Advanced Financial Models Michael TehranchiExample sheet 4 - Michaelmas 2017

Problem 1. (Hull–White extension of Black–Scholes) Consider a market with zero interestrate r = 0 and with a stock price modelled as

dSt = St(µ dt+ σ(t)dWt)

for a given non-random function σ : [0,∞)→ (0,∞).(a) Compute the replication cost of a European call option.(b) A call option is said to be at-the-money if its strike K equals the initial price S0 of theunderlying asset. Explain how you could use your answer to part (a) and quoted time-0prices of at-the-money calls of different maturities T to estimate the function σ.

Solution 1. (a) The replication cost is given by

C(T,K) = EQ[(ST −K)+]

for each maturity T where Q is the unique risk-neutral measure. But, note that d logSt =−1

2σ(t)2dt + σ(t)dWt by Ito’s formula, where dWt = dWt + µ/σ(t)dt defines a Q-Brownian

motion. Since σ is not random, we can conclude from Example sheet 3, Problem 1, that

logSt ∼ N

(logS0 −

1

2

∫ t

0

σ(s)2ds,

∫ t

0

σ(s)2ds

)Hence, the replication cost of the call of the at-the-money call of maturity T is

C(T,K) = S0F

(∫ T

0

σ(s)2ds,K/S0

)where F is the function from Example sheet 2, Problem 9..(b) Note that for K = S0 we have

C(T, S0) = S0

{2Φ

[1

2

(∫ T

0

σ(s)2ds

)1/2]− 1

}.

Assuming that the quoted call prices are their minimal replication costs, we need only chooseσ in such a way that ∫ T

0

σ(s)2ds =

[2Φ−1

(C(T, S0)

2S0

+1

2

)]2by differentiating both sides with respect to T .

Problem 2. * (variance swap) Consider a market a stock with price S, where S be a positiveIto process, and interest rate r = 0. A variance swap is a European contingent claim withpayout

N∑n=1

(log

StnStn−1

)2

.

where 0 ≤ t0 < · · · < tN = T are fixed non-random dates. We know from stochastic calculusthat the payout of the variance swap, for large n, is approximately given by

ξT = 〈logS〉T .The goal of this exercise is to show that ξT can be replicated asymptotically.

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(a) Confirm the identity

log(ST/S0) =

∫ T

0

dStSt− 1

2〈logS〉T .

(b) Confirm the identity

log x = x− 1−∫ 1

0

(k − x)+

k2dk −

∫ ∞1

(x− k)+

k2dk.

(c) Explain how to asymptotically replicate ξT by trading in the stock, cash, and a family ofcall and put options of different strikes but all with maturity date T . Show that the numberof shares of the stock varies dynamically but the portfolio of calls and puts is static. To whatextent is this replication strategy independent of the details of the model for S?

Solution 2. (a) Note that by Ito’s formula:

d logSt =dStSt− d〈S〉t

2S2t

.

and hence

d〈logS〉t =d〈S〉tS2t

.

The conclusion follows.(b) Since (k − x)+ ≥ 0 only if k ≥ x we have∫ 1

0

(k − x)+

k2dk =

∫ 1

x∧1

(1

k− x

k2

)dk

= − log(x ∧ 1)− (1− x)+

Similarly, ∫ ∞1

(x− k)+

k2dk =

∫ x∨1

1

(x

k2− 1

k

)dk

= (x− 1)+ − log(x ∨ 1)

(c) By the construction of quadratic variation in lecture, we have

ξT ≈ 〈logS〉T

= 2

∫ T

0

dStSt− 2 log(ST/S0)

= 2

∫ T

0

dStSt− 2ST + 2(logS0 + 1) + 2

∫ 1

0

PT (T,K)

K2dK + 2

∫ ∞1

CT (T,K)

K2dK

where PT (T,K) = (K − ST )+ is the payout of a put option and CT (T,K) = (ST −K)+ isthe payout of a call. Therefore, a replication strategy is to hold the static position 2

K2dK

puts of strike K ≤ 1 and 2K2dK calls of strike K > 1, and the dynamic position of 2

St− 2

shares of the stock at time t.This replication strategy only requires that S is positive and that Ito’s formula applies, but

is otherwise model-independent. (But don’t forget about notions of admissibility, and evenmore fundamentally, that we need to be in a situation where we can safely ignore bid-askspread, price impact, transaction costs, etc.)

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Problem 3. (implied volatility) Consider a market model with zero interest rate r = 0 andnon-negative stock price (St)t≥0. Suppose the initial call prices are given by

C(T,K) = EQ[(ST −K)+]

where Q is an equivalent measure such that S is a Q- martingale.The implied total variance V (T,K) is defined implicitly as the unique non-negative solu-

tion of the equation

BS(V (T,K), K/S0) = C(T,K)/S0

where

BS(v,m) = Φ(− logm/√v +√v/2)−mΦ(− logm/

√v −√v/2).

The implied volatility is then defined by Σ(T,K) =√V (T,K)/T .

(a) Show for fixed T > 0 the following inequality (due to Roger Lee in 2003)

lim supK→∞

V (T,K)

logK≤ 2.

Is the constant 2 on the right side best possible?Useful fact: ex

2/2Φ(−x)→ 0 as x→ +∞.(b) Challenge: Assume that C(T,K)→ S0 as T →∞ for all K. Show for fixed 0 < K1 < K2

the following inequality (due to Chris Rogers and me in 2009)

lim supT→∞

|V (T,K2)− V (T,K1)|log(K2/K1)

≤ 4.

Is the constant 4 on the right side best possible?Useful fact: xex

2/2Φ(−x)→ 1√2π

as x→ +∞.

Solution 3. (a) By the dominated convergence theorem, we have limK→∞C(T,K) = 0 andhence

(1) limK→∞

BS(V (T,K), K/S0) = 0

On the other hand, we have

(2) limK→∞

S0BS(2 log(K/S0), K/S0) = limK→∞

1

2S0 −KΦ

(−√

2 log(K/S0))

=1

2S0

(We have used the useful fact to assert that the limit of the second term above is zero.)From (1) and (2) combined with the fact that the map v 7→ BS(v,m) is strictly increasing,

we conclude that there must exist a value K0 such that:

V (T,K) < 2 log(K/S0), for all K > K0,

implying the desired conclusion.The constant 2 is best possible. Fix 0 < β < 2 and let

C(T,K) = S0BS(β log(K/S0), K/S0)

= S0Φ(−√

log(K/S0)(1√β−√β2

))−KΦ

(−√

log(K/S0)(1√β

+√β2

))

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for K > S0. We need only show that C(T, ·) is decreasing and concave for all large K sothat we can employ the Breeden–Litzenberger identity to show that there exists an integrablerandom variable ST such that

C(T,K) = E[(ST −K)+]

for sufficiently large K. Note that the corresponding implied total variance V is such that

V (T,K) = β log(K/S0)

and that β can be arbitrarily close to 2.(b) Note that since C(T,K) → S0 we have V (T,K) → ∞. The useful fact shows that√v(1− BS(v,m))ev/8 →

√8πem/2 as v →∞. Hence

V = −8 log(1− C/S0)− 4 log(− log(1− C/S0)) + 4 log(K/S0)− 4 log π + o(1).

Hence

V (T,K2)−V (T,K1) = −8 log1− C(T,K2)/S0

1− C(T,K1)/S0

−4 loglog(1− C(T,K2)/S0)

log(1− C(T,K1)/S0)+4 log(K2/K1)+o(1)

We will show that

(*) 1 ≤ 1− C(T,K2)/S0

1− C(T,K1)/S0

≤ K2/K1

and that

(**)log(1− C(T,K2)/S0)

log(1− C(T,K1)/S0)→ 1.

which will prove the claim. Indeed, note that

C(T,K) = E[(ST −K)+]

= E[ST − ST ∧K]

= S0 − E[ST ∧K].

and that inequality

S ∧K1 ≤ S ∧K2

≤ K2

K1

(S ∧K1)

holds for all S ≥ 0. These combine to show equation (*) Since C(T,K1)→ S0 we have

1 ≤ log(1− C(T,K2)/S0)

log(1− C(T,K1)/S0)≤ 1 +

log(K2/K1)

log(1− C(T,K1)/S0)→ 1

showing equation (**).The constant 4 is best possible. Let U be uniformly distributed on [0, 1], and let St =

1{t≤1} + t1{1<t≤1/U}. This process (St)t≥0 is a martingale in its own filtration, since for1 ≤ s ≤ t,

E[St|Ss = 0] = 0

and

E[St|Ss = s] = tP(t ≤ 1/U |s ≤ 1/U) = tP(U ≤ 1/t)

P(U ≤ 1/s)= s.

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Now note that

E[(ST −K)+] = (1−K/T )+

for T > 1. By the formula established above

V (T,K) = 8 log T − 4 log log T − 4 logK + o(1)

as T →∞. In particular, V (T,K1)− V (T,K2)→ 4 log(K2/K1).

Problem 4. (more implied volatility) Now suppose the stock price dynamics are

dSt = StσtdWt

where σ is independent of the Q-Brownian motion W .(a) Show that there is a family of measures µT on [0,∞) such that

C(T,K) = S0

∫BS(v,K/S0)µT (dv).

(b) If there are constants a ≤ b such that a ≤ σt ≤ b a.s., show that the implied volatilitysatisfies

a ≤ Σ(T,K) ≤ b.

(c) Show the equality Σ(T,K) = Σ(T, S20/K), i.e. the function x 7→ Σ(T, S0e

x) is even. Hint:First prove the identity

BS(v,m) = 1−m+mBS(v, 1/m).

Solution 4. (a) Notice that conditional on the process σ, the distribution of

logST = logS0 −1

2

∫ T

0

σ2sds+

∫ T

0

σsdWs

is normal, since σ and W are independent. Hence

E[(ST −K)+|σ] = S0 BS

(∫ T

0

σ2sds,K/S0

).

The conclusion follows from the tower property of conditional expectations, where the mea-

sure µT is the law of the non-negative random variable∫ T0σ2s ds.

(b) Since v 7→ BS(v,m) is increasing, we have

BS(Ta2, K/S0) ≤∫[a,b]

BS(v,K/S0)µT (dv) ≤ BS(Tb2, K/S0).

But since the middle term is just BS(TΣ(T,K)2, K/S0), we can conclude

a ≤ Σ(T,K) ≤ b.

(c) Notice the Black–Scholes call price function satisfies the following identity:

BS(v,m) = Φ(− logm/√v +√v/2)−mΦ(− logm/

√v −√v/2)

= 1− Φ(logm/√v −√v/2)−m

[1− Φ(logm/

√v +√v/2)

]= 1−m+m

[Φ(logm/

√v +√v/2)− (1/m)Φ(logm/

√v −√v/2)

]= 1−m+m BS(v, 1/m)

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Now use the above calculation:

S0 BS(V (T,K), K/S0) =

∫S0 BS(v,K/S0)µT (dv)

=

∫[S0 −K +K BS(v, S0/K)]µT (dv)

= S0 −K +K BS(V (T, S20/K), S0/K)

= S0 BS(V (T, S20/K), K/S0).

In particular, the implied volatility smile in this model (where the volatility is uncorrelatedwith the driving Brownian motion) is symmetric as a function of log-moneyness log(K/S0).This observation is due to Renault and Touzi in 1996.

Problem 5. (local volatility) Suppose that

dSt = Stσ(t, St)dWt

for a smooth function σ bounded from below and above. Also, suppose that

E[(ST −K)+] = S0 BS(V (T ), K/S0)

for all T,K where T 7→ V (T ) is a given smooth increasing function. Show that σ(t, S) =√V ′(t) for all t, S.

Solution 5. Dupire’s formula says

σ(T,K)2 =2∂C∂T

K2 ∂C2

∂K2

=2V ′(T )∂BS

∂v

(K/S0)2∂BS2

∂m2

.

We will make use of the following identity:

φ(− logm/√v +√v/2) = mφ(− logm/

√v −√v/2).

Now, calculus tells us that

∂BS

∂v=

1

2√vφ(− logm/

√v +√v/2)

∂BS

∂m= −Φ(− logm/

√v −√v/2)

∂2BS

∂m2=

1

m√vφ(− logm/

√v −√v/2)

In particular,

∂BS

∂v=

1

2m2∂

2BS

∂m2

which ought not come as a surprise in light of the Black–Scholes PDE.6

Problem 6. (CEV model) Let the local martingale S satisfy the equation

dSt = cS1−α/2t dWt

for some 0 < α < 2, where c =√

2/α.(a) Let

U(τ, S) =1

Γ(1α

) ∫ Sα/τ

0

z1α−1e−zdz

where Γ is the usual gamma function. For fixed T > 0, show that the process defined byU(T − t, St) is a local martingale. Why is it a true martingale?(b) Use part (a) to conclude that P(ST > 0) = U(T, S0).

Solution 6. (a) Ito’s formula says

dU(T − t, St) = −∂U∂τ

dt+∂U

∂SdS +

1

2

∂2U

∂S2d〈S〉t

=

(−∂U∂τ

+1

2c2S2−α∂

2U

∂S2

)dt+ cS1−α/2∂U

∂SdWt.

Now we calculate some partial derivatives:

∂U

∂τ= − S

Γ(1α

)τ 1+1/α

e−Sα/τ

∂U

∂S=

α

Γ(1α

)τ 1/α

e−Sα/τ

∂2U

∂S2= − α2Sα−1

Γ(1α

)τ 1+1/α

e−Sα/τ

so that∂U

∂τ= α2S2−α∂

2U

∂S2.

and the dt term vanishes from the Ito expansion, leaving only dWt term. Hence U(T − t, St)defines a local martingale. It is actually a true martingale since it is bounded by 0 ≤U(T − t, St) ≤ 1.(b) Note that limτ↓0 U(τ, S) = 1{S>0}. The conclusion follows from the martingale property.

Problem 7. (Hull–White extension of Cox–Ingersoll–Ross) Consider the short rate modelgiven by

drt = λ(r(t)− rt) dt+ σ√rt dWt

for positive constants λ and σ and a deterministic function r : R+ → R. Find the initialforward rate curve f0 = f(0, ·) for this model.

Solution 7. Consider the PDE

∂V

∂t(t, T, r) + λ(r(t)− r)∂V

∂r(t, T, r) +

1

2σ2r

∂2V

∂r2(t, T, r) = rV (t, T, r)

V (T, T, r) = 1.7

As usual we can make the ansatz

V (t, T, r) = e−rA(t,T )−B(t,T )

for some functions A and B which satisfy the boundary conditions A(T, T ) = B(T, T ) = 0.Substituting this into the PDE yields

−∂A∂t

(t, T )r − ∂B

∂t(t, T ))V (t, r)− λ(r(t)− r)A(t, T )V (t, r) +

σ2

2rA(t, T )2V (t, r) = rV (t, r).

which yields the coupled system

∂A

∂t(t, T ) = λA(t) +

σ2

2A(t)2 − 1

∂B

∂t(t, T ) = −λr(t)A(t, T ).

The equation for A is a Riccati equation, whose solution is

A(t, T ) =2(eγ(T−t) − 1)

(γ + λ)eγ(T−t) + (γ − λ)

B(t, T ) =

∫ T

t

λr(s)A(s, T )ds

where γ =√λ2 + 2σ2. Hence, the time 0 forward rates are given by

f0(T ) =4γ2eγT

[(γ + λ)eγT + (γ − λ)]2r0 +

∫ T

0

4γ2eγ(T−s)λr(s)ds

[(γ + λ)eγ(T−s) + (γ − λ)]2.

Problem 8. (Vasicek model) Suppose

drt = λ(r − rt)dt+ σdWt

where W is a Q-Brownian motion. Fix a time horizon S > 0 and let QS be the S-forwardmeasure. Using the fact that the forward rate (f(t, S))t∈[0,S] is a QS-martingale, show thatthe process W S defined by

dW St = dWt +

σ

λ(1− e−λ(S−t))dt

is a QS-Brownian motion.

Solution 8. Recall the formula for the forward rate in the Vasicek model

f(t, S) = e−λ(S−t)rt + (1− e−λ(S−t))r − σ2

2λ2(1− e−λ(S−t))2

for 0 ≤ t ≤ S. Use Ito’s formula to get

df(t, S) = λe−λ(S−t)[rt − r +

σ

λ2(1− e−λ(S−t))

]dt+ e−λ(S−t)drt

= e−λ(S−t)σ[dWt +

σ

λ(1− e−λ(S−t))dt

]= e−λ(S−t)σdW S

t

Since this is a martingale, we must conclude that W S is a QS-Brownian motion. (Actually,we first conclude that W S is a local martingale. But since 〈W S〉t = 〈W 〉t = t, we know thatW S is actually a Brownian motion by Levy’s characterisation theorem.)

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Problem 9. Let P (t, T ) denote the price at time t of the bond of maturity T , and f(t, T )denotes the forward rate at time t for maturity T .

Consider the Gaussian HJM model

df(t, T ) = σ(t, T )

∫ T

t

σ(t, u)du dt+ σ(t, T )dWt

where σ(t, T ) is not random for 0 ≤ t ≤ T.Fix 0 ≤ S ≤ T . By considering the distribution of P (S, T ) under the S-forward measure,

show that the minimal replication cost of a European claim with time-S payout g(P (S, T ))is

P (0, S)F

(P (0, T )

P (0, S),

∫ S

0

∥∥∥∥∫ T

S

σ(t, u)du

∥∥∥∥2 dt)

where

F (c, v) = E[g(ce−v/2+√vZ)]

for Z ∼ N(0, 1).

Solution 9. The replication cost is given by

P (0, S) EQS [g(P (S, T ))].

It remains to find the distribution of P (S, T ) under the S-forward measure. First note

df(t, S) = σ(t, S)

(∫ S

t

σ(t, u)du dt+ dWt

).

Since (f(t, S))0≤t≤S is a martingale under QS we see that the process W S defined by

W St = Wt +

∫ t

s=0

∫ S

u=s

σ(s, u)du ds

is a QS Brownian motion. This means

f(t, T ) = f(0, T ) +

∫ t

s=0

∫ T

u=S

σ(s, T )σ(s, u)du ds+

∫ t

0

σ(s, T )dWs

for all T and hence

− logP (S, T ) =

∫ T

u=S

f(S, u)du

= logP (0, S)/P (0, T ) +1

2

∫ S

0

∥∥∥∥∫ T

S

σ(t, u)du

∥∥∥∥2 dt+

∫ S

0

∫ T

S

σ(t, u)du dWt.

Using Example Sheet 3, Problem 1, the random variable logP (S, T ) under QS is normallydistributed with the mean and variance as desired.(Jamshidian found this formula in 1998.)

Problem 10. (a) Let X1, X2, . . . be a sequence of non-negative random variables such thatE(Xn) = 1 for all n. Use the Borel–Cantelli lemma to show

lim supn→∞

X1/nn ≤ 1 a.s.

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(b) If y(t, T ) is the yield of a bond maturing at time T , the long rate is defined as `t =limT→∞ y(t, T ) whenever the limit exists. Assuming that the bonds are priced by expectation,show that the long rate is non-decreasing, that is

`t ≥ `s a.s. for all 0 ≤ s ≤ t,

a fact first discovered by Dybvig, Ingersoll & Ross in 1996.

Solution 10. (a) For each ε > 0 we have∞∑n=1

P[Xn > (1 + ε)n] ≤∞∑n=1

(1 + ε)−n =1

ε<∞

by Markov’s inequality. The first Borel–Cantelli lemma then says

P(X1/nn > 1 + ε infinitely often) = 0

This shows lim supn→∞X1/nn ≤ 1 as claimed.

(b)Now, let P (t, T ) be the bond price, Bt the bank account, and P (t, T ) = P (t, T )/Bt thediscounted bond price as usual. Suppose the long rate `t exists, so that

`t = − limT→∞

1

TlogP (t, T ) = − lim

T→∞

1

Tlog P (t, T ).

By assumption, the discounted bond prices are given by

P (t, T ) = EQ(B−1T |Ft)each 0 ≤ t ≤ T and a fixed risk-neutral measure Q, and, in particular, P (·, T ) is a martingalefor each T > 0.

Fix 0 ≤ s ≤ t, and let

Xn =P (t, n)

P (s, n).

Note E(Xn) = E[E(Xn|Fs)] = 1 for each n. The first part implies

`s − `t = limn→∞

1

nlogXn ≤ 0 a.s.

as required.

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