Continuous-Time Finance
1 / 46
Self-financing portfolio of stocks and bonds
Discrete time:
At period t, investor holds portfolio of value Vt , consisting of...
∆t shares of the stock, priced at St per unit, and
Vt − ∆t St units of account invested in one-period bonds.
Portfolio value at t + 1 is
Vt+1 = ∆t St+1 + Rf,t (Vt − ∆t St ) .
Subtract Vt from both sides to get
Vt+1 − Vt = ∆t (St+1 − St ) + (Rf,t − 1) (Vt − ∆t St ) .
| {z }
|
{z
} | {z }
≈dVt
≈dSt
≈rt dt
Continuous time analogue:
dVt = ∆t dSt + rt (Vt − ∆t St )dt.
2 / 46
Admissible portfolio
Definition
A self-financing portfolio is admissible if the corresponding wealth process
{Vt } is bounded below, i.e., ∃v̄ < ∞ s.t. Vt ≥ −v̄, ∀ (t, ω) ∈ [0, T ] × Ω.
Intuitively, there must be limit to how much debt can be tolerated.
Without admissibility, can generate any VT > 0 (with probability 1)
from V0 = 0.
Recall that a borrowing limit is required to rule out the discrete-time
doubling strategy.
3 / 46
Replication Pricing & Black-Scholes
4 / 46
Replication pricing in continuous time
Idea is same as in discrete time:
Find self-financing portfolio that perfectly replicates derivative payoff.
Demonstrate idea by pricing options in the Black-Scholes setting:
Return on money market account is constant r, i.e., its value β is
βt = ert .
Stock price, S, follows a geometric BM, i.e.,
dSt = µSt dt + σSt dBt .
5 / 46
Option value
Consider European call option that pays (ST − K)+ at time T .
Conjecture that option price at t should depend on:
Time to expiration, T − t, and
Current price of underlying stock, St .
(Also on parameters, e.g., strike K, but these don’t vary over time)
Why is this reasonable?
Stock price dynamics (dSt ) only depend on itself (St ).
Option payoff depends only on expiration stock price (ST ).
Time matters because of potential for shocks to hit the stock price
before expiration; more things can happen with more time left.
Thus, we write the call value, Ct , as function of t and St :
Ct = c(t, St ).
Note that the function c itself isn’t random, but Ct is, since St is random.
6 / 46
Replicating portfolio
Assets: start with wealth V0 ; make self-financing trades in stock &
risk-free asset.
Goal: want to make sure Vt = Ct for all times 0 < t ≤ T .
Match the terminal value: VT = CT .
Match all the dynamics: dVt = dCt for all t < T .
Result: if so, then we can price the call by V0 = C0 .
Z T
Z T
CT − C0 =
dCt =
dVt
0
0
= VT − V0
= CT − V0
⇒
V 0 = C0 .
(by the same logic Vt = Ct for all times t)
Equivalent method: Match d(Vt /βt ) = d(Ct /βt ) for all t.
Determines the portfolio of stocks and bonds that we need to hold.
Need to derive SDEs for Vt /βt and Ct /βt .
7 / 46
SDE for discounted portfolio value
1. Write Vt /βt = e−rt Vt as f (t, Vt ) for f (t, v) = e−rt v.
2. To use Itô’s lemma, note ft = −re−rt v, fv = e−rt , and fvv = 0, so
1
d(Vt /βt ) = ft dt + fv dVt + fvv Vart [dVt ]
2
−rt
−rt
= −re Vt dt + e dVt
= e−rt (dVt − rVt dt).
3. Since Vt is self-financing, dVt = ∆t dSt + r(Vt − ∆t St )dt, so
e−rt (dVt − rVt dt) = e−rt ∆t (dSt − rSt dt).
4. Finally, use formula for geometric BM, so
h
i
d (Vt /βt ) = e−rt ∆t (µ − r) St dt + σSt dBt .
8 / 46
SDE for discounted option price
1. Use Itô’s lemma on Ct = c(t, St ) and plug in formula for geometric
BM to get
1
dCt = ct dt + cS dSt + cSS Vart [dSt ]
2
1 2 2
= ct + µSt cS + σ St cSS dt + σSt cS dBt .
2
2. For Ct /βt , use Itô’s lemma (exactly as in previous slide) to find
immediately:
d(Ct /βt ) = e−rt (dCt − rCt dt).
3. Then substitute from the above (and substitute Ct = c(t, St )) to get:
h
i
1
d(Ct /βt ) = e−rt (−rc + ct + µSt cS + σ 2 St2 cSS )dt + σSt cS dBt .
2
9 / 46
Deriving the Black-Scholes PDE
Using the previous two slides, the condition d(Vt /βt ) = d(Ct /βt ) becomes
h
i 1
∆t (µ−r)St dt+σSt dBt = −rc+ct +µSt cS + σ 2 St2 cSS dt+σSt cS dBt .
2
Equating the dBt terms, we solve for the option delta:
∆t = cS ,
∀t ∈ [0, T ) .
Remember the formula from the binomial tree, ∆t =
u −C d
Ct+1
t+1
.
u −S d
St+1
t+1
Equating the dt terms, and plugging ∆t , we have
1
rc = ct + rSt cS + σ 2 St2 cSS ,
2
∀t ∈ [0, T ).
In conclusion we seek function c(t, S) satisfying partial differential equation
1
rc(t, S) = ct (t, S) + rScS (t, S) + σ 2 S 2 cSS (t, S), ∀t ∈ [0, T ), S ≥ 0,
2
with the terminal condition c(T, S) = (S − K)+ .
10 / 46
Aside: alternative derivation of Black-Scholes PDE
1. Apply Itô’s lemma on Ct = c(t, St ) as before to get
1
dCt = ct dt + cS dSt + σ 2 St2 cSS dt.
2
2. Rearrange to get
1
= ct + σ 2 St2 cSS dt .
dCt − cS dSt
{z
}
|
2 {z
|
}
capital gains of portfolio
long 1 call, short cS stocks
risk-free return
since no dBt term
3. By no-arbitrage, equate (ct + 12 σ 2 St2 cSS )dt to capital gains of
investing Ct − cS St in risk-free asset, i.e.,
1
r c − cS St dt = ct + σ 2 St2 cSS dt.
2
4. Divide by dt and rearrange to get the same PDE as the previous slide:
1
rc = ct + rSt cS + σ 2 St2 cSS .
2
11 / 46
Black-Scholes formula
How do we solve the PDE to get an expression for c(t, S)?
Typically, we would proceed numerically, by using standard techniques
for solving PDEs.
In this case, there is a nice closed-form solution: (Φ is the standard
normal cdf)
log(S /K) + (r + 1 σ 2 )(T − t) t
2
√
St
Ct = Φ
σ T −t
log(S /K) + (r − 1 σ 2 )(T − t) t
2
√
−Φ
Ke−r(T −t)
σ T −t
It’s a huge pain, but you could check that this satisfies the PDE by
taking derivatives wrt t, S, and SS, and plugging them into the PDE
12 / 46
“The Greeks”
Ct = Φ
log(St /K) + (r + 1 σ 2 )(T − t) log(St /K) + (r − 1 σ 2 )(T − t) 2
2
√
√
St − Φ
Ke−r(T −t)
σ T −t
σ T −t
|
{z
}
|
{z
}
:=x1
:=x2
e.g.,
“Delta”
∆t :=
h
i
∂Ct
1
√
= Φ(x1 ) +
φ(x1 )St − φ(x2 )Ke−r(T −t)
∂St
{z
}
σSt T − t |
=0 (just takes a bit of algebra)
= Φ(x1 )
Some others are
φ(x1 )
∂ 2 Ct
√
=
∂St2
σSt T − t
√
∂Ct
= φ(x1 )St T − t
“Vega” Vt :=
∂σ
h
i
∂Ct
∂Ct
σ
“Theta” Θt :=
=−
= Ke−r(T −t) rΦ(x2 ) + √
φ(x2 )
∂(T − t)
∂t
2 T −t
“Gamma”
Γt :=
13 / 46
Call price
Ct is a convex function of St (because Γt > 0); also Ct > CT (why?)
0.3
0.25
0.2
0.15
0.1
0.05
0
0.75
0.8
0.85
0.9
0.95
1
1.05
1.1
1.15
1.2
1.25
14 / 46
Call price
Ct increases with volatility σ (because Vt > 0)
0.3
0.25
0.2
0.15
0.1
0.05
0
0.75
0.8
0.85
0.9
0.95
1
1.05
1.1
1.15
1.2
1.25
15 / 46
Call price
Ct increases with time-to-maturity T − t (because Θt > 0)
0.3
0.25
0.2
0.15
0.1
0.05
0
0.75
0.8
0.85
0.9
0.95
1
1.05
1.1
1.15
1.2
1.25
16 / 46
Put options
Can recover Black-Scholes for a put option by simple put-call parity:
Ct − Pt + Ke−r(T −t) = St
Thus,
Pt = Φ(x1 )St − Φ(x2 )Ke−r(T −t) +Ke−r(T −t) − St
{z
}
|
=Ct
= [1 − Φ(x2 )]Ke−r(T −t) − [1 − Φ(x1 )]St
= Φ(−x2 )Ke−r(T −t) − Φ(−x1 )St
17 / 46
Risk-Neutral Pricing
18 / 46
Motivation
Goal is to price a financial instrument (e.g., option).
We saw above the PDE approach. But it is sometimes complicated.
From discrete-time, know can write price as risk-neutral expectation
h
i
+
Ct = e−r(T −t) EQ
(S
−
K)
.
T
t
But to calculate Q-expectation, need to know distribution of ST
under Q.
19 / 46
Review: discounted price process is martingale under Q
Remember from the discrete-time dynamic model, we know that for
all t
i
h S i
h
St
T
Q ST
St = EQ
⇒
=
E
.
t
t
Rft
RfT
RfT −t
In our new notation, writing βt = Rft we have
hS i
St
T
= EQ
.
t
βt
βT
In other words, S/β is a martingale under Q.
Note: The above ignores dividends, but see HW3. In general, discounted
value process (i.e., with dividends reinvested) is martingale under Q.
20 / 46
Plan
1. Define Q such that discounted price (or value) process is martingale.
⇒ Under Q, drift of St /βt (or Vt /βt if dividends) is 0.
⇒ Under Q, suppose BtQ is a BM, so that for some αt we have
d(St /βt ) = αt dBtQ .
2. Then we can write SDE for S in terms of Q.
3. Integrate this SDE to find the distribution of ST under Q.
4. Use distribution to compute option price from
h
i
+
.
Ct = e−r(T −t) EQ
(S
−
K)
T
t
21 / 46
Girsanov’s Theorem (how to find Q and BtQ )
Theorem
Assume
{Bt }0≤t≤T is a BM on probability space (Ω, F, P ),
{Ft }0≤t≤T is a filtration for this BM, and
{ηt }0≤t≤T is an adapted process
Define
Zt = exp
n
Z
−
0
t
1
ηs dBs −
2
Z
t
o
ηs2 ds ,
0
RT
and assume that EP [ 0 ηs2 Zs2 ds] < ∞. Then EP [ZT ] = 1, and under
probability measure Q defined by
EQ [X] = EP [ZT X], for all random variables X,
Rt
the process BtQ = Bt + 0 ηs ds is a BM on probability space (Ω, F, Q).
22 / 46
Finding risk-neutral measure - general setup
Risk-free: Consider an adapted interest rate process rt ,
and define compounded return process (money market value):
Rt
βt = e
Let Xt =
Rt
0
0
rs ds
.
rs ds, so βt = eXt . By Itô’s lemma,
dβt = rt βt dt.
Stock: Assume stock price follows generalized geometric BM, i.e., it
satisfies the SDE
dSt = µt St dt + σt St dBt ,
∀0 ≤ t ≤ T .
23 / 46
Finding risk-neutral measure - discounted price process
The discounted stock price process is St /βt . Let f (β, s) = βs . Use
two-dimensional Itô’s lemma:
d(St /βt ) = fs dSt + fβ dβt
1
+
fss Vart [dSt ] + fββ Vart [dβt ] + 2fsβ Covt [dSt , dβt ]
2
Note that fs = 1/β, fss = 0, fβ = −s/β 2 , Vart [dβt ] = 0, and
Covt [dSt , dβt ] = 0, so
St
St dSt St
1
− rt dt.
d(St /βt ) = dSt − 2 dβt =
βt
βt St
βt
βt
Thus, after substituting form of dSt ,
d (St /βt )
dSt
=
− rt dt = (µt − rt )dt + σt dBt
St /βt
St
Define the market price of risk ηt :=
µt −rt
σt ,
and write in terms of ηt :
d (St /βt )
= σt [ηt dt + dBt ].
St /βt
24 / 46
Finding risk-neutral measure - using Girsanov
With Girsanov’s theorem, if we use
Z
n Z t
1 t 2 o
Zt = exp −
ηs dBs −
η ds
2 0 s
0
to define Q, then the process dBtQ = ηt dt + dBt is BM under Q.
Then, under Q the discounted stock dynamics are
d (St /βt )
= σt [ηt dt + dBt ] = σt dBtQ .
St /βt
Notice that St /βt is a Q-martingale (since after integrating, the
right-hand-side is Itô integral with respect to a BM).
Finally, recall
dSt
St
= rt dt +
dSt
St
d(St /βt )
St /βt ,
=
so we have
rt dt + σt dBtQ
|
{z
}
same vol σt ,
changes drift from µt to rt
25 / 46
Risk-neutral measure - discounted portfolio value process
The discounted value of a portfolio, Vt /βt , is also martingale under Q.
To see this, write the self-financing condition
dVt = ∆t dSt + rt (Vt − ∆t St ) dt
and use Itô’s lemma with the P -SDEs for βt and St /βt to find (similar to
step 3 on slide 8):
d (Vt /βt ) = ∆t d (St /βt ) .
Using
d(St /βt )
St /βt
= σt dBtQ , we have
d (Vt /βt ) = ∆t σt (St /βt )dBtQ ,
which is an Itô integral with respect to BM, so martingale.
26 / 46
Risk-neutral pricing
Let CT be random variable representing derivative payoff at T .
We seek initial capital V0 and portfolio process ∆t that replicates
payoff, in particular
VT = CT .
Vt /βt is Q-martingale ⇒ under the ∆t that we are looking for, we
have:
hV i
h
i
Vt
T
Q CT
= EQ
=
E
.
t
t
βt
βT
βT
But V is replicating portfolio value. By no-arbitrage, derivative’s price
is Ct = Vt , so
h RT
i
− t rs ds
Ct = EQ
e
C
T .
t
27 / 46
Simple example - pricing a log contract
Assume
dSt
St
= µdt + σdBtP . Find price of derivative paying log ST at T .
Girsanov Theorem:
dSt
= rdt + σdBtQ .
St
Itô’s lemma:
1 d log St = r − σ 2 dt + σdBtQ .
2
Integrate SDE ⇒ conditional on time-t information,
1 Q
log ST ∼ Normal log St + r − σ 2 (T − t), σ 2 (T − t) .
2
Risk-neutral pricing ⇒ price of log contract is
Lt = e−r(T −t) EQ
t [log ST ]
h
i
1 = e−r(T −t) log St + r − σ 2 (T − t) .
2
28 / 46
Example - pricing an exotic derivative
A bank sells derivative paying
(
(log (ST ))2 + ST − K
(log (ST ))2
if ST > K
if ST ≤ K,
where ST is stock price at expiration and K is strike.
Assumptions:
Stock pays no dividends.
Stock follows geometric BM under true measure P , with µ = 0.1 and
σ = 0.2.
Initial stock price is S0 = 100.
Risk-free rate is r = 0.05.
Derivative expires in half a year, and has strike price K = 100.
Find no-arbitrage price of derivative.
29 / 46
Example - pricing an exotic derivative - solution plan
a. Write down SDE for St , using B P , the BM under true measure.
b. Rewrite SDE using B Q , the BM under risk-neutral measure.
c. Use Itô’s lemma to find SDE for log St under risk-neutral measure.
d. Integrate SDE to find risk-neutral distribution of log ST at t = 0.
e. Express the derivative’s price (at t = 0) as a risk-neutral expectation.
f. Calculate the derivative’s price.
30 / 46
Example - pricing an exotic derivative - solution (a-c)
a. SDE for S under true measure is
dSt
= 0.1dt + 0.2dBtP .
St
b. Girsanov’s theorem ⇒ can change drift to r = 0.05, so under
risk-neutral measure Q, we have
dSt
= 0.05dt + 0.2dBtQ .
St
c. We have log St = f (St ) with f (x) = log x, so Itô’s lemma yields
1
d log St = df (St ) = f 0 (St )dSt + f 00 (St )Vart [dSt ]
2
1
1 1 = dSt −
(0.04St2 dt)
St
2 St2
= 0.03dt + 0.2dBtQ .
31 / 46
Example - pricing an exotic derivative - solution (d-e)
d. Integrating the SDE we have
Z
log ST = log S0 +
T
Z
0.03dt +
0
0
T
0.2dBtQ
= log S0 + 0.03T + 0.2(BTQ − B0Q )
Q
= log(100) + 0.03(0.5) + 0.2B0.5
.
Q
Therefore, log ST ∼ Normal(4.62, 0.02).
e. Derivative’s payoff is sum of two separate payoffs:
(log ST )2 , which we always get, and
ST − K, which we only get if ST > K.
Price of derivative is sum of “price” of these separate payoffs:
h
i
2
+
C0 = e−rT EQ
(log
S
)
+
(S
−
K)
T
T
0
h
i
h
i
2
−rT Q
log ST
+
= e−rT EQ
(log
S
)
+
e
E
(e
−
K)
.
T
0
0
{z
}
|
Black-Scholes call price
32 / 46
Example - pricing an exotic derivative - solution (f)
f. Continuing from above, we have:
n h
i
h
io
2
log ST
C0 = e−rT EQ
+ EQ
− K)+
0 (log ST )
0 (e
m − log K o
n
1 m + s2 − log K − KΦ
,
= e−rT v 2 + m2 + exp m + v 2 Φ
2
v
v
where m = 4.62 and v 2 = 0.02. Here we used fact that if
X ∼ Normal(m, v 2 ) then
1
2
E[max(eX − K, 0)] = em+ 2 v Φ
m + v 2 − log K v
− KΦ
m − log K v
Plugging in the numbers, we have:
n
o
0.02
√
√
C0 = 0.02 + 4.622 + e4.62+ 2 Φ 4.62+0.02−4.605
− 100Φ 4.62−4.605
e−0.05×0.5
0.02
0.02
= 27.7.
33 / 46
Risk-neutral pricing works for very general payoffs
As the previous example illustrates, it doesn’t really matter what the final
payoff looks like. If X is the underlying, and g(XT ) is the final payoff for
some arbitrary function g, then the time-t price is
h RT
i
− t ru du
Pt = EQ
e
g(X
)
T
t
TheR only challenge is trying to figure out the probability distribution of
T
e− t ru du g(XT ) under Q.
Zero-coupon bond: g(x) = 1.
European option: g(x) = max[x − k, 0] and XT = ST .
If the underlying stock St is a GBM, then we showed the distribution of
ST is lognormal, which makes this easy (Black-Scholes formula).
If St is more complicated (e.g., stochastic volatility σt ), then ST will
not be lognormal, so the solution won’t be as pretty.
RT
Asian options: g(x) = max[x − k, 0] and XT = T1 0 St dt.
Barrier options: g(x) = max[x − k, 0] and XT = ST 1{inf t St >b} .
34 / 46
Risk-neutral pricing for dividend-paying stock
Stock follows
dSt
St
= µdt + σdBt and pays dividend δSt dt.
Self-financing strategy: dVt = ∆t dSt + r (Vt − ∆t St ) dt + ∆t δSt dt.
Using Itô’s lemma, we have
Vt
1
dβt + dVt
βt2
βt
Vt
1
= − rdt +
(∆t dSt + r (Vt − ∆t St ) dt + ∆t δSt dt)
βt
β
h t
i
= ∆t (St /βt ) (µ − r + δ) dt + σdBt .
d(Vt /βt ) = −
Girsanov Theorem ⇒ for any ηt , there exists Q such that
dBtQ = ηt dt + dBt is BM, so
i
St h
d(Vt /βt ) = ∆t
(µ − r + δ − σηt )dt + σdBtQ .
βt
Thus, Vt /βt is a Q-martingale (which is required under no-arbitrage)
if and only if ηt = µ−r+δ
, and so we have
σ
dSt /St = (r − δ)dt + σdBtQ .
35 / 46
Extra Topics – not for exam
36 / 46
Risk-neutral measure with multiple stocks
Assume there are m stocks, each with SDE
d
X
dSit = µit Sit dt + Sit
σijt dBjt ,
i = 1, . . . , m.
j=1
Rt
As before, define compounded interest process βt = e
d (Sit /βt )
= (µit − rt ) dt +
Sit /βt
d (Vt /βt ) =
m
X
d
X
0
rs ds
, and find:
σijt dBjt
j=1
∆it d (Sit /βt ) ,
i=1
where ∆it is the adapted portfolio process for each stock i.
Definition
A probability measure Q is risk-neutral if
i) P and Q are equivalent (same null-sets), and
ii) Under Q, the discounted stock price Sβitt is a martingale, ∀i.
37 / 46
Risk-neutral measure with multiple stocks
To make the discounted prices martingales, we would like to find {ηjt } s.t.
d
d (Sit /βt ) X
=
σijt (ηjt dt + dBjt )
Sit /βt
j=1
Then we could use Girsanov’s Theorem to construct the equivalent
measure Q under which dBQ
t = ηt dt + dBt would be a d-dimensional BM,
and write
d
d (Sit /βt ) X
Q
σijt dBjt
,
=
Sit /βt
j=1
so discounted price would be sum of Q-martingales, hence Q-martingale.
Clearly, to find {ηjt } we need to solve:
∀i : µit − rt =
d
X
j=1
σijt ηjt ⇔ µt − rt 1 = Σt ηt
|
{z
}
in vector form
38 / 46
Market prices of risk - like state prices
We want to solve µt − rt 1 = Σt ηt .
|{z} |{z}
|{z}
m×1
m×d d×1
If m = d and Σt is full rank at all times, then the solution is
ηt = Σ−1
t (µt − rt 1).
Analogy to one-period model: with a complete market and no redundant
assets, we solve for a unique state price vector.
If m > d and Σt is full rank at all times, then the candidate solution (which
may not satisfy the original equation) is
ηt = (Σ0t Σt )−1 Σ0t (µt − rt 1).
Analogy to one-period model: with redundant assets, there will exist a state
price vector if and only if assets are priced correctly.
If m < d and Σt is full rank at all times, then there are many solutions, one
of which is
ηt = Σ0t (Σt Σ0t )−1 (µt − rt 1).
Analogy to one-period model: in an incomplete market with no redundant
assets, there are many state price vectors.
39 / 46
American options
Riskless rate is constant at r > 0
Remember – you never exercise the call early if r > 0 (HW3), so we
focus on the put.
Put price satisfies
Pt = max
τ ∈[t,T ]
h
i
−r(τ −t)
EQ
max(K − Sτ , 0)
t e
{z
}
|
present-discounted-value from exercise at τ
|
{z
optimize over possible stopping times τ
}
Then, optimal exercise time is the first time exercising is at least as good
as continuing (and exercising optimally thereafter), i.e.,
n
o
τ ∗ = inf τ : max(K − Sτ , 0) ≥ Pτ
n
o
(by path-continuity) = inf τ : max(K − Sτ , 0) = Pτ
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American options are difficult
In general, it’s hard to price American options (and continuous-time
doesn’t help that much).
We know that the exercise rule
n
o
τ ∗ = inf t : max(K − St , 0) ≥ Pt
boils down to a time-dependent price threshold
n
o
τ ∗ = inf t : St ≤ s∗ (t)
for some function s∗ (t) (the exercise boundary).
But what does s∗ (t) look like? Hard in general.
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Perpetual American options
Simpler setting:
Suppose no expiration date, i.e., T → +∞
Also, suppose dSt = St [µdt + σdBt ], with S0 = s as in Black-Scholes.
Result: the exercise boundary s∗ (t) no longer depends on time t
And for the same reason, the put price no longer depends on time t,
only the current stock price, so there is some function p such that
Pt = p(St )
We want to find time-invariant exercise boundary. To do that, let
s∗ < K be some arbitrary exercise boundary. Let τ be the first time t
that St hits s∗ (or if St never hits s∗ , then we set τ = ∞ and the
option is never exercised). If s∗ describes our exercise strategy, then
the put value is
−rτ
max(K − Sτ , 0)1{τ <∞} ]
p(s; s∗ ) = EQ
0 [e
−rτ
= (K − s∗ )EQ
1{τ <∞} ]
0 [e
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Perpetual American options, continued
−rτ 1
Need to compute EQ
{τ <∞} ]
0 [e
1. Use Girsanov’s theorem: dSt = St [rdt + σdBtQ ]
2. Use Itô and integrate to get the usual St = s exp[(r − 21 σ 2 )t + σBtQ ]
3. At t = τ , we have St = s∗ , so rearrange to get e−rτ =
1
4. Plug this in, and notice that Zt := e− 2 σ
martingale from Girsanov’s theorem, so
−rτ
EQ
1{τ <∞} ] =
0 [e
2 t+σB Q
t
s − 12 σ 2 t+σBtQ
s∗ e
is the same type of
s Q
s
s
E0 [Zt 1{τ <∞} ] = ∗ E∗0 [1{τ <∞} ] = ∗ P∗0 [τ < ∞],
∗
s
s
s
where E∗ is a new expectation operator (and P∗ its associated
probability) under which Bt∗ := BtQ − σt is a BM
5. Using the result of step 2 again, then step 4, note that
P∗0 [τ < ∞] = P∗0 [(r + 12 σ 2 )t + σBt∗ ≤ log(s∗ /s) for some t > 0]
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Perpetual American options, continued
To solve this probability, we use the following fact, which can be proved
using Doob’s optional stopping theorem, with the smart choice of
martingale Mt := exp[−2a(at + Bt )] and clever choice of stopping time
τ := inf{t : Mt = e2ab or Mt = 0}:
Lemma
If B is a BM under P, and if a > 0 and b < 0, then
P0 [at + Bt ≤ b for some t > 0] = e2ab .
Using this lemma, with a = r/σ + 21 σ and b = (1/σ) log(s∗ /s), we get
1
P∗0 [τ < ∞] = P∗0 [(r + σ 2 )t + σBt∗ ≤ log(s∗ /s) for some t > 0]
2
h r
s∗ i s∗ 1+2r/σ2
1
= exp 2 2 +
log
=
σ
2
s
s
Thus,
s∗ 2r/σ2
−rτ
EQ
[e
1
]
=
{τ <∞}
0
s
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Perpetual American options, finish up
Plug in the previous results into the put price to get
s∗ 2r/σ2
p(s; s∗ ) = (K − s∗ )
s
This holds for arbitrary choice of s∗ . To find the optimal exercise, we solve
maxs∗ p(s; s∗ ). The FOC is
s∗ 2r/σ2 2r K − s∗ s∗ 2r/σ2
−
+
= 0.
s
σ2
s∗
s
Solve for s∗ to get
2r
s∗ =
K.
2r + σ 2
as σ increases, s∗ decreases [i.e., wait longer to exercise with more
uncertainty, because more bad things can happen to the stock]
as r decreases, s∗ decreases [i.e., wait longer to exercise under lower
interest rates, because the opportunity cost of waiting declines]
as r & 0, s∗ & 0 [i.e., never exercise with zero interest rates, as you
showed in HW3]
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The End!
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