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Financial Mathematics: A Comprehensive Treatment
(Campolieti-Makarov)Solutions to selected exercises
Alan Marc Watson
July 19, 2016
Contents
1 Chapter 4: Primer on Derivative Securities 2
2 Chapter 10: One-dimensional Brownian motion and related processes. 4
3 Chapter 11: Introduction to continuous-time stochastic calculus 13
4 Chapter 12: Risk-Neutral pricing in the Black-Scholes Economy: one under-lying stock 25
5 Chapter 13: Risk-Neutral pricing in a Multi-Asset Economy 37
6 Chapter 14: American options 50
7 Chapter 15: Interest rate modeling and derivative pricing 53
1
1 Chapter 4: Primer on Derivative Securities
Exercise 1.1. [CM14, Exercise 4.27] A European binary option is a so-called ”all-or-nothing”claim on an underlying asset. For example, one share of a cash-or-nothing binary call has a payoffof exactly one dollar if the asset price ends up above the strike and zero otherwise, i.e. the payofffunction is Λ(S) = 1S≥K(S).
Similarly, a cash-or-nothing binary put has payoff 1S<K . Assume that the asset price process{S(t)}t≥0 is a geometric Brownian motion.
(a) Derive the Black-Scholes exact pricing formulas for both the binary call C(S, T ) and theput P (S, T ).
(b) Give the relationship between the binary call and put price when both options have thesame strike K and maturity T .
(c) Derive the exact formula for the Greek delta of the binary call and put: ∆c = ∂C∂S and
∆p = ∂P∂S .
Solution.
(a) By the risk-neutral pricing formula we have
C(t, S) = e−r(T−t)E [Λ(S(T ))|S(t) = S] (1)
= e−r(T−t)[Λ
(S(t)e(r−σ
2
2)(T−t)+σ
√T−tZ
)|S(t) = S
]= e−r(T−t)E
[1Z≥ 1
σ√T−t
(ln K
S−(r−σ2
2
)(T−t)
)]= e−r(T−t)P
(Z ≥ − 1
σ√T − t
(lnS
K+
(r − σ2
2
)(T − t)
))= e−r(T−t)P (Z ≥ −α)
= e−r(T−t)N (α) (2)
where in the last line we denoted α = 1σ√T−t
(ln S
K +(r − σ2
2
)(T − t)
).
(b) The put-call parity relationship for European options reads
C(t, S)− P (t, S) = S −Ke−r(T−t)
wherebyP (t, S) = C(t, S)− S +Ke−r(T−t) = e−r(T−t)[N (α) +K)− S
(c) ...
�
Exercise 1.2. [CM14, Exercise 4.28] Consider the European-style option with payoff Λ(S) =1K1≤S≤K2(S) and assume the geometric Brownian motion model for the stock price process{S(t)}t≥0.
2
(a) Find the option value by decomposing the payoff in terms of binary options with appropriateindicator functions.
(b) Derive formulas for the following sensitivities of the option value: ∆ = ∂V∂S , Γ = ∂2V
∂S2 and
Θ = ∂V∂t
Solution.
(a) The payoff can be decomposed in terms of binary options
Λ(S) = 1K1≤S≤K2(S) = 1S≥K1 − 1S≥K2
so using the pricing formula (2) from the previous exercise we obtain
V (t, S) = e−r(T−t)E[Λ(S(T ))|S(t) = S] = e−r(T−t)E[1K1≤S(T )≤K2|S(t) = S]
= e−r(T−t)E[1S(T )≥K1|S(t) = S]− e−r(T−t)E[1S(T )≥K2
|S(t) = S]
= CK1(t, S)− CK2(t, S)
= e−r(T−t) [N (α(K1))−N (α(K2))]
where α(Ki) = 1σ√T−t
(ln S
Ki+(r − σ2
2
)(T − t)
).
(b) ...
�
3
2 Chapter 10: One-dimensional Brownian motion and relatedprocesses.
Exercise 2.1. [CM14, Exercise 10.1]
(a) Show that the PDF of a sum of two continuous random variables X and Y is given by theconvolution of the PDF’s fX and fY :
fX+Y (x) =
∫ ∞−∞
fX(x)fY (z − x) dx
(b) Use the result in (a) to show that a sum of two independent standard normal randomvariables results in a normal random variable and find the PDF of such a sum.
(c) Assuming that X1 ∼ N (µ1, σ21) and X2 ∼ N (µ2, σ
22) are correlated with Corr(X1, X2) = ρ,
find the mean and variance of aX1 + a2X2 for a1, a2 ∈ R.
Solution.
(a) We start by computing the CDF of the sum: denoting by fX,Y (x, y) the joint density functionof X and Y we have
P (X + Y ≤ z) = P (X ≤ z − Y ) =
∫ ∞−∞
∫ z−Y
−∞fX,Y (x, y) dx dy
and differentiating with respect to z we obtain
fX+Y (z) =d
dzP (X + Y ≤ z) =
∫ ∞−∞
fX,Y (z − y, y) dy[1]=
∫ ∞−∞
fX(z − y)fY (y) dx
[2]=
∫ ∞−∞
fX(x)fY (z − x) dx
where in [2] we used independence of X and Y and in [2] we changed variables to x = z− y.
(b) We compute the sum of two standard normal variables by convolving their density functions:if X,Y ∼ N (0, 1), then fX(u) = fY (u) = 1√
2πe−u
2/2 so
fX+Y (z) =1
2π
∫ ∞∞
ex2/2e−(z−x)2/2 dx =
1
2πe−z
2/4
∫ ∞∞
e−(x− z2
)2dx
=1
2√
2πe−z
2/4
∫ ∞∞
e−u2/2 du
=1
2√πe−z
2/2
which shows that X + Y is a normal variable with mean 0 and variance 2.
�
Exercise 2.2. [CM14, Exercise 10.3] Suppose that X = (X1, X2, X3)> be a 3-dimensional
Gaussian random vector with mean zero and covariance matrix ΣX =
4 1 21 2 12 1 3
. Set Y =
1 +X1 − 2X2 +X3 and Z = X1 − 2X3.
4
(a) Find the probability distribution of Y .
(b) Find the probability distribution of the vector (Y,Z)>.
Solution. We start by recalling the following facts:
(i) If X = (X1, . . . , Xn) is an n-dimensional vector of random variables with covariance matrixΣX (namely, (ΣX)ij = Cov(Xi, Xj)) and Y = MX where M is an m × n matrix andY = (Y1, . . . , Ym) is an m-dimensional vector of random variables, then
ΣY = MΣXM>
(ii) If Y = b+MX, where Y is an m-dimensional vector of random variables, b is a constantm-dimensional vector and X is an n-dimensional vector of random variables with meanσX and covariance ΣX , then Y has mean vector and covariance matrix given by
µY = b+MµX , ΣY = MΣXM>
(iii) An n-dimensional random vector has the n-dimensional multivariate normal distributionif and only if every combination
∑ni=1 ciXi is a normal random variable. In particular, it
follows from (ii) that if X ∼ N (µX ,ΣX) and Y = b+MX, then
Y ∼ N(b+MµX ,MΣXM
>)
(iv) If Z has the standard n-dimensional normal distribution, Z ∼ N (0, In) and A is a sym-metric positive definite matrix with Cholesky factorization A = U>U , then X = U>Z hasan n-dimensional normal distribution with covariance matrix ΣX = A, so X ∼ N (0, A),
We now solve the exercise.
1. Y has a normal distribution by (iii). If b = (1,−2, 1)>, then
Y ∼ N(
1 + b>0, b>ΣXb)
with
b>ΣXb =(
1 −2 1) 4 1 2
1 2 12 1 3
1−21
= 11
so Y ∼ N(1, 11).
2. Note that
(YZ
)=
(10
)+
not= A︷ ︸︸ ︷(
1 −2 11 0 −2
) X1
X2
X3
so by (iii) the vector W := (Y,Z)> has a bivariate normal distribution with mean µW =
(1, 0)> +A · 0 = (1, 0)> and covariance matrix ΣW = AΣXA> =
(11 −2−2 8
).
5
�
Exercise 2.3. [CM14, Exercise 10.8] For a Brownian motion {W (t)}t≥0 and its naturalfiltration, calculate Es[W
3(t)] for 0 ≤ s ≤ t.
Solution. Method 1. Taking the differential of W 3(t) and using It’s lemma we see that
dW 3(t) = 3W 2(t)dW (t) + 3W (t)dt
since∫ t
0 W2(u) dW (u) is a square-integrable martingale, it follows that that so is W 3(t) −
3∫ t
0 W (u) du, whence
Es[W3(t)] = Es
[W 3(t)− 3
∫ t
0W (u) du
]+ Es
[3
∫ t
0W (u) du
]= W 3(s)− 3
∫ s
0W (u) du+ 3Es
[∫ t
0W (u) du
]As for the last expectation we have
Es
[∫ t
0W (u) du
]= Es
[∫ s
0W (u) du+
∫ t
sW (u) du
]=
∫ s
0W (u) du+
∫ s
tEs[W (u)] du
=
∫ s
0W (u) du+ (t− s)W (s)
so in conclusion
Es[W3(t)] = W 3(s) + 3(t− s)W (s)
Method 2. We can also compute the expectation directly:
Es[W3(t)] = Es
[(W (t)−W (s) +W (s))3
]= Es
[(W (t)−W (s))3
]+ 3Es
[(W (t)−W (s))2W (s)
]+ 3Es
[(W (t)−W (s))W 2(s)
]+ Es
[W 3(s)
]Now note that
Es[(W (t)−W (s))3
]= 0
Es[(W (t)−W (s))2W (s)
]= W (s)Es
[(W (t)−W (s))2W (s)
]= W (s)E
[(W (t)−W (s))2
]= (t− s)W (s)
Es[(W (t)−W (s))W 2(s)
]= W 2(s)Es [W (t)−W (s)] = 0
Es[W 3(s)
]= W 3(s)
In the first equation we used that a normal variable has 0 third central moment (in fact it has0 odd central moments, as the remark below shows). In the second equation, we used the factthat W (s) is Fs-measurable for the first equality (so it can be taken out of the expectation) andalso the fact that W (t)−W (s) is Fs-independent for the second equality, so that the conditionalexpectation Es is in fact unconditional. In the third equation we simply used that W 2(s) isFs-measurable to take it out of the expectation. �
6
Remark 2.1 (Central moments of a normal random variable). The n-th central moment is bydefinition mn = E ((X − E(X))n). Notice that for the normal distribution E(X) = µ, and thatY = X − µ also follows a normal distribution, with zero mean and the same variance σ2 as X.
Therefore, finding the central moment of X is equivalent to finding the raw moment of Y .
mn = E ((X − E(X))n) = E ((X − µ)n) =
∫ ∞−∞
1√2πσ
(x− µ)ne−(x−µ)2
2σ2 dx
y=x−µ=
∫ ∞−∞
1√2πσ
yne−y2
2σ2 dyy=σu
=
∫ ∞−∞
1√2πσ
σnune−u2
2 σdu
= σn∫ ∞−∞
1√2πune−
u2
2 du
The latter integral is zero for odd n as it is the integral of an odd function over a real line.So consider
∫ ∞−∞
1√2πu2ne−
u2
2 du = 2
∫ ∞0
1√2πu2ne−
u2
2 du
u=√
2w=
2√2π
∫ ∞0
(2w)ne−wdw√2w
=2n√π
∫ ∞0
wn−1/2e−wdw =2n√π
Γ
(n+
1
2
)where Γ(x) stands for the Euler’s Gamma function. Using its properties we obtain
m2n = σ2n(2n− 1)!! m2n+1 = 0
Exercise 2.4. [CM14, Exercise 10.9] Find the distribution of W (1) + · · ·+W (n) for n ∈ N.
Solution. A sum of normal random variables is again normal, so it suffices to compute itsmean and variance. Clearly
E [W (1) + · · ·+W (n)] = 0
since each W (k) ∼ N (0, k). As for the variance, recall that
Var[W (1) + · · ·+W (n)] =n∑k=1
Var(W (k)) + 2∑i<j
Cov(W (i),W (j))
and for i < j
Cov(W (i),W (j)) = E[W (i)W (j)] = E[W (i) · (W (j)−W (i))] + E[W 2(i)] = Var(W (i)) = i
We hence observe that
Var[W (1) +W (2)] = (1 + 2) + 2 · 1Var[W (1) +W (2) +W (3)] = (1 + 2 + 3) + 2 · [1 + (1+2)]
Var
(4∑
k=1
W (k)
)= (1 + 2 + 3 + 4) + 2 · [1 + (1 + 2) + (1+2+3)]
...
Var
(n∑k=1
W (k)
)= Var
(n−1∑k=1
W (k)
)+ n+ 2 · (n− 1)n
2= Var
(n−1∑k=1
W (k)
)+ n2
7
So
Var (nk=1W (k)) =n∑k=1
k2 =n(n+ 1)(2n+ 1)
6
and hencen∑k=1
W (k) ∼ N(
0,n(n+ 1)(2n+ 1)
6
)�
Exercise 2.5. [CM14, Exercise 10.10] Let a1, · · · , an ∈ R and 0 < t1 < · · · < tn. Find thedistribution of
∑ni=1 aiW (ti). Note that the choice ak = 1
n , 1 ≤ k ≤ n, leads to Asian options.
Solution. Arguing as in the previous exercise, we conclude that
Var
(n∑i=1
aiW (ti)
)= Var
(n−1∑i=1
aiW (ti)
)+ a2
ntn + 2an
n−1∑k=1
aktk
so by induction
Var
(n∑i=1
aiW (ti)
)=
n∑i=1
[a2i ti + 2ai
i−1∑k=1
aktk
]�
Exercise 2.6. [CM14, Exercise 10.11] Suppose that the processes {X(t)}t≥0 and {Y (t)}t≥0
are respectively given by X(t) = xo + µxt + σxW (t) and Y (t) = yo + µyt + σyW (t), wherex0, y0, µx, µy, σx > 0, σy > 0 are real constants. Find the covariance Cov(X(t), Y (s)), for s, t ≥ 0.
Solution.�
Exercise 2.7. [CM14, Exercise 10.12] Consider the process X(t) = x0 + µt + σW (t) ≥ 0,where x0, µ and σ are real constants. Show that
E[max(X(t)−K, 9)] = (x0 + µt−K)N(x0 + µt−K
σ√t
)+ σ√tn
(x0 + µt−K
σ√t
)Solution.�
Exercise 2.8. [CM14, Exercise 10.13] By directly calculating partial derivatives, verifythat the transition PDF
p0(t;x) =1√2πt
e−x2/2t
of standard Bronwian motion satisfies the diffusion equation
∂
∂tu(t, x) =
1
2
∂2
∂x2u(t, x)
8
Solution. We simply compute the derivatives
∂p0
∂t(t, x) = −1
2
1√2πt3
e−x2
2t +x2
2t21√2πt
e−x2
2t
∂p0
∂x(t, x) =
1√2πt
e−x2
2t
(−xt
)∂2p0
∂x2(t, x) =
1√2πt
e−x2
2tx2
t2− 1
t√
2πte−
x2
2t
whereby clearly
∂
∂tp0(t, x) =
1
2
∂2
∂x2p0(t, x)
�
Exercise 2.9. [CM14, Exercise 10.14] Find the transition PDF of {Wn(t)}t≥0 for n ∈ N.
Solution.
�
Exercise 2.10. [CM14, Exercise 10.15] Find the transition PDF of {Sn(t)}t≥0, whereS(t) = S0e
µt+σW (t), t ≥ 0, S0 > 0, for n ∈ N.
Solution.
�
Exercise 2.11. [CM14, Exercise 10.19] Show that the mean and covariance functions ofthe Brownian bridge from a to b on [0, T ] are, respectively
m(t) = a+b− aT
t,
c(s, t) = s ∧ t− st
T
for s, t ∈ [0, T ].
Solution. Recall that if X(t) ≡ X0→0[0,T ] = W (t)− t
TW (T ) is the Brownian bridge from 0 to 0,ten
Xa→b[0,T ] = a+
b− aT
t+X(t)
so
mXa→b[0,T ]
(t) = a+b− aT
t+mX(t)
cXa→b[0,T ]
(t, s) = cX(t, s)
9
and it suffices to compute mX(t) and cX(t, s).
mX(t) = E[X(t)] = E
[W (t)− t
TW (T )
]= 0,
cX(t, s) = E [X(t)X(s)] = E
[(W (t)− t
TW (T )
)(W (s)− s
TW (T )
)]= E[W (t)W (s)]− s
TE[W (t)W (T )]− t
TE[W (s)W (T )] +
st
T 2W 2(T )
= t ∧ s− st
T
�
Exercise 2.12. [CM14, Exercise 10.21] Consider the GBM; process S(t) = S0eµt+σW (t),
t ≥ 0, S0 > 0. The respective sampled maximum and minimum of this process are defined by
MS(t) = sup0≤u≤t
S(u), mS(t) = inf0≤u≤t
S(u)
and the first hitting time to a level B > 0 is defined by
T SB = inf{t ≥ 0 : S(t) = B}.
Derive expressions for the following:
(a) P (MS(t) ≤ y, S(t) ≤ s), t > 0, 0 < x ≤ y <∞, S0 ≤ y.
(b) P (MS(t) ≤ y), t > 0, S0 ≤ y <∞.
(c) P (T SB ≤ t), t > 0, S0 < B.
Solution.
(a) Let X(t) = µt+ σW (t). Denoting M(t) = sup0≤s≤tW (s), We know that
fMX(t),X(t)(w, y) = e−12µ2+µyfM(t),W (t)(w, y)
fM(t),W (t)(w, y) =2(2w − y)
t√
2πte−(2w−y)2/2t
Note that S(t) ≤ s if and only if X(t) ≤ ln sS0
so
P (MS(t) ≤ y, S(t) ≤ s) = P
(MX(t) ≤ ln
y
S0, X(t) ≤ ln
s
S0
)not= P
(MX(t) ≤ m,X(t) ≤ x
)and
P (MX(t) ≤ m,X(t) ≤ x) =
∫ m
0
∫ x
0fMX(t),X(t)(w, y) dy dw
= e−12µ2t
∫ m
0
∫ x
0eµy
2(2w − y)
t√
2πte−(2w−y)2/2t dy dw
And if you have enough patience you can type the computation of the latter integral...
10
(b) As above note that P (MS(t) ≤ y) = P(MX(t) ≤ ln y
S0
)so it suffices to integrate the joint
density
fMX(t),X(t)(w, y) = e−12µ2+µy 2(2w − y)
t√
2πte−(2w−y)2/2t
over the region −∞ < y ≤ w, 0 ≤ w ≤ ln yS0
, namely
P
(MX(t) ≤ ln
y
S0
)=
∫ ln yS0
0
∫ w
−∞fMX(t),X(t)(w, y) dy dw
=
∫ m
0
∫ m
yfMX(t),X(t)(w, y) dw dy +
∫ 0
−∞
∫ m
0fMX(t),X(t)(w, y) dw dy
where the second equality follows from Fubini’s theorem.
(c) To compute that P (T SB ≤ t), simply note that
P (T SB ≤ t) = P (MS(t) ≥ B) = 1− P (MS(t) ≤ B)
and we computed the latter in part (b).
Remark 2.2. We recall briefly how to obtain
fMX(t),X(t)(w, y) = e−12µ2+µyfM(t),W (t)(w, y) (3)
fM(t),W (t)(w, y) =2(2w − y)
t√
2πte−(2w−y)2/2t (4)
where, recall X(t) = µt + σW (t), MX(t) = sup0≤s≤tX(s) and M(t) = sup0≤s≤tW (t), whereW (t) is a standard Bronwian motion.
The joint density of Bronwian motion and its maximum (4) follows from the reflectionprinciple
P (W (t) ≤ w, T Wm ≤ t) = P (W (t) ≥ 2m− w)
Note on the one hand that for the LHS we have
P (T Wm ≤ t,W (t) ≤ w) = P (M(t) ≥ m,W (t) ≤ w) =
∫ ∞m
∫ w
−∞fM(t),W (t)(x, y) dy dx
As for the RHS, since W (t) ∼ N (0, t) we have
P (W (t) ≥ 2m− w) =
∫ ∞2m−w
1√2πt
e−u2/2t du
Equating both and differentiating with respect to m and w yields (4).
To show (3), assume for simplicity that σ = 1. It then suffices to perform a change ofmeasure that renders the Brownian motion driftless. Assuming that the market price of risk isγ and letting W (t) = W (t) + γ(t) we get
dX(t) = (µ− γ)dt+ dW (t)
11
which is driftless if γ = µ. The change of measure is given by the Radon-Nikodym process
ρt =
(dP
dP
)t
= exp
(−1
2
∫ t
0
µ2
σds−
∫ t
0µds
)= e−
12µ2t−µW (t) = e
12µ2t−µW (t)
Hence
P (MX(t) ≤ m,X(t) ≤ x) = E[IMX(t)≤m,X(t)≤x
]= E
[(dP
dP
)t
IMX(t)≤m,X(t)≤x
]= E
[ρ−1t IMX(t)≤m,X(t)≤x
][1]= e−
12µ2tE
[eµW (t)I
MW (t)≤m,W (t)≤x
][2]= e−
12µ2tE
[eµW (t)IM(t)≤m,W (t)≤x
]= e−
12µ2t
∫ m
0
∫ x
−∞eµyfM(t),W (t)(w, y) dy dw
where in [1] we used that X(t) = W (t) and in [2] we used the fact that the random variables
MW (t) and W (t) under measure P are the same as M(t) and W (t) under measure P . Thisshows (3).
�
12
3 Chapter 11: Introduction to continuous-time stochastic cal-culus
Exercise 3.1. [CM14, Exercise 11.4] Evaluate the following double stochastic integral∫ t
0
(∫ s
0dW (u)
)dW (s)
Solution. The inner integral is simply∫ s
0 dW (u) = W (s) − W (0) = W (s) and then∫ t0 W (s) dW (s) can be computed by applying It’s formula to df(W (t)) with f(x) = x2. In-
deed,
dW 2(s) = 2W (s)dW (t) + 2ds
so integrating over 0 ≤ s ≤ t and re-arranging we obtain∫ t
0W (s) dW (s) =
1
2W 2(t)− t
�
Exercise 3.2. [CM14, Exercise 11.5] Show that∫ t
0W 2(s) dW (s) =
1
3W 3(t)−
∫ t
0W (s) ds
by using an appropriate It formula.
Solution. Applying It’s formula to compute df(W (s)), with f(x) = x3 we obtain
dW 3(s) = 3W 2(s)dW (s) + 3W (s)ds
so re-arranging and integrating over 0 ≤ s ≤ t yields∫ t
0W 2(s) dW (s) =
1
3W 3(t)−
∫ t
0W (s) ds
as claimed. �
Exercise 3.3. [CM14, Exercise 11.6] Use the It isometry property to calculate the variancesof the following It integrals. Also explain why the integrals are well defined.
(a)∫ t
0 |W (s)|1/2 dW (s).
(b)∫ t
0 |W (s) + s|2 dW (s).
(c)∫ t
0 |W (s) + s|3/2 dW (s).
13
Solution. Recall that an It integral∫ t
0 X(s) dW (s) is well defined provided that the integrandX(s) is F(s)-measurable (which is obviously satisfied in all three cases) and provided that itsatisfies the square integrability condition
E
[∫ t
0X2(s) ds
]=
∫ t
0E[X2(s)
]ds <∞
and the It isometry property states that
Var
[∫ t
0X(s) dW (s)
]=
∫ t
0E[X2(s)] ds
(a) By It isometry we have
Var
[∫ t
0|W (s)|1/2 dW (s)
]=
∫ t
0E[W (s)]ds = 0
(b) For the integral∫ t
0 |W (s) + s|2 dW (s) we have that
Var
[∫ t
0|W (s) + s|2 dW (s)
]=
∫ t
0E[|W (s) + s|4
]ds
[1]=
∫ t
0E[W 4(s) + 4W 3(s)s+ 16W 2(s)s2 + 4W (s)s3 + s4
][2]=
∫ t
0
[3s2 + 16s3 + s4
]ds = t3 + 4t4 +
1
5t5
where in [2] we used the result of Exercise 11.9, to the effect that E[W 3(s)] = 0 andE[W 4(s)] = 6
∫ t0 E[W 2(s)] ds = 3t2. This integral is clearly finite, which shows that the
square integrability condition is satisfied. This could have been concluded directly from [1],since we are integrating a continuous function E
[|W (s) + s|4
]over a compact set.
�
Exercise 3.4. [CM14, Exercise 11.8] Using It’s formula, show that the process defined by
X(t) := W 4(t)− 6
∫ t
0W 2(u) du, t ≥ 0
is a martingale with respect to a filtration for Brownian motion.
Solution. Note that X(t) = f(t,W (t)), with f(t, x) = x4 − 6∫ t
0 W2(u) du so by It’s formula
dX(t) = −6W 2(t)dt+ 4W 3(t)dW (t) + 6W 2(t)dt = 4W 3(t)dW (t)
so X(t) is an It integral
X(t) = 4
∫ t
0W 3(s) dW (s)
which is a martingale provided that the usual square integrability is satisfied, which trivially isby continuity of Brownian motion
E
[∫ t
0(W 3(s))2 ds
]= E
[∫ t
0W 6(s) ds
]<∞
�
14
Exercise 3.5. [CM14, Exercise 11.9] Use It’s formula to show that for any integer k ≥ 2
E[W k(t)] =k(k − 1)
2
∫ t
0E[W k−2(s)
]ds
and use this to derive a formula for all the moments of the standard normal distribution.
Solution. By It’s formula we have
dW k(t) = kW k−1(t)dW (t) +1
2k(k − 1)W k−2(t)dt
whence
W k(t) = k
∫ t
0W k−1(s) dW (s) +
k(k − 1)
2
∫ t
0W k−2(s) ds
and taking expectations
E[W k(t)] =k(k − 1)
2
∫ t
0E[W k−2(s)] ds
as claimed. �
Exercise 3.6. [CM14, Exercise 11.10] Show thatM(t) = et/2 sin(W (t)), t ≥ 0 is a martingalewith respect to a filtration for Brownian motion.
Solution. By It’s formula
dM(t) =1
2et/2 sin(W (t)) + et/2 cos(W (t))dW (t)− 1
2et/2 sin(W (t))dt = et/2 cos(W (t))dW (t)
and since M(0) = 0 we have
M(t) =
∫ t
0es/2 cos(W (s))dW (s)
which is an It integral provided that the square integrability condition∫ t
0esE[cos2(W (s))]ds <∞
is satisfied (and this is clearly the case, by continuity). �
Exercise 3.7. [CM14, Exercise 11.11] Use It’s formula to show that for any non-random,continuously differentiable function f(t), the following formula of integration by parts holds:∫ t
0f(s) dW (s) = f(t)W (t)−
∫ t
0f ′(s)W (s) ds
Solution. By It’s formula we have d[f(t)W (t)] = f ′(t)W (t)dt + f(t)dW (t) and integratingwe conclude
f(t)W (t) =
∫ t
0f ′(s) dW (s) +
∫ t
0f(s) dW (s)
as claimed. �
15
Exercise 3.8. [CM14, Exercise 11.15] Let N (x) be the standard normal CDF and considerthe process
X(t)def= N
(W (t)√T − t
), 0 ≤ t < T
Express this process as an It process and show that it is a martingale with respect to anyfiltration of BM. Find the limiting value X(T−) = limt→T− X(t).
Solution. Consider the function
f(t, x) = N(
x√T − t
)=
∫ x√T−t
−∞
1√2πe−u
2/2 du
By It’s formula we have
df(t,W (t)) =∂f
∂t(t,W (t))dt+
∂f
∂x(t,W (t))dW (t) +
1
2
∂2f
∂x2(t,W (t))dt
The derivatives in question are given by
∂f
∂t(t, x) =
1
2
1√2πe−
x2
T−t ,x
(T − t)3/2,
∂f
∂x(t, x) =
1√2πe−
x2
T−t1√T − t
,
∂2f
∂x2(t, x) = − 1√
2πe−
x2
T−tx
(T − t)3/2
whence
dX(t) =1√2πe− W2(t)
2(T−t)1√T − t
dW (t)
This is a martingale provided that the square integrability condition is satisfied, namely∫ t
0E
[1
T − ue−
W2(u)T−u
]du <∞
However
E
[1
T − ue−
W2(u)T−u
]=
1
T − u
∫ ∞−∞
e−uz2
T−u1√2πe−z
2/2 dz =1√2π
1
T − u
∫ ∞−∞
e−z2( 1
2+ uT−u) dz <∞
for u ∈ [0, T ]. Its integral over [0, t] is thus also finite (being the integral of a continuous functionover a compact interval).
Finally, it is clear that
limt→T−
X(t) = N(
limt→T−
W (t)√T − t
)= 1
�
16
Exercise 3.9. [CM14, Exercise 11.16] Suppose that the processes {X(t)}t≥0 and {Y (t)}t≥0
have the log-normal dynamics
dX(t) = X(t) (µX(t)dt+ σXdW (t))
dY (t) = Y (t) (µY (t)dt+ σY dW (t))
Show that the process Z(t) = X(t)Y (t) is also log-normal, with dynamics
dZ(t) = Z(t) (µZ(t)dt+ σZdW (t))
and determine the coefficients µZ and σZ in terms of those of X and Y . Solve the same problemnow assuming that X and Y are governed by two correlated Brownian motions WX and W Y ,respectively, where Corr(WX(t),W Y (t)) = ρt, for a given correlation −1 ≤ ρ ≤ 1.
Solution.�
Exercise 3.10. [CM14, Exercise 11.17] Consider the time-homogeneous diffusion X(t) havea stochastic differential
dX(t) = [3X(t)− 1]dt+ 2√X(t)dW (t), X(t) ≥ 0
Find the stochastic differential for the process Y (t) :=√X(t) and find the generator for Y (t).
Solution. Note that Y (t) = f(X(t)) with f(t, x) =√x. By It’s formula we have
dY (t) = df(X(t)) =∂f
∂x(X(t))dX(t) +
1
2
∂2f
∂x2(X(t))dX(t)dX(t)
=1
2√X(t)
dX(t)− 1
2
1
4X(t)3/2dX(t)dX(t)
=1
2√X(t)
[3X(t)− 1]dt+ dW (t)− 1
2√X(t)
dt
=1
2√X(t)
[3X(t)− 2]dt+ dW (t)
=1
2
[3Y (t)− 2
Y (t)
]dt+ dW (t)
Recall that given an It process
X(t) = µ(t,X(t))dt+ σ(t,X(t))dW (t),
its generator is the operator
Gt,x = µ(t, x)∂
∂x+
1
2σ2(t, x)
∂2
∂x2
which acts on functions f ∈ C1,2. In our case, for the process Y (t) we have
GYt,x =1
2
[3x− 2
x
]∂
∂x+
1
2
∂2
∂x2
�
17
Exercise 3.11. [CM14, Exercise 11.18] Let X(t) = tW 2(t) and Y (t) = eW (t). Find the
stochastic differential of Z(t) = X(t)Y (t) . Compute the mean and variance of Z(t).
Solution. By It’s formula, the differentials of the two processes X(t) and Y (t) are
dX(t) = [W 2(t) + t]dt+ 2tW (t)dW (t),
dY (t) = eW (t)dW (t) +1
2eW (t)dt
To compute dZ(t) = df(X(t), Y (t)) we apply the 2-dimensional It’s formula to the functionf(x, y) = x
y . Since
fx =1
y, fy = − x
y2, fxx = 0, fxy = − 1
y2, fyy =
2x
y3
we have (omitting the dependence (X(t), Y (t)) in the first line)
dZ(t) = fxdX(t) + fydY (t) +1
2fxxdX(t)dX(t) + fxydX(t)dY (t) +
1
2fyydY (t)dY (t)
=1
Y (t)dX(t)− X(t)
Y 2(t)dY (t)− 1
Y 2(t)dX(t)dY (t) +
X(t)
Y 3(t)dY (t)dY (t)
= e−W (t)[(W 2(t) + t)dt+ 2tW (t)dW (t)
]− tW 2(t)e−2W (t)
[1
2eW (t)dt+ eW (t)dW (t)
]−e−2W (t)2tW (t)eW (t)dt+ tW 2(t)e−3W (t)e2W (t)dt
=
[e−W (t)(W 2(t) + t)− 2tW (t)e−W (t) +
1
2tW 2(t)e−W (t)
]dt
+W (t)e−W (t)t [2−W (t)] dW (t)
and I am very likely to have made a mistake somewhere... �
Exercise 3.12. [CM14, Exercise 11.19] Let X(t) be a time-homogeneous diffusion processsolving an SDE dX(t) = cX(t) + σdW (t) with initial condition X(0) = x ∈ R and where c, σare constants. Consider the process defined by Y (t) = X2(t)− 2c
∫ t0 X
2(s) ds− σ2t, t ≥ 0.
(a) Represent Y (t) as an It process and show that it is a martingale with respect to any filtrationfor Brownian motion.
(b) Compute the mean and variance of Y (t) for all t ≥ 0.
Solution.
(a) By It’s formula we have
dY (t) = 2X(t)dX(t) + σ2dt− 2cX2(t)dt− σ2dt
= 2cX2(t)dt+ 2σX(t)dW (t)− 2cX2(t)dt
= 2σX(t)dW (t)
whence
Y (t) = Y (0) + 2σ
∫ t
0X(s) dW (s) = x2 + 2σ
∫ t
0X(s) dW (s)
which is a martingale provided that the square integrability condition∫ t
0 E[X2(s)] ds issatisfied. We verify this in part (b).
18
(b) Clearly E[Y (t)] = x2. In order to compute the variance, we start by solving the SDE forX(t) using an integrating factor. Note that
d(e−ctX(t)) = −ce−ctX(t) + ce−ctX(t)dt+ σe−ctdW (t) = σe−ctdW (t)
whence
X(t) = xect + σ
∫ t
0ec(t−s) dW (s)
The expected value of X(t) is E[X(t)] = xect and by It isometry we have
Var[X(t)] =
∫ t
0σ2e2c(t−s) ds = −σ
2
2ce2c(t−s)
∣∣∣t0
=σ2
2c[e2ct − 1]
Since we also have Var[X(t)] = E[X2(t)]− E[X(t)]2 we conclude that
Var[Y (t)] = 4σ2
∫ t
0E[X2(s)] ds = 4σ2
∫ t
0
[σ2
2c(e2cs − 1) + x2e2cs
]ds
=2σ4
c
[(e2ct − 1)
x2 + 1
2c− t]
�
Exercise 3.13. [CM14, Exercise 11.20] Use the It formula to write down stochastic differ-entials for the following processes.
(a) Y (t) = exp(σW (t)− 1
2σ2t). Also find the expectation and variance of the process X(t) =∫ t
0 Y (s) ds.
(b) Z(t) = f(t)W (t) where f is continuously differentiable.
Solution.
(a) By It’s formula we have
dY (t) = eσW (t)− 12σ2t
[−1
2σ2dt+ σdW (t) +
1
2σ2
]= σeσW (t)− 1
2σ2tdW (t) = σY (t)dW (t)
If A(t) =∫ t
0 Y (s) ds, it is clear that
E[A(t)] =
∫ t
0E[Y (s)] ds =
∫ t
0e−
12σ2sE[eσW (t)] ds =
∫ t
0e−
12σ2se
12σ2s ds = t
As for the variance we have
Var[A(t)] = E[A2(t)]− E[A(t)]2 = E[A2(t)]− t2
and
E[A2(t)] = E
[∫ t
0Y (u) du
∫ t
0Y (v) dv
]=
∫ t
0
∫ t
0E [Y (u)Y (v)] du dv
which is computed as on [CM14, Page 413].
19
(b) By It’s formula we have
dZ(t) = d [f(t)W (t)] = f ′(t)W (t)dt+ f(t)dW (t)
�
Exercise 3.14. [CM14, Exercise 11.22] A time-homogeneous diffusion processX has stochas-tic differential dX(t) = X(t)(1 − X(t))dW (t). Assuming that 0 < X(t) < 1, show that the
process Y (t) = ln(
X(t)1−X(t)
)has a constant diffusion coefficient.
Solution. Note that Y (t) = f(X(t)) with f(x) = ln x1−x . Clearly f ′(x) = 1
x(1−x) . By It’sformula we have
dY (t) = f ′(X(t))dX(t) +1
2f ′′(X(t))X2(t)[1−X(t)]2dt = dW (t) +O(dt)
so that the diffusion term is constant and equal to 1. Since f ′′(x) = 2x−1x2(1−x)2 , the drift term will
be given by1
2f ′′(X(t))X2(t)[1−X(t)]2 =
1
2(2X(t)− 1).
�
Exercise 3.15. [CM14, Exercise 11.23] Let X(t) = (1 − t)∫ t
01
1−s dW (s), where 0 ≤ t < 1.Provide the stochastic differential equation for X(t) and check your answer by solving the SDEobtained subject to the initial condition X(0) = 0.
Solution. Note that X(t) = f(t, Y (t)) where f(t, y) = (1− t)y and Y (t) =∫ t
01
1−s dW (s). ByIt’s formula we have
dX(t) = ft(t, Y (t))dt+ fy(t, Y (t))dY (t) +1
2fyy(t, Y (t))dY (t)dY (t)
= −Y (t)dt+ (1− t)dY (t)
= −[∫ t
0
dW (s)
1− s
]dt+ (1− t)dW (t)
1− t
= −[∫ t
0
dW (s)
1− s
]dt+ dW (t)
= −X(t)
1− tdt+ dW (t)
�
Exercise 3.16. [CM14, Exercise 11.25] Let g(y) be a given function of y, and suppose thatf(x) is a solution of f ′(x) = g(f(x)). Show that X(t) = f(W (t)) is a solution of the SDE
dX(t) =1
2g(X(t))g′(X(t))dt+ g(X(t))dW (t)
20
Solution. Since f ′(x) = g(f(x)), clearly f ′′(x) = g′(f(x))f ′(s). By It’s formula we have that
dX(t) = df(W (t)) = f ′(W (t))dW (t) +1
2f ′′(W (t))dt
= g(f(W (t))dW (t) +1
2g′(f(W (t))f ′(W (t))
= g(X(t))dW (t) +1
2g′(X(t))g(X(t))dt
as claimed. �
Exercise 3.17. [CM14, Exercise 11.26] Use Exercise 11.25 to solve the following non-linearSDE
dX(t) = X3(t)dt+X2(t)dW (t), X(0) = x0
Solution. By Exercise 11.25, the solution is given by
X(t) = f(W (t)), where
{f ′(x) = g(f(x)) = [f(x)]2,f(0) = x0
Separating variables one immediately obtains
f(x) =x0
1− xx0
which is singular at x = 1x0
. �
Exercise 3.18. [CM14, Exercise 11.32] Consider the boundary value problem for the heatequation
∂V
∂t+
1
2
∂2V
∂x2= 0, V (T, x) = f(x)
Show that the solution is given by
V (t, x) =1√
2π(T − t)
∫ ∞−∞
f(y)e−(y−x)2
2(T−t) dy
Solution. This is essentially identical to [CM14, Example 11.10]. By the Feynman-Kactheorem [CM14, Theorem 11.7], if a stochastic process {X(t)}t≥0 satisfies the SDE
dX(t) = µ(t,X(t))dt+ σ(t,X(t))dW (t), (5)
then the C1,2 function V (t, x) solving the PDE
∂V
∂t+
1
2σ2(t, x)
∂2V
∂x2+ µ(t, x)
∂V
∂x= 0, 0 ≤ t ≤ T, x ∈ R+,
V (T, x) = f(x)
admits a representationV (t, x) = E [f(X(T ))|X(t) = x] (6)
provided that E[|f(X(T ))|] <∞.
21
In our case, µ(t, x) ≡ 0 and σ(t, x) = 1 so the SDE (5) becomes dX(t) = dW (t) and has asolution X(T ) = X(t) + [W (T )−W (t)]. The representation (6) then reads
V (t, x) = E [f(X(T ))|X(t) = x] = E [f(X(t) +W (T )−W (t)|X(t) = x]
= E [f(x+W (T )−W (t))][∗]=
∫ ∞−∞
f(y)ϕx,T−t(y) dy
=1√
2π(T − t)
∫ ∞−∞
f(y)e− (y−x)2
2(T−t) dy
where in [∗] we used the fact that E[φ(X)] =∫∞−∞ φ(y)ϕX(y) dy, where ϕX(·) denotes the PDF
of X, and also that x+ [W (T )−W (t)] ∼ N (x, T − t). �
Exercise 3.19. [CM14, Exercise 11.34] Determine f(t, x) satisfying the following boundaryvalue problem
∂V
∂t+
1
2σ2x2∂
2V
∂x2+ µx
∂f
∂x= 0, 0 ≤ t ≤ T, x ∈ R+,
f(T, x) = 1[K1,K2](x)
Solution. Write φ(x) = 1[K1,K2](x) more generally to begin with and we will specialize at theend. By the Feynman-Kac theorem [CM14, Theorem 11.7], if a stochastic process {X(t)}t≥0
satisfies the SDEdX(t) = µ(t,X(t))dt+ σ(t,X(t))dW (t), (7)
then the C1,2 function f(t, x) solving the PDE
∂f
∂t+
1
2σ2(t, x)
∂2f
∂x2+ µ(t, x)
∂f
∂x= 0, 0 ≤ t ≤ T, x ∈ R+,
f(T, x) = φ(x)
admits a representationf(t, x) = E [φ(X(T ))|X(t) = x] (8)
provided that E[|φ(X(T ))|] <∞.In this case, the drift and volatility are µ(t, x) = µx and σ(t, x) = σx, for real constants
µ, σ > 0 and the SDE (7) reads dX(t) = µX(t)dt + σX(t)sW (t). This is a linear SDE withsolution (c.f. [CM14, Equation 11.27])
X(T ) = X(t) exp
[(µ− σ2
2
)(T − t) + σ(W (T )−W (t))
]The representation (8) then reads
f(t, x) = E [φ(X(T ))|X(t) = x]
= E
[φ
(X(t) exp
[(µ− σ2
2
)(T − t) + σ(W (T )−W (t))
])|X(t) = x
]= E
[φ
(x exp
[(µ− σ2
2
)(T − t) + σ(W (T )−W (t))
])]=
∫ ∞−∞
φ
(x exp
[(µ− σ2
2
)(T − t) + σy
])1√
2π(T − t)e− y2
2(T−t) dy (9)
22
We now specialize to φ(x) = 1[K1,K2](x). Note that
K1 ≤ x exp
[(µ− σ2
2
)(T − t) + σy
]≤ K2
if and only if
1
σ
[ln
(K1
x
)−(µ− σ2
2
)(T − t)
]︸ ︷︷ ︸
not= α1
≤ y ≤ 1
σ
[ln
(K2
x
)−(µ− σ2
2
)(T − t)
]︸ ︷︷ ︸
not= α2
so that φ(x exp
[(µ− σ2
2
)(T − t) + σy
])6= 0 if and only if y ∈ [α1, α2] whence the integral in
(9) becomes
f(t, x) =
∫ α2
α1
1√2π(T − t)
e− y2
2(T−t) dy = N (α2)−N (α1)
�
Exercise 3.20. [CM14, Exercise 11.36] Assume that a stock price process {S(t)}t≥0 satisfiesthe SDE
dS(t) = rS(t)dt+ σS(t)dW (t)
with constants r, σ > 0, and where {W (t)}t≥0 is a standard P -BM. By using Girsanov’s theorem,find the explicit expression for the Radon-Nikodym derivative process
ρt =
(dP
dP
)t
such that the process defined by S(t) = ert
S(t) , t ≥ 0 is a P -martingale. Give the SDE satisfied by
the stock price S(t) with respect to the P -BM.
Solution. We start by computing the differential of the process S(t).
dS = d
(ert
S(t)
)= r
ert
S(t)dt− ert
S2(t)dS(t) +
1
2
2ert
S3(t)σ2S2(t)dt
= σS(t)[−σdt+ dW (t)
]Define W (t) := W (t)−σt. By Girsanov’s theorem, W is a P -BM under the measure defined bythe Radon-Nikodym derivative process
ρt =
(dP
dP
)t
= exp
(−1
2
∫ t
0σ ds+
∫ t
0σ dW (s)
)= e−
12σ2t+σW (t)
and the SDE satisfied by the stock price S(t) with respect to the P -BM is
dS(t) = σdW (t)
�
23
Exercise 3.21. [CM14, Exercise 11.38] Consider a one-dimensional general diffusion process{X(t)}t≥0 having a transition PDF p(s, t;x, y), s < t, with respect to a given probability measureP , for all x, y in the state space of the process. Assume a change of measure P→P is definedby the Radon-Nikodym derivative process
ρt =
(dP
dP
)t
= h(t,X(t)), ∀t ≥ 0
Let p(s, t;x, y) denote the PDF with respect to the measure P . Show that the two transitionPDF’s are related by
p(s, t;x, y) =h(t, y)
h(s, x)p(s, t;x, y)
Solution.
P (s, t;x, y) = P (X(t) ≤ y|X(s) = x) = E[1X(t)≤y|X(s) = x
]=
1
ρsE[ρt1X(t)≤y|X(s) = x
]=
1
h(s,X(s))E[h(t,X(t))1X(t)≤y|X(s) = x
]=
1
h(s, x)
∫ ∞−∞
h(t, u)1u≤yp(s, t;x, u) du
=1
h(s, x)
∫ y
−∞h(t, u)p(s, t;x, u) du
so by the first fundamental theorem of calculus
p(s, t;x, y) =∂P
∂y(s, t;x, y) =
h(t, y)
h(s, x)p(s, t;x, y)
as claimed. �
24
4 Chapter 12: Risk-Neutral pricing in the Black-Scholes Econ-omy: one underlying stock
Exercise 4.1. [CM14, Exercise 12.4] Assume the standard Black-Scholes model in an econ-omy with constant continuously compounded interest rate r and with stock price process {S(t)}t≥0
as a GBM with constant volatility σ and constant continuous dividend yield q. Let S(t) = S > 0be the spot at time t < T , where T is the expiry date. Derive the corresponding arbitrage-freetime-t pricing formula V (t, S) for a European option with the following payoffs:
(a) Λ(S(T )) =∑N
n=0 anSn(T ), N ≥ 1, an ∈ R.
(b) Λ(S(T )) = (Sα(T )−K)IS(T )>K , α ∈ R \ 0.
Solution.
(a) Denote as usual τ = T − t. Clearly
V (t, S) = e−rτ Et,S [Λ(S(T ))] = e−rτ Et,S
[N∑n=0
anSn(T )
]= e−rτ
N∑n=0
anEt,S [Sn(T )]
These expectations are computed in [CM14, Equation 12.40]. Writing
S(T ) = S(t)e(r−q− 12σ2)τ+σ
√τZ , Z =
1√τ
(W (T )−W (t)) ∼ N(0, 1)
we have:
Et,S [Sn(T )] = Et,S
[Snen(r−q− 1
2σ2)τ+nσ
√τZ]
= Snen(r−q− 12σ2)τ E
[enσ√τZ]
[∗]= Snen(r−q+ 1
2σ2(n−1))τ
where in [∗] we used [CM14, Formula A.2]. Putting everything together we conclude that
V (t, S) = e−rτN∑n=0
anSnen(r−q+ 1
2σ2(n−1))τ
(b) In this case we have
V (t, S) = e−rτ Et,S[(Sα(T )−K)IS(T )>K
]= e−rτ Et,S
[Sα(T )IS(T )>K
]−Ke−rτ Et,S
[IS(T )>K
]Write as usual
S(T ) = S(t)e(r−q− 12σ2)τ+σ
√τZ , Z =
1√τ
(W (T )−W (t)) ∼ N(0, 1)
25
The second expectation above is simply
Et,S[IS(T )>K
]= P
(Se(r−q− 1
2σ2)τ+σ
√τZ > K
)= P
(Z >
ln KS − (r − q − 1
2σ2)τ
σ√τ
)
= P
(Z < −
ln SK + (r − q − 1
2σ2)τ
σ√τ
)not= N (−d−)
The first expectation is computed in [CM14, Equation 12.41]:
Et,S[Sα(T )IS(T )>K
]= Snen(r−q− 1
2σ2)τ Et,S
[eσ√τZI{Z>
ln KS−(r−q− 1
2σ2)τ
σ√τ
}
]
= Snen(r−q+ 12σ2(n−1)τN
(ln S
K + (r − q + 2n−12 σ2)τ
σ√τ
)
so in conclusion we have:
V (t, S) = e−rτSnen(r−q+ 12σ2(n−1)τN
(ln S
K + (r − q + 2n−12 σ2)τ
σ√τ
)−Ke−rτN (−d−)
d− =ln S
K + (r − q − 12σ
2)τ
σ√τ
�
Exercise 4.2. [CM14, Exercise 12.5] Assume the standard Black-Scholes model in an econ-omy with constant continuously compounded interest rate r and with stock price process {S(t)}t≥0
as a GBM with constant volatility σ and constant continuous dividend yield q. A European callspread has payoff
Λ(S(T ))
0, S(T ) ≤ K,S(T )−K, K < S(T ) < K + ε,ε, S(T ) ≥ K + ε
where K, ε are positive real constants.
(a) Give a sketch of the payoff function.
(b) Derive a formula for the option’s present value V (t, S) and ∆(t, S) = ∂V∂S .
(c) Find limε→0 V (t, S) and limε→∞ V (t, S) and explain your results.
Solution.
(a)
26
(b) The payoff can be decomposed as a combination of call payoffs
Λ(S(T )) = (S(T )−K)+ − (S(T )−K − ε)+
so
V (t, S) = Et,S[(S(T )−K)+
]− Et,S
[(S(T )−K − ε)+
]= e−qτN
(d+(
e−qτS
K, τ)
)− e−rτKN
(d−(
e−qτS
K, τ)
)−e−qτN
(d+(
e−qτS
K + ε, τ)
)+ e−rτKN
(d−(
e−qτS
K + ε, τ)
)(10)
As for the delta we have
∂V
∂S(t, S) = e−qτ
[N(d+(
e−qτS
K, τ)
)−N
(d+(
e−qτS
K + ε, τ)
)](c) Obviously limε→0 V (t, S) = 0 since the terms in the second line of equation (10) then equal
those in the first one and have opposite sign. Note that for ε = 0 the option payoff isidentically zero.
As ε→∞, the option becomes a simple call struck at K whence
limε→∞
V (t, S) = e−qτN(d+(
e−qτS
K, τ)
)− e−rτKN
(d−(
e−qτS
K, τ)
)This can also be seen by taking the limit in equation (10) and noting that
limε→∞
N(d+(
e−qτS
K + ε, τ)
)= N
(limε→∞
d+(e−qτS
K + ε, τ)
)= N (d+(0, τ)) = N (−∞) = 0
since recall d+(x, τ) =lnx−(r+ 1
2σ2)τ
σ√τ
.
�
Exercise 4.3. [CM14, Exercise 12.7]
Solution.�
Exercise 4.4. [CM14, Exercise 12.8]
Solution.�
Exercise 4.5. [CM14, Exercise 12.9] A so-called pay-later European option costs the holdernothing (i.e. zero premium) to set up at present time t = 0. The payoff to the holder is(S(T ) − K)+. Moreover, the holder must pay out X dollars to the writer in the case thatS(T ) ≥ 0. Derive an expression fr the fair value of X. Determine the fair value in the limit ofinfinite volatility limσ→∞X(σ). Assume the standard Black-Scholes model in an economy withconstant continuously compounded interest rate r and with stock price process {S(t)}t≥0 as aGBM with constant volatility σ and constant continuous dividend yield q.
27
Solution. The effective payoff of the pay-later option is
Λeff(S(T )) = (S(T )−K)+ −XI{S(T )≥K}
and we need to determine the fair value of X such that the time-0 expected value of the latteris zero, namely
E0,S
[Λeff(S(T ))
]= E0,S
[(S(T )−K)+ −XI{S(T )≥K}
]= 0
Clearly
E0,S [Λeff(S(T ))] = e−rT E0,S
[(S(T )−K)+
]−Xe−rT E0,S
[I{S(T )≥K}
]= Se−qTN
(d+(
e−qTS
K, T )
)−Ke−rTN
(d−(
e−qTS
K, T )
)−Xe−rTN
(d−(
e−qTS
K, T )
)
where as usual d± =lnx+(r−q± 1
2σ2)T
σ√T
.
Imposing E0,S [Λeff(S(T ))] = 0 and solving for X yields
X = Se(r−q)TN(d+( e
−qTSK , T )
)N(d−( e
−qTSK , T )
) −KNote that
limσ→∞
d±(x, τ) = limσ→∞
lnx+ (r − q ± 12σ
2)T
σ√T
≡ ± limσ→∞
σ√T
2= ±∞
so
limσ→∞
N (d+) = 1, limσ→∞
N (d−) = 0
whence limσ→∞X =∞. �
Exercise 4.6. [CM14, Exercise 12.10] Let C(S, τ) be the Black-Scholes pricing formula ofa standard call option with spot S, strike K, fixed interest rate r, zero stock dividend, constantvolatility σ and time to maturity τ > 0.
(a) Show that the respective limiting values of the call price for vanishing and infinite volatilityare given by
limσ→0
C(S, τ) = (S −Ke−rτ )+, limσ→+∞
C(S, τ) = S
(b) Give a financial interpretation of both limits. Note that the second limit is independent ofthe strike value K; give a financial intuition for this fact.
Solution.
28
(a) We performed an identical computation in the previous exercise. The Black-Scholes formulafor a call option is
C(S, τ) = SN(d+(
S
K, τ)
)−Ke−rτN
(d−(
S
K, τ)
), d±(x, τ) =
lnx+ (r ± 12σ
2)τ
σ√τ
Also note that
limσ→∞
d± ≡ ± limσ→∞
σ√τ
2= ±∞
whence
limσ→∞
N (d+) = 1, limσ→∞
N (d−) = 0
so
limσ→∞
C(S, τ) = limσ→∞
SN(d+(
S
K, τ)
)−Ke−rτN
(d−(
S
K, τ)
)= S
On the other hand,
limσ→0
d±(S
K, τ) ≡ lim
σ→0
ln SK + rτ
σ√τ
=
{+∞, ln S
K + rτ ≥ 0,
−∞, ln SK + rτ < 0
Note further that ln SK + rτ ≥ 0 if and only if S − e−rτ ≥ 0. Hence
limσ→0
C(S, τ) = limσ→0
SN(d+(
S
K, τ)
)−Ke−rτN
(d−(
S
K, τ)
)= 0 · I{ln S
K+rτ<0} + (S −Ke−rτ ) · I{ln S
K+rτ≥0}
= (S −Ke−rτ )+
(b) The more volatile the market is, the more expensive an option is. In the absence of arbitrage,time-t value of a call is bounded above by the spot, so as the volatility grows indefinitely,the call value converges to its no-arbitrage maximum S...
�
Exercise 4.7. [CM14, Exercise 12.11] Consider the value of a European call option writtenby an issuer who only has a fraction 0 ≤ α < 1 of the underlying asset. That is, at expirationtime T the payoff of this type of call is given by
VT = (S(T )−K)+I{αS(T )≥S(T )−K} + αS(T )I{αS(T )<S(T )−K}
Let CL(S, τ ;K,α) denote the value of a European call, where τ = T − t is the time to expiry,K > 0 is the strike, S(t) = S is the spot of the underlying. Show that
CL(S, τ ;K,α) = C(S,K, τ)− (1− α)C
(S,
K
1− α, τ
)
29
Solution. Rewrite thew payoff as
VT = (S(T )−K)+I{ K1−α≥S(T )} + αS(T )I{ K
1−α<S(T )}
so the no-arbitrage price of the options is given by the usual risk-neutral formula
CL(S, τ ;K,α) = e−rτ E [VT ] = e−rτ E[(S(T )−K)+I{ K
1−α≥S(T )} + αS(T )I{ K1−α<S(T )}
]= e−rτ E
[(S(T )−K)I{K≤S(T )≤ K
1−α}+ αS(T )I{ K
1−α<S(T )}
](11)
We seek to show that C(S,K, τ)− (1− α)C(S, K
1−α , τ)
indeed represents the above time-t
value (11) of an option with payoff VT . Indeed:
C(S,K, τ) − (1− α)C
(S,
K
1− α, τ
)= e−rτ Et,S
[(S(T )−K)I{S(T )≥K} − (1− α)
(S(T )− K
1− α
)I{S(T )≥ K
1−α}
]= e−rτ Et,S
[αS(T )I{S(T )≥ K
1−α}+ (S(T )−K)I{S(T )≥K}
−(S(T )− K
1− α+
αK
1− α
)I{S(T )≥ K
1−α}
]= e−rτ Et,S
[αS(T )I{S(T )≥ K
1−α}+ (S(T )−K)
(I{S(T )≥K} − I{S(T )≥ K
1−α}
)]= e−rτ Et,S
[αS(T )I{S(T )≥ K
1−α}+ (S(T )−K)I{K≤S(T )≤ K
1−α}
]and the last expression is precisely (11). �
Exercise 4.8. [CM14, Exercise 12.19] A variant...
Solution.�
Exercise 4.9. [CM14, Exercise 12.20] Assume the stock price {S(t)}t≥0 is a GBM withconstant volatility σ and zero dividend in an economy with constant interest rate r. Let t <T0 < T , i.e. T0 is an arbitrary intermediate time before expiry time T , and consider a European-style option with payoff at time T given by
VT = min{S(T0), S(T )}
(a) Show that the value V at time t ≤ T0 of this option is given by
V = S(t)[N (−d+) + e−r(T−T0)N (d−)
], d± =
(r ± 12σ
2
σ
√T − t
Note: the option value is dependent on T − T0 and does not depend on t.
(b) What is the position held in the stock at time t ≤ T0 in a self-financing replicating strategy?
30
Solution.
(a) We may write the payoff as
VT = min{S(T0), S(T )} = S(T )I{S(T )<S(T0)} + S(T0)I{S(T0)<S(T )}
and the option price is given by the usual risk-neutral pricing formula
V (t, S) = e−r(T−t)Et[VT ] = e−r(T−t)[Et[S(T )I{S(T )<S(T0)}] + E[S(T0)I{S(T0)<S(T )}]
](12)
In order to compute these expectations we condition first on S(T0) and use iterated condi-tioning, namely
Et[S(T )I{S(T )<S(T0)}
]= Et
[E[S(T )I{S(T )<S(T0)}|S(T0)
]][1]= Et
[E[S(T0)e
(r− 12σ2)(T−T0)+σ
√T−T0ZI{Z<−d−} |S(T0)
]]= e(r− 1
2σ2)(T−T0)Et
[S(T0)E
[eσ√T−T0ZI{Z<−d−}
]][2]= er(T−T0)Et [S(T0)N (−d+)]
[3]= er(T−T0)et(T0−t)S(t)N (d+) = er(T−t)S(t)N (d+)
Here [1] follows from the fact that S(T ) < S(T0) if and only if
Z < −(r − 1
2σ2)(T − T0)
σ√T − T0
= −r − 1
2σ2
σ
√T − T0.
In [2] we used [CM14, Formula A.2] to write
E[eσ√T−T0ZI{Z<−d−}
]= e
12σ2(T−T0)N (−d+)
and in [3] we used that e−rtS(t) is a P -martingale. The first summand in equation (12) isthus
e−r(T−t)Et[S(T )I{S(T )<S(T0)}] = S(t)N (−d+)
An identical computation shows that the second expectation in equation (12) is
E[S(T0)I{S(T0)<S(T )}] = Et[E[S(T0)I{S(T0)<S(T )}|S(T0)
]]= Et [S(T0)N (d−)] = er(T0−t)S(t)N (d−)
whence
e−r(T−t)E[S(T0)I{S(T0)<S(T )}] = e−r(T−T0)S(t)N (d−)
Adding up both expressions we conclude that
V = S(t)[N (−d+) + e−r(T−T0)N (d−)
]as claimed.
31
(b) The Delta of this option is
∂V
∂S= N (−d+) + e−r(T−T0)N (d−)
which remains static over time.
�
Exercise 4.10. [CM14, Exercise 12.22]
Solution.
�
Exercise 4.11. [CM14, Exercise 12.25] Consider the discrete geometric averaging of a stockprice process at evenly distributed discrete times tj = t0 + jδt, j = 1, 2, . . . , n, with a time stepδt = T−t0
n ; tn = T is the time of expiration. Define the discretely monitored geometric averagingby
Gk =
k∏j=1
S(tj)
1/k
, k = 1, . . . , n
(a) Assuming that the stock price follows a GBM process, show that Gn is a log-normal randomvariable. Find the mean and variance of lnGn.
(b) Derive the risk-neutral time-t0 prices of the fixed strike Asian call and put options withrespective payoff functions (Gn −K)+ and (K −Gn)+, where K > 0 is a strike price.
Solution.
(a) Since S(t) follows a GBM (assume zero dividends) we have
S(tj) = S(t0)e(r− 12σ2)tj+σW (tj), j = 1, . . . , n
whence
Gn =
n∏j=1
S(tj)
1/n
= S(t0)e(r− 12σ2)( 1
n
∑nj=1 tj)+σ
n
∑nj=1W (tj) (13)
Recall that from [CM14, Exercise 10.10] we have that1
n∑j=1
W (tj) ∼ N (0,n∑j=1
(2j − 1)tn+1−j)d=
√√√√ n∑j=1
(2j − 1)tn+1−j · Z, Z ∼ N (0, 1)
1This follow from the transformation rule: if X,Y are a p-dimensional and m-dimensional random vec-tors, with respective covariance matrices ΣX and SY , and Y = AX, then ΣY = AΣXA
>. In our caseX = (W (t1), . . . ,W (tn))> and A = (1, . . . , 1), so (σX)i,j = Cov(W (ti),W (tj)) = min{i, j} and Var(Y ) =AΣXA
> =∑nj=1(2j − 1)tn+1−j .
32
and therefore
σ
n
n∑j=1
W (tj)d=
σ
n√T
√√√√ n∑j=1
(2j − 1)tn+1−j ·√TZ
not= σ√TZ
with Z ∼ N (0, 1) and T = 1n
∑nj=1 tj .
With regards to the drift term, note that
(r − 1
2σ2)
1
n
n∑j=1
tj
= (r − 1
2σ2)T =
(r − σ2
2+σ2
2− σ2
2
)Tnot=
(r − σ2
2+ δ
)T
so that equation (13) becomes
Gn(T ) = S(t0)e(r−δ− 12σ2)T+σ
√TZ
and clearly
lnGn(T ) ∼ N(
lnS(t0) + (r − δ − 1
2σ2)T, σ2T
)There was no need to do all this to conclude normality, but now we have Gn written as aGBM (with constant dividend δ), so that we can apply the Black-Scholes formula directlyin order to price fixed strike Asian calls and puts as though they were ordinary calls andputs on the underlying Gn(T ).
(b) Since from part (a) we have that Gn(T ) follows a GBM
Gn(T ) = S(t0)e(r−δ− 12σ2)T+σ
√TZ
with
T =1
n
n∑j=1
tj , σ =σ
n√T
√√√√ n∑j=1
(2j − 1)tn+1−j , δ =σ2
2− σ2
2
the Black-Scholes formula yields that time-t0 value of a fixed-strike Asian call with payoff(Gn(T )−K)+ is
C(t0, S) = e−δ(T−t0)SN (d+)− e−r(T−t0)KN (d−)
with
d± = d±
(e−δ·(T−t0)S
K, T − t0
), d±(x, τ) =
lnx− (r − δ ± 12 σ
2)τ
σ√τ
The time-t0 value of a put can be obtained by put-call parity.
�
33
Exercise 4.12. [CM14, Exercise 12.26] Let the stock price process follow the GBM model
dS(t) = S(t)[(r − q)dt+ σdW (t)],
S(T ) = S(t)e(r−q− 12σ2)(T−t)+σ·(W (T )−W (t)) d
= S(t)e(r−q− 12σ2)(T−t)+σ
√T−tZ , Z ∼ N(0, 1)
Define the continuously monitored geometric average of S(t) over a time period [0, t] by
G(t) = exp
(1
t
∫ t
0lnS(u) du
).
(a) Show that the process lnG(t) is Gaussian.
(b) Find the mean and variance of G(T ) conditional on G(t) and S(t), for 0 ≤ t ≤ T .
(c) Show that G(T ) can be written as
G(T ) = G(t)tT S(t)
T−tT eµ+σZ
for some µ, σ ∈ R, and where Z ∼ N(0, 1) under measure P . Find the values of µ and σ.
(d) Derive the risk-neutral time-t pricing functions for the fixed strike Asian call and put optionswith respective payoff functions (G(T )−K)+ and (K −G(T ))+ for 0 ≤ t ≤ T . Express thepricing functions in terms of the spot values G(t) = G > 0, S(t) = S > 0, and times t andT .
(e) Establish the put-call parity relation for the fixed strike Asian call and put options.
Solution.
(a) Since S(t) = S(0)e(r−q− 12σ2)t+σW (t), we have that
lnG(t) =1
t
∫ t
0lnS(u) du =
1
t
∫ t
0
[lnS(0) + (r − q − 1
2σ2)u
]du+
σ
t
∫ t
0W (u) du
From [CM14, Proposition 11.1] we know that∫ t
0 W (u) du ∼ N (0, t3
3 ) whence
lnG(t) ∼ N(
lnS(0) +1
2(r − q − 1
2σ2)t,
σ2t
3
)(b) First note that
lnG(T ) =1
T
∫ T
0lnS(u) du =
t
T
1
t
∫ t
0lnS(u) du+
1
T
∫ T
tlnS(u) du
=t
TlnG(t) +
1
T
∫ T
tlnS(u) du (14)
Now write S(u) = S(t)e(r−q− 12σ2)(u−t)+σ(W (u)−W (t)) whence
1
T
∫ T
tlnS(u) du =
1
T
∫ T
t
[lnS(t) + (r − q − 1
2σ2)(u− t)
]du+
σ
T
∫ T
t[W (u)−W (t)] du
34
For the first integral we have
1
T
∫ T
t
[lnS(t) + (r − q − 1
2σ2)(u− t)
]du = lnS(t)
T − tT
+1
2(r − q − 1
2σ2)
(T − t)2
T
and as for the second one∫ T
t[W (u)−W (t)] du =
∫ T
0W (u) du−
∫ t
0W (u) du−W (t)(T − t)
Note that
Var
(∫ T
0W (u) du−
∫ t
0W (u) du
)=
T 3
3+t3
3− 2Cov
(∫ t
0W (u) du,
∫ T
0W (u) du
)=
T 3
3+t3
3− 2
(t3
3− (T − t) t
2
2
)=
T 3
3− t3
3+ t2(T − t)
where the covariance computation can be found on [CM14, Page 413]. Since W (u), u ∈ [t, T ]is independent of W (t) we conclude that
Var
(∫ T
t[W (u)−W (t)] du
)= Var
(∫ T
tW (u) du
)+ Var (W (t)(T − t))
=T 3
3− t3
3+ t2(T − t) + t(T − t)2
=1
3(T − t)3
Putting everything together, from (14) we obtain
lnG(T ) =t
TlnG(t) + S(t)
T − tT
+1
2(r − q − 1
2σ2)
(T − t)2
T+
1
3σ2 (T − t)3
T 2Z (15)
for Z ∼ N (0, 1). In conclusion
lnG(T )| lnG(t) ∼ N(t
TlnG(t) + lnS(t)
T − tT
+1
2(r − q − 1
2σ2)(T − t)2,
1
3
σ2
T 2(T − t)3
)(c) Exponentiating (15) we obtain
G(T ) = G(t)tT S(t)
T−tT exp
(1
2(r − q − 1
2σ2)(T − t)2 +
1
3
σ2
T 2(T − t)3Z
)not= G(t)
tT S(t)
T−tT eµ+σZ
where
µ =1
2(r − q − 1
2σ2)(T − t)2,
σ =1
3
σ2
T 2(T − t)3
35
(d) In order find the time-t value of an Asian call with payoff (G(T )−K)+ we proceed as usual:
V (t, S(t), G(t)) = e−r(T−t)Et,S,G[(G(T )−K)+
]= e−r(T−t)Et,S,G
[(G(t)
tT S(t)
T−tT eµ+σZ −K
)IG(t)>K
][1]= eµ−r(T−t)G
tT S
T−tT Et,S,G
[eσZIZ>α(t,S,G)
]−Ke−r(T−t)Pt,S,G
(Z > α
)= eµ+ σ2
2−r(T−t)G
tT S
T−tT N
(1
σ
(σ2 + ln
GtT S
T−tT
K+ µ
))
−Ke−r(T−t)N
(1
σ
(lnG
tT S
T−tT
K+ µ
))
where in [1] we denoted
α(t, S,G) =lnK − t
T lnG− T−tT lnS − µ
σ=
1
σ
[−µ+ ln
(K
GtT S
T−tT
)](e) To derive a put-call parity relation (and the time-t value of a put from it) note that
(G(T )−K)+ − (K −G(T ))+ = G(T )−K
Discounting and taking expectations yields
e−r(T−t)Et,S,G[(G(T )−K)+]︸ ︷︷ ︸C(t,S,G)
− e−r(T−t)Et,S,G[(K −G(T ))+]︸ ︷︷ ︸P (t,S,G)
= e−r(T−t)Et,S,G[G(T )]−Ke−r(T−t)
whenceP (t, S,G) = C(t, S,G)− e−r(T−t)Et,S,G[G(T )] +Ke−r(T−t)
and where
e−r(T−t)Et,S,G[G(T )] = eµ−r(T−t)GtT S
T−tT E[eσZ ] = eµ+ 1
2σ2−r(T−t)G
tT S
T−tT
�
36
5 Chapter 13: Risk-Neutral pricing in a Multi-Asset Economy
Exercise 5.1. [CM14, Exercise 13.4] A plain currency call option on a foreign exchange ratehas payoff
CT = (X(T )− κ)+
(a) Derive the pricing function C(t, x), t < T for this call by evaluating the risk-neutral expec-tation formula
C(t, x) = er(T−t)E[(X(T )− κ)+|X(t) = x
](b) Give the BSPDE for the pricing function C(t, x).
Solution.
(a) Suppose that the exchange rate process is a GBM driven by a d-dimensional Brownianmotion, namely (c.f. [CM14, Equation 13.102])
X(T ) = X(t)e(r−rf− 12σ2X)(T−t)+σ(X)·(W (T )−W (t)) d
= X(t)e(r−rf− 12σ2X)(T−t)+σX
√T−tZ
Z =1
σX√T − t
σ(X) · (W (T )− W (t))
where W (t) = (W1(t), . . . ,Wd(t)) and σ(X) = (σX,1, . . . , σX,d).
The pricing function
C(t, x) = er(T−t)E[(X(T )− κ)+|X(t) = x
]= er(T−t)E
[(X(t)e(r−rf− 1
2σ2X)(T−t)+σX
√T−tZ − κ
)+|X(t) = x
]can be obtained using the standard Black-Scholes formula for the price of a call, with interestr − rf , volatility σX , time-to-expiry T − t and strike κ, namely
C(t, x) = er(T−t)[X(t)N
(d+(
X(t)
κ, T − t)
)− κN
(d−(
X(t)
κ, T − t)
)]where as usual
d±(x, τ) =lnx+ (r − rf ± 1
2σ2X)τ
σX√τ
�
Exercise 5.2. [CM14, Exercise 13.5] Assume that a foreign stock price process {Sf (t)}t≥0
and the exchange rate {X(t)}t≥0 are correlated geometric Brownian motions with respectivelog-volatility vectors σ(S) = [σS , 0] and σ(X) = [ρσX ,
√1− ρ2σX ]. Assume the domestic and
foreign interest rates r and rf are constants an that foreign stock pays no dividend.
(a) Derive a formula for the current time t < T price Ct of a call option on foreign stockdenominated in domestic currency with domestic payoff
CT =(X(T )Sf (T )−K
)+
37
(b) Similarly, derive a formula for the current time t < T price Pt of a put option with domesticpayoff
CT =(K −X(T )Sf (T )
)+
(c) Derive a put-call parity formula relating the call and put prices Ct and Pt.
Solution.
(a) We know (c.f. [CM14, Equations 13.1000 and 13.104]) that the process (X(t)Sf (t))t≥0
satisfies the SDE
d[X(t)Sf (t)]
X(t)Sf (t)= (r − qS)dt+ (σ(X) + σ(S)) · dW (t)
with solution
X(T )Sf (T ) = X(t)Sf (t)e(r−qS− 12σ2XS)(T−t)+(σ(X)+σ(S))·(W (T )−W (t))
= X(t)Sf (t)e(r−qS− 12σ2XS)(T−t)+σXS
√T−tX
where
σX = ‖σ(X)‖, σS = ‖σ(S)‖, σXS = ‖σ(X) + σ(S)‖ =√σ2X + σ2
S + 2ρσXσS
and
X =1
σXS√T − t
(σ(X) + σ(S)) · (W (T )− W (t)) ∼ N (0, 1)
The call option can thus be priced by using the standard Black-Scholes formula CBS(S,K, τ ; r, q, σ)with spot S = XSf , strike K, time to maturity τ = T − t, interest rate r, no dividendsq = qS = 0 and volatility σ = σXS , whence
C(t, Sf (t), X(t)) = e−r(T−t)[Sf (t)X(t)N
(d+
(X(t)Sf (t)
K,T − t
))−KN
(d−
(X(t)Sf (t)
K,T − t
))]with
d±(x, τ) =lnx+ (r − rf ± 1
2σ2X)τ
σX√τ
(b) Since we are using the standard Black-Scholes formula, the standard put-call parity relationholds, namely
C(t, Sf (t), X(t))− P (t, Sf (t), X(t)) = Sf (t)X(t)−Ke−r(T−t)
�
Exercise 5.3. [CM14, Exercise 13.6]
38
Solution. This call option can be priced as a standard call using the Black-Scholes formula.�
Exercise 5.4. [CM14, Exercise 13.7] Consider the foreign equity call struck in foreign cur-rency with payoff
CT = X(T )(Sf (T )−Kf
)+.
Assume the foreign stock price is a GBM with constant log-volatility σ(S) and having a dividendyield qS and the exchange rate is a GBM with constant log-volatility σ(X). Derive its pricingfunction C(t, S, S).
Solution. This exercise is identical to [CM14, Example 13.4], with the roles of Sf (T ) andX(T ) exchanged. We will be reproducing most of the argument in [CM14, Exercise 13.9(c)]below, so we don’t duplicate it here. �
Exercise 5.5. [CM14, Exercise 13.9] Assume that a foreign stock price {Sf (t)}t≥0, a foreignexchange rate process {X(t)}t≥0 and a domestic asset price process {A(t)}t≥0 are all geometricBrownian motions with respective constant log-volatility vectors σ(S), σ(X) and σ(A):
dA(t)
A(t)= µAdt+σ(A) · dW (t),
dSf (t)
Sf (t)= µSdt+σ(S) · dW (t),
dX(t)
X(t)= µXdt+σ(X) · dW (t)
W (t) is a 3-dimensional standard P -Bronwian motion in the physical measure P and theassets are correlated, where σS = ‖σ(S)‖, σX = ‖σ(X)‖, σA = ‖σ(A)‖, σ(S) · σ(A) = ρSAσSσA,σ(X) ·σ(A) = ρXAσXσA, σ(X) ·σ(S) = ρXSσXσS . Assume a domestic and foreign economy withrespective interest rates r and rf as constants, zero dividends on all assets, and let A(t) = A,Sf (t) = S and X(t) = X be the spot values.
(a) Derive the time t < T pricing function for a domestic European option with payoff
VT = max{X(T )Sf (T ), A(T )}
(b) Derive the time t < T pricing function for a domestic European-style option with payoff
VT = X(T )Sf (T )IX(T )≥X0+AT IX(T )<X0
where X(0) = X0 is a fixed positive initial exchange rate.
(c) Derive the time t < T pricing function for a domestic European-style option with payoff
VA =(aX(T )Sf (T )− bA(T )
)+
with positive constants a, b.
Solution.
39
(a) Recalling that max{x, y} = (x− y)+ + y we can rewrite the payoff as
VT = max{X(T )Sf (T ), A(T )} =(X(T )Sf (T )−A(T )
)++A(T )
so (see [CM14, Equation 13.67])
V (t, S,X,A) = e−r(T−t)Et,S,X,A
[(X(T )Sf (T )−A(T )
)+]
+ e−r(T−t)Et,S,X,A[A(T )]
and these two expectations have been computed in the exchange option example on [CM14,Section 13.2.2.1] with X(T )Sf (T ) ↔ S2(T ) and A(T ) ↔ S1(T ). We simply need toestablish the correlation coefficient between both processes. Recall that the processesX(t)Sf (t) and A(t) satisfy
X(T )Sf (T ) = X(t)Sf (t)e(r− 12σ2XS)(T−t)+(σ(X)+σ(S))·(W (T )−W (t))
A(T ) = A(t)e(r− 12σ2A)(T−t)+σ(A)·(W (T )−W (t))
where
σXS = ‖σ(X) + σ(S)‖ =√σ2X + σ2
S + 2ρσXσS
The correlation between these two processes is given by
σ(A) · (σ(X) + σ(S))
σA · σXS= (ρAX + ρAS)
σSσXS
not= ρ
so we have the following equivalent expressions:
X(T )Sf (T ) = X(t)Sf (t)e(r− 12σ2XS)(T−t)+σXS
√T−tZXS
A(T ) = A(t)e(r− 1
2σ2A)(T−t)+σA
√T−t
[ρZXS+
√1−ρ2ZA
]
where ZXS and ZA are independent standard normal variables.
Now we are in a position to apply the expression obtained in the aforementioned example:concretely, [CM14, Eqaution 13.70] reads in our case
C(t, S,X,A) = AN(d+
(XS
A, T − t
))+XSN
(d−
(XS
A, T − t
))where
d±(x, τ) =lnx± 1
2ν2τ
ν√τ
, ν = ‖σ(A) − (σ(S) + σ(X))‖ =√σ2A + σ2
XS − 2ρσAσXS
(c) By the risk-neutral pricing formula we have that
V (t,X(t), Sf (t), A(t)) = e−r(T−t)Et
[X(T )Sf (T )IX(T )≥X0
]+ e−r(T−t)Et
[AT IX(T )<X0
]We compute each expectation separately using an appropriate change of numraire. Forthe first one, using X(t)Sf (t) as numraire we are reduced to
e−r(T−t)Et
[X(T )Sf (T )IX(T )≥X0
]= X(t)Sf (t)E
[IX(T )≥X0
]= X(t)Sf (t)P (X(T ) ≥ X0)
40
In order to compute the latter probability it suffices to represent X(T ) as a GBM randomvariable involving the P -BM, exactly as it is done in [CM14, Example 13.4].
Recall that by [CM14, Eqaution 13.102] we have
X(T ) = X(t)e(r−rf− 12σ2X)(T−t)+σ(X)·(W (T )−W (t))
Recall that the Brownian increments in the risk-neutral measure and the measure undernumraire g are related by
W (T )− W (t) = W (g)(T )− W (g)(t) + σ(g) · (T − t) not= W (T )− W (t) + σ(XSf ) · (T − t)
Substituting this in the representation for X(T ) above we obtain
X(T ) = X(t)e(r−rf− 1
2σ2X)(T−t)+σ(X)·
(W (T )−W (t)+σ(XSf )·(T−t)
)= X(t)e(r−rf+σ(X)·σ(S)+ 1
2σ2X)(T−t)+σ(X)·(W (T )−W (t))
= X(t)e(r−rf+σ(X)·σ(S)+ 12σ2X)(T−t)+σX
√T−tZ
where recall that σ(XSf ) = σ(X) + σ(Sf ) and where Z ∼ N (0, 1). Therefore we have that
X(T ) ≥ X0 ⇐⇒ X(t)e(r−rf+σ(X)·σ(S)+ 12σ2X)(T−t)+σX
√T−tZ ≥ X0
⇐⇒ Z ≥ln X0
X(t) − (r − rf + σ(X) · σ(S) + 12σ
2X)(T − t)
σX√T − t
In conclusion, the first expectation we needed to compute is
e−r(T−t)Et
[X(T )Sf (T )IX(T )≥X0
]= X(t)Sf (t)P (X(T ) ≥ X0)
= X(t)Sf (t)N
(ln X(t)
X0+ (r − rf + σ(X) · σ(S) + 1
2σ2X)(T − t)
σX√T − t
)
The second expectation we compute analogously, but now using A(t) as numraire, namely:
e−r(T−t)Et[AT IX(T )<X0
]= A(t)E
(A)t
[IX(T )<X0
] not= A(t)P (X(T ) < X0)
Proceeding as above, we may relate the Brownian motion increments in the risk-neutralmeasure and our new measure P in numraire A(t) to obtain (we skip the details)
X(T ) = X(t)e(r−rf+ 12σ2X)(T−t)+σ(X)·(W (T )−W (t))
whereby
e−r(T−t)Et[AT IX(T )<X0
]= A(t)P (X(T ) < X0)
= A(t)N
(ln X0
X(t) − (r − rf + 12σ
2X)(T − t)
σX√T − t
)
41
In conclusion, adding up both expectation we have shown that
V (t,X(t), Sf (t), A(t)) = e−r(T−t)Et
[X(T )Sf (T )IX(T )≥X0
]+ e−r(T−t)Et
[AT IX(T )<X0
]= X(t)Sf (t)N
(ln X(t)
X0+ (r − rf + σ(X) · σ(S) + 1
2σ2X)(T − t)
σX√T − t
)
+A(t)N
(ln X0
X(t) − (r − rf + 12σ
2X)(T − t)
σX√T − t
)
(d) This exercise is just [CM14, Exercise 13.16] with X(T )Sf (T )↔ S2(T ) and A(T )↔ S1(T )so we may just apply the formulas obtained therein after computing the correlation betweenthe processes X(t)Sf (t) and A(t) as in part (a).
�
Exercise 5.6. [CM14, Exercise 13.11]
Solution.
�
Exercise 5.7. [CM14, Exercise 13.12] Assume as in [CM14, Exercise 13.5] that a foreignstock price process {Sf (t)}t≥0 and the exchange rate {X(t)}t≥0 are correlated geometric Brow-nian motions with respective log-volatility vectors σ(S) = [σS , 0] and σ(X) = [ρσX ,
√1− ρ2σX ]
and with time-0 (spot) values Sf (0) = S, X(0) = X. Assume the domestic and foreign interestrates r and rf are constants an that foreign stock pays no dividend. Derive the time-0 pricingformula as a function of S, T,X, for a domestic European option having payoff at maturityT > 0 given by
VT = X(T )Sf (T )I{M(T )<K}
where M(T ) is the maximum realized value of the exchange rate up to time T :
M(T ) = max0≤t≤T
X(t)
Solution. Using g(t) = X(t)Sf (t) as numraire, we need to compute
e−r(T−t)E[X(T )Sf (T )I{M(T )<K}|F(t)
]= X(t)Sf (t)P (M(T ) < K|F(t))
and it hence suffices to obtain the PDF of the sampled maximum M(T ) in measure P .
Recall that by [CM14, Equation 13.102] we have
X(T ) = X(t)e(r−rf− 12σ2X)τ+σ(X)·(W (T )−W (t))
The Brownian motion increments W (T )− W (t) and W (T )− W (t) are related via
W (T )− W (t) = W (T )− W (t)− σ(XS)(T − t), σ(XS) = σ(X) + σ(S)
42
so the expression for X(t) becomes
X(T ) = X(t)e(r−rf+σ(X)·σ(S)+ 12σ2X)τ+σ(X)·(W (T )−W (t)) d
= X(0)e(r−rf+σ(X)·σ(S)+ 12σ2X)τ+σX
√τZ
where Z = 1σX√τσ(X) · (W (T )− W (t)) ∼ N (0, 1). We can further write
X(t) = X(0)eσXX (t) where X (t) = µt+√tZ, µ =
r − rf + σ(X) · σ(S) + 12σ
2X
σX
andMX(t) = max
0≤u≤tX(u) = X(0)eσM
X (t), MX (t) = max0≤u≤t
X (u)
Similarly, since we will be conditioning on F(t) we have
MX(T ) = max0≤u≤T
X(u) = max{MX(t), X(t)eσM
X (τ)}, MX (τ) = max
t≤u≤TX (u)
We are now in the context of [CM14, Section 12.3.1], whereby the joint density function ofMX(τ) and X(τ) in measure P is given by [CM14, Equation 12.108], namely
fMX(τ),X(τ)(w, x) =2(2w − x)
τ√
2πτe−
12µ2τ+µτ− (2w−x)2
2τ , µ =r − rf + σ(X) · σ(S) + 1
2σ2X
σX
Our expectation is hence either 0 if MX(t) ≥ K or else is given by
X(t)Sf (t)P (M(T ) < K|F(t)) = X(t)Sf (t)P(X(t)eσXM
X (τ) < K|F(t))
= X(t)Sf (t)P
(MX (τ) <
1
σXln
K
X(t)|F(t)
)= X(t)Sf (t)
∫ 1σX
ln KX(t)
0
∫ w
−∞fMX(τ),X(τ)(w, x) dx dw
�
Exercise 5.8. [CM14, Exercise 13.13] Consider a domestic economy with constant interestrate r and two correlated GB stock price processes given by
S1(T ) = S1(t)e(µ1− 12σ2
1)τ+σ1√τZ1 ,
S2(T ) = S2(t)e(µ2− 12σ2
2)τ+σ2√τ(ρZ1+
√1−ρ2Z2)
where Z1, Z2 are independent standard normal random variables. let S1(t) = S1, S2(t) = S2,0 ≤ t ≤ T , be the stock spot values.Derive the pricing function V (t, S1, S2) for a Europeanchooser max call with payoff
VT = (max{S1(T ), S2(T )} −K)+
Solution. Re-write the payoff as
ΛT = (S1(T )−K)+I{S1(T )>S2(T )} + (S2(T )−K)+I{S2(T )>S1(T )}
43
Then
V (t, S1, S2) = e−rτ E[(S1(T )−K)+I{S1(T )>S2(T )} + (S2(T )−K)+I{S2(T )>S1(T )}
]We illustrate how to compute
E[(S1(T )−K)+I{S1(T )>S2(T )}
]We have that
E[(S1(T )−K)+I{S1(T )>S2(T )}
]= E
[(S1(T )−K)I{S1(T )>K,S1(T )>S2(T )}
]= e(r−q1− 1
2σ2
1)τ E[(eσ1
√τZ1 −K)I{Z1>α,f(Z1)>Z2}
]=
∫ ∞α
∫ f(x)
−∞
[eσ1√τx −K
]η2(x, y; ρ) dy dx
= · · ·
where
α =ln K
S1− (r − q1 − 1
2σ21)τ
σ1√τ
f(Z1) = · · ·
η2(x, y; ρ) =1
2π√
1− ρ2e−(x2+y2−2ρxy)/2(1−ρ2)
�
Exercise 5.9. [CM14, Exercise 13.15] Consider three stocks with GBM price dynamics asin [CM14, Exercise 13.1] with constant interest rate r and constant dividend yield qi on eachstock i = 1, 2, 3.
(a) Derive the pricing function V (t, S1, S2, S3), t < T for a European option with payoff
VT = S3(T )I{S3(T )>S1(T ),S3(T )>S2(T )}
(b) Derive the pricing function for a European option with payoff
VT = max{S1(T ), S2(T ), S3(T )}
Solution. For a derivation using the change of numraire technique, see [OW06, Section 6.1].
(b) Part (b) is straightforward once we have established (a), since
max{S1(T ), S2(T ), S3(T )} = S1(T )I{S1(T )>S2(T ),S1(T )>S3(T )}
+ S2(T )I{S2(T )>S1(T ),S2(T )>S3(T )} + S3(T )I{S3(T )>S1(T ),S3(T )>S2(T )}
44
(a) Part (a) is a generalization of [CM14, Equation 13.72], whereby if
S1(T ) = S1(t)e(µ1− 12σ2
1)τ+σ(1)·(W(T )−W(t)), S2(T ) = S2(t)e(µ2− 12σ2
2)τ+σ(2)·(W(T )−W(t))
with
σi = ‖σ(i)‖, σ(1) · σ(2) = ρσ1σ2, ν2 = ‖σ(1) − σ(2)‖2 = σ21 + σ2
2 − 2ρσ1σ2
then
E[S1(T )I{S1(T )>S2(T )}|S1(t) = S1, S2(t) = S2
]= S1e
µ1τN
(ln S1
S2+ (µ1 − µ2 + 1
2ν2)τ
ν√τ
)(16)
We first show this 2-dimensional case and we will then indicate how to extend it to 3 stocks(even though computations are tedious). Start by writing
S1(T ) = S1(t)e(µ1− 12σ2
1)τ+σ1√τZ1 ,
S2(T ) = S2(t)e(µ2− 12σ2
2)τ+σ2√τ(ρZ1+
√1−ρ2Z2)
where Z1, Z2 are independent standard normal random variables. Then we have
E[S1(T )I{S1(T )>S2(T )}|S1(t) = S1, S2(t) = S2
]= S1e
(µ1− 12σ2
1)τE[eσ1√τZ1IZ1>f(Z2)
]where
f(Z2) =ln S2
S1+[µ2 − µ1 − 1
2(σ22 − σ2
1)]τ + σ2
√τ√
1− ρ2Z2
(σ1 − ρσ2)√τ
We now use iterated conditioning
E[S1(T )I{S1(T )>S2(T )}|S1(t) = S1, S2(t) = S2
]= S1e
(µ1− 12σ2
1)τE[E[eσ1√τZ1IZ1>f(Z2)|Z2
]]Recalling that for X ∼ N (0, 1) we have E[eBXIX>A] = eB
2/2N (B −A) we have
E[S1(T )I{S1(T )>S2(T )}|S1(t) = S1, S2(t) = S2
]= S1e
µ1τE[N (σ1
√τ − f(Z2))]
= S1eµ1τE
[N
(ln S1
S2+ (µ1 − µ2 + ν2
2 )τ − σ2√τ√
1− ρ2Z2
(σ1 − ρσ2)√τ
)]
Recalling that for X ∼ N (0, 1) we have E[N (AX + C)] = N(
C√1+A2
)we conclude that
E[S1(T )I{S1(T )>S2(T )}|S1(t) = S1, S2(t) = S2
]= S1e
µ1τN
(ln S1
S2+ (µ1 − µ2 + 1
2ν2)τ
ν√τ
)
In dimension 3 we start by computing the lower Cholesky factorization U of the correlationmatrix 1 ρ12 ρ13
ρ12 1 ρ23
ρ13 ρ23 1
= UU>, U =
1 0 0
ρ12
√1− ρ2
12 0ρ13 α β
45
where
α =ρ23 − ρ12ρ13√
1− ρ212
, β =
√1− ρ2
13 −(ρ23 − ρ12ρ13)2
1− ρ212
Write
S1(T ) = S1(t)e(µ1− 12σ2
1)τ+σ1√τZ1 ,
S2(T ) = S2(t)e(µ2− 12σ2
2)τ+σ2√τ(ρ12Z1+
√1−ρ2
12Z2),
S3(T ) = S3(t)e(µ3− 12σ2
3)τ+σ3√τ(ρ13Z1+αZ2+βZ3)
where Z1, Z2, Z3 are independent standard normal random variables. Now we proceed asin the 2-dimensional case by iterated conditioning
e−rτEt,S1,S2,S3
[S3(T )I{S3(T )>S1(T ),S3(T )>S2(T )}
]= e−rτS3e
(µ3− 12σ2
3)τE[eσ3√τ(ρ13Z1+αZ2+βZ3)I{Z3>f(Z1),Z3>f(Z2)}
]= e−rτS3e
(µ3− 12σ2
3)τE[eσ3√τ(ρ13Z1+αZ2)E
[eβZ3I{Z3>f(Z1),Z3>f(Z2)}|Z1, Z2
]]= e−rτS3e
(µ3− 12σ2
3)τE[eσ3√τ(ρ13Z1+αZ2)E
[eβZ3I{Z3>max{f(Z1),f(Z2)}}|Z1, Z2
]]The inner expectation
E[eβZ3I{Z3>max{f(Z1),f(Z2)}}|Z1, Z2
]can be computed using once again that E[eBXIX>A] = eB
2/2N (B −A) for X ∼ N (0, 1)...
Alternatively: if we instead use S3(t) as numraire, so are reduced to computing
e−r(T−t)Et[S3(T )I{S3(T )>S1(T ),S3(T )>S2(T )}
]= S3(t)P
(1 >
S1(T )
S3(T ), 1 >
S2(T )
S3(T )
)and it now suffices to express S1(T )
S3(T ) and S2(T )S3(T ) as GBM’s generated by two correlated
standard normal variables and to integrate the corresponding bivariate normal densityfunction. Again, by [CM14, Eqaution 13.122] we have
Si(T )
S3(T )=Si(t)
S3(t)e−
12‖σi−σ3‖2(T−t)+(σi−σ3)·(W (T )−W (t)) =
Si(t)
S3(t)e−
12ν2i (T−t)+νi
√T−tZi
for i = 1, 2, where νi = ‖σi − σ3‖ and
Zi =1
νi√T − t
(σi − σ3) · (W (T )− W (t)), ρ12 := Corr(Z1, Z2) =(σ1 − σ3) · (σ2 − σ3)
ν1 · ν2
Moreover note that
Si(T )
S3(T )=Si(t)
S3(t)e−
12ν2i (T−t)+νi
√T−tZi < 1⇐⇒ Zi <
ln S3(t)Si(t)
+ 12ν
2i (T − t)
νi√T − t
not= di, i = 1, 2
46
and hence our expectation is
S3(t)P
(1 >
S1(T )
S3(T ), 1 >
S2(T )
S3(T )
)= S3(t)
∫ d1
−∞
∫ d2
−∞n2(x1, x2; ρ12) dx2 dx1
where
η2(x1, x2; ρ12) =1
2π√
1− ρ212
exp
(−x
21 − 2ρ12x1x2 + x2
2
2(1− ρ212)
)�
Exercise 5.10. [CM14, Exercise 13.16] Consider an exchange option on two stocks havingpayoff
VT = (aS2(T )− bS1(T ))+, a, b > 0.
Assume the stocks are GB; processes
S1(T ) = S1(t)e(µ1− 12σ2
1)τ+σ1√τZ1 ,
S2(T ) = S2(t)e(µ2− 12σ2
2)τ+σ2√τ(ρZ1+
√1−ρ2Z2)
where Z1, Z2 are independent standard normal random variables. Derive the time-t price Vtfor this option by explicitly using one of the stocks as numraire asset and by implementing therisk-neutral pricing [CM14, Formula 13.149]. Note: the derivation is similar to that in [CM14,Example 13.3].
Solution. This exercise is identical to [CM14, Example 13.3]. Working with numraireg(t) = S1(t) and denoting E ≡ ES1 , the time-t value of the option is given by
Vt = S1(t)E
[VT
S1(T )|Ft]
= S1(t)E
[(aS2(T )
S1(T )− b)+
|Ft
]= aS1(t)E
[(S2(T )
S1(T )− b
a
)+
|Ft
]not= aS1(t)E
[(Y (T )− b
a
)+
|Ft
]
where in the last equality we denoted Y (T ) = S2(T )S1(T ) .
It is easy to see2 that the process Y (t) = S2(t)S1(t) is a P -martingale
Y (t) =S2(t)
S1(t)= Y (0)Et
((σ2 − σ1) · W
)As in [CM14, Example 13.3] we can write
Y (t) = Y (0)eX(t)
where
X = −1
2‖σ2 − σ1‖2t+ (σ1 − σ1) · W (t) = −1
2ν2t+ ν
√tZ
2Either see [CM14, Equation 13.122] or directly compute the stochastic differential d(S2(t)S1(t)
).
47
with Z = (σ2−σ1)·W (t)
ν√t
∼ N (0, 1) and ν2 := ‖σ2−σ1‖2 = σ21 +σ2
2−2ρσ1σ2. We can thus compute
Vt = S1(t)E
[(Y (T )− b
a
)+
|Ft
]= S1(t)E
[(Y (0)e−
12ν2T+ν
√T Z − b
a
)+
|Ft
], Z ∼ N (0, 1)
by using the Black-Scholes formula with zero interest rate and dividend, volatility ν, time tomaturity T , spot Y (t) and strike b
a . In other words:
V (T, S1(0), S2(0)) = S1(0)
[Y (0)N (d+(Y (0), T ))− b
aN (d−(Y (0), T )
]= S2(0)N (d+(Y (0), T ))− b
aS1(0)N (d−(Y (0), T )
�
Exercise 5.11. [CM14, Exercise 13.17] Consider a domestic economy with constant interestrate r and two domestic stock price processes
S1(T ) = S1(t)e(µ1− 12σ2
1)τ+σ1√τZ1 ,
S2(T ) = S2(t)e(µ2− 12σ2
2)τ+σ2√τ(ρZ1+
√1−ρ2Z2)
where Z1, Z2 are independent standard normal random variables. Derive the time-0 pricingfunction V0 = V (T, S1, S2) in the spot variables S1(0) = S1, S2(0) = S2, for a European path-dependent option with payoff at maturity T given by
VT = S1(T ) min0≤t≤T
S2(t)
S1(t)
Solution. Recall the numraire invariant form of the risk-neutral pricing formula: for anattainable payoff VT and numraire g(t) we have
Vt = g(t)Et
[VTg(T )
|F(t)
], E
not= E(g)
Picking g(t) = S1(t) as numraire we thus have
Vt = S1(t)E
[min
0≤t≤T
S2(t)
S1(t)|F(t)
]It is easy to see3 that the process Y (t) = S2(t)
S1(t) is a P -martingale
Y (t) =S2(t)
S1(t)= Y (0)Et
((σ2 − σ1) · W
)As in [CM14, Example 13.3] we can write
Y (t) = Y (0)eX(t)
3Either see [CM14, Equation 13.122] or directly compute the stochastic differential d(S2(t)S1(t)
).
48
where
X = −1
2‖σ2 − σ1‖2t+ (σ1 − σ1) · W (t) = −1
2ν2t+ ν
√tZ
with Z = (σ2−σ1)·W (t)
ν√t
∼ N (0, 1) and ν2 := ‖σ2−σ1‖2 = σ21 +σ2
2 − 2ρσ1σ2. We are thus reduced
to computing the expectation
Vt = S1(t)E
[min
0≤t≤TY (t)
]= S1(t)E
[min
0≤t≤TY (0)eX(t)
]= S1(t)E
[min
0≤t≤TY (0)e−
12ν2t+ν
√tZ
]≡ S1(t)E
[min
0≤t≤TY (0)eνZ(t)
]not= S1(t)E
[Y (0)eν·m
Z(t)]
where Z(t) =√tZ −
12ν2t
ν and mZ(t) = min0≤u≤tZ(u) (see [CM14, Section 12.3.1]). The lastexpectation can be computed by integrating against the joint density function of the sampledminimum mZ(t) and Z(t) (c.f. [CM14, Equation 12.109]), namely
E[Y (0)eν·m
Z(t)]
= Y (0)
∫ 0
−∞
∫ ∞w
eνwfmZ(t),Z(t)(w, x) dx dw,
fmZ(t),Z(t)(w) =2(x− 2w)
τ√
2πτe−
12ν2+νx− (x−2w)2
2τ
which is a bit tedious to type. �
49
6 Chapter 14: American options
Exercise 6.1. [CM14, Exercise 14.1] Prove [CM14, Proposition 14.3] for an arbitrary Amer-ican option with a differentiable payoff function Λ. In particular show the following.
(a) At any point (t, S∗t ) of the early-exercise boundary, the American option pricing function Vsatisfies the smooth pasting condition
∂V (t, S)
∂S
∣∣∣∣S=S∗t
= Λ′(S∗t )
(b) The option value V satisfies the zero time-decay condition on the early-exercise domain
∂V (t, S)
∂t= 0, ∀S ∈ Dt = {(t, s) : V (t, S) = Λ(S(t))
Solution.
(a) We proceed as in [CM14, Section 14.2.2]. Suppose the American options has not beenexercised at time t and let B denote the collection of all possible early-exercise boundariesdefined by continuous functions b : [t, T ]→R+, so that for each b ∈ B there is an exercisepolicy Tb. Then
V (t, S) = supb∈B
V (t, S; b), V (t, S; b) = Et,S
[e−r(Tb−t)Λ(S(Tb))
]The total derivative of the function V (t, S; b) along the boundary is given by
dV
db=∂V (t, S; b)
∂S
∂S
∂b
∣∣∣∣S=b
+∂V (t, S; b)
∂b
∣∣∣∣S=b
• Along the curve S = b(t) it is clear that ∂S∂b = 1
• When b = S∗, we have ∂V (t,S;b)∂b
∣∣∣S=b
= 0 (by optimality).
ThereforedV
db=∂V (t, S; b)
∂S
∣∣∣∣S=S∗
(17)
On the other hand, the option value is equal to the payoff function when S = b, that isV (t, b; b) = Λ(b) whence
∂V (t, S)
∂S
∣∣∣∣S=S∗
def=
∂V (t, S;S∗)
∂S
∣∣∣∣S=S∗
by (17)=
dV (t, b; b)
db
∣∣∣∣b=S∗
=dΛ(b)
db
∣∣∣∣b=S∗
= Λ′(S∗)
(b) The total derivative of V (t, S) with respect to time is
dV
dt=∂V (t, S)
∂t+∂V (t, S)
∂S
∂S
∂t
50
On the other hand, on the early-exercise domain Dt we have V (t, S(t)) = Λ(S(t)), so
dV
dt=∂Λ
∂S
∂S
∂t=∂V
∂S
∂S
∂t
and clearly∂V
∂t
∣∣∣∣S∈Dt
= 0
�
Exercise 6.2. [CM14, Exercise 14.2] Let P (S;K, r, q) and C(S;K, r, q), respectively, denotethe price functions of the perpetual American put and call options struck at K. The underlyingasset price follows a geometric Brownian motion
S(t) = S(0)e(r−q− 12σ2)t+σW (t)
Using the closed-form pricing formulas below, show that the option prices satosfy the put-callsymmetry relation
P (K;S, q, r) = C(S;K, r, q)
where (c.f. [CM14, Equations 14.25 and 14.28])
P (S;K, r, q) =
{K
1−λ−
(λ−−1λ−
)λ− (SK
)λ−= − S∗
λ−
(SS∗
)λ−, S∗ < S,
K − S, 0 < S ≤ S∗.
C(S;K, r, q) =
{K
λ+−1
(λ+−1λ+
)λ+ (SK
)λ+= − S∗
λ+
(SS∗
)λ+, 0 < S ≤ S∗,
S −K, S∗ < S.
λ± =−(r − q − σ2
2 )±√
(r − q − σ2
2 )2 + 2σ2r
σ2
Solution. Recall how the pricing formulas for the perpetual American call and put arederived: ... �
Exercise 6.3. [CM14, Exercise 14.3]
Solution.
�
Exercise 6.4. [CM14, Exercise 14.4]
Solution.
�
Exercise 6.5. [CM14, Exercise 14.5]
51
Solution.�
Exercise 6.6. [CM14, Exercise 14.6]
Solution.�
Exercise 6.7. [CM14, Exercise 14.7]
Solution.�
Exercise 6.8. [CM14, Exercise 14.8]
Solution.�
Exercise 6.9. [CM14, Exercise 14.9]
Solution.�
52
7 Chapter 15: Interest rate modeling and derivative pricing
Exercise 7.1. [CM14, Exercise 15.1] Suppose that the continuously compounded spot ratesfor the next three years are
T 1 2 3
y(0, T ) 3% 3.25% 3.5%
Find the forward rates f(0; 1, 2), f(0; 1, 3) and f(0; 2, 3).
Solution. Recall the relation between yield and forward rates. From no-arbitrage, we have
f(t, T, T ′) =1
T ′ − TlnP (t, T )
P (t, T ′)
and by definition of bond yield y(t, T ) we have P (t, T ) = e−y(t,T )(T ′−T ) whence
f(t, T, T ′) =1
T ′ − TlnP (t, T )
P (t, T ′)=
1
T ′ − Tlne−y(t,T )(T ′−T )
e−y(t,T ′)(T ′−T )
= y(t, T ′)T ′ − tT ′ − T
− y(t, T )T − tT ′ − T
so
f(0; 1, 2) = y(0, 2) · 2
1− y(0, 1) · 1
1= 0.0325 · 2− 0.03 · 1,
f(0; 1, 3) = y(0, 3) · 3
2− y(0, 1) · 1
2= 0.035 · 1.5− 0.03 · 0.5,
f(0; 2, 3) = y(0, 3) · 3
1− y(0, 2) · 2
1= 0.035 · 3− 0.0325 · 2
�
Exercise 7.2. [CM14, Exercise 15.3] Consider the Hull-White model
dr(t) = [α(t)− β(t)r(t)] dt+ σ(t)dW (t)
Show that the short rate is a Gaussian process. Find the mean and variance of r(T ) conditionalon r(t) for 0 ≤ t ≤ T .
Solution. Given the market price of risk γ(t), the Hull-White model has risk-neutral dynamics
dr(t) = [α(t)− β(t)r(t)] dt+ σ(t)dW (t), α(t) = α(t)− λ(t)σ(t)
Integrating this SDE (c.f. [CM14, Equation 11.31] or [CM14, Example 11.9]) we obtain
r(T ) = r(t)e−∫ Tt β(s) ds +
∫ T
te−∫ Ts b(u) duα(s) ds+
∫ T
te−∫ Ts β(u) duσ(s) dW (s)
53
The integrand in the It integral is non-random, so r(T ) is indeed Gaussian conditional onr(t) and has mean and variance
E[r(T )|r(t) = r] = re−∫ Tt β(s) ds +
∫ T
te−∫ Ts b(u) duα(s) ds
Var[r(T )|r(t) = r] =
∫ T
te−2
∫ Ts β(u) duσ2(s) ds
where the second equation follows from It isometry. �
Exercise 7.3. [CM14, Exercise 15.5] Find the forward rates f(t, T ) in the Vacisek model
dr(t) = [α− βr(t)] dt+ σdW (t)
Solution. Recall that by definition of the forward rate f(t, T ) we have
ef(t,T,T ′)(T ′−T ) =P (t, T )
P (t, T ′)=⇒ f(t, T, T ′) =
1
T ′ − TlnP (t, T )
P (t, T ′)
and by definition the instantaneous forward rate is
f(t, T ) := limT ′→T
f(t, T, T ′) = − ∂
∂TlnP (t, T )
so it suffices to establish the discount bond P (t, T ) reconstitution formula in the Vacisek model(c.f. [CM14, Equation 15.53]) and differentiate it with respect to maturity T .
We first outline the derivation of the bond reconstitution formula for P (t, T ) following[CM14, Section 15.2.5]. Assuming that the market price of risk is constant γ the Vacisek modelhas risk-neutral dynamics
dr(t) = [α− βr(t)] dt+ σdW (t), α = α− γσ
Integrating this SDE we obtain
r(T ) = e−β(T−t)r(t) +α
β
(1− e−β(T−t)
)+ σ
∫ T
te−β(T−s) dW (s)
...The bond pricing formula for the Vacisek model is hence
P (t, T ) = exp
1− e−β(T−t)
β(y∞ − r)− y∞(T − t)− σ2
4β
(1− e−β(T−t)
β
)2
where y∞ = αβ −
σ2
2β2 . The forward rate is hence given by
f(t, T ) = − ∂
∂TlnP (t, T ) = y∞ +
σ2
2β
(1− e−β(T−t)
β
)e−β(T−t) − (y∞ − r)e−β(T−t)
�
54
Exercise 7.4. [CM14, Exercise 15.7] A d-factor HJM model is given by the SDE
df(t, T ) = α(t, T )dt+ σ(t, T )>dW (t)
where W (t) = (W1(t), . . . ,Wd(t)) is a vector of independent Brownian motions.
(a) Find the SDE for the discounted zero-coupon bond price process P (t, T ) = D(t)P (t, T ) andshow that
P (t, T ) = P (0, T ) exp
(−∫ t
0A(s, T ) ds−
∫ t
0Σ(s, T )>dW (s)
)where
Σ(t, T ) =
∫ T
tσf (t, u) du
A(t, T ) =
∫ T
tα(t, u) du
(b) Show that the no-arbitrage condition is given by
α(t, T ) = σf (t, T )>∫ T
tσf (t, s) ds
Solution. One can argue by extending the discussion in [CM14, Section 15.3.1]. We insteadfollow [AP10, Section 4.4].
(a) In the absence of arbitrage, the deflated bond values are Q-martingales, so by the martin-gale representation theorem there exists a d-dimensional stochastic process σP (t, T ) withσP (T, T ) = 0 such that
dP (t, T ) = −P (t, T )σP (t, T )tdW (t), t ≤ T
By It’s lemma it follows that
dP (t, T )
P (t, T )= r(t)dt− σP (t, T )tdW (t)
By It’s lemma it also follows that
dP (t, T, T + τ)
P (t, T, T + τ)= −[σP (t, T + τ)− σP (t, T )]tσP (t, T )dt− [σP (t, T + τ)− σP (t, T )]tdW (t)
In the T-forward measure, P (t, T, T + τ) is a martingale (by definition), so we must have
dP (t, T, T + τ)
P (t, T, T + τ)= −[σP (t, T + τ)− σP (t, T )]tdW T (t)
where W T (t) is a QT -Brownian motion.
55
Comparison of the last two expressions yields
dW T (t) = dW (t) + σP (t, T )dt
which by Girsanov’s theorem identifies the Radon-Nikodym process for the measure shift
ξ(t) = EQt
[dQT
dQ
]= Et(σP ·W ) = exp
(−∫ t
0||σP (s, T )||2 ds−
∫ t
0σP (s, T ) dW (s)
)
HJM models are traditionally stated in terms of instantaneous forward rates f(t, T ). Onceagain by It’s lemma we have (omitting the drift term)
d lnP (t, T ) = Ø(dt)− σP (t, T )tdW (t)
Differentiating with respect to T on both sides and recalling that f(t, T ) = −∂P (t,T )∂T we have
df(t, T ) = µf (t, T )dt+∂
∂TσP (t, T )t︸ ︷︷ ︸
:=σf (t,T )t
dW (t)
which shows (a).
(b) In order to establish the drift term µf (t, T ) note that since
f(t, T ) = ETt [r(u)]
is a martingale in the T -forward measure we necessarily have
df(t, T ) = σf (t, T )>dW T (t)
Applying the change of measure above
dW T (t) = dW (t) + σP (t, T )dt, σP (t, T ) =
∫ T
tσf (t, u) du
it follows that
df(t, T ) = σf (t, T )tσP (t, T )dt+ σf (t, T )tdW (t)
= σf (t, T )t∫ T
tσf (t, u) dudt+ σf (t, T )tdW (t)
so the drift term in the risk-neutral measure is indeed
µf (t, T ) = σf (t, T )t∫ T
tσf (t, u) du
�
Exercise 7.5. [CM14, Exercise 15.8]
56
Solution.�
Exercise 7.6. [CM14, Exercise 15.9]
Solution.�
Exercise 7.7. [CM14, Exercise 15.12]
Solution.�
Exercise 7.8. [CM14, Exercise 15.13]
Solution.�
Exercise 7.9. [CM14, Exercise 15.14]
Solution.�
57
References
[AP10] L. Andersen, V. Piterbarg, Interest rate modeling. Volumes 1, 2 and 3, Atlantic Financial Press,2010
[CM14] G. Campolieti and R. N. Makarov, Finantial Mathematics: A Comprehensive Introduction,CRC Press, 2014
[Hul09] J. Hull, Options, Futures, and other derivatives, Pearsin, 7th Edition, 09
[OW06] P. Ouwehand and G. West, Pricing rainbow options, Wilmott Magazine 5 74-80, 2006
[Shr04] S. E. Shreve, Stochastic calculus for finance II, Springer, 2004
58
