MATH11150
Stochastic Control and Dynamic Asset Allocation
Solutions and comments Apr 2016
Throughout the examination paper we will assume the existence of a suitable probability space
(Ω,F ,P). Results in the lectures may be used without further justification unless the question is
asking specifically for the proof of a particular result.
1. Given (s, y) ∈ [0, 1)× R, consider the following stochastic control problem
V (s, y) = min
ν
J(s, y; ν)
= min
ν
E
[∫ 1
s
[ (
Xνs,y(t)
)2 − 1
2
ν2(t)
]
dt
]
such that

dXνs,y(r) = ν(r)dW (r), r ∈ [s, T ]
Xνs,y(s) = y
ν(t) ∈ [0, 1] ∀t ∈ [0, 1] and (Ft)t∈[0,T ]-adapted
(a) Let t ∈ [s, 1]. Express E[(Xνs,y(t))2] in terms of the control ν(·) and prove that
E[
(
Xνs,y(t)
)2
] = y2 + E[
(
Xνs,0(t)
)2
].
[4 marks]
(b) Show that V (s, y) can be expressed as V (s, y) = y2(1− s) + g(s) for some function g(s) you
should identify and compute ∂yV (s, y) and ∂yyV (s, y).
[6 marks]
(c) Write down the HJB equation for this stochastic control problem.
[4 marks]
(d) Find a solution to the HJB equation.
[11 marks]
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MATH11150
Stochastic Control and Dynamic Asset Allocation
Solutions and comments Apr 2016
Solution:
(a) Properties of Xν . The solution to the SDE is Xνs,y(t) = y +
∫ t
s
ν(r)dW (r) and hence it
is clear that
Xνs,y(t) = y +X
ν
s,0(t). [2 Marks]
It follows using squares, that E[Xνs,y(t)] = y ∀t ≥ s and Itoˆ’s isometry that
E[
(
Xνs,y(t)
)2
] = y2 + 2yE[
∫ t
s
ν(r)dW (r)] + E[
(∫ t
s
ν(r)dW (r)
)2
] = y2 + E[
∫ t
s
ν2(r)dr]
= y2 + E
[(
Xνs,0(t)
)2]
. [2 Marks]
(b) Properties of the Value function If the value function can effectively be written as
V (s, y) = y2(1−s)+g(s), then it follows immediately that y 7→ V (s, y) is twice continuously
differentiable and we have
∂yV (s, y) = 2y(1− s) [1 Marks]
∂yyV (s, y) = 2(1− s) > 0 ∀(s, y) ∈ [0, 1)× R [1 Marks]
It remains to show the expression V (s, y) = y2(1− s) + g(s) and determine g.
From the property Xνs,y(t) = y +X
ν
s,0(t) ∀t ∈ [s, 1] we have
J(s, y, ν) = E
[∫ 1
s
[ (
Xνs,y(t)
)2 − 1
2
ν2(t)
]
dt
]
= E
[∫ 1
s
[
y2 + 2yXνs,0(t) +
(
Xνs,0(t)
)2 − 1
2
ν2(t)
]
dt
]
= y2(1− s) + 2y
∫ 1
s
E
[
Xνs,0(t)
]
dt+ E
[∫ 1
s
[(
Xνs,0(t)
)2 − 1
2
ν2(t)
]
dt
]
Recall that E[Xνs,0(·)] = 0 then
J(s, y, ν) = y2(1− s) + J(s, 0, ν)
⇒ V (s, y) = y2(1− s) + V (s, 0), hence g(s) = V (s, 0). [4 Marks]
(c) The HJB equation depends on the generator of the diffusion given by dX(t) = ν(t)dW (t),
in this case LνG(s, x) = ∂tG+ 12ν2∂xxG with ν ∈ [0, 1] (and no drift).
Thus, the HJB equation is given by
min
ν∈[0,1]
{
LνV + (x2 − 1
2
ν2
)}
= 0,
or
∂tV + min
ν∈[0,1]
{
1
2
ν2∂xxV − 1
2
ν2
}
+ x2 = 0 for (s, x) ∈ [0, 1)× R, [3 Marks]
V (1, x) = 0 for x ∈ R. [1 Marks]
(d) Find a solution of the HJB equation
Since we have V (s, y) = y2(1− s) + V (s, 0) we have ∂yyV = 2(1− s) which can be replaced
into the HJB equation to yield:
∂tV + min
ν∈[0,1]
{
1
2
ν22(1− s)− 1
2
ν2
}
+ y2 = 0.
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MATH11150
Stochastic Control and Dynamic Asset Allocation
Solutions and comments Apr 2016
We now solve the minimization problem so that we can solve the HJB.
min
ν∈[−1,1]
{
1
2
ν22(1− s)− 1
2
ν2
}
= min
ν∈[−1,1]
{
ν2(
1
2
− s)
}
⇒ ν?(s) =
{
0, if s ∈ [0, 12 )
±1, if s ∈ [ 12 , 1]
[3 Marks] For the minimization
the case ν = −1 ∈ [0, 1] and hence we ignore it.
The minimum reads
min
ν∈[0,1]
{
ν2(
1
2
− s)
}
=
{
0 , if s ∈ [0, 12 )
1
2 − s , if s ∈ [ 12 , 1]
[1 Marks] For the minimimum
We obtain then two branches for the HJB equation depending on the time, s ∈ [0, 12 ] and
s ∈ [ 12 , 1],
∂tV + min
ν∈[0,1]
{
1
2
ν22(1− s)− 1
2
ν2
}
+ y2 = 0 ⇔

∂tV + 0 + y
2 = 0 , if s ∈ [0, 12 )
∂tV + (
1
2 − s) + y2 = 0 , if s ∈ [ 12 , 1]
V (1, y) = 0
[2 Marks] Identify PDE to solve
Since we only have V (1, 0) = 0 we start with the 2nd branch s ∈ [ 12 , 1] and solve the equation
by direct integration. This will also allow to identify V ( 12 , y) to serve as boundary condition
for the 2nd PDE. We have then
V (s, y) = V (1, y)−
∫ 1
s
[(
1
2
− r) + y2]dr
= 0 + y2(1− s) + 1
2
s2 − 1
2
s. [3 Marks] 1st Branch
To solve for the 2nd equation one needs to identify the appropriate boundary condition at
time s = 12 , namely that V (
1
2 , y) =
1
2y
2− 18 . Hence, by solving the PDE by direct integration
over s ∈ [0, 12 ], we have
V (s, y) = V (
1
2
, y)−
∫ 1
2
s
y2dr
=
1
2
y2 − 1
8
+ y2(
1
2
− s) = y2(1− s)− 1
8
. [2 Marks] 2nd Branch
Comment:
(a) Easy; evaluates basic Stochastic Analysis knowledge;
(b) Easy; evaluates basic Stochastic Analysis knowledge;
(c) Easy; Students must identfy the Dynkin generator and write down the HJB equation.
Boundary condition must also be identified
(d) easy to Medium to hard. Solving the minimization problem is easy when the explicit formula
for ∂yyV is injected; The arising HJB is slighlty different from what they have seen as the
PDE has two branches; The branch s ∈ [ 12 , 1] is easy, the other brach is not so easy and
requires more knowledge.
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MATH11150
Stochastic Control and Dynamic Asset Allocation
Solutions and comments Apr 2016
2.
(2.a) A Black-Scholes market is given where there are only one stock (with drift a ∈ R and
volatility σ > 0) and one bank account with interest rate r ∈ R.
In this market an investor, with initial wealth x0 > 0 selects among proportion strategies ν
that are constants and with such a strategy the proportion of wealth invested in the stock is
a constant throughout.
The investor seeks to maximise his expected utility at time T which is a power-type utility
U(x) =
1
γ
xγ , γ ∈ (0, 1).
(2.a.i) Identify explicitly the underlying, show that the SDE expressing the wealth process(
Xν(t)
)
t∈[0,T ] is of Geometric Brownian motion type and write its explicit solution.
[5 marks]
(2.a.ii) Write clearly the optimization problem and then compute the constant optimal
proportion strategy ν? explicitly without applying the stochastic control approach.
You may use without proving that ∀ c ∈ R we have E[ecW (T )] = e 12 c2T .
[7 marks]
(2.b) Let T <∞ and consider the following BSDE with solution (Y (t), Z(t))
t∈[0,T ],
dY (t) =
(
rY (t) + aZ(t)
)
dt+ Z(t)dW (t) , Y (T ) = ξ. (1)
where r, a are constants, ξ is a square-integrable, FT -measurable random variable in a filtered
probability space (Ω, F , {Ft}0≤t≤T ,P), and W is a one-dimensional Brownian Motion.
(2.b.i) Argue that the solution (Y (t), Z(t)) exists and deduce the expression yielding Y (t)
as a map of T , t, r, a and ξ (a so-called closed form solution).
[6 marks]
(2.b.ii) Denote by (Y i, Zi) the solution to BSDE (1) with ξ being replaced by ξi, i = 1, 2
both FT -adapted square-integrable RV. Suppose ξ1 ≥ ξ2 a.s..
Prove that Y 1(t) ≥ Y 2(t) ∀t ∈ [0, T ] a.s.
[7 marks]
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MATH11150
Stochastic Control and Dynamic Asset Allocation
Solutions and comments Apr 2016
Solution:
(2.a) (2.a.i) The wealth process. The stock and riskless process, denoted S and B
respectively, have the following dynamics as postulated by the Black-Scholes market
dS(t) = S(t) [adt+ σdW (t)] and dB(t) = rB(t)dt.
If the strategies are constant proportions of wealth, ν for the proportion of wealth
invested in the Stock and 1 − ν for the proportion of wealth invested in the bank
account, then equation of the wealth is given by
dX(t) =
νX(t)
S(t)
dS(t) +
(1− ν)X(t)
B(t)
dB(t)
= X(t) [(a− r) ν + r] dt+ νσX(t)dW (t), X(0) = x0,
As the controls are constant, X(·) can be computed explicitly as it is a Geometric
Brownian motion. The solution is given by
Xν(t) = x0 · exp
{(
(a− r)ν + r − 1
2
ν2σ2
)
t
}
· exp {νσW (t)} .
(2.a.ii) The optimization For a wealth process Xν and a control ν the optimization
problem can be written as
sup
ν∈R
E
[
U
(
Xν(T )
)]
, U(x) =
1
γ
xγ , γ ∈ (0, 1).
The utility is given by
U
(
Xν(T )
)
=
1
γ
(Xν(T ))
γ
=
xγ0
γ
exp
{
γ(a− r)νT + rγT − 1
2
γν2σ2T
}
· exp {νγσW (T )} .
The expected utility for the strategy ν(t) = ν ∈ R, 0 ≤ t ≤ T is given by
E
[
U
(
Xν(T )
)]
=
xγ0
γ
exp
{
γ(a− r)νT + rγT − 1
2
γν2σ2T +
1
2
ν2γ2σ2T
}
,
or
E
[
1
γ
(Xν(T ))
γ
]
=
xγ0
γ
exp
{(
γ(a− r)ν + rγ − 1
2
ν2σ2γ(1− γ)
)
T
}
, (2)
where we used the fact that (c = νγσ from the question’s statement)
E
[
exp {νγσW (T )}
]
= E
[
exp
{1
2
ν2γ2σ2T
}]
.
The maximum is achieved by simply maximizing the RHS of (2) wrt to ν using
standard analysis techniques, namely, for
f(ν) :=
(
γ(a− r)ν + rγ − 1
2
ν2σ2γ(1− γ)
)
T
d
dν
f(ν) = f ′(ν) = γ(a− r)− νσ2γ(1− γ), f ′′(ν) = σ2γ(1− γ) < 0,
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MATH11150
Stochastic Control and Dynamic Asset Allocation
Solutions and comments Apr 2016
and f ′(ν) = 0 has a unique solution ν? given by
ν? =
γ(a− r)
σ2γ(1− γ) =
a− r
σ2(1− γ) .
for reference’s sake: The optimal utility is given by
1
γ
xγ0 · exp
(
1
2
(
a− r
σ
)2
γ
1− γ T + rγT
)
.
(2.b) (2.b.i) Existence & uniqueness of solutions: The BSDE has a solution because the
terminal condition ξ is a FT -measurable square-integrable RV; the driver function
g is Lipschitz in its spatial variables and g(0, ·, ·) = 0. By the theorem in class there
exists a unique solution to the equation in H2 ×H2.
Closed form solution to the equation: Using integrating factor e(t) := e−rt
and apply Itoˆ’s formula to e(t)Y (t), we have
d
(
e(t)Y (t)
)
= e(t)
(− rY (t) + rY (t) + aZ(t))dt+ e(t)Z(t)dW (t)
= e(t)aZ(t)dt+ e(t)Z(t)dW (t)
= e(t)Z(t)
(
dW (t) + adt
)
.
Define now a new probability measure Q with Radon-Nikodym derivative given by
dQ
dP = E
(−∫ T
0
adWr
)
and under which the process Ŵ (t) = W (t)+adt is a Brownian
motion. Since a is a real number the density dQdP is well-defined (Novikov’s condition
for example).
Changing the measure to Q and integrating over [t, T ] we obtain
e(t)Y (t) = e(T )ξ −
∫ T
t
e(s)Z(s)dŴ (s)
⇒ Y (t) = EQ[ e(T )e−1(t)ξ |Ft] = e−r(T−t)EQ[ ξ |Ft].
(2.b.ii) Method #1: the students recognize that the probability measure Q in the previous
question is the same for both BSDEs and straightforwardly use the closed form
solution to conclude.
Note that the only change in the BSDEs is the terminal condition and not the
coefficients in the driver function. This means that the integrating factor e(·) and
the probability measure Q are the same.
Using the closed form formula for the solution of the previous BSDEs we get that
Y 1(t)− Y 2(t) = e−r(T−t)EQ[ ξ1 − ξ2 |Ft].
Since ξ1 ≥ ξ2 P-a.s. and the measures P and Q are equivalent, then it follows that
EQ[ ξ1 − ξ2 |Ft] ≥ 0 and hence Y 1(t) ≥ Y 2(t).
Method #2: the students compute the difference between the 2 BSDEs and use the
integrating factor+measure change to reach a closed form solution for the difference
Y 1 − Y 2. This would take a bit more time.
Comment:
This question is fairly standard and straightforward; it is the easiest question of the exam.
Tests ability to manipulate the objects discussed in class.
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MATH11150
Stochastic Control and Dynamic Asset Allocation
Solutions and comments Apr 2016
(2.a) Standard optimization without using the big optimization methods learned in class. Relies
on the use of properties of the underlyings models which appear transversally to the whole
course.
Students are supposed to be able to write down the several quantities of interest involved in
the Black-Scholes model.
2.b) This question is standard and is an easy way for students to get some marks. Questions
2.b.i) and 2.b.ii) have been seen in class in some way or the other.
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MATH11150
Stochastic Control and Dynamic Asset Allocation
Solutions and comments Apr 2016
3. In a d-dimensional complete market with zero interest rate, an agent with initial wealth x > 0
trades d stocks and generates wealth process X given by
Xt = x+
∫ t
0
piTs σs(λsds+ dBs) , 0 ≤ t ≤ T.
Here, T denotes transposition, the trading strategy pi is a d-dimensional vector of wealth in
each stock, λ is a d-dimensional vector, σ a d × d invertible matrix, and B a d-dimensional
Brownian motion on a probability space (Ω,F ,P) equipped with the standard augmented filtration
F := (Ft)0≤t≤T , with λ, σ, pi adapted to F.
The agent seeks to maximise E[U(XT )], over the strategies such that the wealth process remains
positive, and with a concave, increasing, differentiable utility function U : (x,∞) → R, for some
x > 0 denoting a constant below which terminal wealth is not permitted to fall. Denote by V the
convex conjugate of U , and by I the inverse of U ′. Denote the maximal expected utility by u(x).
Let Z := E(−λT ·B) and assume Z is a martingale.
(3.a) Derive the dynamics of ZX and deduce that E[ZTXT ] ≤ x.
[3 marks]
(3.b) Show that u(x) ≤ v(y) + xy, where v(y) := E[V (yZT )], for y > 0.
[3 marks]
(3.c) Explain why the optimal terminal wealth, XˆT , is given by XˆT = I(yZT ) , for some y > 0,
and explain how y is fixed.
[3 marks]
(3.d) Suppose U(x) = log(x− x) . Compute a formula for XˆT in terms of x. What is the lowest
value of initial wealth which guarantees that terminal wealth XˆT > x?
[5 marks]
(3.e) By considering ZXˆ, where Xˆ is the optimal wealth process, show that the optimal portfolio
process is given by
pˆit = (Xˆt − x)(σ−1t )Tλt, 0 ≤ t ≤ T.
[5 marks]
(3.f) Suppose now that the agent also receives stochastic income at a rate Y = (Y (t))0≤t≤T per
unit time, where Y is a bounded non-negative adapted process. By considering the dynamics
of X under the unique equivalent martingale measure Q, argue that in this case the wealth
process of any strategy satisfies
E[ZTXT ] ≤ x := x+K,
for some non-negative constant K that you should identify. What is the minimum initial
wealth required for a feasible problem in this case? Interpret the result.
[6 marks]
8
MATH11150
Stochastic Control and Dynamic Asset Allocation
Solutions and comments Apr 2016
Solution:
(3.a) One just applies Itoˆ’s formula to ZtXt which yields
d(ZtXt) = ZtdXt +XtdZt + d[Z,X]t
= Ztpi
T
t σt(λtdt+ dBt)− ZtXtλTt dBt − ZtpiTt σtλtdt
= Zt(pi
T
t σt −XtλTt )dBt.
Since ZX is a local martingale, bounded from below, it is also a super-martingale. Moreover,
X0 = x and Z0 = 1, this implies E[ZTXT ] ≤ Z0X0 = x. [3 marks]
(3.b)
E[U(XT )] ≤ E[U(XT )] + y(x− E[ZTXT ]) (for y > 0)
= E[U(XT )− yZTXT ] + xy
≤ E[V (yZT )] + xy (since U(x)− xy ≤ V (y) ∀y > 0)
=: v(y) + xy.
Maximising the LHS over XT gives u(x) ≤ v(y) + xy [3 marks]
(3.c) One gets equality in part b) if XT = X̂T , such that U
′(X̂T ) = yZT with y fixed by the
constraint E[ZT X̂T ] = x. Inverting the map U ′, we get X̂T = I(yZT ) with y fixed via
E[ZT I(yZT )] = x.
[3 marks]
(3.d) By direct computations
U ′(x) =
1
x− x, so U
′(X̂T ) = yZT gives
1
X̂T − x
= yZT ⇒ X̂T = x+ 1
yZT
.
Substituting this into E[ZT X̂T ] = x gives
E[x¯ZT +
1
y
] = x ⇒ x¯+ 1
y
= x⇔ 1
y
= x− x¯.
Hence
X̂T = x¯+
1
yZT
= x¯+
x− x¯
ZT
.
For X̂T > x¯ we thus require that x > x¯.
[5 marks]
(3.e) ZX̂ is a martingale, so for t ≤ T
ZtX̂t = E[ZT X̂T |Ft]
= E[ZT x¯+ (x− x¯)|Ft]
= x− x¯+ x¯ZT .
Therefore
d(ZtX̂t) = x¯dZt = −x¯ZtλTt dBt
and comparing with the dynamics in part a) gives
piTt σt − X̂tλTt = −x¯λTt
⇒ σTt pit = (X̂t − x¯)λt
⇒ pit = (σ−1t )T(X̂t − x¯)λt
[5 marks]
9
MATH11150
Stochastic Control and Dynamic Asset Allocation
Solutions and comments Apr 2016
(3.f) The wealth dynamics are
dXt = pi
T
t σt(λtdt+ dBt) under P
dXt = pi
T
t σtdB
Q
t + Ytdt, where dB
Q
t = dBt + λtdt B
Q is a Q-BM,
and hence
Xt −
∫ t
0
Ysds = x+
∫ t
0
piTs σsdB
Q
s .
Since Y is bounded, the LHS is bounded from below, so is a Q-super-martingale and it
follows that
EQ[XT −
∫ T
0
Ytdt] ≤ x ⇔ EQ[XT ] ≤ x+ EQ[
∫ T
0
Ytdt]
or equivalently under the measure P
E[ZTXT ] ≤ x+ EQ[
∫ T
0
Ytdt] =: x¯.
By analogy with the problem analysed earlier, we replace x by x˜ in part (d). So for X̂T to
be greater than x¯ we require
x˜ > x¯ ⇔ x+ EQ[
∫ T
0
Ytdt] > x¯ ⇔ x > x¯− EQ[
∫ T
0
Ytdt].
Since the agent is in receipt of income, the initial capital needed to generate a feasible
terminal wealth is reduced by the fair price EQ[
∫ T
0
Ytdt] of the lifetime income.
[6 marks]
Comment:
(3.a) Completely standard
(3.b) Completely standard
(3.c) Completely standard
(3.d) Standard up to the formula for X̂T . This utility was not seen in lectures
(3.e) Fairly standard. Even split marks for computing ZX̂, then using its dynamics to get pi
(3.f) Harder. Students need to realise that X−∫ ·
0
Ysds is a Q-super-martingale, then see that one
can replace x by x˜, and then also interpret the result. This exercise is unseen.
10
MATH11150
Stochastic Control and Dynamic Asset Allocation
Solutions and comments Apr 2016
4. On a filtered probability space (Ω,F ,F = (Ft)0≤t≤T ,P), an agent trades in an incomplete
market, generating wealth process X. A non-attainable claim pays a bounded random variable C
at terminal time T .
An agent with initial capital x and exponential utility function U(x) = −e−x, maximises
expected utility of terminal wealth, with the random endowment of a short position in the claim.
Denote the value function by u(x). You may assume the interest rate is zero and that the wealth
process X is a Q-martingale, for all equivalent local martingale measures (ELMMs) Q with finite
entropy. Denote the density process of such an ELMM Q by Z, and denote the relative entropy
between Q and P by H(Q|P) := E[ZT logZT ].
(4.a) Show that we have the inequality
E[U(XT − C)] ≤ E[V (yZT )− yZTC] + xy, y > 0 (3)
for any trading strategy and any Q, where V is the convex conjugate of U . Hence show
that u(x) ≤ v(y) + xy, where v is the value function of the dual problem, which you should
define.
[4 marks]
(4.b) Suppose that equality is achieved in (3) for the optimal terminal wealth X̂T and an optimal
density ẐT . Show that
X̂T − C = − log(yẐT ) ,
for some y > 0. [2 marks]
(4.c) Explain how y is determined and hence derive a formula for X̂T − C in terms of x and ẐT ,
and some constants that you should identify.
[5 marks]
(4.d) Hence show that the maximal expected utility is given by
u(x) = − exp
{
−x−H(Q̂|P) + EQ̂[C]
}
,
where Q̂ is the ELMM corresponding to Ẑ. [2 marks]
(4.e) Denote by u0 and Z
0 the value function and density of the dual minimiser Q0 when there is
no random endowment in the above utility maximisation problem. The utility indifference
price p of the claim at time zero is defined implicitly by u(x+ p) = u0(x).
Derive a formula for p. [5 marks]
(4.f) Now suppose the market model contains one stock S and one non-traded asset Y , following
the geometric Brownian motions
dSt = σSSt(λSdt+ dB
S
t ), dYt = σY Yt(λY dt+ dB
Y
t ) ,
where BS , BY are correlated Brownian motions with constant correlation ρ ∈ (−1, 1) and
σS , σY , λS , λY are constants. By deriving an expression for H(Q|P), show that in this case
the indifference price has the representation
p = EQ̂
[
C − 1
2
∫ T
0
ψ̂2t dt
]
where ψ̂ is an adapted process. Explain how ψ̂ is related to Q̂. [7 marks]
Solution:
11
MATH11150
Stochastic Control and Dynamic Asset Allocation
Solutions and comments Apr 2016
(4.a) From U(x) ≤ V (y) + xy, so U(XT − C) ≤ V (yZT ) + (XT − C)yZT .
Taking expectations and using that E[ZTXT ] = x gives
E [U(XT − C)] ≤ E [V (yZT )− yZTC] + xy.
Maximising the LHS gives u(x). Defining the dual value function by
v(y) := inf
Q
E[V (yZT )− yZTC]
then we get u(x) ≤ v(y) + xy. [4 marks]
(4.b) We get equality if we choose XT = X̂T , ZT = ẐT such that
U ′(X̂t − C) = yẐT ⇔ e−(X̂T−C) = yẐT ⇔ X̂T − C = − log(yẐT ).
[2 marks]
(4.c) y is fixed via the constraint E[ẐT X̂T ] = x, that is
E
[
ẐT
(
C − log(yẐT )
)]
= x
⇔ E[ẐTC]− (log y)E[ẐT ]−H(Q̂|P) = x
Notice that E[ẐTC] = EQ̂[C] and E[ẐT ] = EQ̂[1] = 1, and hence injecting this is the above
identity yields
⇔ EQ̂[C]− log y −H(Q̂|P) = x ⇔ − log y = x+H(Q̂|P)− EQ̂[C].
Finally,
X̂T − C = − log y − log ẐT
= x+H(Q̂|P)− EQ̂[C]− log ẐT
[5 marks]
(4.d) By directly computing the involved quantity
u(x) = E[−e−(X̂T−C)]
= E
[
− exp
{
−
(
x+H(Q̂|P)− EQ̂[C]
)}
ẐT
]
= − exp
{
−
(
x+H(Q̂|P)− EQ̂[C]
)}
EQ̂ [1]
= − exp
{
−x−H(Q̂|P) + EQ̂[C]
}
[2 marks]
(4.e) We have u0(x) = − exp{−x−H(Q0|P)} and from the identity defining the indifference price
− exp
{
−(x+ p)−H(Q̂|P) + EQ̂[C]
}
= − exp{−x−H(Q0|P)}
Applying logarithms to both sides and simplifying
⇔ −p−H(Q̂|P) + EQ̂[C] = −H(Q0|P) ⇔ p = EQ̂[C]−
(
H(Q̂|P)−H(Q0|P)
)
[5 marks]
12
MATH11150
Stochastic Control and Dynamic Asset Allocation
Solutions and comments Apr 2016
(4.f) Let ZT := E
(
−λSBST −
∫ T
0
ψtdB
S,⊥
t
)
where BS,⊥ is a BM independent of BS (so that
BY = ρBS +
√
1− ρ2BS,⊥) and ψ is an adapted process s.th. ∫ T
0
ψ2t dt < ∞ a.s. (and we
assume that t 7→ ∫ t
0
ψsdB
S,⊥
s is a Martingale); we have by direct computation
logZT = −λSBST −
∫ T
0
ψtdB
S,⊥
t −
1
2
λ2ST −
1
2
∫ T
0
ψ2t dt.
Using Girsanov, we define two new Brownian motions for 0 ≤ t ≤ T
BS,Qt = B
S
t + λSt, B
S,⊥,Q
t = B
S,⊥
t +
∫ t
0
ψsds
which are independent Q-Brownian motions. Then
logZT = −λSBS,QT −
∫ T
0
ψtdB
S,⊥,Q
t +
1
2
λ2ST +
1
2
∫ T
0
ψ2t dt.
Hence
H(Q|P) = E[ZT logZT ] = EQ[logZT ] = 1
2
(λ2ST +
∫ T
0
ψ2t dt).
Denoting by ψ̂ the integrand corresponding to Q̂, we have
H(Q̂|P) = 1
2
(
λ2ST +
∫ T
0
ψ̂2t dt
)
and H(Q0|P) = 1
2
λ2ST,
where we remark that for the H(Q0|P) term, we have ψ0 = 0 since the dual problem for
C = 0 is to purely minimize the entropy, because
E[V (yZT )] = V (y) +H(Q|P) for U(x) = −e−x.
We have then p = EQ̂
[
C − 12
∫ T
0
ψ̂2t dt
]
. [4+3 marks]
Comment:
(4.a) Standard bookwork, relies on knowing properties of V and the definition of v
(4.b) Very standard
(4.c) Standard work, nonetheless needs to be done properly
(4.d) Easy, but students may not be familiar with this representation
(4.e) Easy; students will not have seen this form before for indifference pricing
(4.f) More involved; includes some similar stuff done in the course. Needs students to correctly
compute H(Q|P) ([4 marks]) and then apply it to the indifference price p ([3 marks]),
recognising that ψ0 = 0 for problems without claim.
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