MATH11154
Stochastic Analysis in Finance
Solutions and comments May 2020
1. Let (Wt)t≥0 be a Wiener martingale with respect to a filtration (Ft)t≥0 on a probability space
(Ω,F , P ).
(a) Prove that for any constant λ 6= 0 the process V = (λ−1Wλ2t)t≥0 is a Wiener martingale with
respect to the filtration (Fλ2t)t≥0. [6 marks]
(b) Prove that M = (W 2t )t≥0 is a submartingale with respect to a filtration (Ft)t≥0.
[6 marks]
(c) Consider the process U = (σWt)t∈[0,T ], where σ ∈ R is a constant such that |σ| 6= 1.
(i) Determine [U ]t, the quadratic variation of U over the interval [0, t] for any t ∈ [0, T ].
[7 marks]
(You may use without proof what you know about the quadratic variation of Wiener
processes.)
(ii) Prove there is no probability measure Q, equivalent to P , such that under Q the process
(Ut)t∈[0,T ] is a Wiener process. [6 marks]
Solution:
(a) Vt is Fλ2t-measurable for every t ≥ 0; Vt − Vs = λ−1(Wλ2t −Wλ2s) ∼ N(0, t − s) and it is
independent of Fλ2s for 0 ≤ s ≤ t. [6 marks]
(b) E|Mt| = EW 2t = t <∞ and Mt is Ft-measurable for every t ≥ 0; for 0 ≤ s ≤ t by Jensen’s inequality E(W 2|Fs) ≥ (E(Wt|Fs))2 = W 2s , since W is a martingale with respect to (Ft)t≥0. [6 marks] (c) (i) Clearly, [U ]t = [cW ]t = σ 2[W ]t = σ 2t. [7 marks] (ii) If U is a Wiener process under Q, then for a sequence 0 = tn0 < t n 1 < ... < t n N(n) = t of partitions of [0, t] such that limn→∞maxi |tni+1 − t n i | = 0, we have Sn := N(n)−1∑ i=0 |Utn i+1 − Utn i |2 → t in L2(Ω,F , Q) as n→∞. Hence limk→∞ Snk = t for Q-almost every ω ∈ Ω, for a subsequence nk → ∞. Since P and Q are equivalent, it follows that limk→∞ Snk = t for P -almost every ω ∈ Ω. Consequently, σ2t = [U ]t = t, which means |σ| = 1. [6 marks] 2. Let (Wt)t≥0 be a Wiener martingale with respect to a filtration (Ft)t≥0, let (bt)t∈[0,T ] be an Ft-adapted process such that almost surely∫ T 0 b2t dt <∞ for a given T > 0,
and define the stochastic process
γt(b) = exp
(∫ t
0
bs dWs − 12
∫ t
0
b2s ds
)
for t ∈ [0, T ].
(a) Using Itô’s formula and known properties of stochastic integrals prove that the process
(γt(b))t∈[0,T ] is a local martingale. [5 marks]
(b) Prove that Eγt(b) ≤ 1 for every t ∈ [0, T ].
(Hint: Prove that (γt(b))t∈[0,T ] is a supermartingale.) [6 marks]
1
MATH11154
Stochastic Analysis in Finance
Solutions and comments May 2020
(c) Assume there is a constant K such that |b| ≤ K for all ω ∈ Ω and t ∈ [0, T ]. Prove that for
any constant λ ∈ R
sup
t∈[0,T ]
E exp
(
λ
∫ t
0
bs dWs
)
<∞.
[7 marks]
(Hint: Use that Eγ(λbt) ≤ 1.)
(d) Assume there is a constant K such that |b| ≤ K for all ω ∈ Ω and t ∈ [0, T ]. Prove that
(γt(b))t∈[0,T ] is a martingale with respect to (Ft)t≥0. [7 marks]
Solution:
(a) By Itô’s formula, writing γt in place of γt(b),
dγt = γt(btdWt − 12b
2
t dt) +
1
2
γtb
2
t dt = γtbt dWt, which shows that γ is a local martingale
with respect to (Ft)t≥0.
[5 marks]
(b) By (a) there is an increasing sequence of stopping times (τn)
∞
n=1 such that limn→∞ τn =∞,
and for each n ≥ 1 the stopped process (γt∧τn)t∈[0,T ] is an Ft-martingale. Thus for 0 ≤ s ≤ t
we have E(γt∧τn |Fs) = γs∧τn . Letting here n→∞, by Fatou’s lemma we get
E(γt|Fs) = E( lim
n→∞
γt∧τn |Fs) ≤ lim inf
n→∞
E(γt∧τn |Fs) = lim inf
n→∞
γs∧τn = γs,
which implies Eγt ≤ Eγs for 0 ≤ s ≤ t ≤ T . In particular, Eγt ≤ Eγ0 = 1. [6 marks]
(c) For each t ∈ [0, T ] by (b) we have
E exp
(
λ
∫ t
0
bs dWs
)
= E
(
γt(λb)e
λ2
2
∫
t
0
b2s ds
)
≤ Eγt(λb)eλ
2K2T ≤ eλ
2K2T .
[7 marks]
(d) By (c)
γt = 1 +
∫ t
0
bsγs dWs, t ∈ [0, T ],
where we write γ in place of γ(b). Using (c) we get
E
∫ T
0
b2tγ
2
t dt ≤ K
2
∫ T
0
Ee2
∫
t
0
bs dWs−
∫
t
0
b2s ds ≤ K2T sup
t∈[0,T ]
Ee2
∫
T
0
bs dWs <∞,
i.e., bγ ∈ H([0, T ]), which implies that γ is a martingale. [7 marks]
3. Consider the standard Black-Scholes market with bond price Bt = e
rt and stock price
St = S0 exp(αt+ σWt) at time t ∈ [0, T ], where W is a Wiener process, α is any constant, and S0
and σ are positive constants. Let C(K,T ) and P (K,T ) denote the price at t = 0 of the European
Call and European Put options, respectively, with strike price K > 0 at expiry date T .
(a) Using the Main Theorem on Pricing European type options prove that
C(K,T ) = E(S0e
σWT−Tσ2/2 −Ke−rT )+, P (K,T ) = E(Ke−rT − S0eσWT−Tσ
2/2)+,
where the notation a+ := max(a, 0) is used. [5 marks]
2
MATH11154
Stochastic Analysis in Finance
Solutions and comments May 2020
(b) Prove that the process
Xt = (S0e
σWt−σ2t/2 −Ke−rt)+, t ≥ 0
is a submartingale with respect to the history Ft = σ(Ws, s ≤ t), t ≥ 0, of the Wiener
process W . [5 marks]
(c) Using (a) and (b) prove that C(K,T ) is an increasing function of T . [5 marks]
(d) Calculate P (K,T )− C(K,T ) from (a), and determine limT→∞ P (K,T ) when r > 0.
[5 marks]
(e) Using (c) and (d) prove that P (K,T ) is an increasing function of T if and only if r = 0.
[5 marks]
Solution:
(a) By the Main Theorem on Pricing European we have
C(K,T ) = e−rTEQ(ST −K)+ = EQ(S̃T − e−rTK)+
= EQ(S0e
σW̃T−Tσ2/2 − e−rTK)+ = E(S0eσWT−Tσ
2/2 −Ke−rT )+.
In the same way we get
P (K,T ) = E(Ke−rT − S0eσWT−Tσ
2/2)+.
[5 marks]
(b) Xt is Ft-measurable for every t, E|Xt| ≤ ES0eσWt−σ
2t/2 = S0 and for 0 ≤ s ≤ t by Jensen’s
inequality we have
E(Xt|Fs) ≥
(
E(S0e
σWt−σ2t/2 −Ke−rt|Fs)
)+
=
(
E(S0e
σWt−σ2t/2|Fs)−Ke−rt
)+
=
(
S0e
σWs−σ2s/2 −Ke−rt
)+
≥
(
S0e
σWs−σ2s/2 −Ke−rs
)+
= Xs.
[5 marks]
(c) By (b) C(K,T ) = EXT is an increasing function of T . [5 marks]
(d) By (a) P (K,T ) − C(K,T ) = E(Ke−rT − S0eσWT−Tσ
2/2) = Ke−rT − S0EeσWT−Tσ
2/2 =
Ke−rT − S0. Since P (K,T ) ≥ 0 and P (K,T ) ≤ Ke−rT , we have
0 ≤ lim sup
T→∞
P (K,T ) ≤ lim sup
T→∞
Ke−rT = 0,
which implies limT→∞ P (K,T ) = 0. [5 marks]
(e) If r = 0 then P (K,T ) = C(K,T ) + K − S0, which shows that P (K,T ) is increasing in T
because C(K,T ) is increasing in T . If r > 0 then limT→∞ P (K,T ) = 0, which implies that
P (K,T ) cannot be increasing in T .
[5 marks]
4. Consider again the Black-Scholes market with bond and stock prices as in Question 3. We
want to compute the price V at t = 0 of the European type option with payoff
h :=
{
L if mint∈[0,T ] St ≤ K
0 otherwise
}
at expiry date T , where L > 0 and K > 0 are some constants such that S0 > K.
3
MATH11154
Stochastic Analysis in Finance
Solutions and comments May 2020
(a) Using the main theorem on pricing European type options, show that
V = Le−rTP
(
min
t∈[0,T ]
(Wt + at) ≤ b
)
,
where a := r
σ
− 1
2
σ, b := σ−1 ln(K/S0).
[5 marks]
(b) Using Girsanov’s theorem show that
V = Le−rTE
(
1[mint∈[0,T ]Wt≤b]e
aWT− 12a
2T
)
,
where a and b are the constants defined in (a).
[5 marks]
(c) Denote the event [mT ≤ b,WT ≤ x] by Ax for every x ∈ R, where mT := mint∈[0,T ]Wt.
Using the reflection principle for the Wiener process W , prove that
P (Ax) =
{
P (WT ≤ x) if x < b
P (mT ≤ b)− P (WT ≤ 2b− x) if x ≥ b
}
.
[5 marks]
(d) Find a function g such that
P (Ax) =
∫ x
−∞
g(y) dy for every x ∈ (−∞,∞).
[5 marks]
(e) Using (b), (c) and (d) show that
V = C
∫ b
−∞
(
ea(2b−y) + eay
)
e−
y2
2T dy
with C := Le−(r+
1
2
a2)T /
√
2πT , a = r
σ
− 1
2
σ and b := σ−1 ln(K/S0).
You may use without proof the following fact:
If A ∈ F is an event, X is a random variable and g is a function on R, such that
P (A ∩ [X ≤ x]) =
∫ x
−∞
g(y) dy, for all x ∈ R,
then for every non-negative function f on R we have
E
(
1Af(X)
)
=
∫ ∞
−∞
f(x)g(x) dx.
[5 marks]
Solution:
(a) The payoff is h = L1{mint≤T St≤K}. By the Main Theorem on Pricing European Type
Options
V = e−rTEQh = Le
−rTEQ1{mint≤T St≤K} = Le
−rTQ
(
min
t≤T
St ≤ K
)
,
4
MATH11154
Stochastic Analysis in Finance
Solutions and comments May 2020
where Q is the risk neutral probability measure. Since St = S0 exp(σW̃t + (r − 12σ
2)t) with
a Wiener process (W̃t)t∈[0,T ] under Q, and exp(x) is increasing in x,[
min
t≤T
St ≤ K
]
=
[
min
t≤T
(σW̃t + (r − 12σ
2)t) ≤ ln K
S0
]
=
[
min
t≤T
(W̃t + at) ≤ b
]
with a := r
σ
− 1
2
σ, b := σ−1 ln(K/S0). Hence
V = Le−rTQ
(
min
t∈[0,T ]
(W̃t + at) ≤ b
)
= Le−rTP
(
min
t∈[0,T ]
(Wt + at) ≤ b
)
.
[5 marks]
(b) Set γ := exp(−aWT − 12a
2T ) and define the measure P̄ by dP̄ = γ dP . Then
Eγ = e−
1
2
a2TEe−aWT = 1.
Hence by Girsanov’s theorem P̄ is a probability measure and under P̄ the process
Vt := Wt + at, t ∈ [0, T ]
is a Wiener process. Therefore
P
(
min
t∈[0,T ]
(Wt + at) ≤ b
)
= P
(
min
t∈[0,T ]
Vt ≤ b
)
= EP̄ (γ
−11mint∈[0,T ] Vt≤b).
Notice that γ−1 = exp(aWT +
1
2
a2T ) = exp(aVT − 12a
2T ). Hence
P
(
min
t∈[0,T ]
(Wt + at) ≤ b
)
= EP̄ (e
aVT− 12a
2T1mint≤T Vt≤b) = E(e
aWT− 12a
2T1mint≤T Wt≤b).
Consequently,
V = Le−rTE
(
1mint≤T Wt≤b e
aWT− 12a
2T
)
.
[5 marks]
(c) If x ≤ b then [WT ≤ x] ⊂ [mT ≤ b]. Thus P (Ax) = P (WT ≤ x). Assume x > b. Then
P (Ax) = P (mT ≤ b)− P (mT ≤ b,WT > x),
and by the reflection principle P (mT ≤ b,WT > x) = P (WT ≤ 2b− x). [5 marks]
(d) Since
d
dx
P (Ax) =
1√
2πT
e−
x2
2T if x ≤ b
1√
2πT
e−
(2b−x)2
2T if x > b
=: g(x)
is nonnegative and continuous at every x ∈ R such that
∫∞
−∞ g(x) dx <∞,
P (Ax) =
∫ x
−∞
g(y) dy for all x ∈ (∞,∞).
[3 marks]
(e) From (d)
V =
Le−rT
√
2πT
∫ ∞
−∞
eax−
1
2
a2T g(x) dx = C
{∫ b
−∞
eaxe−
x2
2T dx+
∫ ∞
b
eaxe−
(2b−x)2
2T dx
}
5
MATH11154
Stochastic Analysis in Finance
Solutions and comments May 2020
with C := Le
−rT
√
2πT
e−
1
2
a2T = Le
−(r+1
2
a2)T
√
2πT
. By the change of variable z := 2b− x
∫ ∞
b
eaxe−
(2b−x)2
2T dx =
∫ b
−∞
ea(2b−z)e−
z2
2T dz.
Thus
V = C
∫ b
−∞
(
ea(2b−x) + eax
)
e−
x2
2T dx.
[2 marks]
6