MATH 361: Financial Mathematics for Actuaries I
Albert Cohen
Actuarial Sciences Program
Department of Mathematics
Department of Statistics and Probability
C336 Wells Hall
Michigan State University
East Lansing MI
48823
albert@math.msu.edu
acohen@stt.msu.edu
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
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Course Information
Syllabus to be posted on class page in first week of classes
Homework assignments will posted there as well
Page can be found at https://www.math.msu.edu/classpages/
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Course Information
Many examples within these slides are used with kind permission of
Prof. Dmitry Kramkov, Dept. of Mathematics, Carnegie Mellon
University.
Book for course: Financial Mathematics: A Comprehensive Treatment
(Chapman and Hall/CRC Financial Mathematics Series) 1st Edition.
Can be found in MSU bookstores now
Some examples here will be similar to those practice questions
publicly released by the SOA. Please note the SOA owns the
copyright to these questions.
This book will be our reference, and some questions for assignments
will be chosen from it. Copyright for all questions used from this book
belongs to Chapman and Hall/CRC Press .
From time to time, we will also follow the format of Marcel Finan’s A
Discussion of Financial Economics in Actuarial Models: A Preparation
for the Actuarial Exam MFE/3F. Some proofs from there will be
referenced as well. Please find these notes here
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What are financial securities?
Traded Securities - price given by market.
For example:
Stocks
Commodities
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Financial Mathematics for Actuaries I
MSU Spring 2016
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What are financial securities?
Traded Securities - price given by market.
For example:
Stocks
Commodities
Non-Traded Securities - price remains to be computed.
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Financial Mathematics for Actuaries I
MSU Spring 2016
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What are financial securities?
Traded Securities - price given by market.
For example:
Stocks
Commodities
Non-Traded Securities - price remains to be computed.
Is this always true?
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
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What are financial securities?
Traded Securities - price given by market.
For example:
Stocks
Commodities
Non-Traded Securities - price remains to be computed.
Is this always true?
We will focus on pricing non-traded securities.
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Financial Mathematics for Actuaries I
MSU Spring 2016
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How does one fairly price non-traded securities?
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Financial Mathematics for Actuaries I
MSU Spring 2016
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How does one fairly price non-traded securities?
By eliminating all unfair prices
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Financial Mathematics for Actuaries I
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How does one fairly price non-traded securities?
By eliminating all unfair prices
Unfair prices arise from Arbitrage Strategies
Start with zero capital
End with non-zero wealth
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Financial Mathematics for Actuaries I
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How does one fairly price non-traded securities?
By eliminating all unfair prices
Unfair prices arise from Arbitrage Strategies
Start with zero capital
End with non-zero wealth
We will search for arbitrage-free strategies to replicate the payoff of a
non-traded security
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
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How does one fairly price non-traded securities?
By eliminating all unfair prices
Unfair prices arise from Arbitrage Strategies
Start with zero capital
End with non-zero wealth
We will search for arbitrage-free strategies to replicate the payoff of a
non-traded security
This replication is at the heart of the engineering of financial products
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More Questions
Existence - Does such a fair price always exist?
If not, what is needed of our financial model to guarantee at least one
arbitrage-free price?
Uniqueness - are there conditions where exactly one arbitrage-free
price exists?
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And What About...
Does the replicating strategy and price computed reflect uncertainty
in the market?
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Financial Mathematics for Actuaries I
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And What About...
Does the replicating strategy and price computed reflect uncertainty
in the market?
Mathematically, if P is a probabilty measure attached to a series of
price movements in underlying asset, is P used in computing the
price?
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Notation
Forward Contract:
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Notation
Forward Contract:
A financial instrument whose initial value is zero, and whose final
value is derived from another asset. Namely, the difference of the
final asset price and forward price:
V (0) = 0, V (T ) = S(T ) − F
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
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Notation
Forward Contract:
A financial instrument whose initial value is zero, and whose final
value is derived from another asset. Namely, the difference of the
final asset price and forward price:
V (0) = 0, V (T ) = S(T ) − F
(1)
Value at end of term can be negative - buyer accepts this in exchange
for no premium up front
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Notation
Interest Rate:
The rate r at which money grows. Also used to discount the value
today of one unit of currency one unit of time from the present
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Notation
Interest Rate:
The rate r at which money grows. Also used to discount the value
today of one unit of currency one unit of time from the present
V (0) =
Albert Cohen (MSU)
1
, V (1) = 1
1+r
Financial Mathematics for Actuaries I
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An Example of Replication
Forward Exchange Rate: There are two currencies, foreign and
domestic:
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An Example of Replication
Forward Exchange Rate: There are two currencies, foreign and
domestic:
SAB = 4 is the spot exchange rate - one unit of B is worth SAB of A
today (time 0)
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An Example of Replication
Forward Exchange Rate: There are two currencies, foreign and
domestic:
SAB = 4 is the spot exchange rate - one unit of B is worth SAB of A
today (time 0)
r A = 0.1 is the domestic borrow/lend rate
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
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An Example of Replication
Forward Exchange Rate: There are two currencies, foreign and
domestic:
SAB = 4 is the spot exchange rate - one unit of B is worth SAB of A
today (time 0)
r A = 0.1 is the domestic borrow/lend rate
r B = 0.2 is the foreign borrow/lend rate
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
10 / 161
An Example of Replication
Forward Exchange Rate: There are two currencies, foreign and
domestic:
SAB = 4 is the spot exchange rate - one unit of B is worth SAB of A
today (time 0)
r A = 0.1 is the domestic borrow/lend rate
r B = 0.2 is the foreign borrow/lend rate
Compute the forward exchange rate FAB . This is the value of one unit
of B in terms of A at time 1.
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An Example of Replication: Solution
At time 1, we deliver 1 unit of B in exchange for FAB units of domestic
currency A.
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An Example of Replication: Solution
At time 1, we deliver 1 unit of B in exchange for FAB units of domestic
currency A.
This is a forward contract - we pay nothing up front to achieve this.
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An Example of Replication: Solution
At time 1, we deliver 1 unit of B in exchange for FAB units of domestic
currency A.
This is a forward contract - we pay nothing up front to achieve this.
Initially borrow some amount foreign currency B, in foreign market to
grow to one unit of B at time 1. This is achieved by the initial
SB
amount 1+rA B (valued in domestic currency)
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Financial Mathematics for Actuaries I
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An Example of Replication: Solution
At time 1, we deliver 1 unit of B in exchange for FAB units of domestic
currency A.
This is a forward contract - we pay nothing up front to achieve this.
Initially borrow some amount foreign currency B, in foreign market to
grow to one unit of B at time 1. This is achieved by the initial
SB
amount 1+rA B (valued in domestic currency)
FB
Invest the amount 1+rA A in domestic market (valued in domestic
currency)
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An Example of Replication: Solution
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An Example of Replication: Solution
This results in the initial value
V (0) =
FAB
SAB
−
1 + rA 1 + rB
(3)
Since the initial value is 0, this means
FAB = SAB
Albert Cohen (MSU)
1 + rA
= 3.667
1 + rB
Financial Mathematics for Actuaries I
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Discrete Probability Space
Let us define an event as a point ω in the set of all possible outcomes Ω.
This includes the events ”The stock doubled in price over two trading
periods” or ”the average stock price over ten years was 10 dollars”.
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Discrete Probability Space
Let us define an event as a point ω in the set of all possible outcomes Ω.
This includes the events ”The stock doubled in price over two trading
periods” or ”the average stock price over ten years was 10 dollars”.
In our initial case, we will consider the simple binary space
Ω = {H, T } for a one-period asset evolution. So, given an initial
value S0 , we have the final value S1 (ω), with
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
13 / 161
Discrete Probability Space
Let us define an event as a point ω in the set of all possible outcomes Ω.
This includes the events ”The stock doubled in price over two trading
periods” or ”the average stock price over ten years was 10 dollars”.
In our initial case, we will consider the simple binary space
Ω = {H, T } for a one-period asset evolution. So, given an initial
value S0 , we have the final value S1 (ω), with
S1 (H) = uS0 , S1 (T ) = dS0
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
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MSU Spring 2016
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Discrete Probability Space
Let us define an event as a point ω in the set of all possible outcomes Ω.
This includes the events ”The stock doubled in price over two trading
periods” or ”the average stock price over ten years was 10 dollars”.
In our initial case, we will consider the simple binary space
Ω = {H, T } for a one-period asset evolution. So, given an initial
value S0 , we have the final value S1 (ω), with
S1 (H) = uS0 , S1 (T ) = dS0
(5)
with d < 1 < u. Hence, a stock increases or decreases in price,
according to the flip of a coin.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
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Discrete Probability Space
Let us define an event as a point ω in the set of all possible outcomes Ω.
This includes the events ”The stock doubled in price over two trading
periods” or ”the average stock price over ten years was 10 dollars”.
In our initial case, we will consider the simple binary space
Ω = {H, T } for a one-period asset evolution. So, given an initial
value S0 , we have the final value S1 (ω), with
S1 (H) = uS0 , S1 (T ) = dS0
(5)
with d < 1 < u. Hence, a stock increases or decreases in price,
according to the flip of a coin.
Let P be the probability measure associated with these events:
P[H] = p = 1 − P[T ]
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Arbitrage
Assume that S0 (1 + r ) > uS0
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Arbitrage
Assume that S0 (1 + r ) > uS0
Where is the risk involved with investing in the asset S ?
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Arbitrage
Assume that S0 (1 + r ) > uS0
Where is the risk involved with investing in the asset S ?
Assume that S0 (1 + r ) < dS0
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Arbitrage
Assume that S0 (1 + r ) > uS0
Where is the risk involved with investing in the asset S ?
Assume that S0 (1 + r ) < dS0
Why would anyone hold a bank account (zero-coupon bond)?
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Arbitrage
Assume that S0 (1 + r ) > uS0
Where is the risk involved with investing in the asset S ?
Assume that S0 (1 + r ) < dS0
Why would anyone hold a bank account (zero-coupon bond)?
Lemma Arbitrage free ⇒ d < 1 + r < u
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Derivative Pricing
Let S1 (ω) be the price of an underlying asset at time 1. Define the
following instruments:
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Derivative Pricing
Let S1 (ω) be the price of an underlying asset at time 1. Define the
following instruments:
1
Zero-Coupon Bond : V0B = 1+r
, V1B (ω) = 1
Forward Contract : V0F = 0, V1F = S1 (ω) − F
Call Option : V1C (ω) = max(S1 (ω) − K , 0)
Put Option : V1P (ω) = max(K − S1 (ω), 0)
Albert Cohen (MSU)
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Derivative Pricing
Let S1 (ω) be the price of an underlying asset at time 1. Define the
following instruments:
1
Zero-Coupon Bond : V0B = 1+r
, V1B (ω) = 1
Forward Contract : V0F = 0, V1F = S1 (ω) − F
Call Option : V1C (ω) = max(S1 (ω) − K , 0)
Put Option : V1P (ω) = max(K − S1 (ω), 0)
In both the Call and Put option, K is known as the Strike.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
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Derivative Pricing
Let S1 (ω) be the price of an underlying asset at time 1. Define the
following instruments:
1
Zero-Coupon Bond : V0B = 1+r
, V1B (ω) = 1
Forward Contract : V0F = 0, V1F = S1 (ω) − F
Call Option : V1C (ω) = max(S1 (ω) − K , 0)
Put Option : V1P (ω) = max(K − S1 (ω), 0)
In both the Call and Put option, K is known as the Strike.
Once again, a Forward Contract is a deal that is locked in at time 0 for
initial price 0, but requires at time 1 the buyer to purchase the asset for
price F .
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
15 / 161
Derivative Pricing
Let S1 (ω) be the price of an underlying asset at time 1. Define the
following instruments:
1
Zero-Coupon Bond : V0B = 1+r
, V1B (ω) = 1
Forward Contract : V0F = 0, V1F = S1 (ω) − F
Call Option : V1C (ω) = max(S1 (ω) − K , 0)
Put Option : V1P (ω) = max(K − S1 (ω), 0)
In both the Call and Put option, K is known as the Strike.
Once again, a Forward Contract is a deal that is locked in at time 0 for
initial price 0, but requires at time 1 the buyer to purchase the asset for
price F .
What is the value V0 of the above put and call options?
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Put-Call Parity
Can we replicate a forward contract using zero coupon bonds and put and
call options?
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Put-Call Parity
Can we replicate a forward contract using zero coupon bonds and put and
call options?
Yes: The final value of a replicating strategy X has value
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Put-Call Parity
Can we replicate a forward contract using zero coupon bonds and put and
call options?
Yes: The final value of a replicating strategy X has value
V1C − V1P + (K − F ) = S1 − F = X1 (ω)
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
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Put-Call Parity
Can we replicate a forward contract using zero coupon bonds and put and
call options?
Yes: The final value of a replicating strategy X has value
V1C − V1P + (K − F ) = S1 − F = X1 (ω)
(7)
This is achieved (replicated) by
Purchasing one call option
Selling one put option
Purchasing K − F zero coupon bonds with value 1 at maturity.
all at time 0.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
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Put-Call Parity
Can we replicate a forward contract using zero coupon bonds and put and
call options?
Yes: The final value of a replicating strategy X has value
V1C − V1P + (K − F ) = S1 − F = X1 (ω)
(7)
This is achieved (replicated) by
Purchasing one call option
Selling one put option
Purchasing K − F zero coupon bonds with value 1 at maturity.
all at time 0.
Since this strategy must have zero initial value, we obtain
V0C − V0P =
Albert Cohen (MSU)
F −K
1+r
Financial Mathematics for Actuaries I
(8)
MSU Spring 2016
16 / 161
Put-Call Parity
Can we replicate a forward contract using zero coupon bonds and put and
call options?
Yes: The final value of a replicating strategy X has value
V1C − V1P + (K − F ) = S1 − F = X1 (ω)
(7)
This is achieved (replicated) by
Purchasing one call option
Selling one put option
Purchasing K − F zero coupon bonds with value 1 at maturity.
all at time 0.
Since this strategy must have zero initial value, we obtain
F −K
1+r
Question: How would this change in a multi-period model?
V0C − V0P =
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General Derivative Pricing -One period model
If we begin with some initial capital X0 , then we end with X1 (ω). To price
a derivative, we need to match
X1 (ω) = V1 (ω)
∀ ω∈Ω
(9)
to have X0 = V0 , the price of the derivative we seek.
A strategy by the pair (X0 , ∆0 ) wherein
X0 is the initial capital
∆0 is the initial number of shares (units of underlying asset.)
What does the sign of ∆0 indicate?
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Replicating Strategy
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Replicating Strategy
Initial holding in bond (bank account) is X0 − ∆0 S0
Value of portfolio at maturity is
X1 (ω) = (X0 − ∆0 S0 )(1 + r ) + ∆0 S1 (ω)
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Replicating Strategy
Initial holding in bond (bank account) is X0 − ∆0 S0
Value of portfolio at maturity is
X1 (ω) = (X0 − ∆0 S0 )(1 + r ) + ∆0 S1 (ω)
(10)
Pathwise, we compute
V1 (H) = (X0 − ∆0 S0 )(1 + r ) + ∆0 uS0
V1 (T ) = (X0 − ∆0 S0 )(1 + r ) + ∆0 dS0
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Replicating Strategy
Initial holding in bond (bank account) is X0 − ∆0 S0
Value of portfolio at maturity is
X1 (ω) = (X0 − ∆0 S0 )(1 + r ) + ∆0 S1 (ω)
(10)
Pathwise, we compute
V1 (H) = (X0 − ∆0 S0 )(1 + r ) + ∆0 uS0
V1 (T ) = (X0 − ∆0 S0 )(1 + r ) + ∆0 dS0
Algebra yields
∆0 =
Albert Cohen (MSU)
V1 (H) − V1 (T )
(u − d)S0
Financial Mathematics for Actuaries I
(11)
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Risk Neutral Probability
Let us assume the existence of a pair (p̃, q̃) of positive numbers, and use
these to multiply our pricing equation(s):
p̃V1 (H) = p̃(X0 − ∆0 S0 )(1 + r ) + p̃∆0 uS0
q̃V1 (T ) = q̃(X0 − ∆0 S0 )(1 + r ) + q̃∆0 dS0
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Risk Neutral Probability
Let us assume the existence of a pair (p̃, q̃) of positive numbers, and use
these to multiply our pricing equation(s):
p̃V1 (H) = p̃(X0 − ∆0 S0 )(1 + r ) + p̃∆0 uS0
q̃V1 (T ) = q̃(X0 − ∆0 S0 )(1 + r ) + q̃∆0 dS0
Addition yields
X0 (1 + r ) + ∆0 S0 (p̃u + q̃d − (1 + r )) = p̃V1 (H) + q̃V1 (T )
(12)
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Financial Mathematics for Actuaries I
MSU Spring 2016
If we constrain
0 = p̃u + q̃d − (1 + r )
1 = p̃ + q̃
0 ≤ p̃
0 ≤ q̃
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If we constrain
0 = p̃u + q̃d − (1 + r )
1 = p̃ + q̃
0 ≤ p̃
0 ≤ q̃
then we have a risk neutral probability P̃ where
1
p̃V1 (H) + q̃V1 (T )
Ẽ[V1 ] =
1+r
1+r
1+r −d
p̃ = P̃[X1 (ω) = H] =
u−d
u − (1 + r )
q̃ = P̃[X1 (ω) = T ] =
u−d
V0 = X0 =
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Example: Pricing a forward contract
Consider the case of a stock with
S0 = 400
u = 1.25
d = 0.75
r = 0.05
Then the forward price is computed via
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Example: Pricing a forward contract
Consider the case of a stock with
S0 = 400
u = 1.25
d = 0.75
r = 0.05
Then the forward price is computed via
0=
Albert Cohen (MSU)
1
Ẽ[S1 − F ] ⇒ F = Ẽ[S1 ]
1+r
Financial Mathematics for Actuaries I
(13)
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This leads to the explicit price
F = p̃uS0 + q̃dS0
= (p̃)(1.25)(400) + (1 − p̃)(0.75)(400)
= 500p̃ + 300 − 300p̃ = 300 + 200p̃
3
1 + 0.05 − 0.75
= 300 + 200 ·
= 300 + 200 ·
1.25 − 0.75
5
= 420
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This leads to the explicit price
F = p̃uS0 + q̃dS0
= (p̃)(1.25)(400) + (1 − p̃)(0.75)(400)
= 500p̃ + 300 − 300p̃ = 300 + 200p̃
3
1 + 0.05 − 0.75
= 300 + 200 ·
= 300 + 200 ·
1.25 − 0.75
5
= 420
Homework Question: What is the price of a call option in the case
above,with strike K = 375?
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General one period risk neutral measure
We define a finite set of outcomes Ω ≡ {ω1 , ω2 , ..., ωn } and any
subcollection of outcomes A ∈ F1 := 2Ω an event.
Furthermore, we define a probability measure P̃, not necessarily the
physical measure P to be risk neutral if
P̃[ω] > 0 ∀ ω ∈ Ω
1
X0 = 1+r
Ẽ[X1 ]
for all strategies X .
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General one period risk neutral measure
The measure is indifferent to investing in a zero-coupon bond, or a
risky asset X
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General one period risk neutral measure
The measure is indifferent to investing in a zero-coupon bond, or a
risky asset X
The same initial capital X0 in both cases produces the same
”‘average”’ return after one period.
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General one period risk neutral measure
The measure is indifferent to investing in a zero-coupon bond, or a
risky asset X
The same initial capital X0 in both cases produces the same
”‘average”’ return after one period.
Not the physical measure attached by observation, experts, etc..
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
24 / 161
General one period risk neutral measure
The measure is indifferent to investing in a zero-coupon bond, or a
risky asset X
The same initial capital X0 in both cases produces the same
”‘average”’ return after one period.
Not the physical measure attached by observation, experts, etc..
In fact, physical measure has no impact on pricing
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Example: Risk Neutral measure for trinomial case
Assume that Ω = {ω1 , ω2 , ω3 } with
S1 (ω1 ) = uS0
S1 (ω2 ) = S0
S1 (ω3 ) = dS0
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Example: Risk Neutral measure for trinomial case
Assume that Ω = {ω1 , ω2 , ω3 } with
S1 (ω1 ) = uS0
S1 (ω2 ) = S0
S1 (ω3 ) = dS0
Given a payoff V1 (ω) to replicate, are we assured that a replicating
strategy exists?
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Financial Mathematics for Actuaries I
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Example: Risk Neutral measure for trinomial case
Homework:
Try our first example with
(S0 , u, d, r ) = (400, 1.25.0.75, 0.05)
V1digital (ω) = 1{S1 (ω)>450} (ω).
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Example: Risk Neutral measure for trinomial case
Homework:
Try our first example with
(S0 , u, d, r ) = (400, 1.25.0.75, 0.05)
V1digital (ω) = 1{S1 (ω)>450} (ω).
Now, assume you are observe the price on the market to be
V0digital =
1
Ẽ[V1digital ] = 0.25.
1+r
(14)
Use this extra information to price a call option with strike K = 420.
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MSU Spring 2016
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Solution: Risk Neutral measure for trinomial case
The above scenario is reduced to finding the risk-neutral measure
(p̃1 , p̃2 , p̃3 ). This can be done by finding the rref of the matrix M:


1
1
1
1
420 
M = 500 400 300
1
0
0 0.25(1.05)
(15)
which results in


1 0 0 0.2625
rref (M) = 0 1 0 0.675  .
0 0 1 0.0625
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Solution: Risk Neutral measure for trinomial case
It follows that (p̃1 , p̃2 , p̃3 ) = (0.2625, 0.675, 0.0625), and so
1
Ẽ[(S1 − 420)+ | S0 = 400]
1.05
0.2625
=
× (500 − 420) = 20.
1.05
V0C =
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Solution: Risk Neutral measure for trinomial case
It follows that (p̃1 , p̃2 , p̃3 ) = (0.2625, 0.675, 0.0625), and so
1
Ẽ[(S1 − 420)+ | S0 = 400]
1.05
0.2625
=
× (500 − 420) = 20.
1.05
V0C =
Could we perhaps find a set of digital options as a basis set
n
o
V1d1 (ω), V1d2 (ω), V1d3 (ω) = {1A1 (ω), 1A2 (ω), 1A3 (ω)}
(17)
(18)
with A1 , A2 , A3 ∈ F1 to span all possible payoffs at time 1?
How about (A1 , A2 , A3 ) = ({ω1 } , {ω2 } , {ω3 }) ?
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
28 / 161
Exchange one stock for another
Assume now an economy with two stocks, X and Y . Assume that
(X0 , Y0 , r ) = (100, 100, 0.01)
(19)
and

 (110, 105) : ω = ω1
(100, 100) : ω = ω2
(X1 (ω), Y1 (ω)) =

(80, 95)
: ω = ω3 .
Consider two contracts, V and W , with payoffs
V1 (ω) = max {Y1 (ω) − X1 (ω), 0}
(20)
W1 (ω) = Y1 (ω) − X1 (ω).
Price V0 and W0 .
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Financial Mathematics for Actuaries I
MSU Spring 2016
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Exchange one stock for another
In this case, our matrix M is such that


1
1
1
1
M = 110 100 80 101
105 100 95 101
(21)
which results in

1 0 0

rref (M) = 0 1 0
0 0 1
3
10
6 
10 .
1
10

(22)
It follows that
Ẽ[Y1 ] − Ẽ[X1 ]
= Y0 − X0 = 0
1.01
1
1.5
V0 =
· (15p̃3 ) =
= 1.49.
1.01
1.01
W0 =
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Existence of Risk Neutral measure
Let P̃ be a probability measure on a finite space Ω. The following are
equivalent:
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Existence of Risk Neutral measure
Let P̃ be a probability measure on a finite space Ω. The following are
equivalent:
P̃ is a risk neutral measure
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
31 / 161
Existence of Risk Neutral measure
Let P̃ be a probability measure on a finite space Ω. The following are
equivalent:
P̃ is a risk neutral measure
1
Ẽ S1i
For all traded securities S i , S0i = 1+r
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
31 / 161
Existence of Risk Neutral measure
Let P̃ be a probability measure on a finite space Ω. The following are
equivalent:
P̃ is a risk neutral measure
1
Ẽ S1i
For all traded securities S i , S0i = 1+r
Proof: Homework (Hint: One direction is much easier than others. Also,
strategies are linear in the underlying asset.)
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Complete Markets
A market is complete if it is arbitrage free and every non-traded asset can
be replicated.
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Complete Markets
A market is complete if it is arbitrage free and every non-traded asset can
be replicated.
Fundamental Theorem of Asset Pricing 1: A market is arbitrage free
iff there exists a risk neutral measure
Fundamental Theorem of Asset Pricing 2: A market is complete iff
there exists exactly one risk neutral measure
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
32 / 161
Complete Markets
A market is complete if it is arbitrage free and every non-traded asset can
be replicated.
Fundamental Theorem of Asset Pricing 1: A market is arbitrage free
iff there exists a risk neutral measure
Fundamental Theorem of Asset Pricing 2: A market is complete iff
there exists exactly one risk neutral measure
Proof(s): We will go over these in detail later!
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Optimal Investment for a Strictly Risk Averse Investor
Assume a complete market, with a unique risk-neutral measure P̃.
Characterize an investor by her pair (x, U) of initial capital x ∈ X and
utility function U : X → R+ .
Assume U 0 (x) > 0.
Assume U 00 (x) < 0.
Define the Radon-Nikodym derivative of P̃ to P as the random
variable
P̃(ω)
Z (ω) :=
.
(24)
P(ω)
Note that Z is used to map expectations under P to expectations
under P̃: For any random variable X , it follows that
Ẽ[X ] = E[ZX ].
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Optimal Investment for a Strictly Risk Averse Investor
A strictly risk-averse investor now wishes to maximize her expected utility
of a portfolio at time 1, given initial capital at time 0:
u(x) := max E[U(X1 )]
X1 ∈Ax
(26)
Ax := { all portfolio values at time 1 with initial capital x} .
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MSU Spring 2016
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Optimal Investment for a Strictly Risk Averse Investor
Theorem
Define X̂1 via the relationship
U 0 X̂1 := λZ
(27)
where λ sets X̂1 as a strategy with an average return of r under P̃:
Ẽ[X̂1 ] = x(1 + r ).
(28)
Then X̂1 is the optimal strategy.
Albert Cohen (MSU)
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MSU Spring 2016
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Optimal Investment for a Strictly Risk Averse Investor
Proof.
Assume X1 to be an arbitrary strategy with initial capital x. Then for
f (y ) := E[U(yX1 + (1 − y )X̂1 )]
(29)
h
i
f 0 (0) = E U 0 (X̂1 ) X1 − X̂1
h i
= E λZ X1 − X̂1
h
i
= λẼ X1 − X̂1 = 0
2 00
00
f (y ) = E U (yX1 + (1 − y )X̂1 ) X1 − X̂1
<0
(30)
it follows that
and so f attains its maximum at y = 0. We conclude that
E[U(X1 )] < E[U(X̂1 )] for any admissible strategy X1 .
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Optimal Investment: Example
Assume an investor and economy defined by
U(x) = ln (x)
(S0 , u, d, p, q, r ) = (400, 1.25.0.75, 0.5, 0.5, 0.05).
It follows that
3 2
(p̃, q̃) =
,
5 5
6 4
(Z (H), Z (T )) =
,
.
5 5
(31)
(32)
Since U 0 (x) = x1 , we have
1 1
λ Z (ω)
(33)
1
1
1p
1q
1 1
x = X0 =
Ẽ[X̂1 ] =
p̃ ·
+ q̃ ·
=
.
1+r
1+r
λ p̃
λ q̃
λ1+r
X̂1 (ω) =
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MSU Spring 2016
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Optimal Investment: Example
Combining the previous results, we see that
X̂1 (ω) =
x(1 + r )
Z (ω)
u(x) = p ln X̂1 (H) + (1 − p) ln X̂1 (T )
x(1 + r )
x(1 + r )
= p ln
+ (1 − p) ln
Z (H)
Z (T )
(1 + r )
= ln
x = ln (1.0717x) > ln (1.05x).
Z (H)p Z (T )1−p
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Optimal Investment: Example
In terms of her actual strategy, we see that
ˆ 0 S0
S0 X̂1 (H) − X̂1 (T )
1+r
1
1
∆
=
=
−
π̂0 :=
X0
x S1 (H) − S1 (T )
u − d Z (H) Z (T )
(35)
1+r p 1−p
1.05 5 5
=
−
−
=
= −0.875.
u − d p̃ 1 − p̃
0.5 6 4
Therefore, the optimal strategy is to sell a stock portfolio worth 87.5% of
her initial wealth x and invest the proceeds into a safe bank account.
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Optimal Investment: Example
1+r
In fact, since π̂0 = u−d
strategy involves
p
1−p
p̃ − 1−p̃
, we see that qualitatively, her optimal

 > 0 : p > p̃
= 0 : p = p̃
π̂0 =

< 0 : p < p̃.
This links with her strategy via
1 + r̂1 (ω) :=
S1 (ω)
X̂1 (ω)
= (1 − π̂0 )(1 + r ) + π̂0
X0
S0
(36)
and so for our specific case where (r , u, d, π̂0 ) = (0.05, 1.25, 0.75, −0.875),
we have
1 + r̂1 (H) = (1 − π̂0 )(1 + r ) + π̂0 u = 0.875
1 + r̂1 (T ) = (1 − π̂0 )(1 + r ) + π̂0 d = 1.3125.
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Optimal Investment: U(x) =
√
x
Consider now the same set-up as before, only that the utility function
√
changes to U(x) = x.
It follows that
1 1
p
2 X̂1
1 1
⇒ X̂1 = 2 2
4λ Z
U 0 (X̂1 ) =
(38)
Solving for λ returns
x(1 + r ) = Ẽ[X̂1 ]
= E[Z X̂1 ]
1 1
=E Z 2 2
4λ Z
1
1
= 2E
.
4λ
Z
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Optimal Investment: U(x) =
√
x
Combining the results above, we see that
x(1 + r )
Z 2 E Z1
"s
#
q x(1 + r )
X̂1 = E
⇒ u(x) = E
Z 2 E Z1
s p
1
= x(1 + r ) E
.
Z
X̂1 =
Question: Is it true for all (p, p̃) ∈ (0, 1) × (0, 1) that
s s
1
p 2 (1 − p)2
E
=
+
≥ 1?
Z
p̃
1 − p̃
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(41)
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Optimal Betting at the Omega Horse Track!
Imagine our investor with U(x) = ln (x) visits a horse track.
There are three horses: ω1 , ω2 and ω3 .
She can bet on any of the horses to come in 1st .
The payoff is 1 per whole bet made.
She observes the price of each bet with payoff 1 right before the race
to be
(B01 , B02 , B03 ) = (0.5, 0.3, 0.2).
(42)
Symbolically,
B1i (ω) = 1{ωi } (ω).
(43)
Our investor feels the physical probabilities of each horse winning is
(p1 , p2 , p3 ) = (0.6, 0.35, 0.05).
(44)
How should she bet if the race is about to start?
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Optimal Betting at the Omega Horse Track!
In this setting, we can assume r = 0.
This directly implies that
(p̃1 , p̃2 , p̃3 ) = (0.5, 0.3, 0.2).
(45)
Our Radon-Nikodym derivative of P̃ to P is now
0.5 0.3 0.2
(Z (ω1 ), Z (ω2 ), Z (ω3 )) =
,
,
.
0.6 0.35 0.05
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Optimal Betting at the Omega Horse Track!
Her optimal strategy X̂1 reflects her betting strategy, and satisfies
X̂1 (ω) =
∴
X0
Z (ω)
X̂1 (ω1 ) X̂1 (ω2 ) X̂3 (ω1 )
,
,
X0
X0
X0
!
=
6 7 1
, ,
5 6 4
(47)
.
So, per dollar of wealth, she buys 65 of a bet for Horse 1 to win, 67 of
a bet for Horse 2 to win, and 14 of a bet for Horse 3 to win.
The total price (per dollar of wealth) is thus
6
7
1
· 0.5 + · 0.3 + · 0.2 = 0.6 + 0.35 + 0.05 = 1.
5
6
4
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Dividends
What about dividends? How do they affect the risk neutral pricing of
exchange and non-exchange traded assets? What if they are paid at
discrete times? Continuously paid?
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Dividends
What about dividends? How do they affect the risk neutral pricing of
exchange and non-exchange traded assets? What if they are paid at
discrete times? Continuously paid?
Recall that if dividends are paid continuously at rate δ, then 1 share at
time 0 will accumulate to e δT shares upon reinvestment of dividends into
the stock until time T .
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MSU Spring 2016
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Dividends
What about dividends? How do they affect the risk neutral pricing of
exchange and non-exchange traded assets? What if they are paid at
discrete times? Continuously paid?
Recall that if dividends are paid continuously at rate δ, then 1 share at
time 0 will accumulate to e δT shares upon reinvestment of dividends into
the stock until time T .
It follows that to deliver one share of stock S with initial price S0 at time
T , only e −δT shares are needed. Correspondingly,
Fprepaid = e −δT S0
(49)
F = e rT e −δT S0 = e (r −δ)T S0 .
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MSU Spring 2016
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Binomial Option Pricing w/ cts Dividends and Interest
Over a period of length h, interest increases the value of a bond by a
factor e rh and dividends the value of a stock by a factor of e δh .
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MSU Spring 2016
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Binomial Option Pricing w/ cts Dividends and Interest
Over a period of length h, interest increases the value of a bond by a
factor e rh and dividends the value of a stock by a factor of e δh .
Once again, we compute pathwise,
V1 (H) = (X0 − ∆0 S0 )e rh + ∆0 e δh uS0
V1 (T ) = (X0 − ∆0 S0 )e rh + ∆0 e δh dS0
and this results in the modified quantities
∆0 = e −δh
V1 (H) − V1 (T )
(u − d)S0
e (r −δ)h − d
u−d
u − e (r −δ)h
q̃ =
u−d
p̃ =
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Binomial Models w/ cts Dividends and Interest
For σ, the annualized standard deviation of continuously compounded
stock return, the following models hold:
Futures - Cox (1979)
√
u = eσ h
√
d = e −σ h .
General Stock Model
√
u = e (r −δ)h+σ h
√
d = e (r −δ)h−σ h .
Currencies with rf the foreign interest rate, which acts as a dividend:
√
u = e (r −rf )h+σ h
√
d = e (r −rf )h−σ h .
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1- and 2-period pricing
Consider the case r = 0.10, δ = 0.05, h = 0.01, σ = 0.1, S0 = 10 = K .
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1- and 2-period pricing
Consider the case r = 0.10, δ = 0.05, h = 0.01, σ = 0.1, S0 = 10 = K .
Now price two digital options, using the
1
General Stock Model
2
Futures-Cox Model
with respective payoffs
V1K (ω) := 1{S1 ≥K } (ω)
V2K (ω) := 1{S2 ≥K } (ω).
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Calibration Exercise
Assume table below of realized gains & losses over a ten-period cycle.
Use the adjusted values (r , δ, h, S0 , K ) = (0.02, 0, 0.10, 10, 10).
Calculate binary options from last slide using these assumptions.
Period
1
2
3
4
5
6
7
8
9
10
Albert Cohen (MSU)
Return
S1
S0 = 1.05
S2
S1 = 1.02
S3
S2 = 0.98
S4
S3 = 1.01
S5
S4 = 1.02
S6
S5 = 0.99
S7
S6 = 1.03
S8
S7 = 1.05
S9
S8 = 0.96
S10
S9 = 0.97
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Calibration Exercise: Linear Approximation
We would like to compute σ for the logarithm of returns ln
Si
Si−1
.
Assume the returns per period are all independent.
Q: Can we use a linear (simple) return model instead of a compound
return model as an approximation?
If so, then for our observed simple return rate values:
Calculate the sample variance σ∗2 .
σ∗
Estimate that σ ≈ √
.
h
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Calibration Exercise: Linear Approximation
i
Note that if SSi−1
= 1 + γ for γ 1, then
ln
Si
Si−1
≈γ=
Si − Si−1
.
Si−1
(50)
Approximation: Convert our previous table, using simple interest.
Over small time periods h, define linear return values for i th period:
Xi h :=
Si − Si−1
.
Si−1
(51)
In other words, for simple rate of return Xi for period i:
Si = Si−1 · (1 + Xi h).
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Calibration Exercise: Linear Approximation
Our returns table now looks like
Period
1
2
3
4
5
6
7
8
9
10
sample standard deviation
estimated return deviation
Albert Cohen (MSU)
Return
S1 −S0
= 0.05
S0
S2 −S1
= 0.02
S1
S3 −S2
= −0.02
S2
S4 −S3
= 0.01
S3
S5 −S4
= 0.02
S4
S6 −S5
=
−0.01
S5
S7 −S6
= 0.03
S6
S8 −S7
= 0.05
S7
S9 −S8
= −0.04
S8
S10 −S9
= −0.03
S9
σ∗ = 0.0319
σ ≈ 0.1001
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Calibration Exercise: Linear Approximation
We estimate, therefore, that under the Futures-Cox model
(u, d) = (e 0.0319 , e −0.0319 ) = (1.0324, 0.9686)
p̃ =
Albert Cohen (MSU)
(53)
e 0.002 − e −0.0319
= 0.5234.
e 0.0319 − e −0.0319
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Calibration Exercise: Linear Approximation
For the one-period digital option:
V0 = e −rh Ẽ0 [1{S1 ≥10} ] = e −0.002 · p̃ = 0.5224.
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Calibration Exercise: Linear Approximation
For the one-period digital option:
V0 = e −rh Ẽ0 [1{S1 ≥10} ] = e −0.002 · p̃ = 0.5224.
(54)
For the two-period digital option:
V0 = e −2rh Ẽ0 [1{S2 ≥10} ] = e −0.004 · p̃ 2 + 2p̃q̃ = 0.7698.
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Calibration Exercise: No Approximation
Without the linear approximation, we can directly estimate
√
σY h = 0.03172
(u, d) = (e 0.03172 , e −0.03172 ) = (1.0322, 0.9688)
(56)
e 0.002 − e −0.03172
p̃ = 0.03172
= 0.5246.
e
− e −0.03172
For the one-period digital option:
V0 = e −rh Ẽ0 [1{S1 ≥10} ] = e −0.002 · p̃ = 0.5236.
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Calibration Exercise: No Approximation
Without the linear approximation, we can directly estimate
√
σY h = 0.03172
(u, d) = (e 0.03172 , e −0.03172 ) = (1.0322, 0.9688)
(56)
e 0.002 − e −0.03172
p̃ = 0.03172
= 0.5246.
e
− e −0.03172
For the one-period digital option:
V0 = e −rh Ẽ0 [1{S1 ≥10} ] = e −0.002 · p̃ = 0.5236.
(57)
For the two-period digital option:
V0 = e −2rh Ẽ0 [1{S2 ≥10} ] = e −0.004 · p̃ 2 + 2p̃q̃ = 0.7721.
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1- and 2-period pricing
We can solve for 2-period problems
Albert Cohen (MSU)
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1- and 2-period pricing
We can solve for 2-period problems
on a case-by-case basis, or
by developing a general theory for multi-period asset pricing.
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1- and 2-period pricing
We can solve for 2-period problems
on a case-by-case basis, or
by developing a general theory for multi-period asset pricing.
In the latter method, we need a general framework to carry out our
computations
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Risk Neutral Pricing Formula
Assume now that we have the ”regular assumptions” on our coin flip
space, and that at time N we are asked to deliver a path dependent
derivative value VN . Then for times 0 ≤ n ≤ N, the value of this
derivative is computed via
Albert Cohen (MSU)
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Risk Neutral Pricing Formula
Assume now that we have the ”regular assumptions” on our coin flip
space, and that at time N we are asked to deliver a path dependent
derivative value VN . Then for times 0 ≤ n ≤ N, the value of this
derivative is computed via
Vn = e −rh Ẽn [Vn+1 ]
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
(59)
MSU Spring 2016
58 / 161
Risk Neutral Pricing Formula
Assume now that we have the ”regular assumptions” on our coin flip
space, and that at time N we are asked to deliver a path dependent
derivative value VN . Then for times 0 ≤ n ≤ N, the value of this
derivative is computed via
Vn = e −rh Ẽn [Vn+1 ]
(59)
and so
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
58 / 161
Risk Neutral Pricing Formula
Assume now that we have the ”regular assumptions” on our coin flip
space, and that at time N we are asked to deliver a path dependent
derivative value VN . Then for times 0 ≤ n ≤ N, the value of this
derivative is computed via
Vn = e −rh Ẽn [Vn+1 ]
(59)
X0 = Ẽ0 [XN ]
Vn
Xn := nh .
e
(60)
and so
Albert Cohen (MSU)
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Computational Complexity
Consider the case
p̃ = q̃ =
1
2
(61)
4
3
S0 = 4, u = , d =
3
4
but now with term n = 3.
There are 23 = 8 paths to consider.
However, there are 3 + 1 = 4 unique final values of S3 to consider.
In the general term N, there would be 2N paths to generate SN , but
only N + 1 distinct values.
At any node n units of time into the asset’s evolution, there are n + 1
distinct values.
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Computational Complexity
At each value s for Sn , we know that Sn+1 = 34 s or Sn+1 = 34 s.
Using multi-period risk-neutral pricing, we can generate for
vn (s) := Vn (Sn (ω1 , ..., ωn )) on the node (event) Sn (ω1 , ..., ωn ) = s:
h
4 3 i
vn (s) = e −rh p̃vn+1 s + q̃vn+1 s .
3
4
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An Example:
Assume r , δ, and h are such that
1
9
= q̃, e −rh =
2
10
1
S0 = 4, u = 2, d =
2
V3 := max {10 − S3 , 0} .
p̃ =
(63)
It follows that
v3 (32) = 0
v3 (8) = 2
(64)
v3 (2) = 8
v3 (0.50) = 9.50.
Compute V0 .
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Markov Processes
If we use the above approach for a more exotic option, say a lookback
option that pays the maximum over the term of a stock, then we find
this approach lacking.
There is not enough information in the tree or the distinct values for
S3 as stated. We need more.
Consider our general multi-period binomial model under P̃.
Definition We say that a process X is adapted if it depends only on the
sequence of flips ω := (ω1 , ..., ωn )
Definition We say that an adapted process X is Markov if for every
0 ≤ n ≤ N − 1 and every function f (x) there exists another function g (x)
such that
Ẽn [f (Xn+1 )] = g (Xn ).
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Markov Processes
This notion of Markovity is essential to our state-dependent pricing
algorithm.
Indeed, our stock process evolves from time n to time n + 1, using
only the information in Sn .
We can in fact say that for every f (s) there exists a g (s) such that
g (s) = Ẽn [f (Sn+1 ) | Sn = s] .
(66)
In fact, that g depends on f :
h 4 3 i
g (s) = e −rh p̃f
s + q̃f
s .
3
4
(67)
So, for any f (s) := VN (s), we can work our recursive algorithm
backwards to find the gn (s) := Vn (s) for all 0 ≤ n ≤ N − 1
Albert Cohen (MSU)
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MSU Spring 2016
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Markov Processes
Some more thoughts on Markovity:
Consider the example of a Lookback Option.
Here, the payoff is dependent on the realized maximum
Mn := max0≤i≤n Si of the asset.
Mn is not Markov by itself, but the two-factor process (Mn , Sn ) is.
Why?
Let’s generate the tree!
Homework Can you think of any other processes that are not Markov?
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Call Options on Zero-Coupon Bonds
Assume an economy where
One period is one year
The one year short term interest rate from time n to time n + 1 is rn .
The rate evolves via a stochastic process:
r0 = 0.02
rn+1 = Xrn
1
P̃[X = 2k ] = for k ∈ {−1, 0, 1} .
3
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
(68)
65 / 161
Call Options on Zero-Coupon Bonds
Consider now a zero-coupon bond that matures in 3−years with
common face and redemption value F = 100.
Also consider a call option on this bond that expires in 2−years with
strike K = 97.
Denote Bn and Cn as the bond and call option values, respectively.
Note that we iterate backwards from the values
B3 (r ) = 100
(69)
C2 (r ) = max {B2 (r ) − 97, 0} .
Compute (B0 , C0 ).
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
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Call Options on Zero-Coupon Bonds
Our general recursive formula is
1
Ẽ[Bn+1 (rn+1 ) | rn = r ]
1+r
1
Cn (r ) =
Ẽ[Cn+1 (rn+1 ) | rn = r ].
1+r
Bn (r ) =
(70)
Iterating backwards, we see that at t = 2,
1
B2 (r ) =
1 1 X
B3 (2k r ).
1+r 3
(71)
k=−1
At time t = 2, we have that
r2 ∈ {0.08, 0.04, 0.02, 0.01, 0.005} .
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
(72)
MSU Spring 2016
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Call Options on Zero-Coupon Bonds
Our associated Bond and Call Option values at time 2:
Albert Cohen (MSU)
r2
B2
C2
0.08
0.04
0.02
0.01
0.005
92.59
96.15
98.04
99.01
99.50
0
0
1.04
2.01
2.50
Financial Mathematics for Actuaries I
MSU Spring 2016
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Call Options on Zero-Coupon Bonds
Our associated Bond and Call Option values at time 1:
r1
B1
C1
0.04
0.02
0.01
91.92
95.82
97.87
0.33
1.00
1.83
Our associated Bond and Call Option values at time 0:
r0
B0
C0
0.02
93.34
1.03
Question: What if the delivery time of the option is changed to 3?
Symbolically, what if
C3 (r ) = max {B3 (r ) − 97, 0}?
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
(73)
MSU Spring 2016
69 / 161
Capital Structure Model
As an analyst for an investments firm, you are tasked with advising
whether a company’s stock and/or bonds are over/under-priced.
You receive a quarterly report from this company on it’s return on
assets, and have compiled a table for the last ten quarters below.
Today, just after the last quarter’s report was issued, you see that in
billions of USD, the value of the company’s assets is 10.
There are presently one billions shares of this company that are being
traded.
The company does not pay any dividends.
Six months from now, the company is required to pay off a billion
zero-coupon bonds. Each bond has a face value of 9.5.
Albert Cohen (MSU)
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MSU Spring 2016
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Capital Structure Model
Assume Miller-Modigliani holds with At = Bt + St , where the assets
of a company equal the sum of its share and bond price.
Presently, the market values are (B0 , S0 , r ) = (9, 1, 0.02).
The Merton model for corporate bond pricing asserts that at
redemption time T ,
Bt = e −r (T −t) Ẽ [min {AT , F }]
St = e −r (T −t) Ẽ [max {AT − F , 0}] .
(74)
With all of this information, your job now is to issue a Buy or Sell on
the stock and the bond issued by this company.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
71 / 161
Capital Structure Model
Table of return on assets for Company X, with h = 0.25.
Period
1
2
3
4
5
6
7
8
9
10
sample standard deviation
estimated return deviation
Albert Cohen (MSU)
Return on Assets
A1 −A0
= 0.05
A0
A2 −A1
= 0.02
A1
A3 −A2
=
−0.02
A2
A4 −A3
= 0.01
A3
A5 −A4
= 0.02
A4
A6 −A5
= −0.01
A5
A7 −A6
= 0.03
A6
A8 −A7
= 0.05
A7
A9 −A8
= −0.04
A8
A10 −A9
= −0.03
A9
σ∗ = 0.0319
σ ≈ 0.0638
Financial Mathematics for Actuaries I
MSU Spring 2016
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Capital Structure Model
We scale all of our calculation in terms of billions ($, shares, bonds).
Using the Futures- Cox model, we have
(u, d) = (e 0.0319 , e −0.0319 ) = (1.0324, 0.9686)
p̃ =
(75)
e 0.005 − e −0.0319
= 0.5706.
e 0.0319 − e −0.0319
Using this model, the only time the payoff of the bond is less than the
face is on the path ω = TT .
The price of the bond and stock are thus modeled to be
B̃0 = e −0.02·(2·0.25) p̃ 2 · 9.5 + 2p̃q̃ · 9.5 + q̃ 2 · 9.38
= 9.38 > 9.00
(76)
S˜0 = 10 − 9.38 = 0.62 < 1.00.
It follows that,according to our model, one should Buy the bond as it
is underpriced and one should Sell the stock as it is overpriced.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
73 / 161
The Interview Process
Consider the following scenario:
After graduating, you go on the job market, and have 4 possible job
interviews with 4 different companies.
So sure of your prospects that you know that each company will make
an offer, with an identically, independently distributed probability
attached to the 4 possible salary offers:
P [Salary Offer=50, 000] = 0.1
P [Salary Offer=70, 000] = 0.3
(77)
P [Salary Offer=80, 000] = 0.4
P [Salary Offer=100, 000] = 0.2.
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The Interview Process
Questions:
How should you interview?
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The Interview Process
Questions:
How should you interview?
Specifically, when should you accept an offer and cancel the
remaining interviews?
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
75 / 161
The Interview Process
Questions:
How should you interview?
Specifically, when should you accept an offer and cancel the
remaining interviews?
How does your strategy change if you can interview as many times as
you like, but the distribution of offers remains the same as above?
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
75 / 161
The Interview Process: Strategy
Some more thoughts...
At any time the student will know only one offer, which she can either
accept or reject.
Of course, if the student rejects the first three offers, than she has to
accept the last one.
So, compute the maximal expected salary for the student after the
graduation and the corresponding optimal strategy.
Albert Cohen (MSU)
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MSU Spring 2016
76 / 161
The Interview Process: Optimal Strategy
The solution process {Xk }4k=1 follows an Optimal Stopping Strategy:
n
h
io
Xk (s) = max s, Ẽ Xk+1 | k th offer = s .
(78)
At time 4, the value of this game is X4 (s) = s, with s being the salary
offered.
At time 3, the conditional expected value of this game is
h
i
Ẽ X4 | 3rd offer = s = Ẽ[X4 ]
= 0.1 × 50, 000 + 0.3 × 70, 000
(79)
+ 0.4 × 80, 000 + 0.2 × 100, 000
= 78, 000.
Hence, one should accept an offer of 80, 000 or 100, 000, and reject
the other two.
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MSU Spring 2016
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The Interview Process: Optimal Strategy
This strategy leads to a valuation:
X3 (50, 000) = 78, 000
X3 (70, 000) = 78, 000
(80)
X3 (80, 000) = 80, 000
X3 (100, 000) = 100, 000.
At time 2, similar reasoning using Ẽ2 [X3 ] leads to the valuation
X2 (50, 000) = 83, 200
X2 (70, 000) = 83, 200
(81)
X2 (80, 000) = 83, 200
X2 (100, 000) = 100, 000.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
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The Interview Process: Optimal Strategy
At time 1,
X1 (50, 000) = 86, 560
X1 (70, 000) = 86, 560
(82)
X1 (80, 000) = 86, 560
X1 (100, 000) = 100, 000.
Finally, at time 0, the value of this optimal strategy is
Ẽ0 [X1 ] = Ẽ[X1 ] = 0.8 × 86, 560 + 0.2 × 100, 000 = 89, 248.
(83)
So, the optimal strategy is, for the first two interviews, accept only an
offer of 100, 000. If after the third interview, and offer of 80, 000 or
100, 000 is made, then accept. Otherwise continue to the last interview
where you should accept whatever is offered.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
79 / 161
Review
Let’s review the basic contracts we can write:
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
80 / 161
Review
Let’s review the basic contracts we can write:
Forward Contract Initial Value is 0, because both buyer and seller
may have to pay a balance at maturity.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
80 / 161
Review
Let’s review the basic contracts we can write:
Forward Contract Initial Value is 0, because both buyer and seller
may have to pay a balance at maturity.
(European) Put/Call Option Initial Value is > 0, because both only
seller must pay balance at maturity.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
80 / 161
Review
Let’s review the basic contracts we can write:
Forward Contract Initial Value is 0, because both buyer and seller
may have to pay a balance at maturity.
(European) Put/Call Option Initial Value is > 0, because both only
seller must pay balance at maturity.
(European) ”Exotic” Option Initial Value is > 0, because both only
seller must pay balance at maturity.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
80 / 161
Review
Let’s review the basic contracts we can write:
Forward Contract Initial Value is 0, because both buyer and seller
may have to pay a balance at maturity.
(European) Put/Call Option Initial Value is > 0, because both only
seller must pay balance at maturity.
(European) ”Exotic” Option Initial Value is > 0, because both only
seller must pay balance at maturity.
During the term of the contract, can the value of the contract ever fall
below the intrinsic value of the payoff? Symbolically, does it ever occur
that
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
80 / 161
Review
Let’s review the basic contracts we can write:
Forward Contract Initial Value is 0, because both buyer and seller
may have to pay a balance at maturity.
(European) Put/Call Option Initial Value is > 0, because both only
seller must pay balance at maturity.
(European) ”Exotic” Option Initial Value is > 0, because both only
seller must pay balance at maturity.
During the term of the contract, can the value of the contract ever fall
below the intrinsic value of the payoff? Symbolically, does it ever occur
that
vn (s) < g (s)
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
(84)
MSU Spring 2016
80 / 161
Review
Let’s review the basic contracts we can write:
Forward Contract Initial Value is 0, because both buyer and seller
may have to pay a balance at maturity.
(European) Put/Call Option Initial Value is > 0, because both only
seller must pay balance at maturity.
(European) ”Exotic” Option Initial Value is > 0, because both only
seller must pay balance at maturity.
During the term of the contract, can the value of the contract ever fall
below the intrinsic value of the payoff? Symbolically, does it ever occur
that
vn (s) < g (s)
(84)
where g (s) is of the form of g (S) := max {S − K , 0}, in the case of a Call
option, for example.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
80 / 161
Early Exercise
If σ = 0, and so uncertainty vanishes, then an investor would seek to
exercise early if
rK > δS.
(85)
If σ > 0, then the situation involves deeper analysis.
Whether solving a free boundary problem or analyzing a binomial
tree, it is likely that a computer will be involved in helping the
investor to determine the optimal exercise time.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
81 / 161
For Freedom! (we must charge extra...)
What happens if we write a contract that allows the purchaser to exercise
the contract whenever she feels it to be in her advantage? By allowing this
extra freedom, we must
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
82 / 161
For Freedom! (we must charge extra...)
What happens if we write a contract that allows the purchaser to exercise
the contract whenever she feels it to be in her advantage? By allowing this
extra freedom, we must
Charge more than we would for a European contract that is exercised
only at the term N.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
82 / 161
For Freedom! (we must charge extra...)
What happens if we write a contract that allows the purchaser to exercise
the contract whenever she feels it to be in her advantage? By allowing this
extra freedom, we must
Charge more than we would for a European contract that is exercised
only at the term N.
Hedge our replicating strategy X differently, to allow for the
possibility of early exercise.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
82 / 161
American Options
In the end, the option v is valued after the nth value of the stock
Sn (ω) = s is revealed via the recursive formula along each path ω:
n
h
io
vn (Sn (ω)) = max g (Sn (ω)), e −rh Ẽ v (Sn+1 (ω)) | Sn (ω)
(86)
τ ∗ (ω) = inf {k ∈ {0, 1, .., N} | vk (Sk (ω)) = g (Sk (ω))} .
Here, τ ∗ is the optimal exercise time.
In the Binomial case, we reduce to
n
o
vn (s) = max g (s), e −rh [p̃vn+1 (us) + q̃vn+1 (ds)]
τ ∗ = inf {k | vk (s) = g (s)} .
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
(87)
83 / 161
American Options
Some examples:
”American Bond:” g (s) = 1
”American Digital Option:” g (s) = 1{6≤s≤10}
”American Square Option:” g (s) = s 2 .
Does an investor exercise any of these options early? Consider again the
setting
1
9
= q̃, e −rh =
2
10
1
S0 = 4, u = 2, d = .
2
p̃ =
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
(88)
MSU Spring 2016
84 / 161
American Square Options
Consider the American Square Option. We know via Jensen’s Inequality
that
h
i
h
i
2
e −rh Ẽ g (Sn+1 ) | Sn = e −rh Ẽ Sn+1
| Sn
h
i2 ≥ e −rh Ẽ Sn+1 | Sn
(89)
2
rh
−rh
=e
e Sn
= e rh Sn2 > Sn2 = g (Sn ).
Albert Cohen (MSU)
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MSU Spring 2016
85 / 161
American Square Options
It follows that vN (s) = s 2 and
n
o
vN−1 (s) = max g (s), e −rh Ẽ[vN (SN ) | SN−1 = s]
n
o
= max s 2 , e −rh Ẽ[SN2 | SN−1 = s]
(90)
= e −rh Ẽ[SN2 | SN−1 = s]
= e −rh Ẽ[vN (SN ) | SN−1 = s].
In words, with one period to go, don’t exercise yet!!
The American and European option values coincide. Keep going.
How about with two periods left before expiration?
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
86 / 161
American Options
Let’s return to the previous European Put example, where
1
9
= q̃, e −rh =
2
10
1
S0 = 4, u = 2, d =
2
V3 := max {10 − S3 , 0} .
p̃ =
It follows that S3 (ω) ∈
1
2 , 2, 8, 32
(91)
.
Use this to compute v3 (s) and the American Put recursively.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
87 / 161
Matching Interest Rates to Market Conditions
Consider again a series of coin flips (ω1 , ..., ωn ) where at time n, the
interest rate from n to n + 1 is modeled via
rn = rn (ω1 , ..., ωn )
(92)
and a stochastic volatility σ at time n via
1
rn (ω1 , ..., ωn−1 , ωn = H)
σn = ln
.
2
rn (ω1 , ..., ωn−1 , ωn = T )
(93)
Keep in mind that we will build a recombining binomial tree for this
model. So, for example,
r2 (H, T ) = r2 (T , H).
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
(94)
MSU Spring 2016
88 / 161
Matching Interest Rates to Market Conditions
Futhermore, we can define the yield rate y (t, T , r (t)) for a zero-coupon
bond B(t, T , r (t)) via
B(t, T , r (t)) =
1
(95)
(1 + y (t, T , r (t)))T −t
and the corresponding yield rate volatility
1
y (1, n, r1 (H))
σ̃n = ln
.
2
y (1, n, r1 (T ))
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
(96)
MSU Spring 2016
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Matching Interest Rates to Market Conditions
We can in fact see that
y (1, 2, r1 (H)) = r1 (H)
(97)
y (1, 2, r1 (T )) = r1 (T ).
But, for example, it is not clear how to obtain
(y (1, 3, r1 (H)), y (1, 3, r1 (T ))) .
(98)
Furthermore, how can we match to market conditions and update our
estimates for interest rates rn ?
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
90 / 161
Matching Interest Rates to Market Conditions
One way forward is the Black Derman Toy Model 1
Consider a market with the following observations:
Maturity
1
2
3
4
5
Yield to Maturity y (0, T , r )
0.100
0.110
0.120
0.125
0.130
Yield Volatility σ̃T
N.A.
0.190
0.180
0.150
0.140
(For Daily US Treasury Real Yield Curve Rates click here )
1
Black, Fischer, Emanuel Derman, and William Toy. ”A one-factor model
of interest rates and its application to treasury bond options.” Financial
Analysts Journal 46.1 (1990): 33-39.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
91 / 161
Matching Interest Rates to Market Conditions
In this setting, we match market conditions to a model where at each
time, an interest rate moves up or down with a ”risk-neutral”
probability of 12 .
It follows that we have from time t = 0 to t = 1, with an initial rate
r0 = r ,
1
1
=
1 + y (0, 1, r )
1+r
(99)
⇒ y (0, 1, r ) = r .
Albert Cohen (MSU)
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MSU Spring 2016
92 / 161
Matching Interest Rates to Market Conditions
From time t = 0 to t = 2, again with our initial rate r0 = r ,
Connecting our observed two-year yield with yearly interest rates
returns
1
1
1
1
1
1
+
=
(1 + y (0, 2, r ))2
1 + r 2 1 + r1 (H) 2 1 + r1 (T )
(100)
1
1
1
1
1
1
⇒
+
.
=
1.112
1.10 2 1 + r1 (H) 2 1 + r1 (T )
Also, connecting our one-year yields with yearly interest rates leads to
1
r1 (H)
1
y (1, 2, r1 (H))
σ1 = ln
= ln
2
r1 (T )
2
y (1, 2, r1 (T ))
(101)
= σ̃2 = 0.190.
Solution leads to the pair
(r1 (H), r1 (T )) = (0.1432, 0.0979).
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
(102)
MSU Spring 2016
93 / 161
Matching Interest Rates to Market Conditions
From time t = 0 to t = 3, again with our initial rate r0 = r , we try to
estimate the matching (r2 (H, H), r2 (H, T ), r2 (T , T )).
Connecting our observed two-year yield with yearly interest rates
returns
1
1
1
1
1
=
(1 + y (0, 3, r ))3
1 + r 4 1 + r1 (H) 1 + r2 (H, H)
1
1
1
1
+
1 + r 4 1 + r1 (H) 1 + r2 (H, T )
(103)
1
1
1
1
+
1 + r 4 1 + r1 (T ) 1 + r2 (T , H)
1
1
1
1
+
.
1 + r 4 1 + r1 (T ) 1 + r2 (T , T )
We can now substitute our values
(r , r1 (H), r1 (T ), y (0, 3, r )) = (0.10, 0.1432, 0.0979, 0.120).
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
94 / 161
Matching Interest Rates to Market Conditions
We also know that σ2 6= σ2 (ω1 , ω2 ) and so
1
r2 (H, H)
σ2 = ln
2
r2 (H, T )
1
r2 (T , H)
σ2 = ln
2
r2 (T , T )
p
⇒ r2 (H, T ) = r2 (T , H) = r2 (H, H)r2 (T , T ).
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
(104)
95 / 161
Matching Interest Rates to Market Conditions
Finally, we have one more matching condition via
y (1, 3, r1 (H))
1
0.180 = σ̃3 = ln
2
y (1, 3, r1 (T ))
(105)
where
1
1
=
2
1
+
r
(1 + y (1, 3, r1 (H))
1 (H)
1
1
1
1
+
2 1 + r2 (H, H) 2 1 + r2 (H, T )
1
1
1
1
+
.
2 1 + r2 (T , H) 2 1 + r2 (T , T )
(106)
Now solve for (r2 (H, H), r2 (H, T ), r2 (T , T ))!!
HW1: Price a bond that matures in two years, with the above
observations for term structure, and with F = 100 and coupon rate 10%.
HW2: What about r3 , r4 , r5 ? Can we compute them?
1
1
=
2
1
+
r
(1 + y (1, 3, r1 (T ))
1 (T )
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
96 / 161
Pricing a Two-year Coupon Bond Using Market Conditions
Consider the previous observations for term structure, and a bond with
F = 100 and coupon rate 10%. In this setting, we have
Time
0
1
1
2
2
2
2
Albert Cohen (MSU)
Interest Rate
r0
r1 (H)
r1 (T )
r2 (H, H)
r2 (H, T )
r2 (T , H)
r2 (T , T )
Value
0.100
0.1432
0.0979
0.1942
0.1377
0.1377
0.0976
Financial Mathematics for Actuaries I
MSU Spring 2016
97 / 161
Pricing a Two-year Coupon Bond Using Market Conditions
We can decompose the two-year coupon bond into two component
zero-coupon bonds.
The first is B (1) , which has face 10, maturity T = 1, and initial price
B (1) (0, T , r ) at t = 0.
The first is B (2) , which has face 110, maturity T = 2, and initial price
B (2) (0, T , r ) at t = 0.
The total coupon bond price is thus
B(0, 2, r ) = B (1) (0, 1, r ) + B (2) (0, 2, r ).
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
(107)
98 / 161
Pricing a Two-year Coupon Bond Using Market Conditions
Our component bonds thus have prices
Time
0
1
1
Time
0
1
1
2
2
2
2
Albert Cohen (MSU)
Interest Rate
r0 = 0.10
r1 (H) = 0.1432
r1 (T ) = 0.0979
Interest Rate
r0 = 0.10
r1 (H) = 0.1432
r1 (T ) = 0.0979
r2 (H, H) = 0.1942
r2 (H, T ) = 0.1377
r2 (T , H) = 0.1377
r2 (T , T ) = 0.0976
B 1 (t, T , r )
B (1) (0, 1, r0 ) = 9.09
B (1) (1, 1, r1 (H)) = 10
B (1) (1, 1, r1 (T )) = 10
B 2 (t, T , r )
B (2) (0, 2, r0 ) = 89.28
B (2) (1, 2, r1 (H)) = 96.22
B (2) (1, 2, r1 (T )) = 100.19
B (2) (2, 2, r2 (H, H)) = 110
B (2) (2, 2, r2 (H, T )) = 110
B (2) (2, 2, r2 (T , H)) = 110
B (2) (2, 2, r2 (T , T )) = 110
Financial Mathematics for Actuaries I
MSU Spring 2016
99 / 161
Pricing a Two-year Coupon Bond Using Market Conditions
Finally, we have our coupon bond with price
Time
0
1
1
2
2
2
2
Interest Rate
r0 = 0.10
r1 (H) = 0.1432
r1 (T ) = 0.0979
r2 (H, H) = 0.1942
r2 (H, T ) = 0.1377
r2 (T , H) = 0.1377
r2 (T , T ) = 0.0976
B(t, T , r )
B(0, 2, r0 ) = 98.37
B(1, 2, r1 (H)) = 106.22
B(1, 2, r1 (T )) = 110.19
B(2, 2, r2 (H, H)) = 110
B(2, 2, r2 (H, T )) = 110
B(2, 2, r2 (T , H)) = 110
B(2, 2, r2 (T , T )) = 110
Note: Note that with our coupons, the yield yc (0, 2, r ) = 0.1095, which is
obtained by solving
98.37 =
Albert Cohen (MSU)
10
110
+
.
1 + yc (0, 2, r ) (1 + yc (0, 2, r ))2
Financial Mathematics for Actuaries I
MSU Spring 2016
(108)
100 / 161
Pricing a 1-year Call Option on our 2-year Coupon Bond
Consider the previous term structure, and a European Call Option on the
two year bond with K = 97 and expiry of 1 year.
Compute the initial Call price C0 (r ).
Compute the initial number of bonds to hold to replicate this option.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
101 / 161
Pricing a 1-year Call Option on our 2-year Coupon Bond
In this case, we look at the price of the bond minus the accrued interest:
Time
0
1
1
2
2
2
2
Interest Rate
r0 = 0.10
r1 (H) = 0.1432
r1 (T ) = 0.0979
r2 (H, H) = 0.1942
r2 (H, T ) = 0.1377
r2 (T , H) = 0.1377
r2 (T , T ) = 0.0976
B(t, T , r )
B(0, 2, r0 ) = 98.37
B(1, 2, r1 (H)) = 96.22
B(1, 2, r1 (T )) = 100.19
B(2, 2, r2 (H, H)) = 100
B(2, 2, r2 (H, T )) = 100
B(2, 2, r2 (T , H)) = 100
B(2, 2, r2 (T , T )) = 100
Pricing and hedging is accomplished via
1
1
1
C0 (0.10) =
· 0 + 3.19 = 1.45
1.10 2
2
0 − 3.19
∆0 (0.10) =
= 0.804.
96.22 − 100.19
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
(109)
102 / 161
Asian Options
In times of high volatility or frequent trading, a company may want to
protect against large price movements over an entire time period, using an
average. For example, if a company is looking at foreign exchange markets
or markets that may be subject to stock pinning due to large actors.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
103 / 161
Asian Options
In times of high volatility or frequent trading, a company may want to
protect against large price movements over an entire time period, using an
average. For example, if a company is looking at foreign exchange markets
or markets that may be subject to stock pinning due to large actors.
As an input, the average of an asset is used as an input against a strike,
instead of the spot price.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
103 / 161
Asian Options
In times of high volatility or frequent trading, a company may want to
protect against large price movements over an entire time period, using an
average. For example, if a company is looking at foreign exchange markets
or markets that may be subject to stock pinning due to large actors.
As an input, the average of an asset is used as an input against a strike,
instead of the spot price.
There are two possibilities for the input in the discrete case: h = T
N and
P
Arithmetic Average: IA (T ) := N1 N
k=1 Skh .
1
N
Geometric Average: IG (T ) := ΠN
S
.
k=1 kh
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
103 / 161
Asian Options
In times of high volatility or frequent trading, a company may want to
protect against large price movements over an entire time period, using an
average. For example, if a company is looking at foreign exchange markets
or markets that may be subject to stock pinning due to large actors.
As an input, the average of an asset is used as an input against a strike,
instead of the spot price.
There are two possibilities for the input in the discrete case: h = T
N and
P
Arithmetic Average: IA (T ) := N1 N
k=1 Skh .
1
N
Geometric Average: IG (T ) := ΠN
S
.
k=1 kh
HW: Is there an ordering for IA , IG that is independent of T ?
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
103 / 161
Asian Options: An Example:
Notice that these are path-dependent options, unlike the put and call
options that we have studied until now. Assume r , δ, and h are such that
1
9
S0 = 4, u = 2, d = , e −rh =
2
10
g (I ) = max {I − 2.5, 0} .
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
(110)
MSU Spring 2016
104 / 161
Asian Options: An Example:
Consider an arithmetic average with N = 2. Then
8 + 16
v2 (HH) = max
− 2.5, 0 = 9.5
2
8+4
v2 (HT ) = max
− 2.5, 0 = 3.5
2
2+4
v2 (TH) = max
− 2.5, 0 = 0.5
2
2+1
v2 (TT ) = max
− 2.5, 0 = 0.
2
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
(111)
105 / 161
Asian Options: An Example:
Consider an arithmetic average with N = 2. Then
8 + 16
v2 (HH) = max
− 2.5, 0 = 9.5
2
8+4
v2 (HT ) = max
− 2.5, 0 = 3.5
2
2+4
v2 (TH) = max
− 2.5, 0 = 0.5
2
2+1
v2 (TT ) = max
− 2.5, 0 = 0.
2
(111)
Compute v0 , assuming a European structure.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
105 / 161
Asian Options: An Example:
Consider an arithmetic average with N = 2. Then
8 + 16
v2 (HH) = max
− 2.5, 0 = 9.5
2
8+4
v2 (HT ) = max
− 2.5, 0 = 3.5
2
2+4
v2 (TH) = max
− 2.5, 0 = 0.5
2
2+1
v2 (TT ) = max
− 2.5, 0 = 0.
2
(111)
Compute v0 , assuming a European structure. How about an American
structure?
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
105 / 161
Lognormality
Analyzing returns, we assume:
A probability space Ω, F, P .
Our asset St (ω) has an associated return over any period (t, t + u)
defined as
!
St+u (ω)
rt,u (ω) := ln
.
(112)
St (ω)
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
106 / 161
Lognormality
Partition the interval [t, T ] into n intervals of length h = T n−t , then:
The return over the entire period can be taken as the sum of the
returns over each interval:
!
n
X
ST (ω)
rt,T −t (ω) = ln
=
rtk ,h (ω)
St (ω)
(113)
k=1
tk = t + kh.
We model the returns as being independent and possessing a binomial
distribution.
Employing the Central Limit Theorem, it can be shown that as
n → ∞, this distribution approaches normality.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
107 / 161
Binomial Tree and Discrete Dividends
Another issue encountered in elementary credit and investment theory
is the case of different compounding and deposit periods.
This also occurs in the financial setting where a dividend is not paid
continuously, but rather at specific times.
It follows that the dividend can be modeled as delivered in the middle
of a binomial period, at time τ (ω) < T .
This view is due to Schroder and can be summarized as viewing the
inherent value of St (ω) as the sum of a prepaid forward PF and the
present value of the upcoming dividend payment D:
PFt (ω) = St (ω) − De −r (τ (ω)−t)
√
u = e rh+σ h
(114)
√
d = e rh−σ h .
Now, the random process that we model as having up and down
moves is PF instead of S.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
108 / 161
Back to the Continuous Time Case
Consider the case of a security whose binomial evolution is modeled as an
up or down movement at the end of each day. Over the period of one
year, this amounts to a tree with depth 365. If the tree is not recombining,
then this amounts to 2365 branches. Clearly, this is too large to evaluate
reasonably, and so an alternative is sought.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
109 / 161
Back to the Continuous Time Case
Consider the case of a security whose binomial evolution is modeled as an
up or down movement at the end of each day. Over the period of one
year, this amounts to a tree with depth 365. If the tree is not recombining,
then this amounts to 2365 branches. Clearly, this is too large to evaluate
reasonably, and so an alternative is sought.
Whatever the alternative, the concept of replication must hold. This is
the reasoning behind the famous Black-Scholes-Merton PDE approach.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
109 / 161
Monte Carlo Techniques
Our model for asset evolution is
1
2
√
⇒ St = S0 e (α−δ− 2 σ )t+σ tZ
(115)
Z ∼ N(0, 1).
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
110 / 161
Monte Carlo Techniques
Our model for asset evolution is
1
2
√
⇒ St = S0 e (α−δ− 2 σ )t+σ tZ
(115)
Z ∼ N(0, 1).
Consider now the possibility of simulating the stock evolution by
n
simulating the random variable Z , or in fact an i.i.d. sequence Z (i) i=1 .
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
110 / 161
Monte Carlo Techniques
Our model for asset evolution is
1
2
√
⇒ St = S0 e (α−δ− 2 σ )t+σ tZ
(115)
Z ∼ N(0, 1).
Consider now the possibility of simulating the stock evolution by
n
simulating the random variable Z , or in fact an i.i.d. sequence Z (i) i=1 .
For a European option with time expiry T , we can simulate the expiry
time payoff mulitple times:
√
1 2
(i)
V S (i) , T = G S (i) = G S0 e (α−δ− 2 σ )T +σ T Z .
(116)
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
110 / 161
Monte Carlo Techniques
n on
If we sample uniformly from our simulated values V S (i) , T
we
i=1
can appeal to a sampling-convergence theorem with the appoximation
V (S, 0) = e
Albert Cohen (MSU)
−rT 1
n
n
X
V S (i) , T .
(117)
i=1
Financial Mathematics for Actuaries I
MSU Spring 2016
111 / 161
Monte Carlo Techniques
n on
If we sample uniformly from our simulated values V S (i) , T
we
i=1
can appeal to a sampling-convergence theorem with the appoximation
V (S, 0) = e
−rT 1
n
n
X
V S (i) , T .
(117)
i=1
The challenge now is to simulate our lognormally distributed asset
evolution.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
111 / 161
Monte Carlo Techniques
n on
If we sample uniformly from our simulated values V S (i) , T
we
i=1
can appeal to a sampling-convergence theorem with the appoximation
V (S, 0) = e
−rT 1
n
n
X
V S (i) , T .
(117)
i=1
The challenge now is to simulate our lognormally distributed asset
evolution.
One can simulate the value ST directly by one random variable Z , or a
multiple of them to simulate the path of the evolution until T .
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
111 / 161
Monte Carlo Techniques
n on
If we sample uniformly from our simulated values V S (i) , T
we
i=1
can appeal to a sampling-convergence theorem with the appoximation
V (S, 0) = e
−rT 1
n
n
X
V S (i) , T .
(117)
i=1
The challenge now is to simulate our lognormally distributed asset
evolution.
One can simulate the value ST directly by one random variable Z , or a
multiple of them to simulate the path of the evolution until T .
The latter method is necessary for Asian options.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
111 / 161
Monte Carlo Techniques
There are multiple ways to simulate Z . One way is to find a random
number U taken from a uniform distribution U[0, 1].
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
112 / 161
Monte Carlo Techniques
There are multiple ways to simulate Z . One way is to find a random
number U taken from a uniform distribution U[0, 1].
It follows that one can now map U → Z via inversion of the Nornal cdf N:
Z = N −1 (U).
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
(118)
MSU Spring 2016
112 / 161
Monte Carlo Techniques
There are multiple ways to simulate Z . One way is to find a random
number U taken from a uniform distribution U[0, 1].
It follows that one can now map U → Z via inversion of the Nornal cdf N:
Z = N −1 (U).
(118)
It can be shown that in a sample, the standard deviation of the sample
average σsample is related to the standard deviation of an individual draw
via
σdraw
σsample = √ .
n
(119)
If σdraw = σ, then we can see that we must increase our sample size by
22k if we wish to cut our σsample by a factor of 2k .
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
112 / 161
Black Scholes Pricing using Underlying Asset
In the next course, we will derive the following solutions to the
Black-Scholes PDE:
V C (S, t) = e −r (T −t) Ẽ [(ST − K )+ | St = S]
= Se −δ(T −t) N(d1 ) − Ke −r (T −t) N(d2 )
V P (S, t) = e −r (T −t) Ẽ [(K − ST )+ | St = S]
= Ke −r (T −t) N(−d2 ) − Se −δ(T −t) N(−d1 )
ln KS + (r − δ + 21 σ 2 )(T − t)
√
d1 =
σ T −t
√
d2 = d1 − σ T − t
Z x
z2
1
N(x) = √
e − 2 dz.
2π −∞
(120)
Notice that V C (S, t) − V P (S, t) = Se −δ(T −t) − Ke −r (T −t) .
Question: What underlying model of stock evolution leads to this value?
How can we support such a probability measure?
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
113 / 161
Lognormal Random Variables
We say that Y ∼ LN(µ, σ) is Lognormal if ln(Y ) ∼ N(µ, σ 2 ).
As sums of normal random variables remain normal, products of lognormal
random variables remain lognormal.
Recall that the moment-generating function of
X ∼ N(µ, σ 2 ) ∼ µ + σN(0, 1) is
1
2 2
MX (t) = E[e tX ] = e µt+ 2 σ t
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
(121)
MSU Spring 2016
114 / 161
Lognormal Random Variables
We say that Y ∼ LN(µ, σ) is Lognormal if ln(Y ) ∼ N(µ, σ 2 ).
As sums of normal random variables remain normal, products of lognormal
random variables remain lognormal.
Recall that the moment-generating function of
X ∼ N(µ, σ 2 ) ∼ µ + σN(0, 1) is
1
2 2
1
2 2
MX (t) = E[e tX ] = e µt+ 2 σ t
(121)
If Y = e µ+σZ , then, it can be seen that
E[Y n ] = E[e nX ] = e µn+ 2 σ n
(122)
and
1
fY (y ) = √
exp
σ 2πy
Albert Cohen (MSU)
(ln(y ) − µ)2
−
2σ 2
Financial Mathematics for Actuaries I
!
(123)
MSU Spring 2016
114 / 161
Stock Evolution and Lognormal Random Variables
One application of lognormal distributions is their use in modeling the
evolution of asset prices S. If we assume a physical measure P with α the
expected return on the stock under the physical measure, then
1
St
ln
= N (α − δ − σ 2 )t, σ 2 t
S0
2
(124)
√
(α−δ− 12 σ 2 )t+σ tZ
⇒ St = S0 e
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
115 / 161
Stock Evolution and Lognormal Random Variables
We can use the previous facts to show
E[St ] = S0 e (α−δ)t
!
ln SK0 + (α − δ − 0.5σ 2 )t
√
.
P[St > K ] = N
σ t
(125)
Note that under the risk-neutral measure P̃, we exchange α with r , the
risk-free rate:
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
116 / 161
Stock Evolution and Lognormal Random Variables
We can use the previous facts to show
E[St ] = S0 e (α−δ)t
!
ln SK0 + (α − δ − 0.5σ 2 )t
√
.
P[St > K ] = N
σ t
(125)
Note that under the risk-neutral measure P̃, we exchange α with r , the
risk-free rate:
Ẽ[St ] = S0 e (r −δ)t
(126)
P̃[St > K ] = N(d2 ).
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
116 / 161
Stock Evolution and Lognormal Random Variables
Risk managers are also interested in Conditional Tail Expectations (CTE’s)
of random variables:
h
i
E X 1{X >k}
CTEX (k) := E[X | X > k] =
.
(127)
P[X > k]
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
117 / 161
Stock Evolution and Lognormal Random Variables
In our case,
"
E
√
1 2
S0 e (α−δ− 2 σ )t+σ tZ 1 (α−δ− 1 σ2 )t+σ√tZ 2
S0 e
>K
E[St | St > K ] =
#
h
i
√
1 2
P S0 e (α−δ− 2 σ )t+σ tZ > K
!
S
N
ln K0 +(α−δ+0.5σ 2 )t
√
σ t
N
ln K0 +(α−δ−0.5σ 2 )t
√
σ t
= S0 e (α−δ)t
⇒ Ẽ[St | St > K ] = S0 e (r −δ)t
S
!
N(d1 )
N(d2 )
(128)
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
118 / 161
Stock Evolution and Lognormal Random Variables
In fact, we can use this CTE framework to solve for the European Call
option price in the Black-Scholes framework, where P̃0 [A] = P̃[A | S0 = S]
and
h
i
V C (S, 0) := e −rT Ẽ (ST − K )+ | S0 = S
h
i
= e −rT Ẽ0 ST − K | ST > K · P̃0 [ST > K ]
h
i
= e −rT Ẽ0 ST | ST > K · P̃0 [ST > K ] − Ke −rT P̃0 [ST > K ]
= e −rT Se (r −δ)T
N(d1 )
· N(d2 ) − Ke −rT N(d2 )
N(d2 )
= Se −δT N(d1 ) − Ke −rT N(d2 ).
(129)
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
119 / 161
Black Scholes Analysis: Option Greeks
For any option price V (S, t), define its various sensitivities as follows:
∂V
∂S
∂∆
∂2V
Γ=
=
∂S
∂S 2
∂V
ν=
∂σ
∂V
Θ=
∂t
∂V
ρ=
∂r
∂V
Ψ=
.
∂δ
These are known accordingly as the Option Greeks.
∆=
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
(130)
MSU Spring 2016
120 / 161
Black Scholes Analysis: Option Greeks
Straightforward partial differentiation leads to
∆C = e −δ(T −t) N(d1 )
∆P = −e −δ(T −t) N(−d1 )
(131)
e −δ(T −t) N 0 (d1 )
√
σS T − t
√
ν C = ν P = Se −δ(T −t) T − tN 0 (d1 )
ΓC = ΓP =
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
121 / 161
Black Scholes Analysis: Option Greeks
as well as..
ρC = (T − t)Ke −r (T −t) N(d2 )
ρP = −(T − t)Ke −r (T −t) N(−d2 )
(132)
ΨC = −(T − t)Se −δ(T −t) N(d1 )
ΨP = (T − t)Se −δ(T −t) N(−d1 ).
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
122 / 161
Black Scholes Analysis: Option Greeks
as well as..
ρC = (T − t)Ke −r (T −t) N(d2 )
ρP = −(T − t)Ke −r (T −t) N(−d2 )
(132)
ΨC = −(T − t)Se −δ(T −t) N(d1 )
ΨP = (T − t)Se −δ(T −t) N(−d1 ).
What do the signs of the Greeks tell us?
HW: Compute Θ for puts and calls.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
122 / 161
Option Elasticity
Define
V (S+,t)−V (s,t)
V (S,t)
Ω(S, t) := lim
S+−S
→0
S
S
V (S + , t) − V (s, t)
lim
V (S, t) →0
S +−S
∆·S
=
.
V (S, t)
=
(133)
Consequently,
ΩC (S, t) =
Se −δ(T −t)
∆C · S
=
≥1
V C (S, t)
Se −δ(T −t) − Ke −r (T −t) N(d2 )
∆P · S
Ω (S, t) = P
≤ 0.
V (S, t)
(134)
P
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
123 / 161
Option Elasticity
Theorem
The volatility of an option is the option elasticity times the volatility of the
stock:
σoption = σstock × | Ω | .
(135)
The proof comes from Finan: Consider the strategy of hedging a portfolio
of shorting an option and purchasing ∆ = ∂V
∂S shares.
The initial and final values of this portfolio are
Initally: V (S(t), t) − ∆(S(t), t) · S(t)
Finally: V (S(T ), T ) − ∆(S(t), t) · S(T )
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
124 / 161
Option Elasticity
Proof.
If this portfolio is self-financing and arbitrage-free requirement, then
e r (T −t) V (S(t), t) − ∆(S(t), t) · S(t) = V (S(T ), T ) − ∆(S(t), t) · S(T ).
(136)
It follows that for κ := e r (T −t) ,
"
#
S(T ) − S(t)
V (S(T ), T ) − V (S(t), t)
=κ−1+
+1−κ Ω
V (S(t), t)
S(t)
"
#
"
#
V (S(T ), T ) − V (S(t), t)
S(T ) − S(t)
2
⇒ Var
= Ω Var
V (S(t), t)
S(t)
(137)
⇒ σoption = σstock × | Ω | .
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
125 / 161
Option Elasticity
If γ is the expected rate of return on an option with value V , α is the
expected rate of return on the underlying stock, and r is of course the risk
free rate, then the following equation holds:
γ · V (S, t) = α · ∆(S, t) · S + r · V (S, t) − ∆(S, t) · S .
(138)
In terms of elasticity Ω, this reduces to
Risk Premium(Option) := γ − r = (α − r )Ω.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
(139)
126 / 161
Option Elasticity
If γ is the expected rate of return on an option with value V , α is the
expected rate of return on the underlying stock, and r is of course the risk
free rate, then the following equation holds:
γ · V (S, t) = α · ∆(S, t) · S + r · V (S, t) − ∆(S, t) · S .
(138)
In terms of elasticity Ω, this reduces to
Risk Premium(Option) := γ − r = (α − r )Ω.
(139)
Furthermore, we have the Sharpe Ratio for an asset as the ratio of risk
premium to volatility:
Sharpe(Stock) =
Albert Cohen (MSU)
(α − r )
(α − r )Ω
=
= Sharpe(Call).
σ
σΩ
Financial Mathematics for Actuaries I
MSU Spring 2016
(140)
126 / 161
Option Elasticity
If γ is the expected rate of return on an option with value V , α is the
expected rate of return on the underlying stock, and r is of course the risk
free rate, then the following equation holds:
γ · V (S, t) = α · ∆(S, t) · S + r · V (S, t) − ∆(S, t) · S .
(138)
In terms of elasticity Ω, this reduces to
Risk Premium(Option) := γ − r = (α − r )Ω.
(139)
Furthermore, we have the Sharpe Ratio for an asset as the ratio of risk
premium to volatility:
(α − r )
(α − r )Ω
=
= Sharpe(Call).
(140)
σ
σΩ
HW Sharpe Ratio for a put? How about elasticity for a portfolio of
options? Now read about Calendar Spreads, Implied Volatility, and
Perpetual American Options.
Sharpe(Stock) =
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
126 / 161
Example: Hedging
Under a standard framework, assume you write a 4 − yr European Call
option a non-dividend paying stock with the following:
S0 = 10 = K
σ = 0.2
(141)
r = 0.02.
Calculate the initial number of shares of the stock for your hedging
program.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
127 / 161
Example: Hedging
Recall
∆C = e −δ(T −t) N(d1 )
ln KS + (r − δ + 12 σ 2 )(T − t)
√
d1 =
σ T −t
√
d2 = d1 − σ T − t.
(142)
∆C = N 0.4 = 0.6554.
(143)
It follows that
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
128 / 161
Example: Risk Analysis
Assume that an option is written on an asset S with the following
information:
The expected rate of return on the underlying asset is 0.10.
The expected rate of return on a riskless asset is 0.05.
The volatility on the underlying asset is 0.20.
V (S, t) = e −0.05(10−t) S 2 e S
Compute Ω(S, t) and the Sharpe Ratio for this option.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
129 / 161
Example: Risk Analysis
By definition,
∂V (S,t)
Ω(S, t) =
S · ∂S
∆·S
=
V (S, t)
V (S, t)
d
S dS
(S 2 e S )
S · (2Se S + S 2 e S )
=
(S 2 e S )
S 2e S
= 2 + S.
=
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
(144)
130 / 161
Example: Risk Analysis
By definition,
∂V (S,t)
Ω(S, t) =
S · ∂S
∆·S
=
V (S, t)
V (S, t)
d
S dS
(S 2 e S )
S · (2Se S + S 2 e S )
=
(S 2 e S )
S 2e S
= 2 + S.
=
(144)
Furthermore, since Ω = 2 + S ≥ 2, we have
Sharpe =
Albert Cohen (MSU)
0.10 − 0.05
= 0.25.
0.20
Financial Mathematics for Actuaries I
(145)
MSU Spring 2016
130 / 161
Example: Black Scholes Pricing
Consider a portfolio of options on a non-dividend paying stock S that
consists of a put and a call, both with strike K = 5 = S0 . What is the Γ
for this option as well as the option value at time 0 if the time to
expiration is T = 4, r = 0.02, σ = 0.2.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
131 / 161
Example: Black Scholes Pricing
Consider a portfolio of options on a non-dividend paying stock S that
consists of a put and a call, both with strike K = 5 = S0 . What is the Γ
for this option as well as the option value at time 0 if the time to
expiration is T = 4, r = 0.02, σ = 0.2.
In this case,
V = VC + VP
∂2 C
P
Γ=
V
+
V
= 2ΓC .
∂S 2
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
(146)
MSU Spring 2016
131 / 161
Example: Black Scholes Pricing
√
Consequently, d1 = 0.4 and d2 = 0.4 − 0.2 4 = 0, and so
V (5, 0) = V C (5, 0) + V P (5, 0)
= 5 N(d1 ) + e −4r N(−d2 ) − e −4r N(d2 ) − N(−d1 )
= 5 N(0.4) + e −4r N(0) − e −4r N(0) − N(−0.4)
(147)
= 1.5542
Γ(5, 0) =
2N 0 (0.4)
1
2
√ = N 0 (0.4) = √ e −0.5·(0.4) = 0.4322.
0.2 · 5 · 4
2π
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
132 / 161
Market Making
On a periodic basis, a Market Maker, services the option buyer by
rebalancing the portfolio designed to replicate the payoff written into the
option contract.
Define
Vi = Option Value i periods from inception
∆i = Delta required i periods from inception
(148)
∴ Pi = ∆i Si − Vi
Rebalancing at time i requires an extra (∆i+1 − ∆i ) shares.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
133 / 161
Market Making
On a periodic basis, a Market Maker, services the option buyer by
rebalancing the portfolio designed to replicate the payoff written into the
option contract.
Define
Vi = Option Value i periods from inception
∆i = Delta required i periods from inception
(149)
∴ Pi = ∆i Si − Vi = Cost of Strategy
Rebalancing at time i requires an extra (∆i+1 − ∆i ) shares.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
134 / 161
Market Making
Define
∂Si = Si+1 − Si
∂Pi = Pi+1 − Pi
(150)
∂∆i = ∆i+1 − ∆i
Then
∂Pi = Net Cash Flow = ∆i ∂Si − ∂Vi − rPi
= ∆i ∂Si − ∂Vi − r ∆i Si − Vi
(151)
Under what conditions is the Net Flow = 0?
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
135 / 161
Market Making
For a continuous rate r , we can see that if ∆ := ∂V
∂S , Pt = ∆t St − Vt ,
dV := V (St + dSt , t + dt) − V (St , t)
1
≈ Θdt + ∆ · dSt + Γ · (dSt )2
2
⇒ dPt = ∆t dSt − dVt − rPt dt
1
2
≈ ∆t dSt − Θdt + ∆ · dSt + Γ · (dSt ) − r (∆t St − Vt ) dt
2
!
1
≈ − Θdt + r (∆St − V (St , t))dt + Γ · [dSt ]2 .
2
(152)
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
136 / 161
Market Making
If dt is small, but not infinitessimally small, then on a periodic basis given
the evolution of St , the periodic jump in value from St → St + dSt may be
known exactly and correspond to a non-zero jump in Market Maker profit
dPt .
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
137 / 161
Market Making
If dt is small, but not infinitessimally small, then on a periodic basis given
the evolution of St , the periodic jump in value from St → St + dSt may be
known exactly and correspond to a non-zero jump in Market Maker profit
dPt .
If dSt · dSt = σ 2 St2 dt, then if we sample continuously and enforce a zero
net-flow, we retain the BSM PDE for all relevant (S, t):
∂V
1
∂V
∂2V
+r S
− V + σ2S 2 2 = 0
∂t
∂S
2
∂S
V (S, T ) = G (S)
for final time payoff G (S).
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
(153)
137 / 161
Note: Delta-Gamma Neutrality vs Bond Immunization
In an actuarial analysis of cashflow, a company may wish to immunize
its portfolio. This refers to the relationship between a non-zero value
for the second derivative with respect to interest rate of the
(deterministic) cashflow present value and the subsequent possibility
of a negative PV.
This is similar to the case of market maker with a non-zero Gamma.
In the market makers cash flow, a move of dS in the stock
corresponds to a move 12 Γ(dS)2 in the portfolio value.
In order to protect against large swings in the stock causing non-linear
effects in the portfolio value, the market maker may choose to offset
positions in her present holdings to maintain Gamma Neutrality or she
wish to maintain Delta Neutrality, although this is only a linear effect.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
138 / 161
Option Greeks and Analysis - Some Final Comments
It is important to note the similarities between Market Making and
Actuarial Reserving. In engineering the portfolio to replicate the
payoff written into the contract, the market maker requires capital.
The idea of Black Scholes Merton pricing is that the portfolio should
be self-financing.
One should consider how this compares with the capital required by
insurers to maintain solvency as well as the possibility of obtaining
reinsurance.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
139 / 161
Exam Practice
Consider an economy where :
The current exchange rate is x0 = 0.011 dollar
yen .
A four-year dollar-denominated European put option on yen with a
strike price of 0.008$ sells for 0.0005$.
The continuously compounded risk-free interest rate on dollars is 3%.
The continuously compounded risk-free interest rate on yen is 1.5%.
Compute the price of a 4−year dollar-denominated European call option
on yens with a strike price of 0.008$.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
140 / 161
Exam Practice
Consider an economy where :
The current exchange rate is x0 = 0.011 dollar
yen .
A four-year dollar-denominated European put option on yen with a
strike price of 0.008$ sells for 0.0005$.
The continuously compounded risk-free interest rate on dollars is 3%.
The continuously compounded risk-free interest rate on yen is 1.5%.
Compute the price of a 4−year dollar-denominated European call option
on yens with a strike price of 0.008$.
ANSWER: By put call parity, and the Black Scholes formula, with the
asset S as the exchange rate, and the foreign risk-free rate rf = δ,
V C (x0 , 0) = V P (x0 , 0) + x0 e −rf T − Ke −rT
= 0.0005 + 0.011e −0.015·4 − 0.008e −0.03·4
(154)
= 0.003764.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
140 / 161
Exam Practice
An investor purchases a 1−year, 50− strike European Call option on
a non-dividend paying stock by borrowing at the risk-free rate r .
The investor paid V C (S0 , 0) = 10.
Six months later, the investor finds out that the Call option has
increased in value by one: V C (S0.05 , 0.5) = 11.
Assume (σ, r ) = (0.2, 0.02).
Should she close out her position after 6 months?
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
141 / 161
Exam Practice
An investor purchases a 1−year, 50− strike European Call option on
a non-dividend paying stock by borrowing at the risk-free rate r .
The investor paid V C (S0 , 0) = 10.
Six months later, the investor finds out that the Call option has
increased in value by one: V C (S0.05 , 0.5) = 11.
Assume (σ, r ) = (0.2, 0.02).
Should she close out her position after 6 months?
ANSWER: Simply put, her profit if she closes out after 6 months is
1
11 − 10e 0.02 2 = 0.8995.
(155)
So, yes, she should liquidate her position.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
141 / 161
Exam Practice
Consider a 1−year at the money European Call option on a
non-dividend paying stock.
You are told that ∆C = 0.65, and the economy bears a 1% rate.
Can you estimate the volatility σ?
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
142 / 161
Exam Practice
Consider a 1−year at the money European Call option on a
non-dividend paying stock.
You are told that ∆C = 0.65, and the economy bears a 1% rate.
Can you estimate the volatility σ?
ANSWER: By definition,
∆C = e −δT N(d1 ) = N
=N
Albert Cohen (MSU)
r + 1 σ2 0.01 + 1 σ 2 2
σ
Financial Mathematics for Actuaries I
2
σ
= 0.65
MSU Spring 2016
142 / 161
Exam Practice
Consider a 1−year at the money European Call option on a
non-dividend paying stock.
You are told that ∆C = 0.65, and the economy bears a 1% rate.
Can you estimate the volatility σ?
ANSWER: By definition,
∆C = e −δT N(d1 ) = N
=N
⇒
Albert Cohen (MSU)
r + 1 σ2 0.01 + 1 σ 2 2
σ
2
σ
= 0.65
0.01 + 12 σ 2
= 0.385
σ
Financial Mathematics for Actuaries I
MSU Spring 2016
142 / 161
Exam Practice
Consider a 1−year at the money European Call option on a
non-dividend paying stock.
You are told that ∆C = 0.65, and the economy bears a 1% rate.
Can you estimate the volatility σ?
ANSWER: By definition,
∆C = e −δT N(d1 ) = N
=N
⇒
0.01 + 12 σ 2
σ
r + 1 σ2 0.01 + 1 σ 2 2
σ
2
σ
= 0.65
(156)
= 0.385
⇒ σ ∈ {0.0269, 0.7431} .
More information is needed to choose from the two roots computed above.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
142 / 161
Exam Pointers
When reviewing the material for the exam, consider the following
milestones and examples:
The definition of the Black-Scholes pricing formulae for European
puts and calls.
What are the Greeks? Given a specific option, could you compute the
Greeks?
What is the Option Elasticity? How is it useful? How about the
Sharpe ratio of an option? Can you compute the Elasticity and
Sharpe ration of a given option?
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
143 / 161
Exam Pointers
What is Delta Hedging?
If the Delta and Gamma values of an option are known, can you
calculate the change in option value given a small change in the
underlying asset value?
How does this correspond the Market Maker’s profit?
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
144 / 161
Probability Spaces - Introduction
We define the finite set of outcomes, the Sample Space, as Ω and any
subcollection of outcomes A ⊂ Ω an event.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
145 / 161
Probability Spaces - Introduction
We define the finite set of outcomes, the Sample Space, as Ω and any
subcollection of outcomes A ⊂ Ω an event.
How does this relate to the case of 2 consecutive coin flips
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
145 / 161
Probability Spaces - Introduction
We define the finite set of outcomes, the Sample Space, as Ω and any
subcollection of outcomes A ⊂ Ω an event.
How does this relate to the case of 2 consecutive coin flips
Ω ≡ {HH, HT , TH, TT }
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
145 / 161
Probability Spaces - Introduction
We define the finite set of outcomes, the Sample Space, as Ω and any
subcollection of outcomes A ⊂ Ω an event.
How does this relate to the case of 2 consecutive coin flips
Ω ≡ {HH, HT , TH, TT }
Set of events includes statements like at least one head
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
145 / 161
Probability Spaces - Introduction
We define the finite set of outcomes, the Sample Space, as Ω and any
subcollection of outcomes A ⊂ Ω an event.
How does this relate to the case of 2 consecutive coin flips
Ω ≡ {HH, HT , TH, TT }
Set of events includes statements like at least one head
= {HH, HT , TH}
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
145 / 161
Probability Spaces - Introduction
We define, ∀A, B ⊆ Ω
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
146 / 161
Probability Spaces - Introduction
We define, ∀A, B ⊆ Ω
Ac = {ω ∈ Ω : ω ∈
/ A}
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
146 / 161
Probability Spaces - Introduction
We define, ∀A, B ⊆ Ω
Ac = {ω ∈ Ω : ω ∈
/ A}
A ∩ B = {ω ∈ Ω : ω ∈ A and ω ∈ B}
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
146 / 161
Probability Spaces - Introduction
We define, ∀A, B ⊆ Ω
Ac = {ω ∈ Ω : ω ∈
/ A}
A ∩ B = {ω ∈ Ω : ω ∈ A and ω ∈ B}
A ∪ B = {ω ∈ Ω : ω ∈ A or ω ∈ B}
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
146 / 161
Probability Spaces - Introduction
We define, ∀A, B ⊆ Ω
Ac = {ω ∈ Ω : ω ∈
/ A}
A ∩ B = {ω ∈ Ω : ω ∈ A and ω ∈ B}
A ∪ B = {ω ∈ Ω : ω ∈ A or ω ∈ B}
φ as the Empty Set
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
146 / 161
Probability Spaces - Introduction
We define, ∀A, B ⊆ Ω
Ac = {ω ∈ Ω : ω ∈
/ A}
A ∩ B = {ω ∈ Ω : ω ∈ A and ω ∈ B}
A ∪ B = {ω ∈ Ω : ω ∈ A or ω ∈ B}
φ as the Empty Set
A, B to be Mutually Exclusive or Disjoint if A ∩ B = φ
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
146 / 161
σ−algebras
Given a non-empty set Ω of outcomes, a σ−algebra F is a collection of
subsets of Ω that satisfies
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
147 / 161
σ−algebras
Given a non-empty set Ω of outcomes, a σ−algebra F is a collection of
subsets of Ω that satisfies
∅∈F
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
147 / 161
σ−algebras
Given a non-empty set Ω of outcomes, a σ−algebra F is a collection of
subsets of Ω that satisfies
∅∈F
A ∈ F ⇒ Ac ∈ F
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
147 / 161
σ−algebras
Given a non-empty set Ω of outcomes, a σ−algebra F is a collection of
subsets of Ω that satisfies
∅∈F
A ∈ F ⇒ Ac ∈ F
A1 , A2 , A3 , .... ∈ F ⇒ ∪∞
n=1 An ∈ F
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
147 / 161
σ−algebras
Given a non-empty set Ω of outcomes, a σ−algebra F is a collection of
subsets of Ω that satisfies
∅∈F
A ∈ F ⇒ Ac ∈ F
A1 , A2 , A3 , .... ∈ F ⇒ ∪∞
n=1 An ∈ F
Some Examples
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
147 / 161
σ−algebras
Given a non-empty set Ω of outcomes, a σ−algebra F is a collection of
subsets of Ω that satisfies
∅∈F
A ∈ F ⇒ Ac ∈ F
A1 , A2 , A3 , .... ∈ F ⇒ ∪∞
n=1 An ∈ F
Some Examples
F0 = {∅, Ω}
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
147 / 161
σ−algebras
Given a non-empty set Ω of outcomes, a σ−algebra F is a collection of
subsets of Ω that satisfies
∅∈F
A ∈ F ⇒ Ac ∈ F
A1 , A2 , A3 , .... ∈ F ⇒ ∪∞
n=1 An ∈ F
Some Examples
F0 = {∅, Ω}
F1 = {∅, Ω, {HH, HT }, {TT , TH}}
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
147 / 161
σ−algebras
Given a non-empty set Ω of outcomes, a σ−algebra F is a collection of
subsets of Ω that satisfies
∅∈F
A ∈ F ⇒ Ac ∈ F
A1 , A2 , A3 , .... ∈ F ⇒ ∪∞
n=1 An ∈ F
Some Examples
F0 = {∅, Ω}
F1 = {∅, Ω, {HH, HT }, {TT , TH}}
F2 = {∅, Ω, {HH}, {HT }, {TT }, {TH}, ....}
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σ−algebras
Given a non-empty set Ω of outcomes, a σ−algebra F is a collection of
subsets of Ω that satisfies
∅∈F
A ∈ F ⇒ Ac ∈ F
A1 , A2 , A3 , .... ∈ F ⇒ ∪∞
n=1 An ∈ F
Some Examples
F0 = {∅, Ω}
F1 = {∅, Ω, {HH, HT }, {TT , TH}}
F2 = {∅, Ω, {HH}, {HT }, {TT }, {TH}, ....}
F2 is completed by taking all unions of ∅, Ω, {HH}, {HT }, {TT }, {TH}.
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Power Sets
As a useful example, how many subsets are there of a set containing
n elements?
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Power Sets
As a useful example, how many subsets are there of a set containing
n elements?
Answer: 2n
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Power Sets
As a useful example, how many subsets are there of a set containing
n elements?
Answer: 2n
Proof:
Albert Cohen (MSU)
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Power Sets
As a useful example, how many subsets are there of a set containing
n elements?
Answer: 2n
Proof:
Consider strings of length n where the elements are either 0 or 1....
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Notice that F0 ⊂ F1 ⊂ F2 .
Correspondingly, given an Ω, we define a
Albert Cohen (MSU)
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Notice that F0 ⊂ F1 ⊂ F2 .
Correspondingly, given an Ω, we define a
Filtration as ..
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Notice that F0 ⊂ F1 ⊂ F2 .
Correspondingly, given an Ω, we define a
Filtration as ..
a sequence of σ−algebras F0 , F1 , F2 , ..., Fn , ... such that
Albert Cohen (MSU)
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Notice that F0 ⊂ F1 ⊂ F2 .
Correspondingly, given an Ω, we define a
Filtration as ..
a sequence of σ−algebras F0 , F1 , F2 , ..., Fn , ... such that
F0 ⊂ F1 ⊂ F2 ⊂ ... ⊂ Fn ⊂ ...
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Notice that F0 ⊂ F1 ⊂ F2 .
Correspondingly, given an Ω, we define a
Filtration as ..
a sequence of σ−algebras F0 , F1 , F2 , ..., Fn , ... such that
F0 ⊂ F1 ⊂ F2 ⊂ ... ⊂ Fn ⊂ ...
and F = σ(Ω) as the σ−algebra of all subsets of Ω.
Albert Cohen (MSU)
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Notice that F0 ⊂ F1 ⊂ F2 .
Correspondingly, given an Ω, we define a
Filtration as ..
a sequence of σ−algebras F0 , F1 , F2 , ..., Fn , ... such that
F0 ⊂ F1 ⊂ F2 ⊂ ... ⊂ Fn ⊂ ...
and F = σ(Ω) as the σ−algebra of all subsets of Ω.
Given a pair (Ω, F), we define a Random Variable X (ω) as a mapping
X :Ω→R
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Given a pair (Ω, F), we define a Probability Space as the triple
(Ω, F, P), where
P : F → [0, 1]
P[∅] = 0
For any countable disjoint sets A1 , A2 , ... ∈ F
P [∪∞
n=1 An ] =
Albert Cohen (MSU)
P∞
n=1 P[An ]
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And so
P[A] :=
P
P[ω]
Pω∈A
P
E[X ] := ω X (ω)P[ω] = nk=1 xk P[{X (ω) = xk }]
h
i
with Variance := E (X − E[X ])2
Some useful properties:
P[Ac ] = 1 − P[A]
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And so
P[A] :=
P
P[ω]
Pω∈A
P
E[X ] := ω X (ω)P[ω] = nk=1 xk P[{X (ω) = xk }]
h
i
with Variance := E (X − E[X ])2
Some useful properties:
P[Ac ] = 1 − P[A]
P[A] ≤ 1
Albert Cohen (MSU)
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MSU Spring 2016
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And so
P[A] :=
P
P[ω]
Pω∈A
P
E[X ] := ω X (ω)P[ω] = nk=1 xk P[{X (ω) = xk }]
h
i
with Variance := E (X − E[X ])2
Some useful properties:
P[Ac ] = 1 − P[A]
P[A] ≤ 1
P[A ∪ B] = P[A] + P[B] − P[A ∩ B]
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And so
P[A] :=
P
P[ω]
Pω∈A
P
E[X ] := ω X (ω)P[ω] = nk=1 xk P[{X (ω) = xk }]
h
i
with Variance := E (X − E[X ])2
Some useful properties:
P[Ac ] = 1 − P[A]
P[A] ≤ 1
P[A ∪ B] = P[A] + P[B] − P[A ∩ B]
P[A ∪ B ∪ C ] =
P[A] + P[B] + P[C ] − P[A ∩ B] − P[A ∩ C ] − P[B ∩ C ] + P[A ∩ B ∩ C ]
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MSU Spring 2016
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Example
Consider the case where two dice are rolled separately. What is the Sample
Space Ω here? How about the probability that the dots on the faces of the
pair add up to 3 or 4 or 5?
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
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How Should We Count?
Think about the following problems:
How many pairings of aliens and humans, (aliens, human), can we
have if we can choose from 8 aliens and 9 people?
How many different strings of length 5 can we expect to find of 00s
and 10s ?
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Examples
From a group of 3 aliens and 5 humans, how many alien-people
councils can be formed with 2 of each on the board?
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MSU Spring 2016
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Examples
From a group of 3 aliens and 5 humans, how many alien-people
councils can be formed with 2 of each on the board?
What if two of the humans refuse to serve together?
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Financial Mathematics for Actuaries I
MSU Spring 2016
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Permutations
In general, we are interested in the number of different ways, or
Combinations of ways r objects could be grouped when selected from
a
pool of n total objects. Notationally, for r ≤ n we define this as nr and
the formula can be shown to be
n
n!
=
(n − r )!r !
r
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
155 / 161
Permutations
In general, we are interested in the number of different ways, or
Combinations of ways r objects could be grouped when selected from
a
pool of n total objects. Notationally, for r ≤ n we define this as nr and
the formula can be shown to be
n
n!
=
(n − r )!r !
r
n! = 1 · 2 · ... · n
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
155 / 161
Permutations
In general, we are interested in the number of different ways, or
Combinations of ways r objects could be grouped when selected from
a
pool of n total objects. Notationally, for r ≤ n we define this as nr and
the formula can be shown to be
n
n!
=
(n − r )!r !
r
n! = 1 · 2 · ... · n
(157)
n−1
n−1
n
=
+
r −1
r
r
If order matters when selecting the r objects, then we define the number
of Permutations
n
n!
Pk,n = r ! ·
=
(158)
r
(n − r )!
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
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What if another event has already occured?
Consider the case where two events in our σ−field are under consideration.
In fact, we know that B has already happened. How does that affect the
chances of A happening?
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
156 / 161
What if another event has already occured?
Consider the case where two events in our σ−field are under consideration.
In fact, we know that B has already happened. How does that affect the
chances of A happening?
For example, if you know your friend has one boy, and the chance of a boy
or girl is equal at 0.5, then what is the chance all three of his children are
boys?
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
156 / 161
What if another event has already occured?
In symbols, we seek
P [A | B] ≡
Albert Cohen (MSU)
P [A ∩ B]
P [B]
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What if another event has already occured?
In symbols, we seek
P [A ∩ B]
P [B]
P [ all are boys ∩ first child is a boy]
=
P [ first child is a boy]
P [A | B] ≡
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
157 / 161
What if another event has already occured?
In symbols, we seek
P [A ∩ B]
P [B]
P [ all are boys ∩ first child is a boy]
=
P [ first child is a boy]
P [ all are boys]
=
P [ first child is a boy]
P [A | B] ≡
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
157 / 161
What if another event has already occured?
In symbols, we seek
P [A ∩ B]
P [B]
P [ all are boys ∩ first child is a boy]
=
P [ first child is a boy]
P [ all are boys]
=
P [ first child is a boy]
1 1 1
· ·
= 2 21 2
P [A | B] ≡
2
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
157 / 161
What if another event has already occured?
In symbols, we seek
P [A ∩ B]
P [B]
P [ all are boys ∩ first child is a boy]
=
P [ first child is a boy]
P [ all are boys]
=
P [ first child is a boy]
1 1 1
· ·
= 2 21 2
P [A | B] ≡
(159)
2
1
=
4
Albert Cohen (MSU)
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MSU Spring 2016
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What if another event has already occured?
Now, what if you know your friend has at least one boy. Then what is the
chance all three of his children are boys? This is also known as The
Boy-Girl Paradox.
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
158 / 161
What if another event has already occured?
Now, what if you know your friend has at least one boy. Then what is the
chance all three of his children are boys? This is also known as The
Boy-Girl Paradox.
P [A | C ] ≡
Albert Cohen (MSU)
P [A ∩ C ]
P [C ]
Financial Mathematics for Actuaries I
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What if another event has already occured?
Now, what if you know your friend has at least one boy. Then what is the
chance all three of his children are boys? This is also known as The
Boy-Girl Paradox.
P [A ∩ C ]
P [C ]
P [ all are boys ∩ at least one child is a boy]
=
P [at least one child is a boy]
P [A | C ] ≡
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
158 / 161
What if another event has already occured?
Now, what if you know your friend has at least one boy. Then what is the
chance all three of his children are boys? This is also known as The
Boy-Girl Paradox.
P [A ∩ C ]
P [C ]
P [ all are boys ∩ at least one child is a boy]
=
P [at least one child is a boy]
P [ all are boys]
=
P [ at least one child is a boy]
P [A | C ] ≡
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
158 / 161
What if another event has already occured?
Now, what if you know your friend has at least one boy. Then what is the
chance all three of his children are boys? This is also known as The
Boy-Girl Paradox.
P [A ∩ C ]
P [C ]
P [ all are boys ∩ at least one child is a boy]
=
P [at least one child is a boy]
P [ all are boys]
=
P [ at least one child is a boy]
P [ all are boys]
=
1 − P [all girls]
P [A | C ] ≡
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
158 / 161
What if another event has already occured?
Now, what if you know your friend has at least one boy. Then what is the
chance all three of his children are boys? This is also known as The
Boy-Girl Paradox.
P [A ∩ C ]
P [C ]
P [ all are boys ∩ at least one child is a boy]
=
P [at least one child is a boy]
P [ all are boys]
=
P [ at least one child is a boy]
P [ all are boys]
=
1 − P [all girls]
1 1 1
· ·
= 2 2 12
1− 8
P [A | C ] ≡
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
158 / 161
What if another event has already occured?
Now, what if you know your friend has at least one boy. Then what is the
chance all three of his children are boys? This is also known as The
Boy-Girl Paradox.
P [A ∩ C ]
P [C ]
P [ all are boys ∩ at least one child is a boy]
=
P [at least one child is a boy]
P [ all are boys]
=
P [ at least one child is a boy]
P [ all are boys]
=
1 − P [all girls]
1 1 1
· ·
1
= 2 2 12 =
7
1− 8
P [A | C ] ≡
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
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158 / 161
Total Probability
Notice from our definition of conditional probabiltity that
P[A ∩ B] = P[A | B] · P[B]
Albert Cohen (MSU)
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Total Probability
Notice from our definition of conditional probabiltity that
P[A ∩ B] = P[A | B] · P[B]
(161)
We can expand on this idea: For our sample space Ω, assume we have a
set {A1 , ..., An } where the members are
mutually exclusive - Ai ∩ Aj = φ for all i 6= j
exhaustive - A1 ∪ A2 ∪ ... ∪ An = Ω .
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
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Total Probability
Notice from our definition of conditional probabiltity that
P[A ∩ B] = P[A | B] · P[B]
(161)
We can expand on this idea: For our sample space Ω, assume we have a
set {A1 , ..., An } where the members are
mutually exclusive - Ai ∩ Aj = φ for all i 6= j
exhaustive - A1 ∪ A2 ∪ ... ∪ An = Ω .
Then
P[B] = P[B ∩ A1 ] + P[B ∩ A2 ] + ... + P[B ∩ An ]
Albert Cohen (MSU)
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MSU Spring 2016
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Total Probability
Notice from our definition of conditional probabiltity that
P[A ∩ B] = P[A | B] · P[B]
(161)
We can expand on this idea: For our sample space Ω, assume we have a
set {A1 , ..., An } where the members are
mutually exclusive - Ai ∩ Aj = φ for all i 6= j
exhaustive - A1 ∪ A2 ∪ ... ∪ An = Ω .
Then
P[B] = P[B ∩ A1 ] + P[B ∩ A2 ] + ... + P[B ∩ An ]
= P[B | A1 ]P[A1 ] + P[B | A2 ]P[A2 ] + ... + P[B | An ]P[An ]
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
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159 / 161
Bayes Theorem
For our sample space Ω, assume we have an exhaustive set of events
{A1 , ..., An } with prior probabilities P[Ai ] for i = 1, .., n. Then for any
other event B in our σ−field where P[B] > 0, the posterior probabilty of
Aj given that B has occured is
Albert Cohen (MSU)
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MSU Spring 2016
160 / 161
Bayes Theorem
For our sample space Ω, assume we have an exhaustive set of events
{A1 , ..., An } with prior probabilities P[Ai ] for i = 1, .., n. Then for any
other event B in our σ−field where P[B] > 0, the posterior probabilty of
Aj given that B has occured is
P [Aj | B] =
Albert Cohen (MSU)
P [Aj ∩ B]
=
P [B]
Financial Mathematics for Actuaries I
MSU Spring 2016
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Bayes Theorem
For our sample space Ω, assume we have an exhaustive set of events
{A1 , ..., An } with prior probabilities P[Ai ] for i = 1, .., n. Then for any
other event B in our σ−field where P[B] > 0, the posterior probabilty of
Aj given that B has occured is
P [Aj | B] =
Albert Cohen (MSU)
P [B ∩ Aj ]
P [Aj ∩ B]
=
P [B]
P [B]
Financial Mathematics for Actuaries I
MSU Spring 2016
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Bayes Theorem
For our sample space Ω, assume we have an exhaustive set of events
{A1 , ..., An } with prior probabilities P[Ai ] for i = 1, .., n. Then for any
other event B in our σ−field where P[B] > 0, the posterior probabilty of
Aj given that B has occured is
P [B ∩ Aj ]
P [Aj ∩ B]
=
P [B]
P [B]
P[B | Aj ] · P[Aj ]
=
P[B | A1 ]P[A1 ] + ... + P[B | An ]P[An ]
P [Aj | B] =
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Independent Events
For our sample space Ω, assume we have two events A, B. We say that A
and B are independent if P[A | B] = P[A] and dependent if
P[A | B] 6= P[A]
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
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Independent Events
For our sample space Ω, assume we have two events A, B. We say that A
and B are independent if P[A | B] = P[A] and dependent if
P[A | B] 6= P[A]
In other words, A and B are independent if and only if
P[A ∩ B] = P[A] · P[B].
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
MSU Spring 2016
161 / 161
Independent Events
For our sample space Ω, assume we have two events A, B. We say that A
and B are independent if P[A | B] = P[A] and dependent if
P[A | B] 6= P[A]
In other words, A and B are independent if and only if
P[A ∩ B] = P[A] · P[B].
Generally speaking, for any collection of events {A1 , ..., An } we have for
any subcollection {Ai1 , ..., Ain } ⊆ {A1 , ..., An }
P[Ai1 ∩ ... ∩ Ain ] = P[Ai1 ] · .. · ...P[Ain ]
Albert Cohen (MSU)
Financial Mathematics for Actuaries I
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161 / 161