MATH11154 Stochastic Analysis in Finance
1. Let X and Y be random variables on a probability space (Ω,F , P ) such
that E|X| <∞, and let G be σ-algebra contained in F .
(a) Define precisely the notion of conditional expectations E(X|G)
and E(X|Y ). [6 marks]
(b) For a Wiener process (Wt)t≥0 calculate E(Wt|Ws), E(Wt|2Ws) and
E(Ws|Wt) for 0 < s ≤ t.
[Hint: Note that Wt−Ws is independent of Ws, and find a constant
a such that Ws − aWt is independent of Wt.] [9 marks]
(c) Suppose that X and Y are random variables such that EX2 <∞
and EY 2 <∞. Moreover, suppose that
E(X|Y ) = Y (a.s.) and E(Y |X) = X (a.s.)
Prove X = Y (a.s.). [ 10 marks]
[continued overleaf]
1
MATH11154 Stochastic Analysis in Finance
2. Let W := (Wt)t≥0 be a Wiener process and let (Ft)t≥0 be its history,
i.e., Ft = σ(Ws : 0 ≤ s ≤ t) for every t ≥ 0.
(a) Show that W , X = (W 2t − t)t≥0 and Y = (exp(−
t
2
+ Wt))t≥0 are
martingales with respect to the filtration (Ft)t≥0.
[10 marks]
(b) Consider the equidistant partition {t0, t1, . . . , tn} of the interval
[0, T ] where tj = j
T
n
, j ∈ I := {0, 1, . . . , n} and n ∈ N. Moreover,
consider the stochastic process f (n) := {f (n)t }t≥0 such that
f
(n)
t :=
n−1∑
j=0
Wtj I1(tj ,tj+1](t),
where W := {Wt}t≥0 is a Wiener process, and I1A is the indicator
function of a set A.
(i) Prove that
lim
n→∞
E
∫ T
0
(f
(n)
t −Wt)
2 dt = 0.
[5 marks]
(ii) By observing that
W 2tj+1 −W
2
tj
= (∆Wtj+1)
2 + 2Wtj∆Wtj+1 , for every j ∈ I,
where ∆Wtj+1 := Wtj+1 −Wtj , prove that
W 2T =
n−1∑
j=0
(∆Wtj+1)
2 + 2
n−1∑
j=0
Wtj∆Wtj+1
[5 marks]
(iii) Using the above results, prove that∫ T
0
Wt dWt =
1
2
(W 2T − T ).
[5 marks]
You may use without proof that the quadratic variation of W over
the interval [0, T ] is equal to T .
[continued overleaf]
2
MATH11154 Stochastic Analysis in Finance
3. (a) Consider the stochastic differential equation (SDE):
dXt = µXt dt+ σ dWt, for all t ∈ [0, T ],
for T > 0, where {Wt}t≥0 is a Wiener process, µ and X0 are
constants.
(i) Explain why this stochastic differential equation has a unique
solution. [3 marks]
(ii) Show that the process
Xt = X0e
µt + σeµt
∫ t
0
e−µr dWr, t ∈ [0, T ]
solves the equation. [4 marks]
(iii) Calculate EXt. [2 marks]
(iv) Calculate Cov(Xs, Xt) := E(X̄sX̄t), where X̄r = Xr − EXr
for r ∈ [0, T ].
[6 marks]
(b) Let τ be a a stopping time with respect to the filtration Ft =
σ(Wr : r ≤ t), t ≥ 0, such that Eτ <∞. Show that
(i) Xt = I1{t≤τ} is Ft-measurable, [2 marks]
(ii) EWτ = 0, [4 marks]
(iii) EW 2τ = Eτ . [4 marks]
[continued overleaf]
3
MATH11154 Stochastic Analysis in Finance
4. Assume that the price of a stock at time t is
St = S0 exp(σWt + αt)
for all t ≥ 0, where S0 > 0, σ > 0 and α are constants, andW = (Wt)t≥0
is a Wiener martingale with respect to a filtration (Ft)t≥0.
(a) Prove that S is a martingale with respect to (Ft)t≥0 if and only if
α = −1
2
σ2. [ 5 marks]
(b) State precisely Girsanov’s theorem. [ 5 marks]
(c) Let T > 0 be a fixed time. Using Girsanov’s theorem show the
existence of a probability measureQ such that underQ the process
(St)t∈[0, T ] is a martingale with respect to (Ft)t∈[0,T ].
[ 5 marks]
(d) For a given time T > 0 let τ denote the first time when the process
S = (St)t∈[0, T ] achieves its maximum over [0, T ], i.e,
τ = inf{t ∈ [0, T ] : St = S∗},
where S∗ = maxt∈[0, T ] St. Prove rigorously that τ is not a stopping
time with respect to (Ft)t≥0. (You may use that by Doob’s
theorem on optional sampling EMτ = EM0 for every martingale
(Mt)t∈[0, T ] with respect to a filtration (Ft)t≥0 and stopping times
τ ≤ T with respect to the same filtration.)
[ 5 marks]
(e) Consider a standard Black-Scholes market (B, S) with a stock S
defined above and a bond B = (ert)t≥0. A cash-or-nothing option
in this market pays 1 pound if the stock price exceeds a given
value K at the maturity date T , and it pays nothing otherwise.
Show that the fair price of this option at time t = 0 is
e−rTΦ
( ln(S0/K) + (r − 12σ2)T
σ
√
T
)
,
where Φ is the distribution function of a standard normal random
variable. [ 5 marks]
[continued overleaf]
4
MATH11154 Stochastic Analysis in Finance
5. Consider the standard Black-Scholes model with bond and stock prices,
Bt and St, satisfying
dSt = µSt dt+ σSt dWt, S0 = s > 0,
dBt = rBt dt, B0 = 1,
for t ≥ 0, where r ≥ 0, σ > 0, s > 0 and µ are constants, and (Wt)t≥0
is a Wiener process.
(a) State precisely the Main Theorem on Pricing European Type
Options with payoff h at expiry date T .
[5 marks]
(b) A naive approach to option pricing gives the value
N = e−rTEg(ST )
for the price C at time t = 0 of European type options with pay-off
h = g(ST ) at expiry date T , for non-negative functions g satisfying
the linear growth condition.
(i) Show that N = C if µ = r.
(ii) Assume that g is strictly increasing on (0,∞) i.e., g(x) < g(y) for all 0 < x < y. Compare N and C when µ > r.
[10 marks]
(c) Consider an option with payoff h = | ln(ST )|2 at time T .
(i) Show that EQh
2 <∞, where Q is the risk neutral measure.
(ii) Calculate the price C of the option at t = 0 when r = σ = 2
and S0 = 1.
[10 marks]
[continued overleaf]
5
MATH11154 Stochastic Analysis in Finance
6. Consider a financial market with a riskless asset of the price Bt = e
rt,
and with two risky assets whose prices S
(1)
t and S
(2)
t satisfy
dS
(1)
t = rS
(1)
t dt+ σ1S
(1)
t dW
(1)
t ,
dS
(2)
t = rS
(2)
t dt+ σ2S
(2)
t dW
(2)
t ,
where r, σ1, σ2, S
(1)
0 and S
(2)
0 are positive constants, and (W
(1)
t )t≥0 and
(W
(2)
t )t≥0 are Wiener processes under a ‘risk neutral measure’ Q, such
that W (1) = ρW (2) +
√
1− ρ2W (3) for some constant ρ ∈ [0, 1] and a
Wiener process (W
(3)
t )t≥0, which is independent of (W
(2)
t )t≥0 under Q.
Our aim is to calculate the price C at t = 0 of an exchange option that
gives the holder the right to exchange the asset S(2) for the asset S(1)
at expiry date T . This option has the payoff h =
(
S
(1)
T −S
(2)
T
)+
, and by
a multi-dimensional version of the Main Theorem on Pricing European
type options one knows that C = e−rTEQh, where a
+ := max(a, 0) for
real numbers a.
(a) Prove that
EQh = S
(2)
0 e
rTEQ
{
e−
σ22
2
T+σ2W
(2)
T
(S(1)
T
S
(2)
T
− 1
)+}
.
[2 marks]
(b) Use Girsanov’s theorem to prove that
EQ
{
e−
σ22
2
T+σ2W
(2)
T
(S(1)
T
S
(2)
T
− 1
)+}
= EQ
(
S
(1)
0
S
(2)
0
e−
σ2
2
T+σWT − 1
)+
,
where σ =
√
σ21 − 2ρσ1σ2 + σ22 and (Wt)t≥0 is a Wiener process
under Q. [15 marks]
(c) Prove that
C = S
(1)
0 Φ(d1)− S
(2)
0 Φ(d2),
where Φ is the probability distribution function of a standard
normal random variable, and
d1 =
ln(
S
(1)
0
S
(2)
0
) + σ
2
2
T
σ
√
T
, d2 = d1 − σ
√
T .
[8 marks]
You may use without proof the Black-Scholes European option pricing
formula provided that you state it accurately.
[End of Paper]
6
MATH11154 Stochastic Analysis in Finance
Solutions
1. (a) E(X|G) is a G-measurable random variable Z such that E(1GZ) =
E(1GX) for all G ∈ G, and E(X|Y ) = E(X|σ(Y )).
[6 marks]
(b)
E(Wt|Ws) = E(Wt−Ws+Ws|Ws) = E(Wt−Ws|Ws)+E(Ws|Ws) = Ws
due to the independence of Wt −Ws and Ws.
E(Wt|2Ws) = E(Wt|Ws) = Ws, since σ(2Ws) = σ(Ws).
Note that E{(Ws − aWt)Ws} = s− at. Consequently, Ws − stWt
and Wt are independent. Hence
E(Ws|Wt) = E(Ws −
s
t
Wt +
s
t
Wt|Wt) =
s
t
Wt.
[9 marks]
(c) Note that E(XY ) = EE(XY |Y ) = EY 2. Thus
E(X − Y )2 = EX2 − EY 2,
and by changing the role of X and Y we get
E(X − Y )2 = −E(X − Y )2.
Hence E(X − Y )2 = 0, which implies X = Y (a.s.).
[10 marks]
2. (a) (i) Recall the E|Wt| <∞ for all t ≥ 0, and for 0 ≤ s ≤ t we have
E(Wt|Fs) = E(Wt−Ws +Ws|Fs) = E(Wt−Ws) +Ws = Ws.
[3 marks]
(ii) We note first that W 2t − t is measurable with respect to Ft =
σ(Ws : 0 ≤ s ≤ t). Moreover, observe that
E|W 2t − t| ≤ EW
2
t + t = t+ t = 2t <∞.
For 0 ≤ s ≤ t
E(W 2t − t|Fs) = E((Wt −Ws +Ws)
2 − t|Fs)
= E((Wt −Ws)2 + 2(Wt −Ws)Ws +W 2s − t|Fs)
= E(Wt −Ws)2 + 2WsE(Wt −Ws) +W 2s − t
= t− s+ 0 +W 2s − t = W
2
s − s.
Therefore (W 2t − t)t≥0 is a martingale. [3 marks]
7
MATH11154 Stochastic Analysis in Finance
(iii) We note first that e−
t
2
+Wt is measurable with respect to
Ft = σ(Ws : 0 ≤ s ≤ t) since the exponential function is
a continuous function. Moreover, observe that
Ee−
t
2
+Wt =
1
√
2πt
∫ ∞
−∞
e−
t
2
+xe−
x2
2t dx
=
1
√
2πt
∫ ∞
−∞
e−
x2−2xt+t2
2t dx
=
1
√
2πt
∫ ∞
−∞
e−
(x−t)2
2t dx = 1.
Thus we obtain
E(e−
t
2
+Wt |Fs) = e−
s
2
+WsEe−
t−s
2
+Wt−Ws = e−
s
2
+Ws .
Consequently, (e−
t
2
+Wt)t≥0 is a martingale. [4 marks]
(b) (i)
lim
n→∞
E
∫ T
0
(f
(n)
t −Wt)
2 dt = lim
n→∞
E
n−1∑
j=0
∫ tj+1
tj
(Wtj −Wt)
2 dt
= lim
n→∞
n−1∑
j=0
∫ tj+1
tj
E(Wtj −Wt)
2 dt
= lim
n→∞
n−1∑
j=0
∫ tj+1
tj
(t− tj)dt
= lim
n→∞
n−1∑
j=0
1
2
(tj+1 − tj)2
= lim
n→∞
T 2
2n
= 0,
which implies∫ T
0
Wt dWt = lim
n→∞
∫ T
0
f
(n)
t dWt = lim
n→∞
n−1∑
j=0
Wtj∆Wtj+1 ,
where the limit is taken in mean square sense. [5 marks]
(ii) Since W0 = 0, one obtains that
W 2T =
n−1∑
j=0
(
W 2tj+1 −W
2
tj
)
=
n−1∑
j=0
(∆Wtj+1)
2 + 2
n−1∑
j=0
Wtj∆Wtj+1 .
[5 marks]
8
MATH11154 Stochastic Analysis in Finance
(ii) Thus
∫ T
0
Wt dWt = lim
n→∞
n−1∑
j=0
Wtj∆Wtj+1
= lim
n→∞
1
2
(
W 2T −
n−1∑
j=0
(∆Wtj+1)
2
)
=
1
2
(W 2T − T ).
[5 marks]
3. (a) (i) The linear functions f(t, x) = µx and g(t, x) = σ satisfy the
linear growth and Lipschitz conditions, which ensure that a
unique solution exists.
[3 marks]
Remark: Full mark is given if Itô’s formula is applied to Yt :=
e−µtXt to obtain
d[e−µtXt] = σe
−µtdWt ⇒ eµtXt = X0 + σ
∫ t
0
e−µrdWr,
and thus
Xt = e
µtX0 + σe
µt
∫ t
0
e−µr dWr
for every t ∈ [0, T ].
(ii) By Itô’s formula
dXt = µ(e
µtX0 + σe
µt
∫ t
0
e−µr dWr) dt+ σe
µtd
∫ t
0
e−µr dWr
= µXt dt+ σe
µte−µt dWt = µXt dt+ σ dWt.
[4 marks]
(iii)
EXt = e
µtX0 + σe
µtE
∫ t
0
e−µr dWr = e
µtX0,
since the expectation of stochastic integral is zero due to∫ t
0
e−2µr dr <∞.
[3 marks]
(iv) We may assume that s ≤ t. Then for µ 6= 0
E
(∫ s
0
e−µr dWr
∫ t
0
eµr dWr
)
= E
∫ s
0
e−2µr dr =
1− e−2µs
2µ
,
9
MATH11154 Stochastic Analysis in Finance
and
E(X̄sX̄t) = σ
2eµ(s+t)
1− e−2µs
2µ
.
For µ = 0 we have E(X̄sX̄t) = σ
2s = σ2 min(s, t). [3 marks]
(b) (i) Let us define Xt := I1t≤τ (indicator function of the event
[t ≤ τ ]). Note that
[Xt = 0] = [τ < t] = ∪∞n=1[τ ≤ t−
1
n
] ∈ Ft,
and thus the event [Xt = 1] = [τ < t]
c ∈ Ft. [2 marks]
(ii) Note that
E
∫ ∞
0
X2t dt = E
∫ ∞
0
I1{t≤τ} dt = E
∫ τ
0
1 dt = Eτ <∞.
Therefore
Wτ =
∫ τ
0
1 dWt =
∫ ∞
0
I1{t≤τ} dWt =
∫ τ
0
1 dWt = Wτ ,
and thus EWτ = E
∫∞
0
Xt dWt = 0, since (Xt)t≥0 ∈ H.
[5 marks]
(iii) By Itô’s identity
EW 2τ = E
(∫ ∞
0
Xt dWt
)2
= E
∫ ∞
0
X2t dt = Eτ.
[5 marks]
4. Assume that the price of a stock at time t is
St = S0 exp(σWt + αt),
where S0 > 0, σ > 0 and α are constants, and W = (Wt)t≥0 is a Wiener
martingale with respect to a filtration (Ft)t≥0.
(a) ESt = S0e
αtEeσWt = S0e
αteσ
2t/2 < ∞, and St is Ft-measurable,
since it is a (continuous) function of Wt. For 0 ≤ s < t
E(St|Fs) = Sseα(t−s)Eeσ(Wt−Ws) = Sseα(t−s)+σ
2(t−s)/2,
which shows that E(St|Fs) = Ss if and only if α = −12σ
2.
[5 marks]
10
MATH11154 Stochastic Analysis in Finance
(b) Girsanov’s theorem: On a probability space (Ω,F , P ) let (Wt)t∈[0,T ]
be a Wiener martingale with respect to a filtration (F)t∈[0,T ], and
consider a process
Xt =
∫ t
0
bs ds+Wt, t ∈ [0, T ],
where (bs)s∈[0,T ] ∈ S([0, T ]). Define the measure Q, by
Q(F ) = E(1FγT ), F ∈ F ,
where γT = exp(−
∫ T
0
bs dWs− 12
∫ T
0
b2s ds). Assume that EγT = 1.
Then Q is a probability measure, and under Q the process X is a
Wiener martingale with respect to (F)t∈[0,T ]. [5 marks]
(c) By (a) under a probability measure Q the process
St = S0e
σW̃t−σ2t/2, t ∈ [0, T ]
is a martingale with respect to (Ft)t≥0, if under Q the process
W̃t = Wt + (
α
σ
+ σ
2
)t, t ∈ [0, T ]
is a Wiener martingale with respect to (Ft)t≥0, that by virtue of
Girsanov’s theorem holds for the measure Q with dQ = γTdP ,
γT = exp(−ϑWT − ϑ2/2), ϑ := ασ +
σ
2
,
since by (a) we have EγT = 1 satisfied.
[5 marks]
(d) By (c) there is a probability measure Q such that under Q
the process (St)t∈[0,T ] is a martingale with respect to (Ft)t∈[0,T ].
Assume that τ is a stopping time. Then by Doob’s optional
sampling EQ(Sτ − S0) = 0. Hence Sτ = S0 (a.s.), that gives
σWt ≤ −α (a.s.) for all t ∈ [0, T ], which contradicts the fact that
P (Wt > −α/σ) > 0 for every t > 0.
[5 marks]
(e) The pay-off is 1ST>K , and by the Main Theorem on Pricing
European Type options the price at t = 0 of the option is
N0 = e
−rTEQ(1ST>K),
where Q is the risk neutral probability. Recall that
ST = S0e
(r−σ2/2)T+σW̃T ,
11
MATH11154 Stochastic Analysis in Finance
where W̃T ∼ N(0, T ) under Q. Hence
EQ(1ST>K) = P (S0e
(r−σ2/2)T+σWT > K)
= P
(
WT√
T
>
ln(K/S0)+(σ
2/2−r)T
σ
√
T
)
= Φ
(
ln(S0/K)+(r−σ2/2)T
σ
√
T
)
.
[ 5 marks]
5. (a) Main Theorem on Pricing European Type Options: Let h be a
nonnegative σ(Wr : r ≤ T )-measurable random variable such that
EQh < ∞, where Q is the risk neutral measure. Then there is
a replicable strategy (ψt, φt)t∈[0,T ] for the European type option
with pay-off h at expiry date T , and the price at t ∈ [0, T ] of this
option is the value of the replicating portfolio at t, that equals
e−r(T−t)EQ(h|Ft),
where Ft = σ(Ws : s ≤ t). [5 marks]
(b) (i) By Itô’s formula we can check that St = S0e
(µ−σ2/2)t+σWt) for
t ≥ 0. Hence for the discounted stock price S̃t = e−rtSt we
have
S̃t = S0e
σWt+(µ−r−σ2/2)t
If µ = r, then S̃t = S0e
σWt−σ2t/2 for t ≥ 0, that is a Ft-
martingale under P . Thus P is a risk neutral measure, and
therefore
C = e−rTEh = N.
[5 marks]
(ii) Recall that ST = S0e
r−σ2/2)T+σW̃T where W̃T ∼ N(0, T ) under
Q. Thus
C = e−rTEQg(S0e
(r−σ2/2)T+σW̃T ) = e−rTEg(S0e
(r−σ2/2)T+WT ).
If µ > r then
S0e
(r−σ2/2)T+WT < S0e
(µ−σ2/2)T+WT for all ω ∈ Ω.
Consequently,
C = e−rTEg(S0e
(r−σ2/2)T+WT ) < e−rTEg(S0e
(µ−σ2/2)T+WT ) = N.
[5nmarks]
12
MATH11154 Stochastic Analysis in Finance
(c) (i) Note that ln(ST ) = ln(S0) + (r−σ2/2)T +σW̃T , where W̃T ∼
N(0, T ) under the risk neutral measure Q. Thus
EQh
2 = E| ln(S0) + (r − σ2/2)T + σWT |4
≤ 8| ln(S0) + (r − σ2/2)T |4 + 8σ4EW 4T <∞.
[5 marks]
(ii) If S0 = 1 and r = σ = 2, then h = | ln(ST )|2 = σ2W̃ 2T , and
C = e−rTEQh = e
−rTσ2EW 2T = 4e
−2TT.
[5 marks]
6. (a) Using the notation a+ := max(a, 0) we have
EQ(S
(1)
T − S
(2)
T )
+ = EQ{S
(2)
T (
S
(1)
T
S
(2)
T
− 1)+}
= S
(2)
0 e
rTEQ{e−
σ22
2
T+σ2W
(2)
T (
S
(1)
T
S
(2)
T
− 1)+}.
[2 marks]
(b) Introduce a new probability measure Q̃ by dQ̃ = γdQ,
γ = e−
σ22
2
T+σ2W
(2)
T .
Then by Girsanov’s theorem the process (W̃
(2)
t ,W
(3)
t )t∈[0,T ],
W̃
(2)
t = W
(2)
t − σ2t, W̃
(3)
t = W
(3)
t
is a two-dimensional Wiener process. Thus
S
(1)
T
S
(2)
T
=
S
(1)
0
S
(2)
0
e(σ
2
2−σ
2
1)T+σ1W
(1)
T
−σ2W
(2)
T =
S
(1)
0
S
(2)
0
exp(−σ2T/2 + W̃T )
with
σW̃T := (σ1ρ−σ2)W̃
(2)
T +σ1
√
1− ρ2W̃ (3), σ2 = σ22−2σ1σ2ρ+σ
2
1.
Notice that (W̃t)t∈[0,T ] is a Wiener process under Q̃. Consequently,
EQ{e−
σ22
2
T+σ2W
(2)
T (
S
(1)
T
S
(2)
T
− 1)+} = EQ(
S
(1)
0
S
(2)
0
e−σ
2T/2+σWT − 1)+
where (Wt)t≥0 is a Wiener process under Q. [15 marks]
13
MATH11154 Stochastic Analysis in Finance
(c)
C = e−rTEQh = S
(2)
0 EQ(
S
(1)
0
S
(2)
0
e−σ
2T/2+σWT − 1)+
= EQ(S
(1)
0 e
−σ2/2+σWT − S(2)0 )
+,
that is the same as the price of the European Call option in the
standard (B, S) market with stock price St = S
(1)
0 e
σWt−σ2/2, Bt =
1, strike price K := S
(2)
0 at maturity T , when (W̃t)t∈[0,T ] is a
Wiener process under the probability measure Q. Thus by the
Black-Scholes formula
C = S
(1)
0 Φ(d1)− S
(2)
0 Φ(d2),
where
d1 =
ln(
S
(1)
0
S
(2)
0
) + σ
2
2
T
σ
√
T
, d2 = d1 − σ
√
T ,
and Φ is the probability distribution function of a standard normal
random variable. [8 marks]
14 [End of Solutions]