Prof. Dr. Dr. h.c. Dirk Linowski
Steinbeis University Berlin, Germany
Shanghai Normal University, P.R. of China
Department of Mathematics
linowski@stw.de
Valuation of Financial Options: Part I
Structure:
-
The arbitrage principle
-
Types of derivatives
-
Valuation methods
o Binomial trees
o Black-Scholes formulaes
o Application on currencies
-
Dynamic measures and Hedging strategies
- Application of these concepts on real Options and thus on firms‘ valuations
Literature: (incl. software between 50€ - 70€)
Hull: Options, Futures and other Derivatives
Smithson: Managing Financial Risk
McDonald: Derivatives Markets
Cox / Rubinstein: Options Markets
The arbitrage principle
Arbitrage is a technical term for a risk free profit: Our assumption is that financial
markets are no-arbitrage markets. As long as possibilities exist to make profit without
risk, people will use and let the spread disappear.
We suppose thus there is arbitrage possibility
1
Case a) DCX (Daimler shares): FFM 32,05 Sale
Xetra: 32,00 Purchase
Quotations approach. The spread narrows.
Case b) Gold: Price in USD 590
In Europe: 470 €
Exchange rate 1,28USD/€
Equivalent to 461 Euro
Thus purchase in America, sale Europe
Case c) Money Market Approach
Porsche has revenues of USD 1.000 million in exactly 1 year
t=0
t =1
+1.000 USD (Sales)
956.94 USD
iUS = 4.5%
- 1.000 USD (loan)
1€ = 1,28USD
=0
747.61 €
Futures on currencies: You can sell and purchase the major currencies forward (up to
years).
t=0
956.94 USD
t=1
1.045 (4.5%)
1.28USD = 1€ (spot price)
747.61 €
1.03 (3%)
1000 USD
1.29864 USD/€
770.04 €
Spot exchange rate: sUSD/€ = 1.28
Futures rate in 1 year: f(1) USD/€ = 1.29864
2
Suppose now f(1) USD/€ = 1.29. You sell forward (in 1 year) 1.000 €. You deliver in 1
year thus 1,000 € and get 1,290 USD in 1 year. You obtain thus “too less” USD, i.e.
the € is forward too cheap or the USD is forward too expensive.
Possibility a) Buy € forward,
Possibility b) Sell $ forward.
Look on b)
t=0
t=1
+1.000.000 €
Future
- 1,290,000 USD
i € = 0.03,
+970,874 €
i$ = 0.045
- 1,000,000€ (Payback of bank loan in €)
- 970,874€
+1,290,000USD (account in USD)
+1,242,718 USD
=0
- 1,234,450 USD
1.290.000/1,045
= 8,269 USD = 6.460 €
(Exchange at spot price: 1.28 USD/€)
The biggest volumes of all markets are on currencies: One change affects all other
ratios!
€/USD
1.28
€/¥
140
€/£
0.69 and
For instance: £/$ = 1/0.69*1,28 = 1.855. This must be the exchange rate. Otherwise
is arbitrage possible!
Suppose now the € rises on 1.30 USD ?! What is the impact of this change?
d) Other cases: Arbitrage in expectations (CAPM or APT valuation based)
3
Options I:
In contrast to futures, options are rights for the holder, not obligations.
Applications in
-
Hedging
-
Speculation
-
Arbitrage
Call = Right to buy a well-defined quantity of a well-defined asset at (until) a defined
future date for a given price. Example: Call on Daimler; maturity 30.06.2009;
Exercise price 35 €.
Put = Right to sell a well-defined quantity of a well-defined asset at (until) a defined
future date for a given price.
American style: Exercise possible at any time until maturity
European style: Exercise possible only at maturity.
What is the value of these right?
At maturity:
CT = max (0, ST-E)
PT = max (0, E - ST)
S 0 = 32 .
ST1 = 20, E = 30 CT = 0, PT = 10
ST2 = 40, E = 30 CT = 10, PT = 0
American Put: We can sell a DCX-share for 35 € until 30/12/2005. Thus holds P>3 =
35-32.
Call:
CT = max (0, ST – E)
4
Seen from the buyer or holder.
T – Maturity, option’s life’s end: Here 30.12.2005
ST – Spot price at maturity, here DCX
E – Exercise price, here 35 €
30
S
Put:
35
PT = max (0, E - ST)
35
In these graphs, or pay-off diagrams, no prices to be paid today were considered.
E
Co
E + C0
Put-Call-Parity for European options ∀t ≤ T : Pt – Ct + St = E / (1+i)T-t
At maturity, i.e. t = T, there are two possibilities:
(I)
ST > E : Then do we have
max( 0, E − ST ) − max( 0, ST − E ) = 0 − ( ST − E ) + ST = E
(II)
ST < E : E − ST + ST = E .
Often, 1/(1+i)T-t is replaced by e-i(T-t) (continuously compounded return)
5
Variables influencing the values of put and call options
Variable
Call
Put
S+
+
-
E+
-
+
Time to maturity T - t +
+
?
Risk-free interest rate i+
+
-
Volatility σ +
+
+
Dividends +
-
+
Classification of options depending on the relation of actual price of the
underlying and exercise price
Call
Put
S>E
in the money
out of the money
S~E
at the money
at the money
S<E
Out of the money
in the money
C
S
E
At the money
Out of the money
In the money
A Call which is in the money has a value which is the sum of inner (or intrinsic) value
(S – E) and time value. Time value decreases with the option nearing to maturity.
6
Beside Put-Call-parities, the most important formulae in option pricing is BlackScholes formulae to value a European call on a stock paying no dividends before the
option’s maturity.
ln
C ( S , E , T − t , σ , i ) = SΦ (d 1 ) − e − i (T − t ) EΦ (d 2 ); d 1 =
Φ(d ) =
1
d
∫
2π − ∞
−
S
σ2
+(
+ i ) ⋅ (T − t )
E
2
; d 2 = d1 − σ T − t
σ T −t
x2
e 2 dx is the probability distribution function of the Standard Normal
Distribution.
Here, interest rate and volatility are assumed to be constant over time. To compute
the Black-Scholes formulae for valuing a European put option with the same
underlying, same maturity, same exercise price, we can use Put-Call-Parity!
P – C + S = e-i(T-t) E; C = S Φ(d1) - e-i(T-t) E Φ(d2)
Pt = e-i(T-t) E – St + S Φ(d1) - e-i(T-t) E Φ(d2)
= S (Φ(d1)-1) + e-i(T-t) E (1-Φ(d2))
How does the Black-Scholes formula come up?
We show here the derivation about the binomial model. Other models, especially
using partial stochastic differential equations, are considered later. We start with a 1period model.
Example: Given a stock with today’s price of 250 €. At the end of a period, we
assume two possible outcomes, namely
7
Su(1) = 400, Sd(1) = 200
(u = up, d = down).
Suppose further the exercise price of a call maturing at the end of a period is E =
250.
SU(1) = 400 = u S(0) = 1.6*250
(u = 1.6)
Sd(1) = 200 = d S(0) = 0.8*250
(d = 0.8)
S(0) = 250
t=0
t=1
CU = max(0,U S(0) - E) = max(0,400-250) = 150
?=C
Cd = max(0,d S(0) - E) = max(0,200-250) = 0
Question: What is the value of this call today, given a risk-free asset with a return of
12%, if E = 250 and, alternatively, S can quote to either 400 or 200 at the end of the
period? We did not mention any probabilities of the two future outcomes: Let’s make
a first attempt: Suppose the probability for an upward movement is estimated to be
30 percent. Consequently, the probability for a downward movement is 70 percent.
The discounted expected cash flow is thus:
C0 =
p ⋅ Cu + (1 − p) ⋅ Cd 0.3⋅150+ 0.7 ⋅ 0 45
=
=
= 40.18
(1 + i)
1.12
1.12
Suppose the following:
Action
t=0
t =1
Up (S = 400)
Down (S =200)
Purchase 100 Calls - 4.018
+ 15.000
0
Sale of 75 shares
+ 18.750
- 30.000
- 15.000
Account
-13.393
+ 15.000
+15.000
Sum
+1.339
0
0
8
We assume here we hold enough shares. We purchase today 100 calls for 40.18
each, sell 75 shares for today’s price of 250 each. At the end of the period, we will
either exercise the Call (if S = 400) or forget about it if there is no inner value.
Further, we repurchase the 75 shares in t = 1, independent on the outcome. To get a
zero cash-flow in t = 1, we need in both cases 15.000. This will be the payment of the
bank, if we invest today at the risk free interest rate. To get 15.000 after 1 period, we
need today 13.393 (with interest rate 12 %) to be put on a bank account. However,
our call must be wrongly priced: It can not work so because we can make a risk free
profit in t = 0! Consequently, the call is too cheap!
This leads us to the idea of pricing options: To combine the risk-free asset with a
number of shares such that the option’s payoff-profile is duplicated!
Stock:
Call:
400
250
150
Diff. 200
?
200
Difference is 150
0
In our example, we need ¾ of a share to duplicate a call (or 75 shares for 100 calls).
1,6 * ¾ * 250 = 300
187,5 = ¾ * 250
Difference is 150
0,8 * ¾ * 250 = 150
If we can further take a risk free position of 150/1.12. Our portfolio position will be
t=0
t=1
¾ *u * S – 150 = 150
¾ * S – 150/1,12
¾ *d * S – 150 = 0
9
This is he same payoff at t = 1 like for the option.
Action
t=0
t =1
Up (S = 400)
Purchase a Call
-C
Down (S =200)
+ 150
0
Sale of 0.75 shares + 187.5
- 300
- 150
Account
150/1.12 = 133.93
+ 150
+150
Sum
0
0
0
The sum of the columns must be equal to zero, too. Consequently, C=53.57. This is
the „fair” price in an arbitrage free market under the assumptions made above.
Su(1) = 400
S0 = 250
Sa(1) = 200
C0 = 53.57 with a risk-free interest rate of 12 %, u =1.6; d = 0.8, E = 250.
Derivation of Black-Scholes formulae for a European Call with no dividends
until maturity
The following assumptions hold:
a) Changes in quotations are not continuously but happen at discrete times (like
in practice)
b) The price of a stock can take only two different values at any time (that is often
the case but not always)
10
c)
Su(1)
S0
SU
closer to reality:
Sd(1)
S0
Sd
SM
Either bid or ask quote ist usually he next price. Sometimes are averages between bid and
a ask quotation the the new price.
d) There are neither taxes nor transaction cost are to be taken into account.
e) Assets are divisible until infinity, there are limitations of short sales
f) The risk-free interest rate is constant over time, the rates of borrowers and
lenders are equal.
g) There is only one possibility to exercise the call
h) There are no dividends to be paid until maturity of the call
i) It holds u > 1 + i > , otherwise is arbitrage possible
In the example, we had u = 1.6; 1+i = 1.12; d = 0.8.
Suppose u > d > 1 + i ( “Take loans until infinity to purchase shares)
Suppose 1 + i > u > d ( Sell shares (short) and invest in the risk-free asset)
j) The duplicating portfolio consists from ∆ parts of a share and the risk free
amount B
t=0
t=1
∆uS - rB
∆S - B
∆dS - rB
For duplication we must choose ∆ and B such that
∆uS – rB = Cu
∆dS – rB = Cd
⇒ ∆=
Cu − Cd
d ⋅ Cu − u ⋅ Cd
; B=
(u − d ) ⋅ S
(u − d ) ⋅ r
In the example we have
11
∆=
150 − 0
0,8 ⋅ 150 − 1,6 ⋅ 0
= 0.75; B =
= 133.93
(1.6 − 0.8) ⋅ 250
(1.6 − 0.8) ⋅ 1.12
∆ is also called Hedge-Ratio: The bank would purchase 75 shares if it sells
r −d
u−r
⋅ Cu +
⋅ Cd
Cu − Cd
dCu − uCd (Cu − Cd ) ⋅ r − dCu + uCd u − d
u
−
d
∆S − B =
⋅S =
=
=
r
(u − d ) ⋅ S
(u − d ) ⋅ r
(u − d ) ⋅ r
100 calls to a customer to generate a risk-free position.
We define
p :=
r−d
r−d
⇒ 1− p =1−
u−d
u−d
r−d u−r
+
=1
u−d u−d
and obtain
In our example we have
1 .12 − 0 .8
1 .6 − 1 .12
⋅ 150 +
⋅0
−
1
.
6
0
.
8
1
.
6
0
.
8
−
= 53 .57
C=
1 .12
p und (1 – p) are called “Pseudo-probabilities“!
2-period-model:
In these models, we use recombining trees, i.e. u, d and the step length are
assumed to be constant.
12
u2 S
uS
S
udS
dS
d2 S
Cuu
Cu
C
Cud
Cd
Cdd
Computation happens backward:
Cu =
p ⋅ C uu + (1 − p ) ⋅ C ud
p ⋅ C ud + (1 − p ) ⋅ C dd
; Cd =
r
r
640
400
250
320
200
160
At T = t, there are three values possible:
- max (u2S-E,0) = 640 – 250 = 390
- max (udS-E,0) = 320 – 250 = 70
- max (d2S-E,0) = 0
p ⋅ [ p ⋅ Cuu + (1 − p ) ⋅ Cud ] (1 − p ) ⋅ [ p ⋅ Cud + (1 − p ) ⋅ Cdd ]
+
r
r
C=
r
p 2Cuu + 2 p(1 − p ) ⋅ Cud + (1 − p ) 2 ⋅ Cdd
=
r2
13
Cu =
0.4 ⋅ 390 + 0.6 ⋅ 70
0.4 ⋅ 70 + 0.6 ⋅ 0
= 25;
= 176.78; Cd =
1.12
1.12
0.4 2 ⋅ 390 + 2 ⋅ 0.4 ⋅ 0.6 ⋅ 25 + 0.62 ⋅ 0
= 76.5
C=
1.12
Remark: In t = 1, the portfolio must be restructured to keep it risk free. The change of
∆ describes how many shares must be sold or purchased at t = 1. We discuss this in
more detail when speaking about hedging strategies.
Excursus: Valuing Real Options modelled by Binomial trees
3-period model
1024
Cuuu = max(0,1024 - 250) = 774
512
Cuud = max(0,512 - 250) = 262
256
Cudd = max(0,256 - 250) = 6
128
Cddd = 0
640
400
250
320
200
160
C = 1 / (r3) * [p3 Cuuu + 3 p2 (1-p) Cuud + 3p(1-p) 2 Cudd + (1-p)3 Cddd]
3
0
3
1
3
2
3
3
1 3
3!
= 3 ∑
⋅ p j (1 − p ) 3− j max(ujd3-jS - E, 0)
r j = 0 j! (3 − j )!
14
For N steps:
1 N
N!
C= N ∑
⋅ p j (1 − p ) N − j max(ujdN-jS - E, 0)
r j = 0 j! ( N − j )!
There is a minimum number of upward movements necessary such that the call can
be exercised at maturity. In other words, the call needs a minimum number of upward
movements to have an inner value at t = T. All knots with a lower number of upmoves are equal to zero at t = T.
N N
C = S ⋅ ∑   ⋅ p
j = a j 
j
j N− j
N− j u d
(1 − p )
−
N
r
E N N j
  p (1 − p ) N − j
N ∑ j 
r j = a 
Black-Scholes formulae was
C ( S , E , T − t , σ , i ) = SΦ ( d1 ) − e −i (T − t ) EΦ ( d 2 );
S σ2
ln + (
+ i ) ⋅ (T − t )
E
2
d1 =
; d 2 = d1 − σ T − t
σ T −t
If we look on our binomial model, we face a complementary binomial distribution:
Φ (d : µ , σ 2 ) =
1
2
d (x−µ)
2
e 2σ dx
∫
2πσ 2 − ∞
Complementary value is 1 - Φ(d) = Φcomp(d). We look thus on the probability over d to
+ infinity.
Name a = minimum number of upward movements such that C is strictly positive at
maturity:
15
That is equivalent to uadN- a S > E.
p⋅u
r
N N
E N N j
j
N− j
C = S ⋅ ∑   ⋅ ( p ' ) (1 − p )
− N ∑   p (1 − p ) N − j
r j = a j 
j = a j 
Define p ' :=
p j (1 − p ) N − j u j d N − j
= ( p ' ) j (1 − p ' ) N − j
N
r
Assertion:
We rewrite the left side:
N− j
p ju j
r−d u
N− j d
(1 − p )
⋅ N − j ; 1 − p' = 1 −
⋅
j
u−d r
r
r
p ju j
= ( p' ) j
rj
r−d
=p
u−d
It remains to show that
(1 − p ) N − j
dN− j
r
(1 − p )
N− j
= (1 − p ) N − j
d
r − d d u − d − r + d d ud − rd (u − r )d
= (1 −
)⋅ =
⋅ =
=
r
u−d r
u−d
r ur − dr (u − d ) r
Bcomp is used to name the complementary binomial distribution function. We have
three types of parameters: the number of steps (N), the pseudo-probability (p or p’),
the number of necessary upward steps (a).
C = S Bcomp(a, N,p’) – E / rN Bcomp(a, N,p)
What is the value of a? Look on
S u a dN-a > E (u / d)a > E /(SdN)
16
E
E
ln
N
N
u
E
a ⋅ ln ≥ ln N ⇒ a ≥ Sd ⇒ a ' = Sd
u
u
d
Sd
ln
ln
d
d
ln
a’ is the integer up-rounded value of a.
It remains to show that Bcomp(a,N,p´) Φ(d1) and Bcamp(a´,N,p) Φ(d2)
This, however, will not be done here.
Interpretation of Black-Scholes-Formulae
The first term is the product of the discounted expected value of the stock’s price at
maturity, under condition that the price of the stock is bigger than the exercise price
at maturity and the probability that the price of the stock exceeds the exercise price at
maturity. The second term is the product of the discounted exercise price and the
probability that the stock price exceeds the exercise price at maturity. The value of
Black-Scholes-formulae for a call option is in the interval between 0 (for options very
far out of the money) and the difference between the stock price and the discounted
exercise price. (for calls being very deep in the money).
Extreme case 1:
S T << E
ln
⇒
S
σ 2
ln
+ (
+ i ) ⋅ (T − t )
E
2
d1 =
;
σ T − t
S
→ −∞ ; Φ ( −∞ ) → 0
E
Extreme case 2:
17
S T >> E ⇒ d1 =
ln
S
σ2
+(
+ i ) ⋅ (T − t )
S
E
2
; ln → ∞; Φ ( ∞) → 1
E
σ T −t
Numerical example for a European Call and Put:
Time to maturity: 3 months; risk-free interest rate 10%; E = 200; S = 190;
σ = 40% = 0,4
190 0.42
ln
+(
+ 0.1) ⋅ 0,25
200
2
d1 =
= 0.1818;
0,4 0,25
d 2 = d1 − σ T − t = 0.1818 − 0,4 ⋅ 0,25 = −0.0182
Note: Volatility time to maturity and interest rates have to put in the formulae on an
annual base!
Φ(d1) = 0.575
Φ(d2) = 0.493 C ≈ 10
Interpretation: We pay 10 today for acquiring the right to purchase an asset costing
190 today in 3 months for 200. From a speculator’s perspective, our stock must rise
about 10 percent in these three months to have no loss.
C
1. Leverage
2. Elasticity
Elasticity =
relative change of depending var iable
relative change of independent var iable
10
E = 210
E = 200
Sales price
∆C
Leverage
18
∆C C1 − C0
∆C S
∂C S
C0
ΩC = C =
=
⋅ ⇒
⋅
∆S
S1 − S0 ∆S C
∂S C
S
S0
To compute ∆, we apply chain and product rule of derivations
∂ C ∂ ( S ⋅ Φ (d1 ) − e −i (T − t ) E ⋅ Φ (d 2 ))
=
∂S
∂S
(cf. below)
Arbitrage tables for Puts and Calls in the 1-period model
In the example, we had S 0 = 250; S (1) u = 400; S (1) d = 200; i = 12% , E = 250
t=0
t=1
(Su=400)
(Sd=200)
“3 unknown values“
u
+150
d
0
C, ∆, B
Purchase Call
-C
Short sale of
∆ C share
∆c * 250
-300
(-∆c*400)
-150
(-∆c*200)
Risk-free asset
150/1.12
=133.5
0
+150
+150
0
0
Put
t=0
t=1
d
u
Purchase Put
-P
+50
0
Purchase of
∆P shares
-∆P*250
(=-∆P*S0)
+∆P*200
+∆P*400
Loan
+100/1.12
= 89.28
-100
-100
(B = risk-free amount) B =
C = 53.5
|∆P| = ¼
P = 26,78
B = 100
u ⋅ C d − d ⋅ Cu
C − Cd
; ∆C = u
(u − d )(1 + i )
(u − d ) ⋅ S
Graphical „motivation“
19
Call:
Profit
in t = 1
„Long-position“ of a share
50
250
300
=
S0 = E
S(1)
Short-position (Short sale)
Profit
in t = 1
Call
250
= S0
If the bank sells calls, ∆C tells us the how many shares to purchase to obtain a risk
free portfolio. It is the number of calls multiplied by delta.
Derivation of the ∆-factor:
C = SΦ (d1 ) − e −i (T −t ) EΦ ( d 2 )
∆C =
S σ2
ln + (
+ i ) (T − t )
E
2
with d1 =
,
σ T −t
∂d
∂d
∂C
= Φ ( d 1 ) + S ⋅ ϕ ( d 1 ) ⋅ 1 − e −i ( T − t ) E ⋅ ϕ ( d 2 ) 2
∂S
∂S
∂S
d 2 = d1 − σ T − t
ϕ density function of SND
20
ln
d1 =
σ2
S
+(
+ i ) ⋅ (T − t )
E
2
σ T −t
∂d1
∂d
1
=
= 2
∂S Sσ T − t
∂S
⇒
S σ2
ln + (
+ i ) ⋅ (T − t )
∂d1
∂d
1
2
d1 = E
⇒
=
= 2
∂S Sσ T − t
∂S
σ T −t
1
∆ C = Φ (d 1 ) +
( S ϕ ( d 1 ) − e − i ( T − t ) Eϕ ( d 2 ) )
Sσ T − t
e
d 2
− 2
2
( d −σ T −t ) 2
2
=
1 − 1
e
2π
=
1 − 21 + d1σ T −t +
e
2π
d2
=S⋅
1
e
2π
−
d1
2
2
− e − i (T − t ) E ⋅ e
−
d 12
σ 2 (T − t )
+ d 1σ T − t −
2
2
σ 2 ( T −t )
2
1
2π
2
=
1
e
2π
−
d1
σ 2 (T − t )
d 1σ T − t −
2 ( S − e − i (T − t ) E ⋅ e
2
)
d1
=
ln
2
1 −2
e
( S − e − i (T − t ) E ⋅ e
2π
S σ2
+(
+ i )( T − t )
σ 2 T −t
2
E
⋅σ T − t −
2
σ T −t
)
d2
1 − 21
S
e ( S − e − i ( T −t ) E ⋅ ⋅ e i ( T −t ) )
E
2π
=
∂C
= Φ ( d1 )
∂S
∈ (0,1]
=0
∂C
>0
∂S
Computation of the ∆ of a put option
P − C + S = e −i (T −t ) E ⇒ P = e −i (T −t ) E + C − S
∂P
∆P =
= ∆ C − 1 = Φ (d1 ) − 1 ∈ ( −1,0) < 0
∂S
Alternatively, the ∆-factor can be derived via the binomial model.
21
Thumb rules for options with “reasonable” time to maturity
In the money
At the money
Out of the money
∆C
∆P
∈ (½,1)
∈ (-1,-½)
~ -0.5
∈ (-½,0)
~ 0.5
∈ (0,½)
Question: How to manage a portfolio to keep it risk-free?
∆ is influenced by the price of the underlying and (decreasing) time to maturity.
Tiny example: Bank A sells to customer B 1.000 Calls, each costing 5 €. The price of
the underlying is assumed to be 100 € per share. ∆C = 0,5.
The bank gets thus a premium of 1.000*5 € =5.000 €.
The bank buys now ∆ * #Calls = 0.5*1.000= 500 shares at 100 € each, i.e the bank
buys shares for 50,000 €.
Suppose now the share price increases and ∆C turns to 0.7.
The bank purchases 200 new shares at the higher price. With decreasing ∆,
shares are sold.
More relevant in practice is Γ − Hedging [Gamma]: Γ =
∂ 2C
∂S 2
Volatility σ
Historical volatilities are computed from former quotes:
Week
1
Quote Sj
S1
2
S2
3
...
26
S26
Sj/ Sj-1
Ln(Sj / Sj-1)
Discrete
returns
 S j  S j − S j −1
≈
ln
S 
S j −1
j
−
1


1 26
Estimate for the expected return: µˆ =
∑ ln( S j / S j −1 )
25 j =1
1 25
2
Variance on weekly base: Vaˆr =
∑ ln( S j / S j −1 ) − µˆ )
24 j =1
Variance on year’s base: σˆ
2
= Vaˆr ⋅ 52
22
Historical volatility: σˆ = σˆ 2 = 52 ⋅ Vaˆr
Volatility is the annualized standard deviation of the return distribution
1. The numerical result (historical volatility) depends on the data base.
2. The historical volatility provides not necessarily a good forecast for future
volatilities.
3. The implicit volatility is the numerical value satisfying the price. There is exactly
one implicit volatility in the relevant range.
How to use Black and Scholes formulae
- All variables entering in B&S-formulae are annualized variables.
- The historical volatility depends on several items.
- The length and the time window to be examined determine partly its numerical
value.
quote
At the beginning, lower deviations of the underlying
time
What is here the „right“ estimate for the volatility? historical volatility must be an
interval.
In contrast, the implicit volatility is the volatility put by the issuer in the derivative! It
shows the issuer’s expected deviations of the underlying until maturity.
σ = σ (t ) is a function of time.
S = SΦ (d1 ) − e − i (T − t ) EΦ ( d 2 )
S σ2
ln + (
+ i ) (T − t )
E
2
with d1 =
,
σ T −t
d 2 = d1 − σ T − t
S – known from the stock exchange
E – part of the contract
i – risk free interest rate, known from the central bank
23
T – t – time until maturity, known
C – Actual observed price of the derivative: It physically exists!
There exists one and only one σ in the relevant range satisfying the B&Sformulae. That is the implicit volatility.
Dependency of the call price of σ , Vega =
⇒
∂C
∂σ
∂C
∂d
∂d
= S ⋅ ϕ ( d1 ) ⋅ 1 − e − i (T − t ) E ⋅ ϕ (d 2 ) ⋅ 2 = S T − tϕ ( d1 ) > 0
∂σ
∂σ
∂σ
∂Pt
= e − i (T − t ) E + Ct − St = Vega (Call ) = St T − tϕ ( d1 )
∂σ
(
)
What about the impact of a change in interest rates?
∂C
= ϑ = (T − t )e − i (T − t ) ΕΦ ( d 2 ) > 0
δi
With a rise of the risk-free interest rate, the price of a call rises, everything else being
constant, too. Consequently, P decreases.
Fundamental principle is the absence of arbitrage.
Ε[ ST | St ] = S0 ⋅ ei (T − t ) ; i.e. the expected return of an investment in S is the risk-free
return.
Estimates for u and d in the binomial model
^
σ T −t
u=e
^
;
d = e −σ T − t ; u =
1
d
What about the impact of the time to maturity for options’ prices? Intuitively,
we expect higher prices with longer times to maturity.
24
ϑCall =
δC
δ (T − t )
= SΦ (d1 ) − iEe −i (T −t ) Φ (d 2 ) −
Ee − i (T −t )ϕ (d 2 )σ
2 T −t
The farer the maturity, the higher (everything else being constant) the value of a call
option. However, this does not hold anymore for all put options.
ϑ Put = ϑCall + iEe −i (T −t )
How does ∆ − Hedging work? An example:
We hold 100 shares and we aim to insure our portfolio with put options. Given the
following data:
^
S0 = 200; E = 200; i = 0,075; σ = 0,15 ( = 15%)
Time to maturity
90
70
50
40
Share’s price
200
212
198
Value of the shares 20.000 21.200 19.800
…
…
188
181
18.800 18.100
Put- ∆
…
-1
-1
…
…
100
12.83
100
19
…
20,083 20,000
-0.391 -0.129 -0.49
(1)
#Puts
Put price
256
4.29
Portfolio value
21,098 21,920
(1)
(2)
775
0.93
204
4.40
(2)
30
20
10
5
B&S formulae
Purchase of 519 Puts at 0.93 each
Sale of 571 Puts at 4.40 each
Graphical motivation:
Profit /
Loss
in t = T
Put
Long position share
Portfolio insurance with Puts
ST
E = S0
25
Call holder
Bank insures itself by selling calls
Call seller
Γ -Hedging:
Portfolios are hedged by buying puts and selling calls.
 ∂S

∆ of a share is always 1 
= 1
 ∂S

 ∂2S

Γ of a share is always 0  2 = 0 
 ∂S

Γ changes of long and short positions cancel themselves.
-
-
Time to maturity
90
70
Quote
200
Long Put- ∆
-0,391
Short Call- ∆
-0,609
Positions’ delta -1
Put price
4.29
Call price
7.82
#Long Puts
100
#Short Calls
100
212
-0.129
-0.871
-1
0.93
15.68
446
446
Cashflow: 353
Dividends and option pricing
We look on a call option on a stock which pays out a dividend before the option
matures. At t1 ∈ (0,T), a non-stochastic dividend D will be paid out. We regard now
the time value of the dividend.
D
e −( t − t ) D
1
t0
0
t1
26
In B&S-formulae, we substitute S t → S t − e
−i ( t1 − t o )
D
Example:
S = 41; E = 40; σ = 0,3; i = 0.08; T = 3 months = 0,25 years
The share pays out a dividend D = 3 after 1 month.
Time value of the dividend: 3e
− 0 ,8 (
1
)
12 = 2.980067 .
A forward price of the share at t = 0 ís thus S0 – 2.98 = 38.02 = 41 – 2.98.
C t = ( S t − e − i ( t 1 − t ) D ) ⋅ Φ ( d1 ) − e − i ( T − t ) E Φ ( d 2 )
 S − e − i ( t1 − t ) D   σ 2

+
ln  t
+ i (T − t )


  2
E

 
d1 = 
; d 2 = d1 − σ T − t
σ T −t
Example : d1 = −0.138; d 2 = −0.288 ⇒ C = 1.763
Note that the call’s price without dividend is 3.399.
Continuous dividends: We „subtract“ at any time a tiny bit. (Reverse of
continuously compounding)
C t = S t e − q (T −t ) Φ (d 1 ) − E ⋅ e − i (T −t1 ) Φ (d 2 )
q = Discount factor for
continuous dividends.
σ2
S
Ln( ) + (i − q +
)(T − t )
E
2
d1 =
; d 2 = d1 − σ T − t
σ T −t
27
d1 can be rewritten as
 S ⋅ e − q (T − t )  σ 2
+
Ln
(T − t )
− i (T − t ) 
2
E
e
⋅

d1 = 
σ T −t
Note that the dividend yield q discounts only the price of the stock while the risk free
interest rate i discounts only the exercise price.
We can use this Garman-Kohlhagen-formulae to price options on currencies. The
dividend yield q becomes the foreign risk-free interest rate iF. (f: foreign); iD=
domestic risk-free interest rate .
Garman-Kohlhagen-formulae for the value of a Call on currencies:
C = C ( S , E , T − t , σ ; iF , iD ) = St ⋅ e −i (T −t )Φ (d1 ) − e −i (T −t ) EΦ (d 2 )
F
D
S
1
Ln( t ) + (iD − iF + σ 2 )(T − t )
E
2
d1 =
; d 2 = d1 − σ T − t
σ T −t
Example:
Valuation of a USD-denominated call on Euro. The USD is our domestic currency.
S = 1.20 USD / Euro; E1 = 1.25 USD / Euro; σ = 0,1; iF = 2.25%; iD =3.75% ; T = 1
year.
C = 0.0225 USD
The right to purchase 100 Euro for 125 USD next year costs thus USD 2.25.
E2 = 1.20 USD / Euro C2 = 0.03997
E3 = 1.15 USD / Euro C3 = 0.0654
28
In the money
6.5
Out of the money
4
2
1.15
1.20
1.25
E in USD / Euro
Put-Call-Parity: Pt = C t + E ⋅ e
P1 = 0.0837
−i D (T − t )
− S t e −i (T − t ) P3 = 0.0277; P2 = 0.0517;
F
A hedging European Exporter in USD-currency countries can use futures / MMA –
and calls on Euro or puts on USD.
Example: Porsche has revenues of 3 billion USD next year, Porsche uses ∆ hedging.
∆=e
− i F (T − t )
Ln(
Φ ( d1 );
d1 =
S
E
) + (i D − i F +
σ2
2
)(T − t )
σ T −t
∆ C = 0.31
1
* 3 billion * 0,0225 = 217.846.109 USD, for this hedge: That
∆C
corresponds to 7 % of the revenues!
Porsche needs thus
A USD-denominated call on Euro, giving the holder the right to pay USD for Euro, is
equivalent to a Euro-denominated put on in USD, representing the right to sell USD
for Euro.
Example: S0 = 0.8 Euro / USD (corresponds to 1.25 USD / Euro)
1) We look on a USD-denominated call on Euro with E = 1.30 USD / Euro.
At maturity, we have thus the right to purchase 1 Euro by selling 1,30 USD
At maturity, CT = max (0, ST – 1.30)
29
2) A Euro-denominated put on USD with an exercise price of
1
1.30 USD / Euro
= 0.77 Euro /USD gives the holder the right to sell one USD at maturity for
0.77 Euro zu erhalten. The holder of this put option will exercise this right, if a USD
Is worth less than 0.77 Euro. PT = max (0, 1/1.30 – 1/ST )
ST > 1.30 USD/Euro if and only if 1/1.3 > 1/ST. Call on Euro and put on USD are
thus either exercised at maturity or worthless.
Further Applications: Real options
The “Real Options Approach” assesses the value of managerial flexibility in
responding to new information. Managers have many options to adapt and revise
decisions in response to new and unexpected developments. Such flexibility is clearly
valuable and should be accounted for in the valuation of a project or firm. Often,
managers can expand or contract production in response to changes in demand. The
firm would be less valuable if they had to choose a fixed production level before
knowing the level of demand. There are two steps in Real Options Analysis:
Problem of Identification: Are there real options imbedded in a given project?
What type of options? Are they important?
Problem of Valuation: How do we value the (important) options? How do we value
different types of options? Why can’t we just use NPV? Are there twin securities?
30