Advanced Financial Models Problem 1. R1 R W 1 Lmike/AFM/example3-solution.pdf · Example sheet 3 -...

10

Click here to load reader

Transcript of Advanced Financial Models Problem 1. R1 R W 1 Lmike/AFM/example3-solution.pdf · Example sheet 3 -...

Page 1: Advanced Financial Models Problem 1. R1 R W 1 Lmike/AFM/example3-solution.pdf · Example sheet 3 - Michaelmas 2017 Problem 1. Let f: [0;1) !R be a continuous (non-random) ... Since

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).

1

Page 2: Advanced Financial Models Problem 1. R1 R W 1 Lmike/AFM/example3-solution.pdf · Example sheet 3 - Michaelmas 2017 Problem 1. Let f: [0;1) !R be a continuous (non-random) ... Since

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

Page 3: Advanced Financial Models Problem 1. R1 R W 1 Lmike/AFM/example3-solution.pdf · Example sheet 3 - Michaelmas 2017 Problem 1. Let f: [0;1) !R be a continuous (non-random) ... Since

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

Page 4: Advanced Financial Models Problem 1. R1 R W 1 Lmike/AFM/example3-solution.pdf · Example sheet 3 - Michaelmas 2017 Problem 1. Let f: [0;1) !R be a continuous (non-random) ... Since

(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

Page 5: Advanced Financial Models Problem 1. R1 R W 1 Lmike/AFM/example3-solution.pdf · Example sheet 3 - Michaelmas 2017 Problem 1. Let f: [0;1) !R be a continuous (non-random) ... Since

(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

Page 6: Advanced Financial Models Problem 1. R1 R W 1 Lmike/AFM/example3-solution.pdf · Example sheet 3 - Michaelmas 2017 Problem 1. Let f: [0;1) !R be a continuous (non-random) ... Since

∂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

6

Page 7: Advanced Financial Models Problem 1. R1 R W 1 Lmike/AFM/example3-solution.pdf · Example sheet 3 - Michaelmas 2017 Problem 1. Let f: [0;1) !R be a continuous (non-random) ... Since

(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

Page 8: Advanced Financial Models Problem 1. R1 R W 1 Lmike/AFM/example3-solution.pdf · Example sheet 3 - Michaelmas 2017 Problem 1. Let f: [0;1) !R be a continuous (non-random) ... Since

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.

8

Page 9: Advanced Financial Models Problem 1. R1 R W 1 Lmike/AFM/example3-solution.pdf · Example sheet 3 - Michaelmas 2017 Problem 1. Let f: [0;1) !R be a continuous (non-random) ... Since

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

Page 10: Advanced Financial Models Problem 1. R1 R W 1 Lmike/AFM/example3-solution.pdf · Example sheet 3 - Michaelmas 2017 Problem 1. Let f: [0;1) !R be a continuous (non-random) ... Since

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.

10