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.) Prove X = Y (a.s.).
and E(Y |X) = X (a.s.)
[ 10 marks]
1
[continued overleaf]
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)foreveryt≥0.
(a) Show that W, X = (Wt2 − t)t≥0 and Y = (exp(−2t + 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 = jTn, j ∈ I := {0,1,...,n} and n ∈ N. Moreover,
consider the stochastic process f(n) := {f(n)}
t t≥0
n−1
f(n):=W 1I (t),
such that
t
tj (tj,tj+1]
j=0
where W := {Wt}t≥0 is a Wiener process, and 1IA is the indicator
function of a set A. (i) Prove that
(ii) By observing that
Wt2 −Wt2 =(∆Wtj+1)2+2Wtj∆Wtj+1,
j+1 j
where ∆Wtj+1 := Wtj+1 − Wtj , prove that
n−1 n−1
WT2 = (∆Wtj+1 )2 + 2 Wtj ∆Wtj+1
lim E n→∞ 0
(ft − Wt) dt = 0.
T(n) 2
j=0 j=0
[5 marks] You may use without proof that the quadratic variation of W over
(iii) Using the above results, prove that T12
the interval [0,T] is equal to T.
2
[continued overleaf]
WtdWt =2(WT −T). 0
[5 marks]
foreveryj∈I,
[5 marks]
MATH11154 Stochastic Analysis in Finance
3. (a) Consider the stochastic differential equation (SDE):
dXt =μXtdt+σdWt, forallt∈[0,T],
for T > 0, where {Wt}t≥0 is a Wiener process, μ and X0 are constants.
(i) (ii)
(iii) (iv)
Explain why this stochastic differential equation has a unique
solution.
Show that the process
Xt =X0eμt+σeμt
solves the equation. Calculate EXt.
t 0
e−μrdWr,
[3 marks]
t∈[0,T]
[4 marks]
[2 marks] Calculate Cov(Xs, Xt) := E(X ̄sX ̄t), where X ̄r = Xr − EXr
for r ∈ [0,T].
τ be a a stopping time with respect to the filtration Ft =
(b) Let
σ(Wr : r ≤ t), t ≥ 0, such that Eτ < ∞. Show that
(i) Xt = 1I{t≤τ} is Ft-measurable, (ii) EWτ = 0,
[2 marks] [4 marks] [4 marks]
(iii) EWτ2 = Eτ.
3
[6 marks]
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MATH11154 Stochastic Analysis in Finance
4. Assume that the price of a stock at time t is St = S0 exp(σWt + αt)
forallt≥0,whereS0 >0,σ>0andαareconstants,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 = −21σ2. [ 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 measure Q such that under Q 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
−rT ln(S0/K)+(r−21σ2)T eΦ√,
σT
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 =μStdt+σStdWt, S0 =s>0,
dBt =rBtdt, 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 .
(b) A naive approach to option pricing gives the value N = e−rT Eg(ST )
[5 marks]
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) ShowthatN=Cifμ=r.
(ii) Assume that g is strictly increasing on (0, ∞) i.e., g(x) < g(y) forall0
[10 marks]
(c) Consider an option with payoff h = | ln(ST )|2 at time T .
(i) Show that EQh2 < ∞, where Q is the risk neutral measure.
(ii) CalculatethepriceC oftheoptionatt=0whenr=σ=2 and S0 = 1.
5
[10 marks]
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MATH11154 Stochastic Analysis in Finance
6. Consider a financial market with a riskless asset of the price Bt = ert,
and with two risky assets whose prices S(1) and S(2) satisfy tt
dS(1) = rS(1)dt + σ S(1)dW (1), tt1tt
dS(2) = rS(2)dt + σ S(2)dW (2), tt2tt
where r, σ , σ , S(1) and S(2) are positive constants, and (W(1)) and 120 0 tt≥0
(W(2)) are Wiener processes under a ‘risk neutral measure’ Q, such t t≥0
that W(1) = ρW(2) +1−ρ2W(3) for some constant ρ ∈ [0,1] and a Wiener process (W(3)) , which is independent of (W(2)) under Q.
t t≥0 t t≥0
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) −S(2)+, and by TT
a multi-dimensional version of the Main Theorem on Pricing European type options one knows that C = e−rT EQh, where a+ := max(a, 0) for real numbers a.
(a) Prove that
σ2 (2)S(1) E h=S(2)erTE e− 2 T+σ2WT T
Q0Q S(2) T
(b) Use Girsanov’s theorem to prove that
+ −1 .
σ2 (2)S(1) + S(1)
E e−2T+σ2WT T −1 =E 0 e−2T+σWT−1 ,
σ2
where σ = σ12 − 2ρσ1σ2 + σ2 and (Wt)t≥0 is a Wiener process
Q S(2) Q S(2) T0
[2 marks]
+
under Q. (c) Prove that
[15 marks]
C = S(1)Φ(d ) − S(2)Φ(d ), 0102
where Φ is the probability distribution function of a standard normal random variable, and
[8 marks]
S(1) 2 ln(0 )+σT
S(2)2 √ d1= 0√ ,d2=d1−σT.
σT
You may use without proof the Black-Scholes European option pricing formula provided that you state it accurately.
6
[End of Paper]
MATH11154 Stochastic Analysis in Finance Solutions
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] 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 − st Wt and Wt are independent. Hence
E ( W s | W t ) = E ( W s − st W t + st W t | W t ) = st W t .
[9 marks]
(c) NotethatE(XY)=EE(XY|Y)=EY2. 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) RecalltheE|Wt|<∞forallt≥0,andfor0≤s≤twehave E(Wt|Fs) = E(Wt −Ws +Ws|Fs) = E(Wt −Ws)+Ws = Ws.
[3 marks] (ii) We note first that Wt2 − t is measurable with respect to Ft =
σ(Ws : 0 ≤ s ≤ t). Moreover, observe that
E|Wt2 −t|≤EWt2 +t=t+t=2t<∞.
For 0 ≤ s ≤ t
E(Wt2 − t|Fs) = E((Wt − Ws + Ws)2 − t|Fs)
=E((Wt −Ws)2 +2(Wt −Ws)Ws +Ws2 −t|Fs)
=E(Wt −Ws)2 +2WsE(Wt −Ws)+Ws2 −t =t−s+0+Ws2 −t=Ws2 −s.
Therefore (Wt2 − t)t≥0 is a martingale. [3 marks] 7
1. (a)
(b)
MATH11154 Stochastic Analysis in Finance
(iii) We note first that e−2t+Wt is measurable with respect to Ft = σ(Ws : 0 ≤ s ≤ t) since the exponential function is a continuous function. Moreover, observe that
t 1∞tx2 Ee−2+Wt =√ e−2+xe−2tdx
2πt −∞
1 ∞ x2−2xt+t2 =√ e− 2t dx
(b) (i)
[4 marks]
(W −W)2dt tj t
E(Wtj − Wt)2 dt (t−tj)dt
=lim (tj+1−tj)2 n→∞ j=0 2
T2
= lim = 0,
Thus we obtain
Consequently, (e− 2t +Wt )t≥0 is a martingale. T n−1 t
limE
n→∞ t t n→∞
0
n→∞ 2n
T T n−1
2πt −∞
1 ∞ (x−t)2
e− 2t dx=1. 2s222
=√
E(e−t+Wt|F )=e−s+WsEe−t−s+Wt−Ws =e−s+Ws.
2πt −∞
(f(n)−W)2dt= limE
n−1 t
which implies
W dW = lim f(n) dW = lim W ∆W ,
j+1 j=0 tj
j+1 n→∞ j=0 tj
n−1 t j+1
= lim = lim
n→∞ j=0 tj n−1 1
t t n→∞ t t n→∞ tj tj+1 0 0 j=0
where the limit is taken in mean square sense. [5 marks] (ii) Since W0 = 0, one obtains that
n−1
WT2 =Wt2 j+1
j=0
n−1 n−1 −Wt2=(∆Wtj+1)2 +2Wtj∆Wtj+1.
j
8
j=0 j=0
[5 marks]
MATH11154 (ii)
3. (a) (i)
Stochastic Analysis in Finance
Thus
(ii)
(iii)
(iv)
t t e−μrdWr)dt+σeμtd
T n−1
Wt dWt = lim Wtj ∆Wtj+1
0 n→∞ j=0
1n−1 1
= lim WT2 −(∆Wtj+1)2 = (WT2 −T). n→∞ 2 j=0 2
[5 marks]
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 Itoˆ’s formula is applied to Yt :=
e−μtXt to obtain d[e−μtXt] = σe−μtdWt
and thus
Xt = eμtX0 + σeμt for every t ∈ [0,T].
By Itoˆ’s formula
dXt =μ(eμtX0+σeμt
t 0
e−μrdWr,
t e−2μr dr < ∞. 0
⇒ eμtXt = X0 + σ
t 0
e−μr dWr
e−μrdWr =μXtdt+σeμte−μtdWt =μXtdt+σdWt.
00
EXt = eμtX0 + σeμtE
since the expectation of stochastic integral is zero due to
Wemayassumethats≤t. Thenforμ̸=0
s t s 1−e−2μs
E e−μr dWr eμr dWr = E e−2μr dr = 2μ , 000
9
t
e−μr dWr = eμtX0,
[3 marks]
0
[4 marks]
MATH11154
(b) (i)
(ii)
and
Stochastic Analysis in Finance
̄ ̄ 2 μ(s+t)1−e−2μs E(XsXt) = σ e 2μ
(iii)
By Itoˆ’s identity EWτ2 =E
[5 marks]
∞
=E Xt2dt=Eτ.
.
For μ = 0 we have E(X ̄sX ̄t) = σ2s = σ2 min(s, t). [3 marks]
Let us define X := 1I (indicator function of the event t t≤τ
[t ≤ τ]). Note that
[Xt =0]=[τ
(a) ESt = S0eαtEeσWt = S0eαteσ2t/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)+σ2(t−s)/2,
which shows that E(St|Fs) = Ss if and only if α = −21σ2.
[5 marks]
10
[5 marks]
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
t 0
where (bs)s∈[0,T] ∈ S([0,T]). Define the measure Q, by Q(F)=E(1FγT), F∈F,
whereγT =exp(−TbsdWs−1Tb2sds).AssumethatEγT =1. 020
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 = S0eσW ̃t−σ2t/2, t ∈ [0,T]
is a martingale with respect to (Ft)t≥0, if under Q the process W ̃t=Wt+(ασ+σ2)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 = γT dP ,
γT=exp(−θWT−θ2/2), θ:=ασ+σ2, 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)>0foreveryt>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+σW ̃T ,
11
Xt =
bsds+Wt, t∈[0,T],
MATH11154 Stochastic Analysis in Finance where W ̃T ∼ N(0,T) under Q. Hence
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),
whereFt =σ(Ws :s≤t). [5marks]
EQ(1ST >K) = P(S0e(r−σ2/2)T+σWT > K)
= PWT > ln(K/S0)+(σ2/2−r)T = Φln(S0/K)+(r−σ2/2)T .
[ 5 marks]
√√√
TσTσT
(b) (i) By Itˆo’s formula we can check that St = S0e(μ−σ2/2)t+σWt) for t ≥ 0. Hence for the discounted stock price S ̃t = e−rtSt we have
S ̃t = S0eσWt+(μ−r−σ2/2)t
Ifμ=r,thenS ̃t =S0eσWt−σ2t/2 fort≥0,thatisaFt- martingale under P. Thus P is a risk neutral measure, and therefore
C = e−rT Eh = N.
(ii) Recall that ST = S0er−σ2/2)T+σW ̃T where W ̃T ∼ N(0,T) under
Q. Thus
C = e−rT EQg(S0e(r−σ2/2)T +σW ̃ T ) = e−rT Eg(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−rT Eg(S0e(r−σ2/2)T +WT ) < e−rT Eg(S0e(μ−σ2/2)T +WT ) = N.
12
[5 marks]
[5nmarks]
MATH11154 Stochastic Analysis in Finance
(c) (i) Note that ln(ST) = ln(S0)+(r−σ2/2)T +σW ̃T, where W ̃T ∼
6. (a)
Using the notation a+ := max(a, 0) we have
(1) (2) + (2) S(1) +
N(0,T) under the risk neutral measure Q. Thus EQh2 =E|ln(S0)+(r−σ2/2)T +σWT|4
≤8|ln(S0)+(r−σ2/2)T|4 +8σ4EWT4 <∞.
[5 marks]
(ii) IfS0 =1andr=σ=2,thenh=|ln(ST)|2 =σ2W ̃T2,and C = e−rT EQh = e−rT σ2EWT2 = 4e−2T T.
[5 marks]
E(S−S)=E{S(T −1)} QT T QTS(2)
σ2 (2) (1) =S(2)erTE {e− 2 T+σ2WT (ST −1)+}.
0Q S(2) T
[2 marks] (b) Introduce a new probability measure Q ̃ by dQ ̃ = γdQ,
σ2 (2) γ=e−2T+σ2WT .
Then by Girsanov’s theorem the process (W ̃ (2), W (3)) , t t t∈[0,T]
W ̃(2) =W(2) −σ t, W ̃(3) =W(3) tt2tt
is a two-dimensional Wiener process. Thus
(1) (1) (1)
S S 2 2 (1) (2) S T = 0 e(σ2−σ1)T+σ1WT −σ2WT = 0
exp(−σ2T/2 + W ̃T )
S(2) S(2) S(2) T00
with
σW ̃ := (σ ρ−σ )W ̃ (2)+σ 1 − ρ2W ̃ (3),
T
σ2 = σ2−2σ σ ρ+σ2. T12T1 2121
Notice that (W ̃ t)t∈[0,T ] is a Wiener process under Q ̃. Consequently,
σ2 (2) (1) E {e− 2 T+σ2WT (ST
(1)
− 1)+} = E (S0 e−σ2T/2+σWT − 1)+
Q S(2) T0
Q S(2)
where (Wt)t≥0 is a Wiener process under Q. [15 marks]
13
MATH11154 (c)
Stochastic Analysis in Finance
−rT (2) S(1) −σ2T/2+σW + C=e Eh=SE(0 e T−1)
Q 0QS(2) 0
= E (S(1)e−σ2/2+σWT − S(2))+, Q00
that is the same as the price of the European Call option in the standard (B, S) market with stock price S = S(1)eσWt−σ2/2, B =
t0t
1, strike price K := S(2) at maturity T, when (W ̃ ) is a
0 t t∈[0,T] Wiener process under the probability measure Q. Thus by the
Black-Scholes formula
C = S(1)Φ(d ) − S(2)Φ(d ),
where
0102
S(1) 2 ln(0 )+σT
S(2)2 √ d1= 0√ ,d2=d1−σT,
σT
and Φ is the probability distribution function of a standard normal random variable. [8 marks]
14 [End of Solutions]