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Advanced Financial Models Problem 1. R1 R W 1 Lmike/AFM/example3-solution.pdf · Example sheet 3 -...
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Advanced Financial Models Michael TehranchiExample sheet 3 - Michaelmas 2017
Problem 1. Let f : [0,∞)→ R be a continuous (non-random) function and W a Brownianmotion, and let σ2 =
∫∞0f(s)2ds and assume σ2 < ∞. By considering the L2 construction
of the stochastic integral, show that∫∞0f(s)dWs is a normal random variable with mean
zero and variance σ2.
Solution 1. Approximate f by piece-wise constant left-continuous non-random functions fnsuch that ∫ ∞
0
(f(s)− fn(s))2ds→ 0.
The existence of such a sequence is a consequence of the density of step functions in L2(Leb).But for a concrete construction, let
fM,N =MN∑i=1
f(tNi−1)1(tNi−1,tNi ]
where tNi = i/N . Note∫ ∞0
(f(s)− fM,N(s))2ds =
∫ ∞M
f(s)2ds+M sup{|f(s)− f(t)| : 0 ≤ s, t ≤M, |s− t| ≤ 1N}.
For each n ≥ 1, by square-integrability of f we can find Mn such that∫ ∞Mn
f(s)2ds <1
n
and by uniform continuity of f on the compact [0,Mn] we can find Nn such that
sup{|f(s)− f(t)| : 0 ≤ s, t ≤Mn, |s− t| ≤ 1Nn} ≤ 1
nMn
.
Setting fn = fMn,Nn does the job sinceNow ∫ ∞
0
fn(t)dWt =∑i
f(ti−1)(Wti −Wti−1)
is the sum of independent mean-zero normal random variables, and hence is mean-zeronormal with variance ∑
i
f(ti−1)2(ti − ti−1)→
∫ ∞0
f(t)2dt.
Since∫∞0fn(t)dWt →
∫∞0f(t)dWt in L2(P) by the definition of the stochastic integral, we
will be done once we appeal to the following standard fact:Fact. Suppose Xn ∼ N(µn, σ
2n) and µn → µ and σ2
n → σ2. If Xn → X in L2 thenX ∼ N(µ, σ2).
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Proof. Since Xn → X in L2, then Xn → X in distribution. Let Fn and F be the distributionfunctions of Xn and X, respectively. At the points of continuity of continuity of F we have
F (x) = limnFn(x)
= limn
Φ
(x− µnσn
)= Φ
(x− µσ
)by the continuity of the standard normal distribution function Φ.
Problem 2. * (Ornstein–Uhlenbeck process) Let W be a Brownian motion, and let
Xt = eatx+ b
∫ t
0
ea(t−s)dWs
for some a, b, x ∈ R.(a) Verify that (Xt)t≥0 satisfies the following stochastic differential equation:
dXt = aXt dt+ b dWt, X0 = x.
(b) Show that
Xt ∼ N
(eatx,
b2
2a(e2at − 1)
).
(c) What is the distribution of the random variable∫ T0Xt dt?
Solution 2. (a) Since
Xt = eat(x+ b
∫ t
0
e−as dWs
)we can apply Ito’s formula
dXt = eat(be−atdWt
)+
(x+ b
∫ t
0
e−as dWs
)aeatdt
= b dWt + aXt dt
(b) Since ∫ t
0
(ea(t−s))2ds =e2at − 1
2a
this part follows from Problem 1.(c) Method 1: Note that by rearranging the stochastic differential equation we have∫ T
0
Xt dt =1
a(XT − x− bWT )
2
and hence∫ T0Xt dt is normally distributed with mean (eaT−1)x/a. To compute the variance,
first note that
Cov(XT ,WT ) = Cov
(b
∫ T
0
ea(T−t), dWt
∫ T
0
dWt
)= b
∫ T
0
ea(T−t)dt
=b
a(eaT − 1)
by Ito’s isometry. Hence
Var
(∫ T
0
Xt dt
)=
1
a2(Var(XT )− 2b Cov(XT ,WT ) + b2Var(WT )
)=
b2
2a3(e2aT − 4eaT + 3 + 2aT ).
Method 2: ∫ T
0
Xt dt =
∫ T
0
eaTx dt+
∫ T
0
∫ t
0
ea(t−s)b dWs dt
=
∫ T
0
eaTx dt+
∫ T
0
∫ T
s
ea(t−s)b dt dWs
=eaT − 1
ax+
∫ T
0
ea(T−s) − 1
ab dWs
Hence∫ T0Xt dt is normally distributed with mean (eaT − 1)x/a and variance
b2
a2
∫ T
0
(ea(T−s) − 1)2ds =b2
2a3(e2aT − 4eaT + 3 + 2aT )
This calculation is useful in the study of the Vasicek interest rate model.
Problem 3. Let W be a Brownian motion. Show that if Yt = W 3t − 3tWt then Y is a
martingale (1) by hand, and (2) by Ito’s formula.
Solution 3. (1) By hand: Since Gaussian random variables have finite moments of all orders,Y is integrable. Indeed, we have
E(|Yt|) ≤ E(|W 3t |) + 3tE(|Wt|) = Ct3/2 <∞
where C = 5√
2/π. Therefore, using the independence of the increments of W we have
E(Yt|Fs) =E(W 3t − 3tWt|Fs)
=E[(Wt −Ws +Ws)3 − 3t(Wt −Ws +Ws)|Fs]
=E[(Wt −Ws)3] + 3E[(Wt −Ws)
2]Ws + 3E(Wt −Ws)W2s +W 3
s
− 3tE(Wt −Ws)− tWs
=0 + 3(t− s)Ws + 0 +W 3s + 0− tWs
=Ys
for 0 ≤ s < t.3
(2) By Ito’s rule:
dYt = d(W 3t − 3tWt)
= (3W 2t dWt + 3Wt dt)− 3(t dWt +Wt dt)
= 3(W 2t − t)dWt
and hence Y is a local martingale. Recall that if E(∫ t
0α2s ds
)< ∞ for all t ≥ 0 then
the process(∫ t
0αs dWs
)t≥0
is a martingale. Again, it’s clear that the integrand is square
integrable in this case since Guassian random variables have finite moments of all orders.But, just to be explicit,
E∫ t
0
[3(W 2s − s)]2ds = 9
∫ t
0
E(W 4s − 2W 2
s s+ s2)ds = 9
∫ t
0
2s2 ds = 6t3 <∞
and hence (Yt)t≥0 is a martingale.
Problem 4. (Heat equation) Let W be a scalar Brownian motion, and let g : [0, T ]×R→ Rbe a smooth function that satisfy the partial differential equation
∂g
∂t+
1
2
∂2g
∂x2= 0
with terminal condition
g(T, x) = G(x).
(a) Show that (g(t,Wt))t∈[0,T ] is a local martingale.(b) If the function g is bounded, deduce the formula
g(t, x) =
∫ ∞−∞
G(x+√T − tz)
e−z2/2
√2π
dz.
(c) Use Problem 2 to find explicitly the unique bounded solution to the PDE
∂h
∂t+ x
∂h
∂x+
1
2
∂2h
∂x2= 0
with terminal condition
h(T, x) = cos x.
Useful fact: If Z ∼ N(µ, σ2) then E(cosZ) = e−σ2/2 cosµ.
Solution 4. (a) By Ito’s formula:
dg(t,Wt) =
(∂g
∂t+
1
2
∂2g
∂x2
)dt+
∂g
∂xdWt
=∂g
∂xdWt
and hence (g(t,Wt))t∈[0,T ] is a local martingale.4
(b) Recall a bounded local martingale is a true martingale. In particular, by the independenceof the increments of Brownian motion, we have
g(t,Wt) = E[g(T,WT )|Ft]= E[G(WT )|Ft]= E[G(Wt +WT −Wt)|Ft]
=
∫ ∞−∞
G(Wt +√T − tz)
e−z2/2
√2π
dz
since WT −Wt ∼ N(0, T − t). Since the above formula holds identically, we have the desiredintegral representation of the solution of the heat equation.(c) Let X be the Ornstein–Uhlenbeck process
dXt = Xtdt+ dWt.
By Ito’s formula, the process (h(t,Xt))t∈[0,T ] is a local martingale if and only if h is solutionto the PDE. Now if h is bounded, then (h(t,Xt))t∈[0,T ] is a true martingale. Using problem2, we have
h(t,Xt) = E[cosXT |Ft]
= E[cos
(eT−tXt +
∫ T
t
eT−t−sdWs
)|Ft]
= cos(eT−tXt) exp(−(e2(T−t) − 1)/4)
since the conditional distribution of XT given Ft is normal with mean eT−tXt and variance(e2(T−t) − 1)/2. In particular, the unique solution is
h(t, x) = cos(eT−tx) exp(−14(e2(T−t) − 1))
Problem 5. (Strictly local martingale) This is a technical exercise to exhibit a local martin-gale that is not a true martingale. Let W = (W 1,W 2,W 3) be a three-dimensional Brownianmotion and let u = (1, 0, 0). It is a fact that P(Wt 6= u for all t ≥ 0) = 1.(a) Let Xt = |Wt − u|−1. Use Ito’s formula and Levy’s characterisation of Brownian motionto show that
dXt = X2t dZt, X0 = 1
where Z is a Brownian motion. In particular, show that X is a positive local martingale.(b) By directly evaluating the integral or otherwise, show that
E(Xt) = 2Φ(t−1/2)− 1
for all t > 0, where Φ is the distribution function of a standard normal random variable.Why does this imply that X is a strictly local martingale?
Solution 5. (a) Let f(x1, x2, x3) = ((x1 − 1)2 + x22 + x23)−1/2 so that(
∂f
∂x1,∂f
∂x2,∂f
∂x3
)= −[f(x1, x2, x3)]
3 (x1 − 1, x2, x3)
5
∂2f
∂x21+∂2f
∂x22+∂2f
∂x23= −f 3 + 3f 5(x1 − 1)2 +−f 3 + 3f 5x22 − f 3 + 3f 5x23
= 0.
In particular, Ito’s formula yields
dXt = −X3t [(W 1
t − 1)dW 1t +W 2
t dW2t +W 3
t dW3t ].
Since X can be written as a stochastic integral of a three dimensional Brownian motion, itis a local martingale. Now let Z be the local martingale such that Z0 = 0 and
dZt = −Xt[(W1t − 1)dW 1
t +W 2t dW
2t +W 3
t dW3t ].
Since
d〈Z〉t = X2t [(W 1
t − 1)2 + (W 2t )2 + (W 3
t )2]dt
= dt
by construction, the process Z is a Brownian motion by Levy’s characterisation theorem.(b) Switch to spherical coordinates:
E(Xt) = (2π)−3/2∫∫∫
e−x21/2−x22/2−x23/2√
(√tx1 − 1)2 + tx22 + tx23
dx1 dx2 dx3
= (2π)−3/2∫ ∞r=0
∫ π
θ=0
∫ 2π
φ=0
r2 sin θe−r2/2√
tr2 − 2√t cos θ + 1
dφ dθ dr
= (2π)−1/2∫ ∞r=0
∫ π
θ=0
r2 sin θe−r2/2√
tr2 − 2√t cos θ + 1
dθ dr
= (2πt)−1/2∫ ∞r=0
re−r2/2
√tr2 − 2
√t cos θ + 1
∣∣∣∣πθ=0
dr
= (2πt)−1/2∫ ∞r=0
2(r1{r>t−1/2} +√tr21{r≤t−1/2})e
−r2/2dr
= 2
∫ t−1/2
0
e−r2/2
√2π
dr
Note that E(Xt) < X0 for all t > 0, so X is a strictly local martingale.
Problem 6. (strictly local martingales again) (a) Suppose that X is positive martingalewith X0 = 1. Fix T > 0 and let
dQdP
= XT .
Let Yt = 1/Xt for all t ≥ 0. Use Girsanov’s theorem to show that (Yt)0≤t≤T is a positivemartingale under Q.(b) Continuing from part (a), now suppose that X has dynamics
dXt = XtσtdWt
where W is a Brownian motion under P. Show that there exists a Q-Brownian motion Wsuch that
dYt = YtσtdWt
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(c) Let X be a positive local martingale with X0 = 1 and dynamics
dXt = X2t dWt.
Our goal is to show that X is a strictly local martingale. For the sake of finding a contra-diction, suppose X is a true martingale. In the notation of parts (a) and (b), show that
P(Yt > 0) = 1 but Q(Yt > 0) = Φ(t−1/2).
Why does this contradict the assumption that X is a true martingale?
Solution 6. (a) Since P and Q are equivalent and
P(Xt > 0 for all t) = 1 then Q(Yt > 0 for all t) = 1.
Now to show that Y is a Q-martingale, note that
EQ(YT |Ft) =EP(XTYT |Ft)EQ(XT |Ft)
=1
Xt
(b) By Ito’s formula,
dYt = dX−1t
= −X−2t dXt +X−3t d〈X〉t= −Ytσt(dWt − σtdt)
Now by Girsanov’s theorem, the process dWt = dWt − σtdt defines a Q Brownian motion.And of course W = −W is a Brownian motion also.(c) Now assuming X is a true martingale, then Girsanov’s theorem applies and hence
dYt = YtσtdWt = dWt
since σt = Xt. Hence Q(Yt > 0) = Q(Wt > −1) = Φ(t−1/2) < 1. But since
P(Yt > 0) = P(Xt > 0) = 1.
Therefore P and Q are not equivalent after all.
Problem 7. Consider a three asset market with prices given by
dBt
Bt
= 2 dt
dS(1)t
S(1)t
= 3 dt+ dW(1)t − 2 dW
(2)t
dS(2)t
S(2)t
= 5 dt− 2 dW(1)t + 4 dW
(2)t .
Construct an absolute arbitrage.
Solution 7. If the pure investment strategy is decomposed as (φ, π) a good choice for theholding is stock is given by
πt =
(2
S(1)t
,1
S(2)t
)7
but, of course, it is not unique. It remains to find the holding in the bank account φ. Notethat the wealth X evolves as
dXt = r(Xt − πt · St)dt+ πt · dSt= (2Xt + 5)dt
so the unique solution with X0 = 0 is
Xt =5
2(e2t − 1).
Since Xt > 0 a.s. for t > 0, this is an arbitrage with the holding in the bank account givenby
φt =Xt − πt · St
Bt
=5− 11e−2t
2B0
.
Problem 8. * Consider a Black–Scholes market with two assets with dynamics given by
dBt = Bt r dt
dSt = St(µ dt+ σdWt)
Find a replicating strategy H and the associated wealth process for a claim with payout
(1) ξT = SpT for some p ∈ R(2) ξT = (logST )2
(3) ξT =∫ T0Ssds
Show that your answer to part (3) is unchanged if we only assume that S is a positive Itoprocess.
Solution 8. Note that ST = Ste(r−σ2/2)(T−t)+σ(WT−Wt) where Wt = Wt + (µ− r)t/σ defines a
Brownian motion for the equivalent martingale measure Q.
(1) We could solve the Black–Scholes PDE with boundary condition V (T, S) = Sp, butit is easier to compute conditional expections.
ξt = EQ[e−r(T−t)SpT |Ft] = Spt e(p−1)(r+pσ2/2)(T−t)
so that
V (t, S) = Spe(p−1)(r+pσ2/2)(T−t).
The hedging portfolio is
πt =∂V
∂S(t, St) = pSp−1t e(p−1)(r+pσ
2/2)(T−t).
(2) Similarly, since
ξt = EQ[e−r(T−t)(logST )2|Ft] = e−r(T−t){[logSt + (r − σ2/2)(T − t)]2 + σ2(T − t)}we have
πt = 2e−r(T−t)[logSt + (r − σ2/2)(T − t)]/St.(3) Finally,
ξt = EQ[e−r(T−t)
∫ T
0
Ssds|Ft]
= e−r(T−t)∫ t
0
Ssds+ St1− e−r(T−t)
r.
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Unfortunately, the payout is not of the form g(ST ) so our theorem for calculating thereplication portfolio doesn’t apply. Still, let
πt =1− e−r(T−t)
r
and
φt =ξt − πtSt
Bt
=1
B0
e−rT∫ t
0
Ssds.
Notice that φtBt + πtSt = ξt by construction and that by a calculation
dξt = φtdBt + πtdSt,
so (φ, π) is a self-financing replication strategy. To check the self-financing conditionwe did not need to use the specific dynamics of S. Indeed, we need only assume thatwe can apply Ito’s formula.
Problem 9. (Black–Scholes formula) Let X ∼ N(0, 1) be a standard normal random vari-able, and v and m be positive constants. Express the expectation
F (v,m) = E[(e−v/2+√vX −m)+]
in terms of Φ, the distribution function of X.
Solution 9.
E[(e−v/2+√vX −m)+] =
∫ ∞−∞
(e−v/2+√vx −m)+
e−x2/2
√2π
dx
=
∫ ∞logm/
√v+√v/2
(e−v/2+√vx −m)
e−x2/2
√2π
dx
=
∫ ∞logm/
√v+√v/2
e−v/2+√vx−x2/2
√2π
dx−m∫ ∞logm/
√v+√v/2
e−x2/2
√2π
dx
=
∫ − logm/√v+√v/2
−∞
e−s2/2
√2π
ds−m∫ − logm/
√v−√v/2
−∞
e−t2/2
√2π
dt
= Φ
(− logm√
v+
√v
2
)−m Φ
(− logm√
v−√v
2
)Problem 10. (strictly local martingale in finance) Consider a market with zero interest rater = 0 and stock price with dynamics
dSt = S2t dWt.
Consider a European claim with payout ξT = ST .(a) Show that there exists a trading strategy which replicates the claim with correspondingwealth ξt = V (t, St) where
V (t, S) = S
[2Φ
(1
S√T − t
)− 1
].
(b) Consider the strategy of buying S0 claims and selling ξ0 shares. The time 0 wealth isV0 = 0 and the time T wealth is VT = (S0−ξ0)ST > 0. Is this strategy an absolute arbitrage?
9
Solution 10. (a) It is straight-forward, if a bit tedious, to verify
∂V
∂t+
1
2S4∂
2V
∂S2= 0
and limt↑T V (t, S) = S. The replication strategy is given by πt = ∂V∂S
(t, St) as usual.(b) This candidate arbitrage is not admissible. Indeed, let Vt = S0ξt− ξ0St. Note that V is alocal martingale, as it is the linear combination of two local martingales. Now if the strategywere admissible, the wealth V would be bounded from below and hence a supermartingale.Therefore we would have to conclude
0 = V0 ≥ E(VT ) = E(S0ξT − ξ0ST ) = (S0 − ξ0)E(ST )
But this contradicts S0 > ξ0. So in this market it is impossible to lock in the sure futureprofit at zero initial cost, because doing so leaves open the possibility that the wealth is goesnegative between times t = 0 and t = T .
Note, however, that there is an admissible arbitrage relative to the asset with price S.Indeed, the strategy of holding one share of the claim is a relative arbitrage, since the initialdiscounted wealth is ξ0/S0 < 1 and the terminal discounted wealth is ξT/ST = 1. This isonly a relative arbitrage since you do not short S, and hence must start with positive initialwealth.
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