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Chapter 11
American Options
In contrast with European option which have fixed maturities, the holder of
an American option is allowed to exercise at any given (random) time. This
transforms the valuation problem into an optimization problem in which one
has to find the optimal time to exercise in order to maximize the payoff of
the option. As will be seen in the first section below, not all random times
can be considered in this process, and we restrict ourselves to stopping times
whose value at time t be can decided based on the historical data available.
11.1 Filtrations and Information Flow
Let (Ft )t∈R+ denote the filtration generated by a stochastic process (Xt )t∈R+ .
In other words, Ft denotes the collection of all events possibly generated by
{Xs : 0 6 s 6 t} up to time t. Examples of such events include the event
{Xt0 6 a0 , Xt1 6 a1 , . . . , Xtn 6 an }
for a0 , a1 , . . . , an a given fixed sequence of real numbers and 0 6 t1 < · · · <
tn < t, and Ft is said to represent the information generated by (Xs )s∈[0,t]
up to time t.
By construction, (Ft )t∈R+ is an increasing family of σ-algebras in the sense
that we have Fs ⊂ Ft (information known at time s is contained in the information known at time t) when 0 < s < t.
One refers sometimes to (Ft )t∈R+ as the increasing flow of information
generated by (Xt )t∈R+ .
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N. Privault
11.2 Martingales, Submartingales, and Supermartingales
Let us recall the definition of martingale (cf. Definition 5.5) and introduce in
addition the definitions of supermartingale and submartingale.∗
Definition 11.1. An integrable stochastic process (Zt )t∈R+ is a martingale
(resp. a supermartingale, resp. a submartingale) with respect to (Ft )t∈R+ if
it satisfies the property
Zs = IE[Zt | Fs ],
0 6 s 6 t,
Zs > IE[Zt | Fs ],
0 6 s 6 t,
Zs 6 IE[Zt | Fs ],
0 6 s 6 t.
resp.
resp.
Clearly, a process (Zt )t∈R+ is a martingale if and only if it is both a supermartingale and a submartingale.
A particular property of martingales is that their expectation is constant.
Proposition 11.2. Let (Zt )t∈R+ be a martingale. We have
IE[Zt ] = IE[Zs ],
0 6 s 6 t.
The above proposition follows from the “tower property” (17.37) of conditional expectations, which shows that
IE[Zt ] = IE[IE[Zt | Fs ]] = IE[Zs ],
0 6 s 6 t.
(11.1)
Similarly, a supermartingale has a decreasing expectation, while a submartingale
has a increasing expectation.
Proposition 11.3. Let (Zt )t∈R+ be a supermartingale, resp. a submartingale.
Then we have
IE[Zt ] 6 IE[Zs ],
0 6 s 6 t,
resp.
IE[Zt ] > IE[Zs ],
0 6 s 6 t.
Proof. As in (11.1) above we have
IE[Zt ] = IE[IE[Zt | Fs ]] 6 IE[Zs ],
The proof is similar in the submartingale case.
0 6 s 6 t.
“This obviously inappropriate nomenclature was chosen under the malign influence of
the noise level of radio’s SUPERman program, a favorite supper-time program of Doob’s
son during the writing of [Doo53]”, cf. [Doo84], historical notes, page 808.
∗
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Independent increments processes whose increments have negative expectation give examples of supermartingales. For example, if (Xt )t∈R+ is such a
process then we have
IE[Xt | Fs ] = IE[Xs | Fs ] + IE [Xt − Xs | Fs ]
= IE[Xs | Fs ] + IE[Xt − Xs ]
6 IE[Xs | Fs ]
= Xs ,
0 6 s 6 t.
Similarly, a process with independent increments which have positive expectation will be a submartingale. Brownian motion Bt + µt with positive drift
µ > 0 is such an example, as in Figure 11.1 below.
5
drifted Brownian motion
drift
4.5
4
3.5
3
2.5
2
1.5
1
0.5
0
-0.5
0
2
4
6
8
10
12
14
16
18
20
Fig. 11.1: Drifted Brownian path.
The following example comes from gambling.
Fig. 11.2: Evolution of the fortune of a poker player vs number of games played.
A natural way to construct submartingales is to take convex functions of
martingales.
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Proposition 11.4. Given (Mt )t∈R+ a martingale and φ : R −→ R a convex
function we have
φ(Ms ) 6 IE[φ(Mt ) | Fs ],
0 6 s 6 t,
i.e. (φ(Mt ))t∈R+ is a submartingale.
Proof. By Jensen’s inequality we have
φ(IE[Mt | Fs ]) 6 IE[φ(Mt ) | Fs ],
0 6 s 6 t,
(11.2)
which shows that
φ(Ms ) = φ(IE[Mt | Fs ]) 6 IE[φ(Mt ) | Fs ],
0 6 s 6 t.
More generally, the proof of Proposition 11.4 shows that φ(Mt )t∈R+ remains a submartingale when φ is convex nondecreasing and (Mt )∈R+ is
a submartingale. Similarly, (φ(Mt ))t∈R+ will be a supermartingale when
(Mt )∈R+ is a martingale and the function φ is concave.
Other examples of (super, sub)-martingales include geometric Brownian
motion
2
St = S0 e rt+σBt −σ t/2 ,
t ∈ R+ ,
which is a martingale for r = 0, a supermartingale for r 6 0, and a
submartingale for r > 0.
11.3 Stopping Times
Next, we turn to the definition of stopping time.
Definition 11.5. A stopping time is a random variable τ : Ω −→ R+ ∪{+∞}
such that
{τ > t} ∈ Ft ,
t ∈ R+ .
(11.3)
The meaning of Relation (11.3) is that the knowledge of the event {τ > t}
depends only on the information present in Ft up to time t, i.e. on the knowledge of (Xs )06s6t .
In other words, an event occurs at a stopping time τ if at any time t it
can be decided whether the event has already occured (τ 6 t) or not (τ > t)
based on the information Ft generated by (Xs )s∈R+ up to time t.
For example, the day you bought your first car is a stopping time (one
can always answer the question “did I ever buy a car”), whereas the day you
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will buy your last car may not be a stopping time (one may not be able to
answer the question “will I ever buy another car”).
Proposition 11.6. Let τ and θ be stopping times.
i) Every constant time is a stopping time.
ii) The minimum τ ∧ θ = min(τ, θ) of τ and θ is also a stopping time.
iii) The maximum τ ∨ θ = max(τ, θ) of τ and θ is also a stopping time.
Proof. Point 1. is easily checked. We have
{τ ∧ θ > t} = {τ > t and θ > t} = {τ > t} ∩ {θ > t} ∈ Ft ,
t ∈ R+ .
On the other hand, we have
{τ ∨ θ > t} = {τ > t and θ > t} = {τ > t} ∩ {θ > t} ∈ Ft ,
t ∈ R+ ,
which implies
{τ ∨ θ < t} = {τ ∨ θ > t}c ∈ Ft ,
t ∈ R+ .
Hitting times provide natural examples of stopping times. The hitting time
of level x by the process (Xt )t∈R+ , defined as
τx = inf{t ∈ R+ : Xt = x},
is a stopping time, as we have (here in discrete time)
∗
{τx > t} = {Xs 6= x for all s ∈ [0, t]}
= {X0 6= x} ∩ {X1 6= x} ∩ · · · ∩ {Xt 6= x} ∈ Ft ,
t ∈ N.
In gambling, a hitting time can be used as an exit strategy from the game.
For example, letting
τx,y := inf{t ∈ R+ : Xt = x or Xt = y}
(11.4)
defines a hitting time (hence a stopping time) which allows a gambler to exit
the game as soon as losses become equal to x = −10, or gains become equal
to y = +100, whichever comes first.
However, not every R+ -valued random variable is a stopping time. For
example the random time
(
)
τ = inf
t ∈ [0, T ] : Xt = sup Xs
,
s∈[0,T ]
∗
As a convention we let τ = +∞ in case there exists no t ∈ R+ such that Xt = x.
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which represents the first time the process (Xt )t∈[0,T ] reaches its maximum
over [0, T ], is not a stopping time with respect to the filtration generated by
(Xt )t∈[0,T ] . Indeed, the information known at time t ∈ (0, T ) is not sufficient
to determine whether {τ > t}.
Given (Zt )t∈R+ a stochastic process and τ : Ω −→ R+ ∪ {+∞} a stopping
time, the stopped process (Zt∧τ )t∈R+ is defined as

 Zt if t < τ,
Zt∧τ =

Zτ if t > τ,
Using indicator functions we may also write
Zt∧τ = Zt 1{t<τ } + Zτ 1{t>τ } ,
t ∈ R+ .
The following Figure 11.3 is an illustration of the path of a stopped process.
0.065
0.06
0.055
0.05
0.045
0.04
0.035
0.03
0.025
0
2
4
6
τ 8
10
12
14
16
18
20
t
Fig. 11.3: Stopped process.
Theorem 11.7 below is called the stopping time (or optional sampling, or
optional stopping) theorem, it is due to the mathematician J.L. Doob (19102004). It is also used in Exercise 11.4 below.
Theorem 11.7. Assume that (Mt )t∈R+ is a martingale with respect to
(Ft )t∈R+ . Then the stopped process (Mt∧τ )t∈R+ is also a martingale with
respect to (Ft )t∈R+ .
Proof. We only give the proof in discrete time by applying the martingale
transform argument of Proposition 2.6. Writing the telescoping sum
Mn = M0 +
n
∞
X
X
(Ml − Ml−1 ) = M0 +
1{l6n} (Ml − Ml−1 ),
l=1
l=1
we have
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Mτ ∧n = M0 +
τX
∧n
(Ml − Ml−1 ) = M0 +
l=1
∞
X
1{l6τ ∧n} (Ml − Ml−1 ),
l=1
and for k 6 n,
IE[Mτ ∧n | Fk ] = M0 +
∞
X
IE[1{l6τ ∧n} (Ml − Ml−1 ) | Fk ]
l=1
= M0 +
k
X
IE[1{l6τ ∧n} (Ml − Ml−1 ) | Fk ]
l=1
+
∞
X
IE[1{l6τ ∧n} (Ml − Ml−1 ) | Fk ]
l=k+1
= M0 +
k
X
(Ml − Ml−1 ) IE[1{l6τ ∧n} | Fk ]
l=1
+
∞
X
IE[IE[(Ml − Ml−1 )1{l6τ ∧n} | Fl−1 ] | Fk ]
l=k+1
= M0 +
k
X
(Ml − Ml−1 )1{l6τ ∧n}
l=1
+
∞
X
IE[1{l6τ ∧n} IE[(Ml − Ml−1 ) | Fl−1 ] | Fk ]
l=k+1
= M0 +
τ ∧n∧k
X
(Ml − Ml−1 )1{l6τ ∧n}
l=1
= M0 +
τ ∧k
X
(Ml − Ml−1 )1{l6τ ∧n}
l=1
= Mτ ∧k ,
k = 0, 1, . . . , n,
because the martingale property of (Ml )l∈N implies
IE[(Ml − Ml−1 ) | Fl−1 ] = IE[Ml | Fl−1 ] − IE[Ml−1 | Fl−1 ]
= IE[Ml | Fl−1 ] − Ml−1
= 0,
l > 1.
Since the stopped process (Mτ ∧t )t∈R+ is a martingale by Theorem 11.7 we
find that its expectation is constant by Proposition 11.2. More generally, if
(Mt )t∈R+ is a supermartingale with respect to (Ft )t∈R+ , then the stopped
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process (Mt∧τ )t∈R+ remains a supermartingale with respect to (Ft )t∈R+ .
As a consequence, if τ is a stopping time bounded by T > 0, i.e. τ 6 T
almost surely, we have
IE[Mτ ] = IE[Mτ ∧T ] = IE[Mτ ∧0 ] = IE[M0 ].
(11.5)
In case τ is finite with probability one but not bounded we may also write
h
i
IE[Mτ ] = IE lim Mτ ∧t = lim IE[Mτ ∧t ] = IE[M0 ],
(11.6)
t→∞
t→∞
provided that
|Mτ ∧t | 6 C,
a.s.,
t ∈ R+ .
(11.7)
More generally, (11.6) will hold provided that the limit and expectation signs
can be exchanged, and this can be done using e.g. the Dominated Convergence Theorem.
In case P(τ = +∞) > 0, (11.6) will hold under the above conditions,
provided that
M∞ := lim Mt
(11.8)
t→∞
exists with probability one.
In addition, if τ and ν are two a.s. bounded stopping times such that τ 6 ν,
a.s., we have
IE[Mτ ] > IE[Mν ]
(11.9)
if (Mt )t∈R+ is a supermartingale, and
IE[Mτ ] 6 IE[Mν ]
(11.10)
if (Mt )t∈R+ is a submartingale, cf. Exercise 11.4 below for a proof in discrete
time. From (11.5)), if τ and ν are two bounded stopping times, we have
IE[Mτ ] = IE[Mν ]
(11.11)
if (Mt )t∈R+ is a martingale.
As a counterexample to (11.11), the random time
n
o
τ := inf t ∈ [0, T ] : Mt = sup Ms ,
s∈[0,T ]
which is not a stopping time, will satisfy
IE[Mτ ] > IE[MT ],
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although τ 6 T almost surely.
Similarly,
n
o
τ := inf t ∈ [0, T ] : Mt = inf Ms ,
s∈[0,T ]
is not a stopping time and satisfies
IE[Mτ ] < IE[MT ].
Relations (11.9), (11.10) and (11.11) can be extended to unbounded stopping
times along the same lines and conditions as (11.6), such as (11.7) applied to
both τ and ν. Dealing with unbounded stopping times can be necessary in
the case of hitting times.
In general, for all a.s. finite (bounded or unbounded) stopping times τ it
remains true that
IE[Mτ ] = IE lim Mτ ∧t 6 lim IE Mτ ∧t 6 lim IE[M0 ] = IE[M0 ], (11.12)
t→∞
t→∞
t→∞
provided that (Mt )t∈R+ is a nonnegative supermartingale, where we used Fatou’s Lemma 17.1.∗ As in (11.6), the limit (11.8) is required to exist with
probability one if P(τ = +∞) > 0.
When (Mt )t∈[0,T ] is a martingale, e.g. a centered random walk with independent increments, the message of the stopping time Theorem 11.7 is that
the expected gain of the exit strategy τx,y of (11.4) remains zero on average
since
IE Mτx,y = IE[M0 ] = 0,
if M0 = 0. This shows that, on average, this exit strategy does not increase
the average gain of the player. More precisely we have
0 = M0 = IE[Mτx,y ] = xP(Mτx,y = x) + yP(Mτx,y = y)
= −10 × P(Mτx,y = −10) + 100 × P(Mτx,y = 100),
which shows that
P(Mτx,y = −10) =
10
11
and P(Mτx,y = 100) =
1
,
11
provided that the relation P(Mτx,y = x) + P(Mτx,y = y) = 1 is satisfied, see
below for further applications to Brownian motion.
∗
IE[limn→∞ Fn ] 6 limn→∞ IE[Fn ] for any sequence (Fn )n∈N of nonnegative random
variables, provided that the limits exist.
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Similar arguments are applied in the examples below. In the next table we
summarize some of the results of this section for bounded stopping times.
Mt
bounded stopping times τ 6 ν






 supermartingale IE[Mτ ] > IE[Mν ] if τ 6 ν.

martingale







submartingale
IE[Mτ ] = IE[Mν ].
IE[Mτ ] 6 IE[Mν ] if τ 6 ν.
Examples of application
In this section we note that, as an application of the stopping time theorem,
a number of expectations can be computed in a simple and elegant way.
Brownian motion hitting a barrier
Given a, b ∈ R, a < b, let the hitting∗ time τa,b : Ω −→ R+ be defined by
τa,b = inf{t > 0 : Bt = a or Bt = b},
which is the hitting time of the boundary {a, b} of Brownian motion (Bt )t∈R+ ,
a, b ∈ R, a < b.
Recall that Brownian motion (Bt )t∈R+ is a martingale since it has independent increments, and those increments are centered:
IE[Bt − Bs ] = 0,
0 6 s 6 t.
Consequently, (Bτa,b ∧t )t∈R+ is still a martingale and by (11.6) we have
IE[Bτa,b | B0 = x] = IE[B0 | B0 = x] = x,
as the exchange between limit and expectation in (11.6) can be justified since
|Bt∧τa,b | 6 max(|a|, |b|),
t ∈ R+ .
Hence we have

 x = IE[Bτa,b | B0 = x] = a × P(Bτa,b = a | B0 = x) + b × P(Bτa,b = b | B0 = x),

∗
P Xτa,b = a | X0 = x + P(Xτa,b = b | X0 = x) = 1,
A hitting time is a stopping time
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which yields
P(Bτa,b = b | B0 = x) =
x−a
,
b−a
a 6 x 6 b,
b−x
,
b−a
a 6 x 6 b.
which also shows that
P(Bτa,b = a | B0 = x) =
Note that the above result and its proof actually apply to any continuous
martingale, and not only to Brownian motion.
Drifted Brownian motion hitting a barrier
Next, let us turn to the case of drifted Brownian motion
Xt = x + Bt + µt,
t ∈ R+ .
In this case the process (Xt )t∈R+ is no longer a martingale and in order to
use Theorem 11.7 we need to construct a martingale of a different type. Here
we note that the process
Mt := e σBt −σ
2
t/2
,
t ∈ R+ ,
is a martingale with respect to (Ft )t∈R+ . Indeed, we have
i
h
2
2
IE[Mt | Fs ] = IE e σBt −σ t/2 Fs = e σBs −σ s/2 ,
0 6 s 6 t,
cf. e.g. Example 3 page 174. By Theorem 11.7 we know that the stopped
process (Mτa,b ∧t )t∈R+ is a martingale, hence its expectation is constant by
Proposition 11.2, and (11.6) gives
1 = IE[M0 ] = IE[Mτa,b ],
as the exchange between limit and expectation in (11.6) can be justified since
|Mt∧τa,b | 6 max( e σ|a| , e σ|b| ),
t ∈ R+ .
Next, we note that letting σ = −2µ we have
e σXt = e σx+σBt +σµt = e σx+σBt −σ
2
t/2
= e σx Mt ,
or Mt = e −σx e σXt , hence
1 = IE[Mτa,b ]
= e −σx IE[ e σXτa,b ]
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= e σ(a−x) P Xτa,b = a | X0 = x + e σ(b−x) P(Xτa,b = b | X0 = x),
under the additional condition
P Xτa,b = a | X0 = x + P(Xτa,b = b | X0 = x) = 1.
Finally this gives

e σx − e σb
e −2µx − e −2µb



P(Xτa,b = a | X0 = x) = σa
= −2µa

σb

e
−
e
e
− e −2µb






P(X
τa,b
= b | X0 = x) =
e −2µa − e −2µx
,
e −2µa − e −2µb
(11.13a)
(11.13b)
a 6 x 6 b. Letting b tend to infinity in the above equalities shows that the
probability P(τa = +∞) of escape to infinity of Brownian motion started
from x ∈ [a, ∞) is equal to

 1 − P(Xτa,∞ = a | X0 = x) = 1 − e −2µ(x−a) , µ > 0,
P(τa = +∞) =

0, µ 6 0.
(11.14)
Similarly for x ∈ (−∞, b], letting a tend to infinity we have

 1 − P(Xτ−∞,b = b | X0 = x) = 1 − e −2µ(x−b) ,
P(τb = +∞) =

0, µ > 0.
µ < 0,
(11.15)
Mean hitting time for Brownian motion
The martingale method also allows us to compute the expectation IE[Bτa,b ],
after checking that (Bt2 − t)t∈R+ is also a martingale. Indeed we have
IE[Bt2 − t | Fs ] = IE[(Bs + (Bt − Bs ))2 − t | Fs ]
= IE[Bs2 + (Bt − Bs )2 + 2Bs (Bt − Bs ) − t | Fs ]
= IE[Bs2 − s | Fs ] − (t − s) + IE[(Bt − Bs )2 | Fs ] + 2 IE[Bs (Bt − Bs ) | Fs ]
= Bs2 − s − (t − s) + IE[(Bt − Bs )2 | Fs ] + 2Bs IE[Bt − Bs | Fs ]
= Bs2 − s − (t − s) + IE[(Bt − Bs )2 ] + 2Bs IE[Bt − Bs ]
= Bs2 − s,
0 6 s 6 t.
Consequently the stopped process (Bτ2a,b ∧t − τa,b ∧ t)t∈R+ is still a martingale
by Theorem 11.7 hence the expectation IE[Bτ2a,b ∧t − τa,b ∧ t] is constant in
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t ∈ R+ , hence by (11.6) we get∗
x2 = IE[B02 − 0 | B0 = x]
= IE[Bτ2a,b − τa,b | B0 = x]
= IE[Bτ2a,b | B0 = x] − IE[τa,b | B0 = x]
= b2 P(Bτa,b = b | B0 = x) + a2 P(Bτa,b = a | B0 = x) − IE[τa,b | B0 = x],
i.e.
IE[τa,b | B0 = x] = b2 P(Bτa,b = b | B0 = x) + a2 P(Bτa,b = a | B0 = x) − x2
b−x
x−a
+ a2
− x2
= b2
b−a
b−a
= (x − a)(b − x),
a 6 x 6 b.
Mean hitting time for drifted Brownian motion
Finally we show how to recover the value of the mean hitting time IE[τa,b |
X0 = x] of drifted Brownian motion Xt = x + Bt + µt. As above, the process
Xt − µt is a martingale the stopped process (Xτa,b ∧t − µ(τa,b ∧ t))t∈R+ is still
a martingale by Theorem 11.7. Hence the expectation IE[Xτa,b ∧t − µ(τa,b ∧ t)]
is constant in t ∈ R+ .
Since the stopped process (Xτa,b ∧t − µt)t∈R+ is a martingale, we have
x = IE[Xτa,b − µτa,b | X0 = x],
which gives
x = IE[Xτa,b − µτa,b | X0 = x]
= IE[Xτa,b | X0 = x] − µ IE[τa,b | X0 = x]
= bP(Xτa,b = b | X0 = x) + aP(Xτa,b = a | X0 = x) − µ IE[τa,b | X0 = x],
i.e. by (11.13a),
µ IE[τa,b | X0 = x] = bP(Xτa,b = b | X0 = x) + aP(Xτa,b = a | X0 = x) − x
e −2µa − e −2µx
e −2µx − e −2µb
+ a −2µa
−x
−2µa
−2µb
e
−e
e
− e −2µb
−2µa
−2µx
−2µx
−2µb
b( e
−e
) + a( e
−e
) − x( e −2µa − e −2µb )
=
,
−2µa
−2µb
e
−e
=b
hence
∗
Here we note that it can be showed that IE[τa,b ] < ∞ in order to apply (11.6).
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IE[τa,b | X0 = x] =
b( e −2µa − e −2µx ) + a( e −2µx − e −2µb ) − x( e −2µa − e −2µb )
,
µ( e −2µa − e −2µb )
a 6 x 6 b.
11.4 Perpetual American Options
The price of an American put option with finite expiration time T > 0 and
strike price K can be expressed as the expected value of its discounted payoff:
i
h
f (t, St ) =
sup
IE∗ e −r(τ −t) (K − Sτ )+ St ,
t6τ 6T
τ stopping time
under the risk-neutral probability measure P∗ , where the supremum is taken
over stopping times between t and a fixed maturity T . Similarly, the price of
a finite expiration American call option with strike price K is expressed as
i
h
f (t, St ) =
sup
IE∗ e −r(τ −t) (Sτ − K)+ St .
t6τ 6T
τ stopping time
In this section we take T = +∞, in which case we refer to these options as
perpetual options, and the corresponding put and call are respectively priced
as
i
h
f (t, St ) =
sup
IE∗ e −r(τ −t) (K − Sτ )+ St ,
τ >t
τ stopping time
and
f (t, St ) =
sup
τ >t
τ stopping time
i
h
IE∗ e −r(τ −t) (Sτ − K)+ St .
Two-choice optimal stopping at a fixed price level for perpetual
put options
In this section we consider the pricing of perpetual put options. Given L ∈
(0, K) a fixed price, consider the following choices for the exercise of a put
option with strike price K:
1. If St 6 L, then exercise at time t.
2. Otherwise if St > L, wait until the first hitting time
τL := inf{u > t : Su 6 L}
(11.16)
of the level L > 0, and exercise the option at time τL if τL < ∞.
Note that by definition of (11.16) we have τL = t if St 6 L.
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In case St 6 L, the payoff will be
(K − St )+ = K − St
since K > L > St , however in this case one would buy the option at price
K − St only to exercise it immediately for the same amount.
In case St > L, the price of the option will be
i
h
fL (t, St ) = IE∗ e −r(τL −t) (K − SτL )+ St
i
h
= IE∗ e −r(τL −t) (K − L)+ St
i
h
= (K − L) IE∗ e −r(τL −t) St .
(11.17)
We note that the starting date t does not matter when pricing perpetual
options, hence fL (t, x) is actually independent of t ∈ R+ , and the pricing of
the perpetual put option can be performed by taking t = 0 and in the sequel
we will work under
fL (t, x) = fL (x),
x > 0.
Recall that the underlying asset price is written as
St = S0 e rt+σB̃t −σ
2
t/2
,
t ∈ R+ ,
(11.18)
where (B̃t )t∈R+ is a standard Brownian motion under the risk-neutral probability measure P∗ , r is the risk-free interest rate, and σ > 0 is the volatility
coefficient.
Proposition 11.8. Assume that r > 0. We have
h
i
fL (x) = IE∗ e −r(τL −t) (K − SτL )+ St = x

K − x,
0 < x 6 L,


=
2

 (K − L) x −2r/σ , x > L.
L
Proof. We take t = 0 without loss of generality.
i) The result is obvious for S0 = x 6 L since in this case we have τL = t and
SτL = S0 = x, so that we only focus on the case x > L.
ii) Next, we consider the case S0 = x > L. By the relation
h
h
i
i
IE∗ e −rτL (K − SτL )+ S0 = x = IE∗ 1{τL <∞} e −rτL (K − SτL )+ S0 = x
h
i
= IE∗ 1{τL <∞} e −rτL (K − L)S0 = x
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i
h
= (K − L) IE∗ e −rτL S0 = x ,
(11.19)
∗
we note that it suffices to compute IE e −rτL S0 = x . For this, we note that
(λ)
from (11.18), for all λ ∈ R the process Zt t∈R+ defined as
(λ)
Zt
:= Stλ e −t(rλ−λ(1−λ)σ
2
/2)
= S0λ e λσB̃t −λ
2
σ 2 t/2
,
t ∈ R+ ,
is a martingale under the risk-neutral probability measure P∗ . Choosing λ
such that
σ2
r = rλ − λ(1 − λ) ,
(11.20)
2
we have
(λ)
Zt = Stλ e −rt ,
t ∈ R+ .
The equation (11.20) rewrites as
0 = λ2 σ 2 /2 + λ(r − σ 2 /2) − r =
with solutions
σ2
(λ + 2r/σ 2 )(λ − 1),
2
(11.21)
λ+ = 1 and λ− = −2r/σ 2 .
Choosing the negative solution∗ λ− = −2r/σ 2 < 0, we have
(λ− )
0 6 Zt
λ
= St − e −rt 6 Lλ− ,
0 6 t < τL ,
(11.22)
(λ )
limt→∞ Zt −
since r > 0. Next, we note that
= 0, and
Lλ− IE∗ e −rτL = IE∗ SτλL− e −rτL 1{τL <∞}
h
i
= IE∗ Zτ(λL− ) 1{τL <∞}
h
i
(λ )
= IE∗ lim ZτL−
∧t
t→∞
h
i
(λ )
= lim IE∗ ZτL−
∧t
t→∞
(λ ) = IE∗ Z0 −
(11.23)
(11.24)
λ
= S0 − ,
where by (11.22) we used the dominated convergence theorem from (11.23)
to (11.24), hence we find
Note that P(τL = ∞) > 0 since (St )t∈R+ is a submartingale, cf. (11.14), and the bound
(11.22) does not hold for the positive solution λ+ = 1.
∗
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x −2r/σ2
,
IE∗ e −rτL S0 = x =
L
x > L,
(11.25)
and we conclude by (11.19), which shows that
IE∗ e −rτL (K − SτL )+ S0 = x = (K − L) IE∗ e −rτL S0 = x
x −2r/σ2
,
= (K − L)
L
when S0 = x > L.
We note that taking L = K would yield a payoff always equal to 0 for the
option holder, hence the value of L should be strictly lower than K. On the
other hand, if L = 0 the value of τL will be infinite almost surely, hence
the option price will be 0 when r > 0 from (11.17). Therefore there should
be an optimal value L∗ , which should be strictly comprised between 0 and K.
Figure 11.4 shows for K = 100 that there exists an optimal value L∗ =
85.71 which maximizes the option price for all values of the underlying.
35
L=75
L=L*=85.71
L=98
(K-x)+
30
Option price
25
20
15
10
5
0
70
80
90
100
110
120
Underlying x
Fig. 11.4: Graphs of the option price by exercise at τL for several values of L.
In order to compute L∗ we observe that, geometrically, the slope of fL (x) at
x = L∗ is equal to −1, i.e.
2
fL0 ∗ (L∗ ) = −
i.e.
or
"
2r
(L∗ )−2r/σ −1
(K − L∗ )
= −1,
2
σ
(L∗ )−2r/σ2
2r
(K − L∗ ) = L∗ ,
σ2
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L∗ =
2r
K < K.
2r + σ 2
The same conclusion can be reached by the vanishing of the derivative of
L 7−→ fL (x):
x −2r/σ
∂fL (x)
2r K − L x −2r/σ
=−
+ 2
= 0,
∂L
L
σ
L
L
2
2
cf. page 351 of [Shr04]. The next Figure 11.5 is a 2-dimensional animation
that also shows the optimal value L∗ of L.
Fig. 11.5: Animated graph of the option price depending on the values of L.∗
The next Figure 11.6 gives another view of the put option prices according
to different values of L, with the optimal value L∗ = 85.71.
(K-x)+
fL(x)
fL*(x)
K-L
30
25
20
15
10
5
0
70
75
80
75
85
70
90
65
Underlying x 95 100 105
110 60
80
85
90
100
95
L
Fig. 11.6: Option price as a function of L and of the underlying asset price.
∗
The animation works in Acrobat Reader on the entire pdf file.
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In Figure 11.7, which is based on the stock price of HSBC Holdings (0005.HK)
over year 2009, the optimal exercise strategy for an American put option with
strike price K=$77.67 would have been to exercise whenever the underlying
price goes above L∗ = $62, i.e. at approximately 54 days, for a payoff of
$38. Note that waiting a longer time, e.g. until 85 days, would have yielded a
higher payoff of at least $65. This is due to the fact that, here, optimization
is done based on the past information only and makes sense in expectation
(or average) over all possible future paths.
Payoff (K-x)+
American put price
Option price path
L*
80
70
60
50
40
30
20
10
0
0
50
100
150
Time in days
200
100
90
80
70
50
60
underlying HK$
40
30
Fig. 11.7: Path of the American put option price on the HSBC stock.
PDE approach
We can check by hand calculations that the function

2r


K − x,
0 < x 6 L∗ =
K,


2r + σ 2

fL∗ (x) :=
−2r/σ2


Kσ 2
2r + σ 2 x
2r


, x > L∗ =
K,

2
2r + σ
2r K
2r + σ 2
(11.26)
satisfies the PDE

0 < x 6 L∗ < K,
 −rK < 0,
1 2 2 00
0
∗
− rfL (x) + rxfL∗ (x) + σ x fL∗ (x) =

2
0,
x > L∗ .
(11.27)
in addition to the conditions

0 < x 6 L∗ < K,
 fL∗ (x) = K − x,

fL∗ (x) > (K − x)+ , x > L∗ ,
which can be checked from (11.26).
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The above statements can be summarized in the following set of differential
inequalities, or variational differential equation:


(11.28a)
fL∗ (x) > (K − x)+ ,










σ 2 2 00


rxfL0 ∗ (x) +
(11.28b)
x fL∗ (x) 6 rfL∗ (x),
2






σ 2 2 00

0

∗ (x) − rxf ∗ (x) −
x
f
(x)
(fL∗ (x) − (K − x)+ ) = 0, (11.28c)
rf

∗
L
L
L


2



which admits an interpretation in terms of absence of arbitrage, as shown
below.
By (11.27) and Itô’s formula the discounted portfolio price f˜L∗ (St ) =
e −rt fL∗ (St ) satisfies
d(f˜L∗ (St ))
1
= − rfL∗ (St ) + rSt fL0 ∗ (St ) + σ 2 St2 fL00∗ (St ) e −rt dt + e −rt σSt fL0 ∗ (St )dB̃t
2
= −1{St 6L∗ } rK + e −rt σSt fL0 ∗ (St )dB̃t
= −1{fL∗ (St )6(K−St )+ } rK + e −rt σSt fL0 ∗ (St )dB̃t .
(11.29)
In other words, from Equation (11.28c), f˜L∗ (St ) is a martingale when
fL∗ (St ) > (K − St )+ ,
i.e.
St > L∗ ,
and the expected rate of return of the option price fL∗ (St ) then equals the
rate r of the risk-free asset as
d fL∗ (St ) = d e rt f˜L∗ (St ) = rfL∗ (St )dt + e rt df˜L∗ (St ),
and the investor prefers to wait.
On the other hand if fL∗ (St ) = (K − St )+ , i.e. 0 < St < L∗ , it is not
worth waiting as (11.28b) and (11.28c) show that the return of the option is
lower than that of the risk-free asset, i.e.:
1
−rfL∗ (St ) + rSt fL0 ∗ (St ) + σ 2 St2 fL00∗ (St ) = −rK < 0,
2
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and exercise becomes immediate since the process f˜L∗ (St ) becomes a (strict)
supermartingale. In this case, (11.28c) implies fL∗ (x) = (K − x)+ .
In view of the above derivation it should make sense to assert that fL∗ (St )
is the price at time t of the perpetual American put option. The next proposition shows that this is indeed the case, and that the optimal exercise time
is τ ∗ = τL∗ .
Proposition 11.9. The price of the perpetual American put option is given
for all t > 0 by
fL∗ (St ) = sup
τ >t
τ stopping time
i
h
IE∗ e −r(τ −t) (K − Sτ )+ St
i
h
= IE∗ e −r(τL∗ −t) (K − SτL∗ )+ St

K − St , 0 < St 6 L∗ ,



−2r/σ2
=

2r + σ 2 St
Kσ 2


,
S t > L∗ .
2r + σ 2
2r K
Proof. i) Since the drift
1
−rfL∗ (St ) + rSt fL0 ∗ (St ) + σ 2 St2 fL00∗ (St )
2
in Itô’s formula (11.29) is nonpositive by the inequality (11.28b), the discounted portfolio price
u 7−→ e −ru fL∗ (Su ),
u ∈ [t, ∞),
is a supermartingale. As a consequence, for all (a.s. finite) stopping times
τ ∈ [t, ∞) we have, by (11.12),
i
i
h
h
e −rt fL∗ (St ) > IE∗ e −rτ fL∗ (Sτ )St > IE∗ e −rτ (K − Sτ )+ St ,
from (11.28a), which implies
e −rt fL∗ (St ) >
sup
τ >t
τ stopping time
i
h
IE∗ e −rτ (K − Sτ )+ St .
(11.30)
ii) The converse inequality is obvious by Proposition 11.8 as
i
h
fL∗ (St ) = IE∗ e −r(τL∗ −t) (K − SτL∗ )+ St
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6
sup
τ >t
τ stopping time
i
h
IE∗ e −r(τ −t) (K − Sτ )+ St ,
(11.31)
since τL∗ is a stopping time larger than t ∈ R+ . The inequalities (11.30) and
(11.31) allow us to conclude to the equality
i
h
fL∗ (St ) =
sup
IE∗ e −r(τ −t) (K − Sτ )+ St .
τ >t
τ stopping time
Remark. Note that the converse inequality (11.31) can also be obtained
from the variational PDE (11.28a)-(11.28c) itself, without relying on Proposition 11.8. For this, taking τ = τL∗ we note that the process
u 7−→ e −ru∧τL∗ fL∗ (Su∧τL∗ ),
u > t,
is not only a supermartingale, it is also a martingale until exercise at time
τL∗ by (11.27) since Su∧τL∗ > L∗ , hence we have
i
h
e −rt fL∗ (St ) = IE∗ e −r(u∧τL∗ ) fL∗ (Su∧τL∗ )St ,
u > t,
hence after letting u tend to infinity we obtain
i
h
e −rt fL∗ (St ) = IE∗ e −rτL∗ fL∗ (SτL∗ )St
i
h
= IE∗ e −rτL∗ fL∗ (L∗ )St
i
h
= IE∗ e −rτL∗ (K − SτL∗ )+ St
i
h
6
sup
IE∗ e −rτL∗ (K − SτL∗ )+ St ,
τ >t
τ stopping time
which shows that
e −rt fL∗ (St ) 6
sup
τ >t
τ stopping time
i
h
IE∗ e −rτ (K − Sτ )+ St ,
t ∈ R+ .
Two-choice optimal stopping at a fixed price level for perpetual
call options
In this section we consider the pricing of perpetual call options. Given L > K
a fixed price, consider the following choices for the exercise of a call option
with strike price K:
1. If St > L, then exercise at time t.
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2. Otherwise, wait until the first hitting time
τL = inf{u > t : Su = L}
and exercise the option and time τL .
In case St > L, the payoff will be
(St − K)+ = St − K
since K < L 6 St .
In case St < L, the price of the option will be
h
i
fL (St ) = IE∗ e −r(τL −t) (SτL − K)+ St
h
i
= IE∗ e −r(τL −t) (L − K)+ St
h
i
= (L − K) IE∗ e −r(τL −t) St .
Proposition 11.10. We have
fL (x) =


 x − K,
x > L > K,

 (L − K) x ,
L
0 < x 6 L.
(11.32)
Proof. We only need to consider the case S0 = x < L. Note that for all λ ∈ R
we have
Zt := Stλ e −rλt+λσ
2
t/2−λ2 σ 2 t/2
2
= S0λ e λσB̃t −λ
σ 2 t/2
,
t ∈ R+ ,
is a martingale under the risk-neutral probability measure P̃. Hence the
stopped process (Zt∧τL )t∈R+ is a martingale and it has constant expectation, i.e. we have
IE∗ [Zt∧τL ] = IE∗ [Z0 ] = S0λ ,
t ∈ R+ .
(11.33)
Choosing λ such that
r = rλ − λσ 2 /2 + λ2 σ 2 /2,
i.e.
0 = λ2 σ 2 /2 + λ(r − σ 2 /2) − r =
σ2
(λ + 2r/σ 2 )(λ − 1),
2
Relation (11.33) rewrites as
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IE∗ (St∧τL )λ e −r(t∧τL ) = S0λ ,
(11.34)
t ∈ R+ .
Choosing the positive solution λ+ = 1 yields the bound
∗
0 6 St∧τL e −r(t∧τL ) 6 L,
(11.35)
t ∈ R+ ,
since S0 = x < L. Hence by letting t go to infinity in (11.34) and from (11.35)
and the dominated convergence theorem we get
L IE∗ e −rτL = IE∗ SτL e −rτL 1{τL <∞}
h
i
= IE∗ lim SτL ∧t e −r(τL ∧t)
t→∞
= lim IE∗ SτL ∧t e −r(τL ∧t)
t→∞
= S0 ,
which yields
S0
.
IE∗ e −rτL =
L
(11.36)
One sees from Figure 11.8 that the situation completely differs from the
perpetual put option case, as there does not exist an optimal value L∗ that
would maximize the option price for all values of the underlying.
450
L=150
L=250
L=400
(x-K)+
x
400
350
Option price
300
250
200
150
100
50
0
0
50
100
150
200
250
Underlying
300
350
400
450
Fig. 11.8: Graphs of the option price by exercising at τL for several values of L.
The intuition behind this picture is that there is no upper limit above which
one should exercise the option, and in order to price the American perpetual
call option we have to let L go to infinity, i.e. the “optimal” exercise strategy
is to wait indefinitely.
Note that P(τL = ∞) = 0 since (St )t∈R+ is a submartingale, cf. (11.15), and the bound
(11.35) does not hold for the negative solution λ− = −2r/σ 2 .
∗
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+
(x-K)
fL(x)
K-L
180
160
140
120
100
80
60
40
20
0
300
150
250
200
L
150
250
200
Underlying x
100
100
Fig. 11.9: Graphs of the option prices parametrized by different values of L.
We check from (11.32) that
lim fL (x) = x − lim K
L→∞
L→∞
x
= x,
L
x > 0.
(11.37)
As a consequence we have the following proposition.
Proposition 11.11. Assume that r > 0. The price of the perpetual American
call option is given by
IE∗ e −r(τ −t) (Sτ − K)+ St = St ,
sup
t ∈ R+ .
(11.38)
τ >t
τ stopping time
Proof. For all L > K we have
fL (St ) = IE∗ e −r(τL −t) (SτL − K)+ St
6
sup
IE∗ e −r(τ −t) (Sτ − K)+ St ,
t ∈ R+ ,
τ >t
τ stopping time
hence from (11.37), taking the limit as L → ∞ yields
St 6
sup
IE∗ e −r(τ −t) (Sτ − K)+ St .
(11.39)
τ >t
τ stopping time
On the other hand, since u 7−→ e −r(u−t) Su is a martingale, by (11.12) we
have, for all stopping times τ ∈ [t, ∞),
IE∗ e −r(τ −t) (Sτ − K)+ St 6 IE∗ e −r(τ −t) Sτ St 6 St ,
t ∈ R+ ,
hence
sup
IE∗ e −r(τ −t) (Sτ − K)+ St 6 St ,
t ∈ R+ ,
τ >t
τ stopping time
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N. Privault
which shows (11.38) by (11.39).
We may also check that since ( e −rt St )t∈R+ is a martingale, the process t 7−→
( e −rt St −K)+ is a submartingale since the function x 7−→ (x−K)+ is convex,
hence for all bounded stopping times τ such that t 6 τ we have
h
h
i
i
(St − K)+ 6 IE∗ ( e −r(τ −t) Sτ − K)+ St 6 IE∗ e −r(τ −t) (Sτ − K)+ St ,
t ∈ R+ , showing that it is always better to wait than to exercise at time t,
and the optimal exercise time is τ ∗ = +∞. This argument does not apply to
American put options.
11.5 Finite Expiration American Options
In this section we consider finite expirations American put and call options
with strike price K, whose prices can be expressed as
h
i
f (t, St ) =
sup
IE∗ e −r(τ −t) (K − Sτ )+ St ,
t6τ 6T
τ stopping time
and
f (t, St ) =
sup
h
i
IE∗ e −r(τ −t) (Sτ − K)+ St .
t6τ 6T
τ stopping time
Two-choice optimal stopping at fixed times with finite expiration
We start by considering the optimal stopping problem in a simplified setting
where τ ∈ {t, T } is allowed at time t to take only two values t (which corresponds to immediate exercise) and T (wait until maturity).
Call options
Since x 7−→ (x − K)+ is a nondecreasing convex function and the process
( e −rt St − e −rt K)t∈R+ is a submartingale under P∗ , we know that t 7−→
( e −rt St − e −rt K)+ is a submartingale by the Jensen inequality (11.2), hence
by (11.12), for any stopping time τ 6 t we have
(St − K)+ = e rt ( e −rt St − e −rt K)+
6 e rt IE∗ [( e −rτ Sτ − e −rτ K)+ | Ft ]
6 IE∗ [ e −r(τ −t) (Sτ − K)+ | Ft ],
i.e.
(x − K)+ 6 IE∗ [ e −r(τ −t) (Sτ − K)+ |St = x],
x, t > 0.
In particular for τ = T > t a deterministic time we get
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(x − K)+ 6 e −r(T −t) IE∗ [(ST − K)+ |St = x],
x, t > 0.
as illustrated in Figure 11.10 using the Black-Scholes formula for European
call options.
In other words, taking x = St , the payoff (St −K)+ of immediate exercise at
time t is always lower than the expected payoff e −r(T −t) IE∗ [(ST − K)+ |St =
x] given by exercise at maturity T . As a consequence, the optimal strategy
for the investor is to wait until time T to exercise an American call option,
rather than exercising earlier at time t.
Black-Scholes European call price
Payoff (x-K)+
80
70
60
50
40
30
20
10
0
140
120
underlying 100
HK$
80
60
10 9
8
7
2 1 0
5 4to 3maturity T-t
6 Time
Fig. 11.10: Expected Black-Scholes European call price vs (x, t) 7−→ (x − K)+ .
More generally it can be in fact shown that the price of an American call option equals the price of the corresponding European call option with maturity
T , i.e.
i
h
f (t, St ) = e −r(T −t) IE∗ (ST − K)+ St ,
i.e. T is the optimal exercise date, cf. e.g. §14.4 of [Ste01] for a proof.
Put options
For put options the situation is entirely different. The Black-Scholes formula
for European put options shows that the inequality
(K − x)+ 6 e −r(T −t) IE∗ [(K − ST )+ |St = x],
does not always hold, as illustrated in Figure 11.11.
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Black-Scholes European put price
Payoff (K-x)+
16
14
12
10
8
6
4
2
0
0
1
2
3
4
5
Time to maturity T-t 6
7
8
9
10
90
100
110 underlying HK$
120
Fig. 11.11: Black-Scholes put price function vs (x, t) 7−→ (K − x)+ .
As a consequence, the optimal exercise decision for a put option depends on
whether (K − St )+ 6 e −r(T −t) IE∗ [(K − ST )+ |St ] (in which case one chooses
to exercise at time T ) or (K − St )+ > e −r(T −t) IE∗ [(K − ST )+ |St ] (in which
case one chooses to exercise at time t).
A view from above of the graph of Figure 11.11 shows the existence of an
optimal frontier depending on time to maturity and on the value of the underlying asset instead of being given by a constant level L∗ as in Section 11.4,
cf. Figure 11.12:
0
1
2
3
4
5
T-t
6
7
8
9
10
120
115
110
100
105
underlying HK$
95
90
Fig. 11.12: Optimal frontier for the exercise of a put option.
At a given time t, one will choose to exercise immediately if (St , T −t) belongs
to the blue area on the right, and to wait until maturity if (St , T − t) belongs
to the red area on the left.
PDE characterization of the finite expiration American put price
Let us describe the PDE associated to American put options. After discretization {0 = t0 < t1 < . . . < tN = T } of the time interval [0, T ], the optimal
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exercise strategy for the American put option can be described as follow at
each time step:
If f (t, St ) > (K − St )+ , wait.
If f (t, St ) = (K − St )+ , exercise the option at time t.
Note that we cannot have f (t, St ) < (K − St )+ .
If f (t, St ) > (K − St )+ the expected return of the option equals that of
the risk-free asset. This means that f (t, St ) follows the Black-Scholes PDE
rf (t, St ) =
∂f
∂f
1
∂2f
(t, St ) + rSt (t, St ) + σ 2 St2 2 (t, St ),
∂t
∂x
2
∂x
whereas if f (t, St ) = (K − St )+ it is not worth waiting as the return of the
option is lower than that of the risk-free asset:
rf (t, St ) >
∂f
∂f
1
∂2f
(t, St ) + rSt (t, St ) + σ 2 St2 2 (t, St ).
∂t
∂x
2
∂x
As a consequence, f (t, x) should solve the following variational PDE:


f (t, x) > (K − x)+ ,









 ∂f
∂f
σ2 2 ∂ 2 f


(t, x) + rx (t, x) +
x
(t, x) 6 rf (t, x),


∂x
2
∂x2
 ∂t




∂f
∂f
σ2 2 ∂ 2 f


(t,
x)
+
rx
(t,
x)
+
x
(t,
x)
−
rf
(t,
x)


∂x
2
∂x2
 ∂t




× (f (t, x) − (K − x)+ ) = 0,



(11.40a)
(11.40b)
(11.40c)
subject to the terminal condition f (T, x) = (K − x)+ . In other words, equality holds either in (11.40a) or in (11.40b) due to the presence of the term
(f (t, x) − (K − x)+ ) in (11.40c).
The optimal exercise strategy consists in exercising the put option as soon
as the equality f (u, Su ) = (K − Su )+ holds, i.e. at the time
τ ∗ = inf{u > t : f (u, Su ) = (K − Su )+ },
after which the process f˜L∗ (St ) ceases to be a martingale and becomes a
(strict) supermartingale.
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A simple procedure to compute numerically the price of an American put
option is to use a finite difference scheme while simply enforcing the condition
f (t, x) > (K − x)+ at every iteration by adding the condition
f (ti , xj ) := max(f (ti , xj ), (K − xj )+ )
right after the computation of f (ti , xj ).
The next figure shows a numerical resolution of the variational PDE
(11.40a)-(11.40c) using the above simplified (implicit) finite difference scheme,
see also [Jac91] for properties of the optimal boundary function. In comparison with Figure 11.7, one can check that the PDE solution becomes timedependent in the finite expiration case.
Finite expiration American put price
Immediate payoff (K-x)+
L*=2r/(2r+sigma2)
16
14
12
10
8
6
4
2
0
0
2
4
Time to maturity T-t
90
6
8
110
10
100
underlying
120
Fig. 11.13: Numerical values of the finite expiration American put price.
In general, one will choose to exercise the put option when
f (t, St ) = (K − St )+ ,
i.e. within the blue area in Figure (11.13). We check that the optimal threshold L∗ = 90.64 of the corresponding perpetual put option is within the exercise region, which is consistent since the perpetual optimal strategy should
allow one to wait longer than in the finite expiration case.
The numerical computation of the put price
i
h
f (t, St ) =
sup
IE∗ e −r(τ −t) (K − Sτ )+ St
t6τ 6T
τ stopping time
can also be done by dynamic programming and backward optimization using the Longstaff-Schwartz (or Least Square Monte Carlo, LSM) algorithm
[LS01], as in Figure 11.14.
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Longstaff-Schwartz algorithm
+
Immediate
payoff (K-x)
L*=2r/(2r+sigma2)
16
14
12
10
8
6
4
2
0
0
2
4
Time to maturity T-t
90
6
8
110
10
100
underlying
120
Fig. 11.14: Longstaff-Schwartz algorithm for the finite expiration American put price.
In Figure 11.14 above and Figure 11.15 below the optimal threshold of the
corresponding perpetual put option is again L∗ = 90.64 and falls within the
exercise region. Also, the optimal threshold is closer to L∗ for large time to
maturities, which shows that the perpetual option approximates the finite
expiration option in that situation. In the next Figure 11.15 we compare the
numerical computation of the American put price by the finite difference and
Longstaff-Schwartz methods.
10
Longstaff-Schwartz algorithm
Implicit finite differences
+
Immediate payoff (K-x)
Time to maturity T-t
8
6
4
2
0
90
100
110
120
underlying
Fig. 11.15: Comparison between Longstaff-Schwartz and finite differences.
It turns out that, although both results are very close, the Longstaff-Schwartz
method performs better in the critical area close to exercise at it yields the
expected continuously differentiable solution, and the simple numerical PDE
solution tends to underestimate the optimal threshold. Also, a small error
in the values of the solution translates into a large error on the value of the
optimal exercise threshold.
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The finite expiration American call option
In the next proposition we compute the price of a finite expiration American
call option with an arbitrary convex payoff function φ.
Proposition 11.12. Let φ : R −→ R be a nonnegative convex function such
that φ(0) = 0. The price of the finite expiration American call option with
payoff function φ on the underlying asset (St )t∈R+ is given by
i
i
h
h
f (t, St ) =
sup
IE∗ e −r(τ −t) φ(Sτ )St = e −r(T −t) IE∗ φ(ST )St ,
t6τ 6T
τ stopping time
i.e. the optimal strategy is to wait until the maturity time T to exercise the
option, and τ ∗ = T .
Proof. Since the function φ is convex and φ(0) = 0 we have
φ(px) = φ((1 − p) × 0 + px) 6 (1 − p) × φ(0) + pφ(x) = pφ(x),
for all p ∈ [0, 1] and x > 0. Hence the process s 7−→ e −rs φ(St+s ) is a
submartingale since taking p = e −r(τ −t) we have
e −rs IE∗ [φ (St+s ) | Ft ] > e −rs φ (IE∗ [St+s | Ft ])
> φ e −rs IE∗ [St+s | Ft ]
= φ(St ),
where we used Jensen’s inequality (11.2) applied to the convex function φ.
Hence by the optional stopping theorem for submartingales, cf (11.10), for
all (bounded) stopping times τ comprised between t and T we have,
IE∗ [ e −r(τ −t) φ(Sτ ) | Ft ] 6 e −r(T −t) IE∗ [φ(ST ) | Ft ],
i.e. it is always better to wait until time T than to exercise at time τ ∈ [t, T ],
and this yields
i
i
h
h
sup
IE∗ e −r(τ −t) φ(Sτ )St 6 e −r(T −t) IE∗ φ(ST )St .
t6τ 6T
τ stopping time
The converse inequality
i
h
e −r(T −t) IE∗ φ(ST )St 6
sup
t6τ 6T
τ stopping time
i
h
IE∗ e −r(τ −t) φ(Sτ )St ,
being obvious because T is a stopping time.
As a consequence of Proposition 11.12 applied to the convex function φ(x) =
(x − K)+ , the price of the finite expiration American call option is given by
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f (t, St ) =
sup
t6τ 6T
τ stopping time
i
h
IE∗ e −r(τ −t) (Sτ − K)+ St
i
h
= e −r(T −t) IE∗ (ST − K)+ St ,
i.e. the optimal strategy is to wait until the maturity time T to exercise the
option.
In the following table we summarize the optimal exercise strategies for the
pricing of American options.
Option
type
Perpetual
Finite expiration

K − St , 0 < St 6 L∗ ,



Put
−2r/σ2
option 

St
 (K − L∗ )
, St > L∗ .
L∗
τ ∗ = τL∗
Call
option
St ,
τ ∗ = +∞.
Solve the PDE (11.40a)-(11.40c) for f (t, x)
or use Longstaff-Schwartz [LS01].
τ ∗ = T ∧ inf{u > t : f (u, Su ) = (K − Su )+ }.
e −r(T −t) IE∗ [(ST − K)+ | St ],
τ∗ = T.
Exercises
Exercise 11.1
i.e. B0 = 0.
Let (Bt )t∈R+ be a standard Brownian motion started at 0,
a) Is the process t 7−→ (2 − Bt )+ a submartingale, a martingale or a supermartingale?
b) Is the process t 7−→ e Bt a submartingale, a martingale, or a supermartingale?
c) Consider the random time ν defined by
ν := inf{t ∈ R+ : Bt = B2t },
which represents the first time the curves (Bt )t∈R+ and (B2t )t∈R+ intersect.
Is ν a stopping time?
d) Consider the random time τ defined by
τ := inf{t ∈ R+ : e Bt = (α + βt) e t/2 },
which represents the first time geometric Brownian motion Bt crosses the
straight line t 7−→ α + βt. Is τ a stopping time?
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N. Privault
e) If τ is a stopping time, compute IE[τ ] by the Doob stopping time theorem
in the cases (α > 1 and β < 0) or (α < 1 and β > 0).
Exercise 11.2 Stopping times. Let (Bt )t∈R+ be a standard Brownian motion
started at 0.
a) Consider the random time ν defined by
ν := inf{t ∈ R+ : Bt = B1 },
which represents the first time Brownian motion Bt hits the level B1 . Is
ν a stopping time?
b) Consider the random time τ defined by
τ := inf t ∈ R+ : e Bt = α e −t/2 ,
which represents the first time the exponential of Brownian motion Bt
crosses the path of t 7−→ α e −t/2 , where α > 1.
Is τ a stopping time? If τ is a stopping time, compute IE[ e −τ ] by the
stopping time theorem.
c) Consider the random time τ defined by
τ := inf t ∈ R+ : Bt2 = 1 + αt ,
which represents the first time the process (Bt2 )t∈R+ crosses the straight
line t 7−→ 1 + αt, with α ∈ (−∞, 1).
Is τ a stopping time? If τ is a stopping time, compute IE[τ ] by the Doob
stopping time theorem.
Exercise 11.3 Consider a standard Brownian motion (Bt )t∈R+ started at
B0 = 0, and let
τL = inf{t ∈ R+ : Bt = L}
denote the first hitting time of level L > 0.
a) Compute the Laplace transform IE[ e −rτL ] of τL for all r > 0.
√
Hint: Use the stopping time theorem and the fact that e 2rBt −rt t∈R+
is a martingale when r > 0.
b) Find the optimal level stopping strategy depending on the value of r > 0
for the maximization problem
sup IE e −rτL BτL .
L>0
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Exercise 11.4 (Doob-Meyer decomposition in discrete time). Let (Mn )n∈N
be a discrete-time submartingale with respect to a filtration (Fn )n∈N , with
F−1 = {∅, Ω}.
a) Show that there exists two processes (Nn )n∈N and (An )n∈N such that
i) (Nn )n∈N is a martingale with respect to (Fn )n∈N ,
ii) (An )n∈N is nondecreasing, i.e. An 6 An+1 a.s., n ∈ N,
iii) (An )n∈N is predictable in the sense that An is Fn−1 -measurable,
n ∈ N, and
iv) Mn = Nn + An , n ∈ N.
Hint: Let A0 := 0,
An+1 := An + IE[Mn+1 − Mn | Fn ],
n > 0,
and define (Nn )n∈N in such a way that it satisfies the four required properties.
b) Show that for all bounded stopping times σ and τ such that σ 6 τ a.s.,
we have
IE[Mσ ] 6 IE[Mτ ].
Hint: Use the stopping time Theorem 11.7 for martingales and (11.11).
Exercise 11.5 Consider a two-period binomial model (Sk )k=0,1,2 with interet
rate r = 0% and risk-neutral measure (p∗ , q ∗ ):
/3
p∗ = 2
/3
p∗ = 2
S1 = 1.2
q∗ = 1
/3
/3
p∗ = 2
S0 = 1
q∗ = 1
/3
S2 = 1.44
S2 = 1.08
S1 = 0.9
q∗ = 1
/3
S2 = 0.81
a) At time t = 1, would you exercise the American put with strike price
K = 1.25 if S1 = 1.2? If S1 = 0.9?
b) What would be your strategy at time t = 0?∗
∗
Download the corresponding discrete-time IPython notebook that can be run here.
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Exercise 11.6
Let r > 0 and σ > 0.
2
a) Show that for every C > 0, the function f (x) := Cx−2r/σ solves the
differential equation

1 2 2 00

0

 rf (x) = rxf (x) + σ x f (x),
2


 lim f (x) = 0.
x→∞
b) Show that for every K > 0 there exists a unique level L∗ ∈ (0, K) and
constant C > 0 such that f (x) also solves the smooth fit conditions
f (L∗ ) = K − L∗ and f 0 (L∗ ) = −1.
Exercise 11.7 American binary options. An American binary (or digital)
call (resp. put) option with maturity T > 0 can be exercised at any time
t ∈ [0, T ], at the choice of the option holder.
The call (resp. put) option exercised at time t yields the payoff 1[K,∞) (St )
(resp. 1[0,K] (St )), and the option holder wants to find an exercise strategy
that will maximize his payoff.
a) Consider the following possible situations at time t:
i) St > K,
ii) St < K.
In each case (i) and (ii), tell whether you would choose to exercise the
call option immediately or to wait.
b) Consider the following possible situations at time t:
i) St > K,
ii) St 6 K.
In each case (i) and (ii), tell whether you would choose to exercise the
put option immediately or to wait.
c) The price CdAm (t, St ) of an American binary call option is known to satisfy
the Black-Scholes PDE
rCdAm (t, x) =
∂CdAm
∂C Am
1
∂ 2 CdAm
(t, x) + rx d (t, x) + σ 2 x2
(t, x).
∂t
∂x
2
∂x2
Based on your answers to Question (a), how would you set the boundary
conditions CdAm (t, K), 0 6 t < T , and CdAm (T, x), 0 6 x < K?
d) The price PdAm (t, St ) of an American binary put option is known to satisfy
the same Black-Scholes PDE
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rPdAm (t, x) =
∂PdAm
∂P Am
1
∂ 2 PdAm
(t, x). (11.41)
(t, x) + rx d (t, x) + σ 2 x2
∂t
∂x
2
∂x2
Based on your answers to Question (b), how would you set the boundary
conditions PdAm (t, K), 0 6 t < T , and PdAm (T, x), x > K?
e) Show that the optimal exercise strategy for the American binary call option with strike price K is to exercise as soon as the underlying reaches
the level K, at the time
τK = inf{u > t : Su = K},
starting from any level St 6 K, and that the price CdAm (t, St ) of the
American binary call option is given by
CdAm (t, x) = IE[ e −r(τK −t) 1{τK <T } | St = x].
f) Show that the price CdAm (t, St ) of the American binary call option is equal
to
(r + σ 2 /2)(T − t) + log(x/K)
x
√
CdAm (t, x) = Φ
K
σ T −t
x −2r/σ2 −(r + σ 2 /2)(T − t) + log(x/K) √
+
,
0 6 x 6 K.
Φ
K
σ T −t
Show that this formula is consistent with your answer to Question (c).
g) Show that the optimal exercise strategy for the American binary put option with strike price K is to exercise as soon as the underlying reaches
the level K, at the time
τK = inf{u > t : Su = K},
starting from any level St > K, and that the price PdAm (t, St ) of the
American binary put option is
PdAm (t, x) = IE[ e −r(τK −t) 1{τK <T } | St = x],
x > K.
h) Show that the price PdAm (t, St ) of the American binary put option is equal
to
x
−(r + σ 2 /2)τ − log(x/K)
√
PdAm (t, x) = Φ
K
σ τ
x −2r/σ2 (r + σ 2 /2)τ − log(x/K) √
+
Φ
,
x > K,
K
σ τ
and that this formula is consistent with your answer to Question (d).
i) Does the call-put parity hold for American binary options?
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Exercise 11.8 American forward contracts. Consider (St )t∈R+ an asset price
process given by
dSt
= rdt + σdBt ,
St
where r > 0 and (Bt )t∈R+ is a standard Brownian motion.
a) Compute the price
f (t, St ) =
sup
t6τ 6T
τ stopping time
i
h
IE∗ e −r(τ −t) (K − Sτ )St ,
and optimal exercise strategy of a finite expiration American type short
forward contract with strike price K on the underlying asset (St )t∈R+ ,
with payoff K − ST .
b) Compute the price
i
h
f (t, St ) =
sup
IE∗ e −r(τ −t) (Sτ − K)St ,
t6τ 6T
τ stopping time
and optimal exercise strategy of a finite expiration American type long
forward contract with strike price K on the underlying asset (St )t∈R+ ,
with payoff ST − K.
c) How are the answers to Questions (a) and (b) modified in the case of
perpetual options?
Exercise 11.9
Consider an underlying asset price process written as
St = S0 e rt+σB̃t −σ
2
t/2
,
t ∈ R+ ,
where (B̃t )t∈R+ is a standard Brownian motion under the risk-neutral probability measure P∗ , with σ, r > 0.
a) Show that the processes (Yt )t∈R+ and (Zt )t∈R+ defined as
−2r/σ 2
Yt := e −rt St
and Zt := e −rt St ,
t ∈ R+ ,
are both martingales under P∗ .
b) Let τL denote the hitting time
τL = inf{u ∈ R+ : Su = L}.
By application of the stopping time theorem to the martingales (Yt )t∈R+
and (Zt )t∈R+ , show that
∗
IE
e
−rτL
| S0 = x =
0 < x 6 L,

 x/L,

2
(x/L)−2r/σ ,
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American Options
c) Compute the price IE∗ [ e −rτL (K − SτL )] of a short forward contract under
the exercise strategy τL .
d) Show that for every value of S0 = x there is an optimal value L∗x of L
that maximizes L 7−→ IE[ e −rτL (K − SτL )].
e) Would you use the strategy
τL∗x = inf{u ∈ R+ : Su = L∗x }
as an optimal exercise strategy for the short forward contract with payoff
K − Sτ ?
Exercise 11.10 Let p > 1 and consider a power put option with payoff

 (κ − Sτ )p if Sτ 6 κ,
((κ − Sτ )+ )p =

0
if Sτ > κ,
when exercised at time τ , on an underlying asset whose price St is written as
St = S0 e rt+σBt −σ
2
t/2
,
t ∈ R+ ,
where (Bt )t∈R+ is a standard Brownian motion under the risk-neutral probability measure P∗ , r > 0 is the risk-free interest rate, and σ > 0 is the
volatility coefficient.
Given L ∈ (0, κ) a fixed price, consider the following choices for the exercise
of a put option with strike price κ:
i) If St 6 L, then exercise at time t.
ii) Otherwise, wait until the first hitting time τL := inf{u > t : Su = L},
and exercise the option at time τL .
a) Under the above strategy, what is the option payoff equal to if St 6 L ?
b) Show that in case St > L, the price of the option is equal to
i
h
fL (St ) = (κ − L)p IE∗ e −r(τL −t) St .
c) Compute the price fL (St ) of the option at time t.
−2r/σ 2
Hint. Recall that by (11.25) we have IE∗ [ e −r(τL −t) | St = x] = (x/L)
,
x > L.
d) Compute the optimal value L∗ that maximizes L 7−→ fL (x) for all fixed
x > 0.
Hint. Observe that, geometrically, the slope of x 7−→ fL (x) at x = L∗ is
equal to −p(κ − L∗ )p−1 .
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e) How would you compute the American option price
f (t, St ) =
sup
IE∗ e −r(τ −t) ((κ − Sτ )+ )p St ?
τ >t
τ stopping time
Exercise 11.11 Same questions as in Exercise 11.10, this time for the option
with payoff κ − (Sτ )p when exercised at time τ , with p > 0.
Exercise 11.12 American put options with dividends, cf. Exercise 8.5 in
[Shr04]. Consider a dividend-paying asset priced as
St = S0 e (r−δ)t+σB̃t −σ
2
t/2
,
t ∈ R+ ,
where r > 0 is the risk-free interest rate, δ > 0 is the continuous dividend rate,
(B̃t )t∈R+ is a standard Brownian motion under the risk-neutral probability
measure P∗ , and σ > 0 is the volatility coefficient. Consider the american put
option with payoff

 κ − Sτ if Sτ 6 κ,
(κ − Sτ )+ =

0
if Sτ > κ,
when exercised at the stopping time τ > 0. Given L ∈ (0, κ) a fixed level,
consider the following exercise strategy for the above option:
- If St 6 L, then exercise at time t.
- If St > L, wait until the hitting time τL := inf{u > t : Su = L}, and
exercise the option at time τL .
a) Give the intrinsic option payoff at time t = 0 in case S0 6 L.
In the sequel we work with S0 = x > L.
(λ) b) Show that for all λ ∈ R the process Zt t∈R+ defined as
(λ)
:=
Zt
St
S0
λ
e −t((r−δ)λ−λ(1−λ)σ
2
/2)
is a martingale under the risk-neutral probability measure P∗ .
(λ) c) Show that Zt t∈R+ can be rewritten as
(λ)
Zt
=
St
S0
λ
e −rt ,
t ∈ R+ ,
for two values λ− 6 0 6 λ+ of λ that can be computed explicitly.
d) Choosing the negative solution λ− , show that
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American Options
(λ− )
0 6 Zt
=
St
S0
λ−
e −rt 6
L
S0
λ−
,
0 6 t < τL .
e) Let τL denote the hitting time
τL = inf{u ∈ R+ : Su 6 L}.
By application of the stopping time theorem to the martingale (Zt )t∈R+ ,
show that
λ−
S0
IE∗ e −rτL =
,
(11.42)
L
with
λ− :=
−(r − δ − σ 2 /2) −
p
(r − δ − σ 2 /2)2 + 4rσ 2 /2
.
σ2
(11.43)
f) Show that for all L ∈ (0, K) we have
h
i
IE∗ e −rτL (K − SτL )+ S0 = x

0 < x 6 L,

 K − x,

√
=
2 /2)2 +4rσ 2 /2
−(r−δ−σ2 /2)− (r−δ−σ


σ2
 (K − L) x
, x > L.
L
g) Show that the value L∗ of L that maximizes
h
i
fL (x) := IE∗ e −rτL (K − SτL )+ S0 = x
for all x is given by
L∗ =
λ−
K.
λ− − 1
h) Show that
fL∗ (x) =



K − x,



0 < x 6 L∗ =
λ−
K,
λ− − 1
λ −1 λ−


1 − λ− −
x
λ−



, x > L∗ =
K,
K
−λ−
λ− − 1
i) Show by hand computation that fL∗ (x) satisfies the variational differential
equation
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N. Privault


fL∗ (x) > (K − x)+ ,










1 2 2 00

0

 (r − δ)xfL∗ (x) + σ x fL∗ (x) 6 rfL∗ (x),
2





1



rfL∗ (x) − (r − δ)xfL0 ∗ (x) − σ 2 x2 fL00∗ (x)


2




× (f ∗ (x) − (K − x)+ ) = 0.
(11.44a)
(11.44b)
(11.44c)
L
j) By Itô’s formula, check that the discounted portfolio price
t 7−→ e −rt fL∗ (St )
is a supermartingale.
k) Show that we have
fL∗ (S0 ) >
τ
sup
IE∗ e −rτ (K − Sτ )+ S0 .
stopping time
l) Show that the stopped process
s 7−→ e −r(s∧τL∗ ) fL∗ (Ss∧τL∗ ),
s ∈ R+ ,
is a martingale, and that
fL∗ (S0 ) 6
τ
sup
IE∗ e −rτ (K − Sτ )+ .
stopping time
m) Fix t ∈ R+ and let τL∗ denote the hitting time
τL∗ = inf{u > t : Su = L∗ }.
Conclude that the price of the perpetual American put option with dividend is given for all t ∈ R+ by
fL∗ (St ) = IE∗ e −r(τL∗ −t) (K − SτL∗ )+ St

λ−

 K − St ,
K,
0 < St 6


λ− − 1

= λ −1 λ−


1 − λ− −
St
λ−



,
St >
K,
K
−λ−
λ− − 1
where λ− < 0 is given by (11.43), and
τL∗ = inf{u > t : Su 6 L}.
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Exercise 11.13 American call options with dividends, cf. § 9.3 of [Wil06].
2
Consider a dividend-paying asset priced as St = S0 e (r−δ)t+σB̃t −σ t/2 , t ∈ R+ ,
where r > 0 is the risk-free interest rate, δ > 0 is the continuous dividend
rate, and σ > 0.
2
(λ)
a) Show that for all λ ∈ R the process Zt := (St )λ e −t((r−δ)λ−λ(1−λ)σ /2)
is a martingale under P∗ .
(λ)
b) Show that we have Zt = (St )λ e −rt for two values λ− 6 0, 1 6 λ+ of λ
satisfying a certain condition.
(λ )
c) Show that 0 6 Zt + 6 Lλ+ , 0 6 t < τL := inf{u > t : Su = L}, and
compute IE∗ e −rτL (SτL − K)+ S0 = x , with S0 = x < L and K < L.
Exercise 11.14
Optimal stopping for exchange options [GS96].
We consider two risky assets S1 and S2 modeled by
2
S1 (t) = S1 (0) e σ1 Wt +rt−σ2 t/2
2
S2 (t) = S2 (0) e σ2 Wt +rt−σ2 t/2 ,
(11.45)
t ∈ R+ , with σ2 > σ1 > 0, and the perpetual optimal stopping problem
sup
τ stopping time
and
IE[ e −rτ (S1 (τ ) − S2 (τ ))+ ],
where (Wt )t∈R+ is a standard Brownian motion under P.
a) Find α > 1 such that the process
Zt := e −rt S1 (t)α S2 (t)1−α ,
t ∈ R+ ,
(11.46)
is a martingale.
b) For some fixed L > 1, consider the hitting time
τL = inf t ∈ R+ : S1 (t) > LS2 (t) ,
and show that
IE[ e −rτL (S1 (τL ) − S2 (τL ))+ ] = (L − 1) IE[ e −rτL S2 (τL )].
c) By an application of the stopping time theorem to the martingale (11.46),
show that we have
IE[ e −rτL (S1 (τL ) − S2 (τL ))+ ] =
L−1
S1 (0)α S2 (0)1−α .
Lα
d) Show that the price of the perpetual exchange option is given by
sup
τ stopping time
"
IE[ e −rτ (S1 (τ ) − S2 (τ ))+ ] =
L∗ − 1
S1 (0)α S2 (0)1−α ,
(L∗ )α
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N. Privault
where
L∗ =
α
.
α−1
e) As an application of Question (d), compute the perpetual American put
option price
sup
IE[ e −rτ (κ − S2 (τ ))+ ]
τ stopping time
when r =
σ22 /2.
Exercise 11.15
Consider an underlying asset whose price is written as
St = S0 e rt+σBt −σ
2
t/2
,
t ∈ R+ ,
where (Bt )t∈R+ is a standard Brownian motion under the risk-neutral probability measure P∗ , σ > 0 denotes the volatility coefficient, and r ∈ R is the
(λ)
risk-free rate. For any λ ∈ R we consider the process (Zt )t∈R+ defined by
(λ)
Zt
2
:= e −rt (St )λ = (S0 )λ e λσBt −λ
σ 2 t/2+(λ−1)(λ+2r/σ 2 )σ 2 t/2
,
t ∈ R+ .
(11.47)
(λ)
a) Assume that r > −σ 2 /2. Show that, under P∗ , the process (Zt )t∈R+ is
a supermartingale when −2r/σ 2 6 λ 6 1, and that it is a submartingale
when λ ∈ (−∞, −2r/σ 2 ] ∪ [1, ∞).
(λ)
b) Assume that r 6 −σ 2 /2. Show that, under P∗ , the process (Zt )t∈R+ is
2
a supermartingale when 1 6 λ 6 −2r/σ , and that it is a submartingale
when λ ∈ (−∞, 1] ∪ [−2r/σ 2 , ∞).
c) From now on we assume that r < 0. Given L > 0, let τL denote the hitting
time
τL = inf{u ∈ R+ : Su = L}.
(λ)
By application of the stopping time theorem to (Zt )t∈R+ , show that
∗
IE
e
−rτL
1{τL <∞} | S0 = x 6

2
 (x/L)min(1,−2r/σ ) ,

2
(x/L)max(1,−2r/σ ) ,
x > L,
0 < x 6 L.
d) Deduce an upper bound on the price
IE∗ e −rτL (K − SτL )+ | S0 = x
of a European put option exercised in finite time under the strategy τL
when L ∈ (0, K) and x > L.
e) Show that when r 6 −σ 2 /2, the upper bound of Question (d) increases
and tends to +∞ when L decreases to 0.
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American Options
f) Find an upper bound on the price
IE∗ e −rτL (SτL − K)+ 1{τL <∞} | S0 = x
of a European call option exercised in finite time under the strategy τL
when L > K and x 6 L.
g) Show that when −σ 2 /2 6 r < 0, the upper bound of Question (f) increases
in L > K and tends to S0 as L increases to +∞.
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