MATH0094:Market Risk Measures and Portfolio Theory
Solutions to LN Exercises
2021-2022
1 Fundamentals of market theory
Exercise 1.1. The following table summarises all the questions.
Predictable Adapted Markovian Martingale
1 N Y Y N
2 N Y Y N
3 N Y Y Y
4 N Y Y N
5 N Y Y N
6 Y Y N N
7 N N N N
Table 1: Summary of answers for this question
Let us see the details for some of these:
1. The process is clearly adapted but not predictable (for the latter it suffices to see that Xˆ1
is not deterministic).
Since Xt ⊥ Ft−1, Xt ⊥ Xˆt, so
E[Xˆt∣Ft−1] = t − 1
t
Xˆt−1 + 1
t
E[Xt] = E[Xˆt∣Xˆt−1]. (1)
A simple induction together with the chain rule then shows that the process is Markovian.
Equation (1) also shows that the process is not a Martingale in general, since
Xˆt−1 ≠ E[Xˆt∣Ft−1].
2. If Y is the process in 1., this process is just exp(Y ), so it is also not predictable, but it is
adapted and Markovian, since
E[Xˆt∣Ft−1] = (Xˆt−1) t−1t E[exp(Xt
t
)] = E[Xˆt∣Xˆt−1]
where we used independence to separate the two terms in the conditional expectation.
Also, the process is not in general a martingale.
1
3. This is a special modification of 2: if we call the process solution of 2 Z, we get
Xˆt = exp(−tE[X1] − 1
2
tvar[X1])(Zt)t.
Note that the first term is deterministic. Thus, it inherits adaptability, non-predictability
and Markovianity. Hence, we just need to verify if this is a martingale. Since for a random
variable N ∼ N (m,σ2) we get that
E[exp(N)] = exp(m + 1
2
σ2),
it follows that
E[Xˆt∣Ft−1] = Xˆt−1 exp(E[X1] + 1
2
var(X1)) exp(−E[X1] − 1
2
var[X1]) = Xˆt−1
Hence, it is a martingale.
4. It follows easily that the process is not predictable but is adapted. Moreover,
E[ξt∣Ft−1] = (E[Xˆt∣Ft−1]E[σt∣Ft−1]) = ( E[σtXt∣Ft−1]α0 + α1Xˆ2t−1 + βtσ2t−1)
= ((α0 + α1Xˆ2t−1 + βtσ2t−1)1/2E[X1]
α0 + α1Xˆ2t−1 + βtσ2t−1 ) = E[ξt∣ξt−1]
where we used once more independence and the fact that the expression before the last one
only depends on the values of ξt−1. Also, the same expression shows that in general it will
not be a martingale.
5. The process Xˆ is in this case adapted but not predictable, since the event {Xt > max
0≤sis Ft but not Ft−1 measurable.
Since
Xˆt = max{Xˆt−1,Xt},
it is clear that the process is Markovian. Note also that
E[Xˆt∣Ft−1] = E[max{Xˆt−1,Xt}∣Xˆt−1]
and so in general it will not be a martingale.
6. We can write
Xˆt = ⎧⎪⎪⎨⎪⎪⎩Xˆt−1 if XXˆt−1 > 1t otherwise ,
which is Ft−1 measurable. Hence, it is predictable and adapted, and in particular E[Xˆt∣Ft−1] =
Xˆt. Now, note that the value of Xˆt is not completely known by conditioning by Xˆt−1. For
instance,
E[Xˆt∣Xˆt−1 = t − 1] = ⎧⎪⎪⎨⎪⎪⎩t − 1 if Xt−1 > 1t otherwise. .
Hence, it follows that this is not a Markovian process.
7. The process Xˆ in this case is not even adapted as it uses information for Fs for s > t.
2
Exercise 1.2. 1. pi is a valid strategy if it denotes the actions of investors, and it is a pre-
dictable process.
pi is deterministic, so that it is trivially predictable. Moreover, both pi+ and pi− are pre-
dictable processes since pi+,−1 are both deterministic while pi+,−2 are F1 measurable.
2. Calling w0 the initial investment, we have that:
Spi1 = Spi+1 = Spi−1 = w02 (1 +R11)
Spi2 = w04 (1 +R11)(1 +R12)
Spi
+
2 = w04 (1 +R11) [1 +R12 + 2δ(R12 − 1)sign(R11 − u + d2 )]
Spi
−
2 = w04 (1 +R11) [1 +R12 − 2δ(R12 − 1)sign(R11 − u + d2 )] ,
where the function sign(x) = ⎧⎪⎪⎨⎪⎪⎩∣x∣x
−1 if x ≠ 0
0 otherwise
.
Table 2 contains the values of the risky asset under different outcomes. Replacing the
values, we obtain Table 3 (note the sense of inequalities marked in yellow, which come
from the assumptions).
(0,0) (1,1) (0,1) (1,0)
S12 S
1
0d
2 S10u
2 S10du S
1
0du
Table 2: Values of risky asset under different outcomes.
Spi
−
2 S
pi
2 S
pi+
2(0,0) w0
4
(1 + d)[1 + d − 2δ(1 − d)] w0
4
(1 + d)2 w0
4
(1 + d)[1 + d + 2δ(1 − d)](1,1) w0
4
(1 + u)[1 + u − 2δ(u − 1)] w0
4
(1 + u)2 w0
4
(1 + u)[1 + u + 2δ(u − 1)](0,1) w0
4
(1 + d)[1 + u + 2δ(u − 1)] w0
4
(1 + d)(1 + u) w0
4
(1 + d)[1 + u − 2δ(u − 1)](1,0) w0
4
(1 + u)[1 + d + 2δ(1 − d)] w0
4
(1 + d)(1 + u) w0
4
(1 + u)[1 + d − 2δ(1 − d)]
Table 3: Values of portfolios under different outcomes: in the cases ω ∈ {(0,0), (1,1)} the values
increase from left to right; in the cases ω ∈ {(0,1), (1,0)} the values decrease from left to right.
Now, if we want Spi
+
2 > Spi−2 , we can take both P[S11 = u],P[S11 = u] > 0; while setting
P[S12 = u∣S11 = u] = P[S12 = d∣S11 = d] = 1.
We can easily verify the inequality by observing the table 3, since we do not have the cases(0,1) or (1,0).
3. Similar to previous point but for Spi
+
2 < Spi−2 we choose P[S12 = u∣S11 = d] = P[S12 = d∣S11 =
u] = 1.
3
4. By using again the table, we can see that for Spi2 > Spi−2 we would need
P[S12 = u∣S11 = d] = P[S12 = d∣S11 = u] = 0,
but to get Spi2 > Spi+2 we require
P[S12 = d∣S11 = d] = P[S12 = u∣S11 = u] = 0.
All these conditions cannot be imposed simultaneously, so it is impossible.
Exercise 1.3. We show that we can deduce the risk premia from the SDF. From the properties
of covariance, conditional expectation and (1.14) in the Lecture Notes, we have
Covt (Mt+1
Mt
,Rit+1) = Et [Mt+1Rit+1Mt ] −Et [Mt+1Mt ]Et[Rit+1] = 1 − Et[Mt+1]Et[Rit+1]Mt .
On the other hand, turning our attention to the money market account, we deduce from
(1.14) in the LN and the predictable property of R0 that
Mt = Et[Mt+1R0t+1] = R0t+1Et[Mt+1].
Thus,
Covt (Mt+1
Mt
,Rit+1) = 1 − Et[Rit+1]R0t+1
which after reordering concludes the proof.
Exercise 1.4. (i) Is there any risk-less asset in this market?
Yes, note that the first entry of S1(ω1), S1(ω2), S1(ω3) are all equal to 1. Hence, it is
deterministic and thus risk-less. Its rate is R01 = 1/1 = 1.
(ii) The return of the risky assets:
R1 = [S11(ω1)/p1,S11(ω2)/p2,S11(ω3)/p3] (2)= [3
2
,
1
2
,
5
2
] (3)
i.e., R1(ω1) = 32 , R1(ω2) = 12 , R1(ω3) = 52 . And R2 = [ 97 , 57 , 107 ].
If probability is uniform, we have that the risk premia are:
E[R1] −R0 = 3
2
P[{ω1}] + 1
2
P[{ω2}] + 5
2
P[{ω3}] − 1 (4)
= 1
3
[3
2
+ 1
2
+ 5
2
] − 1 = 3
2
− 1 = 1
2
(5)
E[R2] −R0 = 9
7
P[{ω1}] + 5
7
P[{ω2}] + 10
7
P[{ω3}] − 1 (6)
= 1
3
[9
7
+ 5
7
+ 10
7
] − 1 = 8
7
− 1 = 1
7
(7)
(iii) We now check if the market model is complete and arbitrage free. This kind of problem
can be solved in several ways using the matrix properties we reviewed, or exposing explicitly
arbitrage opportunities or non-replicable profiles.
4
We follow a method based on linear algebra. Since the matrix is square, we can use the
determinant to check if the matrix MS1 is invertible. Remember that if this is the case, then
there is a unique solution to any linear system associated to it.
We obtain that
det(MS1) = det(M⊺S1) = (1)(10 − 25) − (1)(30 − 45) + (1)(15 − 9) = −15 + 15 + 6 = 6.
Hence, any linear system has a unique solution, so MS1 has full range. The market is therefore
complete.
It can also be deduced that there is at most one set of AD prices. We need to check that
they are all positive. By solving the linear system
S0 =MS1pAD
We get ⎛⎜⎝
1 1 1 1
3 1 5 2
9 5 10 7
⎞⎟⎠→
⎛⎜⎝
1 1 1 1
0 −2 2 −1
0 −4 1 −2
⎞⎟⎠→
⎛⎜⎝
1 0 0 1/2
0 1 −1 1/2
0 0 −3 0
⎞⎟⎠
Note that from the last equation, we have that pAD3 = 0. As this is the only solution, we get that
there is no strictly positive AD prices (or SDF or equivalent risk neutral probability).
This means that the market has an arbitrage.
Exercise 1.5. (a) We know that absence of arbitrage is equivalent to the existence of a positive
SDF, positive AD prices, or an equivalent risk-neutral probability. Using the latter, there is no-
arbitrage if and only if there exists Q equivalent to the original probability such that EQ[R] = R0.
Define Q{ω1} = q1, Q{ω2} = q2. We solve the system:
R0 = q1R1(ω1) + q2R1(ω2) = q1Ru + q2Rd
and
q1 + q2 = 1(from probability properties)
Moreover, from no arbitrage we require, q1 > 0, q2 > 0. Solving, we get:
q2 = Ru −R0
Ru −Rd q1 = R0 −RdRu −Rd (8)
But since Ru > Rd, then q2 > 0 and q1 > 0 if and only if Ru > R0 and R0 > Rd. Therefore, the
condition is Ru > R0 > Rd.
(b) We obtained the risk neutral probabilities as a by product of the previous exercise, as
given in (8)
(c) There are two Arrow-Debreu securities, one for each {ω1, ω2}.
pAD1 = EQ[1ω1R0 ] = q11{ω1}(ω1)R0 + q21{ω1}(ω2)R0 = q1R0 = R0 −Rd(Ru −Rd)R0 (9)
pAD2 = EQ[1ω2R0 ] = q2R0 = Ru −R0(Ru −Rd)R0 (10)
(d) If a new asset paying (S1 −K)+ is introduced, we can use the risk neutral probabilities
to find popt, the price of this asset.
5
popt = EQ[(S1 −K)+
R0
] = q1(Rup −K)+
R0
+ q2(Rdp −K)+
R0
(11)
= 1
R0(Ru −Rd)[(Rup −K)+(R0 −Rd) + (Rdp −K)+(Ru −R0)] (12)
but since Ru > R0 > Rd, and K = R0p
popt = p
R0(Ru −Rd)[(Ru −R0)(R0 −Rd)]
Exercise 1.6. (a) Arbitrage-free ⇒ law of one price. Take θ,ξ such that θ ⋅S1 = ξ ⋅S1. If there
is no-arbitrage, there exists a positive SDF. Hence, E[MSi1] = pi.
But then, by linearity:
θ ⋅ p = θ ⋅E[MS1] = E[M(θ ⋅S1)] (13)= E[M(ξ ⋅S1)] = ξ ⋅E[MS1] = ξ ⋅ p (14)
(b) Example of market with arbitrage and such that the law of one price is satisfied. Take
the market in question 2.1. We showed it admitted an arbitrage. Let θ, ξ such that θ ⋅S1 = ξ ⋅S1.
This means that
[θ0, θ1, θ2]⎡⎢⎢⎢⎢⎢⎣
1 1 1
3 1 5
9 5 10
⎤⎥⎥⎥⎥⎥⎦´udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¸udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¶
MS1
= [ξ1, ξ2, ξ3]⎡⎢⎢⎢⎢⎢⎣
1 1 1
3 1 5
9 5 10
⎤⎥⎥⎥⎥⎥⎦. (15)
Since we showed MS1 is invertible, because det(MS1) = 6. Therefore, we conclude that θ = ξ , so
trivially the law of one price holds.
Exercise 1.7. Recall that M is an SDF if for each market instrument, i = 0, ..., n, we have
Si0 = E[Si1M]. Let Mα1 = αM1 + (1 − α)M˜1 for two SDFs M1, M˜1. Then,
E[(αM1 + (1 − α)M˜1)Si1] = αE[M1Si1] + (1 − α)E[M˜1Si1] (16)= αSi0 + (1 − α)Si0 = Si0 (17)
Moreover, if M1 > 0 and M˜1 > 0, we have that for 0 ≤ α ≤ 1, αM1 > 0 and (1 − α)M˜1 > 0 and
thus Mα1 > 0.
Exercise 1.8. We show the two applications: Assume that the price for the asset satisfies
pnew = E[Snew1 M] for some strictly positive SDF M . Then, M is a strictly positive SDF for the
market with n + 2 assets, and by the FTAP-1, there is no arbitrage.
On the other hand, if pnew ≠ E[DnewM] for every M strictly positive SDF on the market
with n + 1 assets, then we will not be able to find an SDF on the market with n + 2 assets. By
the FTAP-1, this implies there exists an arbitrage opportunity.
Exercise 1.9. By Exercise 1.8, we have that p˜ = E[DnewM˜1] and pˆ = E[DnewMˆ1] for two strictly
positive SDFs M˜1, Mˆ1. But since
αp˜ + (1 − α)pˆ = E[Dnew{αM˜1 + (1 − α)Mˆ1}] (18)= E[DnewMα1 ], (19)
and Mα1 is a strictly positive SDF by Exercise 1.7. Hence, it is an arbitrage free price.
6
2 Utility functions
Exercise 2.1.
1. The investor chooses from the two alternatives the one with larger expected utility. Hence,
we need to compare the expected utility of a final wealth W1 = w0 = 200 and a final wealth
W2 = ⎧⎪⎪⎨⎪⎪⎩w0 + 100 = 300 with probability
1
2
w0 − 100 = 100 with probability 12 .
Now, note that E[W2] = 12300+ 12100 = 200 = w0 =W1, and also that uA(x) = xγγ is concave
(in fact it is a CRRA utility function). Thus, by Jensen’s inequality we have that
E[uA(W2)] ≤ uA[E[W2]] = uA(W1) = E[uA(W1)]
Therefore, the investor prefers W1.
The argument is the same for uB(x) = − exp(−x), since it is also a concave function.
2. In this case we have that
W1 = ⎧⎪⎪⎨⎪⎪⎩11 = with probability
1
3
31 = with probability 2
3
; W2 = ⎧⎪⎪⎨⎪⎪⎩21 with probability 0.91 with probability 0.1 .
We can numerically make the calculation to get, E[log(W1)] = 13 log(11)+ 23 log(31) ≈= 3.09,
while E[log(W2)] = 0.93 log(21) + 0.1 log(1) ≈= 2.74. Clearly, the investor will choose the
first investment.
Exercise 2.2. Since the functions to be considered are twice differentiable, it suffices to look at
the sign of the second derivative: we need to verify that the second derivative is negative on the
domain of definition.
1. f1(x) = log(x) for x > 0: We get
f ′(x) = 1
x
; and f ′′(x) = −1
x
< 0.
2. f2(x) = a − bx2 for b > 0: We get
f ′(x) = −2bx; and f ′′(x) = −2b < 0.
Hence, both functions are concave. Recall though that the quadratic function needs a restricted
domain of definition to be considered a utility function.
Exercise 2.3. To evaluate the expected utility, we can either use integral methods or Monte
Carlo methods.
7
# Import modules
import numpy as np
import s c ipy . s t a t s as s t
# We use two approaches , e i t h e r us ing the ’ s t a t s ’ module
# and the ’ expec t ’ method , or us ing Monte Carlo wi th a l a r g e
# number o f s imu la t i on s
# We i n i t i a l i s e the func t i on
myf = lambda x : np . l og ( x )
mc number = 1000000
# Pareto wi th shape a lpha=2 and s c a l e ( or mode) xm = 0.5
# Method 1
u = s t . pareto . expect (myf , args =(2 ,) , s c a l e =0.5)
print ( ’ Pareto , method 1 : ’ , u )
# Method 2
sample p = (np . random . pareto (2 , mc number ) + 1) ∗ 0 .5
u = myf ( sample p ) . mean ( )
print ( ’ Pareto , method 2 : ’ , u )
Pareto, method 1: -0.19314718055994637
Pareto, method 2: -0.1932406330316995
# Exponent ia l wi th lambda = 1
# Method 1
u = s t . expon . expect (myf , s c a l e =1)
print ( ’ Exponential , method 1 : ’ , u )
# Method 2
sample e = np . random . exponent i a l ( s i z e =(mc number , ) )
u = myf ( sample e ) . mean ( )
print ( ’ Exponential , method 2 : ’ , u )
Exponential, method 1: -0.5772156649008394
Exponential, method 2: -0.5804589164442893
# Log normal ( s tandard )
# Method 1
u = s t . lognorm . expect (myf , args =(1 ,))
print ( ’ Exponential , method 1 : ’ , u )
# Method 2
sample ln = np . random . lognormal ( s i z e =(mc number , ) )
u = myf ( sample ln ) . mean ( )
print ( ’ Exponential , method 2 : ’ , u )
8
Exponential, method 1: 4.844603335998414e-17
Exponential, method 2: -0.00023813580546780955
Note that there is a good agreement between the two methods.
Exercise 2.4. An affine function on R is any function g ∶ R → R of the type g(x) = ax + b for
a, b ∈ R. Clearly, if a > 0, we have that
x < y⇒ ax < ay⇒ ax + b < ay + b⇒ g(x) < g(y)
so that an affine function with a > 0 preserves the order. To show the converse, note that if a = 0,
for all x < y we have g(x) = g(y). And similarly, if a < 0, we get
x < y⇒ ax > ay⇒ ax + b > ay + b⇒ g(x) > g(y),
so that the order is reversed. Hence, the only affine functions that preserve the order are the
ones with a > 0.
Exercise 2.8. (i) We take the case u(x) = log(x). Take w initial wealth and wX additional
gamble.
E[u(w(1 +X))] = E[log(w(1 +X))] (20)= E[log(w) + log(1 +X)] (21)= log(w) +E[log(1 +X)] (22)
On the other hand,
u(E[w(1 +X)] −wη(X)) = log(w(E[1 +X] − η(X))) (23)= log(w) + log(E[1 +X] − η(X)) (24)
⇒ η(X) solves: log(E[1 +X] − η(X)) = E[log(1 +X)] (25)⇒ η(X) = E[1 +X] + exp(E[log(1 +X)]), (26)
which is independent of ω. Similar development for the case u(x) = 1
1−γx1−γ with γ ∈ (0,1)⋃(1,∞).
(ii) E[R] = E[exp(Z)] with Z ∼ N (−σ2
2
, σ).
E[R] = exp(E[Z] + 1
2
var(Z)) = exp(−σ2
2
+ σ2
2
) = 1 (27)
Since we have that u is CRRA, we can use the results of the previous exercise, and solve the
problem for initial wealth 1 If the initial wealth is w0 ≠ 1 we simply multiply by w0. Let us first
focus on the case u(x) = log(x). We get,
log(E[R] − η(R)) = E[log(R)] = E[Z] = −σ2
2
, (28)
log(1 − η(R)) = −σ2
2
⇒ η(R) = 1 − exp(−σ2
2
) > 0. (29)
9
Hence, η(w0R) = w0(1 − exp(−σ22 )).
In the case of u(x) = 1
1−γx1−γ ,
1
1 − γ (1 − η(R))1−γ = E[ 11 − γ [exp(Z)]1−γ] (30)= E[ 1
1 − γ exp((1 − γ)Z)] (31)
= 1
1 − γ exp((1 − γ)(−σ22 ) + σ22 (1 − γ)2) (32)
= 1
1 − γ exp(−σ22 (1 − γ)γ) (33)
= 1
1 − γ [exp(−σ22 γ)]1−γ (34)
⇔ (1 − η(R)) = exp(−σ2
2
γ) (35)
η(R) = 1 − exp(−σ2
2
γ) (36)
η(w0R) = w0(1 − exp(−σ2
2
γ)) (37)
We see that the solution is the same independently of the relative risk aversion coefficient.
Exercise 2.9. (i) Since W ∼ N (w,σ2w), we calculate
E[u(W )] = E[−exp(−αW )] = −exp(−αw + 1
2
α2σ2w). (38)
On the other hand,
E[u(W +D − pBID)] = E[−exp(−α(W +D − pBID))] (39)= −exp(−α(w +m − pBID) + 1
2
α2(σ2w + σ2 + 2σσwρ)) (40)
Hence, − αw + 1
2
α2σ2w = −α(w +m − pBID) + 12α2(σ2w + σ2 + 2σσwρ) (41)⇒ pBID =m − 1
2
α(σ2 + 2σσwρ) (42)
where in (40) we use the expression for the mean of the exponential of a Gaussian distributed
random variable and the fact that if the joint vector (A,B)is Gaussian,
A +B ∼ N (E[A +B], var(A +B)) (43)= N (E[A] +E[B], var(A) + var(B) + 2cov(A,B)) (44)
(ii) By similar arguments as before,
10
E[u(W −D + pask)] = −exp(−α(w −m + pask) + 1
2
α2(σ2w + σ2 − 2σσwρ)) (45)
Hence, − αw + 1
2
α2σ2w = −α(w −m + pask) + 12α2(σ2w + σ2 − 2σσwρ) (46)⇔ pask =m + 1
2
α(σ2 − 2σσwρ) (47)
Note that this simple model proposes an explanation to the fact that there is a bid-ask spread
around the average value of the asset m. This simple model captures the fact that an investor
perceives an asset to be worth more than its average price if they want to sell, and to be worth
less if they want to buy.
Exercise 2.11.
i. Following the hint, we take W such that
P[W = 2i] = 2−i; for all i = 1,2, . . . .
We easily verify
E[W ] =∞.
ii. Let us now modify the above to get the required condition. Take for example W˜ = exp(−W ),
i.e.,
P[W = e1/2i] = 2−i; for all i = 1,2, . . . .
Clearly, W˜ is always positive. Moreover,
E[log(W˜ )] = E[−W ] = −E[W ] = −∞.
11
3 Portfolio choice
One period case
Exercise 3.4. In this setting, we have only two possible assets. Let φ be the amount to be
invested in the risky asset. Note that if we set w0 − φ the amount to be invested in the risk-free
asset, the budget constraint on the initial time, is satisfied, since w0 = φ + (w0 − φ). Hence, we
can write the optimisation problem as
max
φ ∈ R ψ(θ) ∶= E[u(W (φ))]
s.t. W (φ) = (w0 − φ)R0 + φR1 (48)
In this case, we take u(x) = log(x), and we have,
E[u(W )] = 1
2
u((w0 − φ)R0 + φu) + 1
2
u((w0 − φ)R0 + φd) (49)
Using the first order conditions, we have that
d
dφ
E[u(W (φ))]∣
φ=φ∗ = 0 (50)
⇒ u′((w0 − φ∗)R0 + φ∗u)1
2
(u −R0) + u′((w0 + φ∗)R0 + φ∗d)1
2
(d −R0) = 0 (51)
⇒ u −R0(w0 − φ∗)R0 + φ∗u = R0 − d(w0 − φ∗)R0 + φ∗d (52)⇒ φ∗(u −R0)(d −R0) + (u −R0)w0R0 = φ∗(R0 − d)(u −R0) + (R0 − d)(w0R0) (53)
φ∗ = w0R0(u − 2R0 + d)
2(R0 − d)(u −R0) (54)
Compare with φ¯ = w0R0(K−1)
u−R0+K(R0−d) (Eq. 5.15 from lecture notes). Given that ρ = 1, we have
that
K = u −R0
R0 − d ⇒ φ¯ = w0R0(u −R0 −R0 − d)(u −R0)(R0 − d) + (R0 − d)(u −R0) . (55)
Thus, it follows that the expression is unchanged.
Exercise 3.5. The main difference between this problem and Exercise (3.4) is the additional
constraint on having no short positions. In terms of the variable φ (the amount invested in the
risky asset), this means that
0 ≤ φ ≤ w0. (56)
Indeed, φ cannot be negative nor be greater than the total (otherwise we would need to short
the risk-free asset). Hence, the optimisation problem now reads,
max
φ ∈ [0,w0] ψ(θ) ∶= E[u(W (φ))]
s.t. W (φ) = (w0 − φ)R0 + φR1 (57)
12
For the rest of the solution of this exercise, let us call φ¯ to the solution to the problem (48)
(i.e., the solution with allowed short positions), and let φ∗ be the solution to (57), i.e., with no
short positions. Let us remark that since [0,w0] ⊂ R, we have that
ψ(φ¯) ≥ ψ(φ¯∗).
Clearly, from this inequality we deduce that if φ¯ ∈ [0,w0], we have equality and φ¯ = φ∗. Let
us explicitly write this condition in terms of the data of the problem. From (55) we get:
0 ≤ φ¯ ≤ w0 ⇔ 0 ≤ w0R0(u − 2R0 + d)
2(R0 − d)(u −R0) ≤ w0 ⇔ 0 ≤ R0(u − 2R0 + d) ≤ 2(R0 − d)(u −R0) (58)⇔ 0 ≤ R0((u −R0) − (R0 − d)) ≤ 2(u −R0)(R0 − d) (59)
Recall that we are working under the assumption 0 < d < R0 < u. We now consider the cases
where (59) does not hold (i.e. if either the left or the right inequality fail).
• If (u −R0) < (R0 − d): we would have as optimal φ∗ = 0. To verify this, let us check the
directional derivative version of the first order conditions. We obtain
∂φE[u(W (φ))]∣φ=0 = u1(w0R0)(u −R0)
2
+ u1(w0R0)(d −R0)
2= u1(w0R0)
2
((u −R0) − (R0 − d)) < 0
Thus, ∂φE[u(W (φ))]∣φ=0(φ − 0) ≤ 0 ∀φ ∈ [0,w0]. This means that 0 is a local (and thanks
to concavity) global minimum of our problem. Note also that in this case, φ¯ < φ∗, so as a
result of not being able to short we invest more on the risky asset (or more appropriately,
we do not short it).
• If R0((u −R0) − (R0 − d)) ≥ 2(u −R0)(R0 − d) , we claim that φ∗ = w0. The verification
follows the same lines as above. In this case, φ¯∗ < φ¯, so we invest less on the risky asset (in
order not to short the risk-free one).
Exercise 3.6. The optimisation problem reads
max
φ ∈ Rn E[u(W (φ))]
s.t. w0 = φ ⋅ 1,
W (φ) = φ ⋅R1
(60)
where u(x) ∶= − exp(−αx). We can replace directly the second constraint on the utility function.
Moreover, recalling that since R1 is a Gaussian vector, (φ ⋅R1 is a Gaussian random variable
with mean φ ⋅µ and variance φ⊺Σ¯φ, it follows that
E[u(φ ⋅R1))] = E[−exp(−αφ ⋅R1)] = −exp(−αφ ⋅µ + 1
2
α2φ⊺Σ¯φ). (61)
The problem (60) then reads
max
φ ∈ Rn − exp(−αφ ⋅µ + 12α2φ⊺Σ¯φ)
s.t. w0 = φ ⋅ 1 (62)
13
The main difference of (62) with respect to section 5.5. is that the control variable is an
element of Rn, which corresponds to the fact that there is no risk-free asset. However, as before,
observing that − exp(−x) is increasing, we get that φ∗ solves (62) if and only if it also solves
max
φ ∈ Rn αφ ⋅µ − 12α2φ⊺Σ¯φ
s.t. w0 = φ ⋅ 1 (63)
We solve (63) by using the Lagrange multipliers technique1, i.e. we set
L(φ,λ) ∶= αφ ⋅µ − 1
2
α2φ⊺Σ¯φ − λ(φ ⋅ 1 −w0) (64)
and applying the first order conditions, to get
▽φL(φ∗, λ∗) = −αµ + α2Σ¯φ∗ − λ∗1 = 0 (65)
and also,
∂λL(φ∗, λ∗) = φ∗ ⋅ 1 −w0 = 0 (66)
Solving in (65) for φ∗ gives
φ∗ = 1
α2
Σ¯
−1(λ∗1 + αµ) (67)
On the other hand, multiplying (inner product) by 1 and using (66) we can write
w0 = φ∗ ⋅ 1 = ( 1
α2
Σ¯
−1(λ∗1 + αµ)) ⋅ 1 (68)
= λ∗
α2
1⊺Σ¯−11 + 1
α
1⊺Σ¯−1µ (69)
⇒ λ∗ = α2(w0 − 1α1⊺Σ¯−1µ)
1⊺Σ¯−11 = α
2w0 − α1⊺Σ¯−1µ
1⊺Σ¯−11 (70)
which gives us the value of λ∗. Replacing in (67) we conclude that
φ∗ = 1
α
Σ¯
−1
µ + ⎛⎝αw0 − 1⊺Σ¯
−1
µ
α1⊺Σ¯−11
⎞⎠ Σ¯−11 (71)
as claimed. It is interesting to compare this equation with the one in the case where a risk-free
asset is available. Note first that the optimal portfolio now does depend on the initial wealth!
The point is that because there is no risk-free investment and all the initial wealth has to be
invested, all alternatives are risky. So the small perturbations around risk-free alternatives are
not possible.
To have more insight, let us call ϕ the optimal portfolio when there is a risk-free asset. From
Section 5.5 we have that ϕ = α−1Σ¯−1µ. By replacing and reordering terms, we get
φ∗ = ϕ + (w0 − 1 ⋅ϕ) Σ¯−11
1⊺Σ¯−11
1It is possible to use the same technique as in the section 5.5., but this time is slightly more involved to write
this time
14
In other words, we are just dividing the amount that would be assigned to the risk-free asset
among the remaining ones. This is done proportionally to the sum of all covariances of each
asset.
Exercise 3.7. In this case, we find that the optimisation problem can be written
max
c0 ∈ R,φ ∈ R E[u(c0) + u(C1)]
s.t. w0 = c0 +φ ⋅ 1,
C1 = φ ⋅R
(72)
where R = [R0
Rˆ
]. Note that using the constraints, we have:
E[u(c0) + u(C1)] = u(c0) +E[u(C1)] = u(c0) +E[u(φ0R0 + φˆ ⋅ Rˆ)] (73)= u(c0) +E[u((w0 − c0 − φˆ ⋅ 1)R0 + φˆ ⋅ Rˆ)] (74)
Once again, we see that C1 = (w0 − c0 − φˆ ⋅ 1)R0 + φˆ ⋅ Rˆ is Gaussian with mean E[C1] =(w0 − c0 − φˆ ⋅ 1)R0 + φˆ ⋅µ and variance var[C1] = φˆ⊺Σ¯φ. Hence, we have that
E[u(c0) + u(C1)] = −exp(−αc0) − exp(−αE[C1] + 1
2
α2var(C1)) (75)
= −exp(−αc0) − exp(−α[(w0 − c0 − φˆ ⋅ 1)R0 + φˆ ⋅ µ] + 1
2
α2φˆ
⊺
Σ¯φˆ´udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¸udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¶
ϕ
) (76)
Let us apply the first order conditions. We get
∂c0E[u(c∗0) + u(c∗1)] = αexp(−αc∗0) − αR0exp(ϕ) = 0 (77)▽φˆE[u(c∗0) + u(c∗1)] = αexp(−αϕ)(µ − 1R0 − αΣ¯φˆ∗) = 0. (78)
Since the exponential function is always positive, from (78), we obtain
φˆ∗ = 1
α
Σ¯
−1(µ − 1R0) (79)
Let us highlight that this is exactly as in the case of the pure investment problem! The
intuition is that riskiness comes only from investment (as the consumption at initial time is
deterministic), so that this investor prioritises his decision on how to invest on the risky assets.
Also, note that if we express this problem in terms of the number of shares, we have
θi,∗ = φi,∗
Si0
= 1
αSi0
Σ¯
−1(µ − 1R0)
It remains to find c∗0 and the amount to be invested in the risk-free asset. Using (77), we
have by reordering, dividing by α and taking log that
−αc∗0 = −α[(w0 − c∗0 − φˆ∗ ⋅ 1)R0 + φˆ∗ ⋅µ] + 12α2 ˆφ∗⊺Σ¯φˆ∗] + log(R0) (80)⇒ c∗0 = 11 +R0 [(w0 − φˆ∗ ⋅ 1)R0 + φˆ∗ ⋅µ − 12α ˆφ∗⊺Σ¯φˆ∗ − 1α log(R0)] (81)
15
Noting that φˆ∗ ⋅ 1 = 1
α
1⊺Σ¯−1(µ − 1R0); φˆ∗ ⋅µ = 1
α
µ⊺Σ¯−1(µ − 1R0) and
ˆφ∗⊺Σ¯φ = 1
α2
(µ − 1R0)⊺Σ¯−1(µ − 1R0)
we finally get
c∗0 = 11 +R0 [w0R0 + 12α(µ − 1R0)⊺Σ¯−1(µ − 1R0) − 1α log(R0)].
Using that φ0,∗ = w0 − c∗0 − φˆ∗, we also get the initial amount invested in the risky-free asset.
Exercise 3.8. The problem can be written
max
θ,η
[u (ηω0) + δE [u (Ww0(1−η),φ)] = max
η
[u(ηw0) + δmax
φ
E [u (Ww0(1−η),φ)]]
By Example 3.1.3 we have that
φ∗ = w0(1 − η)R0(x − 1)
u −R0 +K (R0 − d)
solves the max problem with initial wealth w0(1 − η).
We now find the value of expected utility at this maximum. We look first at Ww0(1−eta),φ∗ .
We have that if R1 = u,
Ww0(1−η),Φ∗(ω = 1) = w0(1 − η)R0 ⎛⎝1 + (K − 1) (u −R0)u −R0 +K (R0 − d)⎞⎠
but since Kρ = u−R0
R0−d , we have
Ww0(1−η),Φ∗ = w0(1 − η)R0 (1 + K − 1
1 +K1−ρ ) .
By a similar procedure, we get that if R1 = d
Ww0(1−η),Φ∗(ω = 0) = w0(1 − η)R0 (1 + 1 −K
Kρ +K ) .
Hence,
E [u (Ww0(u−v)1φ2)] = (w0R0(1 − η))1−ρ
2(1 − ρ)
⎡⎢⎢⎢⎢⎣(K +K
1−ρ
1 +K1−ρ )1−ρ + ( Kρ + 1Kρ +K )
⎤⎥⎥⎥⎥⎦´udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¸udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¶=C
Finally, we solve
max
η
⎡⎢⎢⎢⎢⎣(w0η)
1−ρ
1 − ρ + (w0R0(1 − η))
1−ρ
2(1 − ρ) Cδ
⎤⎥⎥⎥⎥⎦
which by a first order condition gives
16
(w0η)−ρw0 − (w0R0(1−η))−ρ2 (Cδ) (w0R0) = 0⇒ η = R0(1−η)21/ρ(Cδ)1/ρ(R0)1/ρ⇒ η = (R0) ρ−1ρ 21/ρ(Cδ)1/ρ+(R0)(ρ−1)/ρ21/ρ
Some checks are in order: note that η ∈ (0,1) as expected. Moreover, when δ → 0, η → 1.
Exercise 3.9. Since we have only one riskless asset, we have that the only uncertainty on this
case comes from the random endowment. The decision variables are c0 and φ (the amount to be
invested in the risk-free asset), and the problem reads
max
c0, φ ∈ R E[u(c0) + δu(C1)]
s.t. w0 = c0 + φ,
C1 = φR0 + Y
(82)
We can introduce both restrictions on the utility as follows: from the first restriction, we have
φ = (w0 − c0) and replacing on the end of period constraint, we get that C1 = (w0 − c0)R0 + Y .
Thus, we can solve the unconstrained problem
max
c0 ∈ R u(c0) + δE[u((ω0 − c0)R0 + Y )] (83)
which solves the first question. For the second claim, because u′′ < 0, we know that u is strictly
concave. Using first order conditions, we get
u′(c∗0) − δE[u′((ω0 − c∗0)R0 + Y )]R0 = 0. (84)
On the other hand, using u′′′ > 0, we get by Jensen’s inequality that
E[u′((ω0 − c∗0)R0 + Y )] ≥ u′(E[(ω0 − c∗0)R0 + Y ]) = u′((ω0 − c∗0)R0) (85)
where we used the fact that E[Y ] = 0. Equations (84) and (85) imply that
u′(c∗0) ≥ δu′((ω0 − c∗0)R0)R0. (86)
Assuming now that Y = 0 (denoting cˆ0 the optimal in this case), we have
u′(cˆ0) = δu′((ω0 − cˆ0)R0)R0. (87)
We claim now that (86) and (87) deduce that cˆ0 ≥ c∗0. Assume by contradiction that cˆ0 < c∗0, and
observe that the left hand of (86) is monotonously decreasing as function of c∗0 (due to concavity)
while the right-hand side is increasing with c∗0 (also due to concavity). Hence, we would get from
(87) and cˆ0 < c∗0 that u′(c∗0) < δu′((ω0 − c∗0)R0), which would contradict (86). It follows that
cˆ0 ≥ c∗0 as desired.
Multi-period case
Exercise 3.10. By definition 1.28, we have that a measure Q is risk neutral if and only if the
Radon-Nikodym density with respect to P can be written
dQ
dP
∶=MTS0T
17
for some SDF M . Hence, it follows that for any s = 0, . . . , T
EQ [Cs
S0s
] = E [MTS0T CsS0s ] = E [Es[MTS0T ]CsS0s ]
where we used the definition of the RN density and of conditional expectation, given that both
Cs, S
0
s are Fs measurable. But by the definition of SDF, Es[MTS0T ] =MsS0s . Hence,
EQ [Cs
S0s
] = E [MsCs] .
Exactly the same arguments show that
EQ [ Is
S0s
] = E [MsIs] .
Hence, the claim follows by Lemma 3.16.
Exercise 3.12. As per the question, let us call pi∗ ∈ Rn+1 the portfolio achieving the maximum
in the log-utility maximisation for one period.
Let pi ∈ Rn+1 be other non-optimal strategy (that we keep fixed), and we need to compare
St(pi) and St(pi∗) for t large enough. Let
∆ ∶= E[log(pi∗R1)] −E[log(piR1)].
By assumption, we have that 0 < ∆ <∞.
Let us now recall that for a dividend-reinvested portfolio without consumption, we have that
S0(pi) = w0; St(pi) = w0 t∏
s=1(pi⊺Rs),
Clearly the strict monotonicity of log implies that
St(pi) < St(pi∗) if and only if 1
t
log(St(pi)) < 1
t
log(St(pi∗)).
So we focus in showing a statement on the right-hand side. Indeed, note that
1
t
log(St(pi)) = log(w0)
t
+ 1
t
t∑
s=1 log(pi⊺Rs);
and, similarly for pi∗. Thus,
1
t
log(St(pi∗)) − 1
t
log(St(pi)) = 1
t
t∑
s=1{log((pi∗)⊺Rs) − log(pi⊺Rs)}. (88)
Now, the law of large numbers implies that 1
t
log(St(pi∗)) → E[log(St(pi∗))] almost surely (and
similarly for pi). Their difference converges almost surely then to ∆. Hence, there exists T > 0
such that for all t > T ,
P(∣1
t
log(St(pi∗)) − 1
t
log(St(pi)) −∆∣ ≤ ∆
2
) = 1
From where we deduce that
P(1
t
log(St(pi∗)) − 1
t
log(St(pi)) ≥ ∆
2
> 0) = 1
for all t > T , as wanted.
18

























































is Ft but not Ft−1 measurable.
Since
Xˆt = max{Xˆt−1,Xt},
it is clear that the process is Markovian. Note also that
E[Xˆt∣Ft−1] = E[max{Xˆt−1,Xt}∣Xˆt−1]
and so in general it will not be a martingale.
6. We can write
Xˆt = ⎧⎪⎪⎨⎪⎪⎩Xˆt−1 if XXˆt−1 > 1t otherwise ,
which is Ft−1 measurable. Hence, it is predictable and adapted, and in particular E[Xˆt∣Ft−1] =
Xˆt. Now, note that the value of Xˆt is not completely known by conditioning by Xˆt−1. For
instance,
E[Xˆt∣Xˆt−1 = t − 1] = ⎧⎪⎪⎨⎪⎪⎩t − 1 if Xt−1 > 1t otherwise. .
Hence, it follows that this is not a Markovian process.
7. The process Xˆ in this case is not even adapted as it uses information for Fs for s > t.
2
Exercise 1.2. 1. pi is a valid strategy if it denotes the actions of investors, and it is a pre-
dictable process.
pi is deterministic, so that it is trivially predictable. Moreover, both pi+ and pi− are pre-
dictable processes since pi+,−1 are both deterministic while pi+,−2 are F1 measurable.
2. Calling w0 the initial investment, we have that:
Spi1 = Spi+1 = Spi−1 = w02 (1 +R11)
Spi2 = w04 (1 +R11)(1 +R12)
Spi
+
2 = w04 (1 +R11) [1 +R12 + 2δ(R12 − 1)sign(R11 − u + d2 )]
Spi
−
2 = w04 (1 +R11) [1 +R12 − 2δ(R12 − 1)sign(R11 − u + d2 )] ,
where the function sign(x) = ⎧⎪⎪⎨⎪⎪⎩∣x∣x
−1 if x ≠ 0
0 otherwise
.
Table 2 contains the values of the risky asset under different outcomes. Replacing the
values, we obtain Table 3 (note the sense of inequalities marked in yellow, which come
from the assumptions).
(0,0) (1,1) (0,1) (1,0)
S12 S
1
0d
2 S10u
2 S10du S
1
0du
Table 2: Values of risky asset under different outcomes.
Spi
−
2 S
pi
2 S
pi+
2(0,0) w0
4
(1 + d)[1 + d − 2δ(1 − d)] w0
4
(1 + d)2 w0
4
(1 + d)[1 + d + 2δ(1 − d)](1,1) w0
4
(1 + u)[1 + u − 2δ(u − 1)] w0
4
(1 + u)2 w0
4
(1 + u)[1 + u + 2δ(u − 1)](0,1) w0
4
(1 + d)[1 + u + 2δ(u − 1)] w0
4
(1 + d)(1 + u) w0
4
(1 + d)[1 + u − 2δ(u − 1)](1,0) w0
4
(1 + u)[1 + d + 2δ(1 − d)] w0
4
(1 + d)(1 + u) w0
4
(1 + u)[1 + d − 2δ(1 − d)]
Table 3: Values of portfolios under different outcomes: in the cases ω ∈ {(0,0), (1,1)} the values
increase from left to right; in the cases ω ∈ {(0,1), (1,0)} the values decrease from left to right.
Now, if we want Spi
+
2 > Spi−2 , we can take both P[S11 = u],P[S11 = u] > 0; while setting
P[S12 = u∣S11 = u] = P[S12 = d∣S11 = d] = 1.
We can easily verify the inequality by observing the table 3, since we do not have the cases(0,1) or (1,0).
3. Similar to previous point but for Spi
+
2 < Spi−2 we choose P[S12 = u∣S11 = d] = P[S12 = d∣S11 =
u] = 1.
3
4. By using again the table, we can see that for Spi2 > Spi−2 we would need
P[S12 = u∣S11 = d] = P[S12 = d∣S11 = u] = 0,
but to get Spi2 > Spi+2 we require
P[S12 = d∣S11 = d] = P[S12 = u∣S11 = u] = 0.
All these conditions cannot be imposed simultaneously, so it is impossible.
Exercise 1.3. We show that we can deduce the risk premia from the SDF. From the properties
of covariance, conditional expectation and (1.14) in the Lecture Notes, we have
Covt (Mt+1
Mt
,Rit+1) = Et [Mt+1Rit+1Mt ] −Et [Mt+1Mt ]Et[Rit+1] = 1 − Et[Mt+1]Et[Rit+1]Mt .
On the other hand, turning our attention to the money market account, we deduce from
(1.14) in the LN and the predictable property of R0 that
Mt = Et[Mt+1R0t+1] = R0t+1Et[Mt+1].
Thus,
Covt (Mt+1
Mt
,Rit+1) = 1 − Et[Rit+1]R0t+1
which after reordering concludes the proof.
Exercise 1.4. (i) Is there any risk-less asset in this market?
Yes, note that the first entry of S1(ω1), S1(ω2), S1(ω3) are all equal to 1. Hence, it is
deterministic and thus risk-less. Its rate is R01 = 1/1 = 1.
(ii) The return of the risky assets:
R1 = [S11(ω1)/p1,S11(ω2)/p2,S11(ω3)/p3] (2)= [3
2
,
1
2
,
5
2
] (3)
i.e., R1(ω1) = 32 , R1(ω2) = 12 , R1(ω3) = 52 . And R2 = [ 97 , 57 , 107 ].
If probability is uniform, we have that the risk premia are:
E[R1] −R0 = 3
2
P[{ω1}] + 1
2
P[{ω2}] + 5
2
P[{ω3}] − 1 (4)
= 1
3
[3
2
+ 1
2
+ 5
2
] − 1 = 3
2
− 1 = 1
2
(5)
E[R2] −R0 = 9
7
P[{ω1}] + 5
7
P[{ω2}] + 10
7
P[{ω3}] − 1 (6)
= 1
3
[9
7
+ 5
7
+ 10
7
] − 1 = 8
7
− 1 = 1
7
(7)
(iii) We now check if the market model is complete and arbitrage free. This kind of problem
can be solved in several ways using the matrix properties we reviewed, or exposing explicitly
arbitrage opportunities or non-replicable profiles.
4
We follow a method based on linear algebra. Since the matrix is square, we can use the
determinant to check if the matrix MS1 is invertible. Remember that if this is the case, then
there is a unique solution to any linear system associated to it.
We obtain that
det(MS1) = det(M⊺S1) = (1)(10 − 25) − (1)(30 − 45) + (1)(15 − 9) = −15 + 15 + 6 = 6.
Hence, any linear system has a unique solution, so MS1 has full range. The market is therefore
complete.
It can also be deduced that there is at most one set of AD prices. We need to check that
they are all positive. By solving the linear system
S0 =MS1pAD
We get ⎛⎜⎝
1 1 1 1
3 1 5 2
9 5 10 7
⎞⎟⎠→
⎛⎜⎝
1 1 1 1
0 −2 2 −1
0 −4 1 −2
⎞⎟⎠→
⎛⎜⎝
1 0 0 1/2
0 1 −1 1/2
0 0 −3 0
⎞⎟⎠
Note that from the last equation, we have that pAD3 = 0. As this is the only solution, we get that
there is no strictly positive AD prices (or SDF or equivalent risk neutral probability).
This means that the market has an arbitrage.
Exercise 1.5. (a) We know that absence of arbitrage is equivalent to the existence of a positive
SDF, positive AD prices, or an equivalent risk-neutral probability. Using the latter, there is no-
arbitrage if and only if there exists Q equivalent to the original probability such that EQ[R] = R0.
Define Q{ω1} = q1, Q{ω2} = q2. We solve the system:
R0 = q1R1(ω1) + q2R1(ω2) = q1Ru + q2Rd
and
q1 + q2 = 1(from probability properties)
Moreover, from no arbitrage we require, q1 > 0, q2 > 0. Solving, we get:
q2 = Ru −R0
Ru −Rd q1 = R0 −RdRu −Rd (8)
But since Ru > Rd, then q2 > 0 and q1 > 0 if and only if Ru > R0 and R0 > Rd. Therefore, the
condition is Ru > R0 > Rd.
(b) We obtained the risk neutral probabilities as a by product of the previous exercise, as
given in (8)
(c) There are two Arrow-Debreu securities, one for each {ω1, ω2}.
pAD1 = EQ[1ω1R0 ] = q11{ω1}(ω1)R0 + q21{ω1}(ω2)R0 = q1R0 = R0 −Rd(Ru −Rd)R0 (9)
pAD2 = EQ[1ω2R0 ] = q2R0 = Ru −R0(Ru −Rd)R0 (10)
(d) If a new asset paying (S1 −K)+ is introduced, we can use the risk neutral probabilities
to find popt, the price of this asset.
5
popt = EQ[(S1 −K)+
R0
] = q1(Rup −K)+
R0
+ q2(Rdp −K)+
R0
(11)
= 1
R0(Ru −Rd)[(Rup −K)+(R0 −Rd) + (Rdp −K)+(Ru −R0)] (12)
but since Ru > R0 > Rd, and K = R0p
popt = p
R0(Ru −Rd)[(Ru −R0)(R0 −Rd)]
Exercise 1.6. (a) Arbitrage-free ⇒ law of one price. Take θ,ξ such that θ ⋅S1 = ξ ⋅S1. If there
is no-arbitrage, there exists a positive SDF. Hence, E[MSi1] = pi.
But then, by linearity:
θ ⋅ p = θ ⋅E[MS1] = E[M(θ ⋅S1)] (13)= E[M(ξ ⋅S1)] = ξ ⋅E[MS1] = ξ ⋅ p (14)
(b) Example of market with arbitrage and such that the law of one price is satisfied. Take
the market in question 2.1. We showed it admitted an arbitrage. Let θ, ξ such that θ ⋅S1 = ξ ⋅S1.
This means that
[θ0, θ1, θ2]⎡⎢⎢⎢⎢⎢⎣
1 1 1
3 1 5
9 5 10
⎤⎥⎥⎥⎥⎥⎦´udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¸udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¶
MS1
= [ξ1, ξ2, ξ3]⎡⎢⎢⎢⎢⎢⎣
1 1 1
3 1 5
9 5 10
⎤⎥⎥⎥⎥⎥⎦. (15)
Since we showed MS1 is invertible, because det(MS1) = 6. Therefore, we conclude that θ = ξ , so
trivially the law of one price holds.
Exercise 1.7. Recall that M is an SDF if for each market instrument, i = 0, ..., n, we have
Si0 = E[Si1M]. Let Mα1 = αM1 + (1 − α)M˜1 for two SDFs M1, M˜1. Then,
E[(αM1 + (1 − α)M˜1)Si1] = αE[M1Si1] + (1 − α)E[M˜1Si1] (16)= αSi0 + (1 − α)Si0 = Si0 (17)
Moreover, if M1 > 0 and M˜1 > 0, we have that for 0 ≤ α ≤ 1, αM1 > 0 and (1 − α)M˜1 > 0 and
thus Mα1 > 0.
Exercise 1.8. We show the two applications: Assume that the price for the asset satisfies
pnew = E[Snew1 M] for some strictly positive SDF M . Then, M is a strictly positive SDF for the
market with n + 2 assets, and by the FTAP-1, there is no arbitrage.
On the other hand, if pnew ≠ E[DnewM] for every M strictly positive SDF on the market
with n + 1 assets, then we will not be able to find an SDF on the market with n + 2 assets. By
the FTAP-1, this implies there exists an arbitrage opportunity.
Exercise 1.9. By Exercise 1.8, we have that p˜ = E[DnewM˜1] and pˆ = E[DnewMˆ1] for two strictly
positive SDFs M˜1, Mˆ1. But since
αp˜ + (1 − α)pˆ = E[Dnew{αM˜1 + (1 − α)Mˆ1}] (18)= E[DnewMα1 ], (19)
and Mα1 is a strictly positive SDF by Exercise 1.7. Hence, it is an arbitrage free price.
6
2 Utility functions
Exercise 2.1.
1. The investor chooses from the two alternatives the one with larger expected utility. Hence,
we need to compare the expected utility of a final wealth W1 = w0 = 200 and a final wealth
W2 = ⎧⎪⎪⎨⎪⎪⎩w0 + 100 = 300 with probability
1
2
w0 − 100 = 100 with probability 12 .
Now, note that E[W2] = 12300+ 12100 = 200 = w0 =W1, and also that uA(x) = xγγ is concave
(in fact it is a CRRA utility function). Thus, by Jensen’s inequality we have that
E[uA(W2)] ≤ uA[E[W2]] = uA(W1) = E[uA(W1)]
Therefore, the investor prefers W1.
The argument is the same for uB(x) = − exp(−x), since it is also a concave function.
2. In this case we have that
W1 = ⎧⎪⎪⎨⎪⎪⎩11 = with probability
1
3
31 = with probability 2
3
; W2 = ⎧⎪⎪⎨⎪⎪⎩21 with probability 0.91 with probability 0.1 .
We can numerically make the calculation to get, E[log(W1)] = 13 log(11)+ 23 log(31) ≈= 3.09,
while E[log(W2)] = 0.93 log(21) + 0.1 log(1) ≈= 2.74. Clearly, the investor will choose the
first investment.
Exercise 2.2. Since the functions to be considered are twice differentiable, it suffices to look at
the sign of the second derivative: we need to verify that the second derivative is negative on the
domain of definition.
1. f1(x) = log(x) for x > 0: We get
f ′(x) = 1
x
; and f ′′(x) = −1
x
< 0.
2. f2(x) = a − bx2 for b > 0: We get
f ′(x) = −2bx; and f ′′(x) = −2b < 0.
Hence, both functions are concave. Recall though that the quadratic function needs a restricted
domain of definition to be considered a utility function.
Exercise 2.3. To evaluate the expected utility, we can either use integral methods or Monte
Carlo methods.
7
# Import modules
import numpy as np
import s c ipy . s t a t s as s t
# We use two approaches , e i t h e r us ing the ’ s t a t s ’ module
# and the ’ expec t ’ method , or us ing Monte Carlo wi th a l a r g e
# number o f s imu la t i on s
# We i n i t i a l i s e the func t i on
myf = lambda x : np . l og ( x )
mc number = 1000000
# Pareto wi th shape a lpha=2 and s c a l e ( or mode) xm = 0.5
# Method 1
u = s t . pareto . expect (myf , args =(2 ,) , s c a l e =0.5)
print ( ’ Pareto , method 1 : ’ , u )
# Method 2
sample p = (np . random . pareto (2 , mc number ) + 1) ∗ 0 .5
u = myf ( sample p ) . mean ( )
print ( ’ Pareto , method 2 : ’ , u )
Pareto, method 1: -0.19314718055994637
Pareto, method 2: -0.1932406330316995
# Exponent ia l wi th lambda = 1
# Method 1
u = s t . expon . expect (myf , s c a l e =1)
print ( ’ Exponential , method 1 : ’ , u )
# Method 2
sample e = np . random . exponent i a l ( s i z e =(mc number , ) )
u = myf ( sample e ) . mean ( )
print ( ’ Exponential , method 2 : ’ , u )
Exponential, method 1: -0.5772156649008394
Exponential, method 2: -0.5804589164442893
# Log normal ( s tandard )
# Method 1
u = s t . lognorm . expect (myf , args =(1 ,))
print ( ’ Exponential , method 1 : ’ , u )
# Method 2
sample ln = np . random . lognormal ( s i z e =(mc number , ) )
u = myf ( sample ln ) . mean ( )
print ( ’ Exponential , method 2 : ’ , u )
8
Exponential, method 1: 4.844603335998414e-17
Exponential, method 2: -0.00023813580546780955
Note that there is a good agreement between the two methods.
Exercise 2.4. An affine function on R is any function g ∶ R → R of the type g(x) = ax + b for
a, b ∈ R. Clearly, if a > 0, we have that
x < y⇒ ax < ay⇒ ax + b < ay + b⇒ g(x) < g(y)
so that an affine function with a > 0 preserves the order. To show the converse, note that if a = 0,
for all x < y we have g(x) = g(y). And similarly, if a < 0, we get
x < y⇒ ax > ay⇒ ax + b > ay + b⇒ g(x) > g(y),
so that the order is reversed. Hence, the only affine functions that preserve the order are the
ones with a > 0.
Exercise 2.8. (i) We take the case u(x) = log(x). Take w initial wealth and wX additional
gamble.
E[u(w(1 +X))] = E[log(w(1 +X))] (20)= E[log(w) + log(1 +X)] (21)= log(w) +E[log(1 +X)] (22)
On the other hand,
u(E[w(1 +X)] −wη(X)) = log(w(E[1 +X] − η(X))) (23)= log(w) + log(E[1 +X] − η(X)) (24)
⇒ η(X) solves: log(E[1 +X] − η(X)) = E[log(1 +X)] (25)⇒ η(X) = E[1 +X] + exp(E[log(1 +X)]), (26)
which is independent of ω. Similar development for the case u(x) = 1
1−γx1−γ with γ ∈ (0,1)⋃(1,∞).
(ii) E[R] = E[exp(Z)] with Z ∼ N (−σ2
2
, σ).
E[R] = exp(E[Z] + 1
2
var(Z)) = exp(−σ2
2
+ σ2
2
) = 1 (27)
Since we have that u is CRRA, we can use the results of the previous exercise, and solve the
problem for initial wealth 1 If the initial wealth is w0 ≠ 1 we simply multiply by w0. Let us first
focus on the case u(x) = log(x). We get,
log(E[R] − η(R)) = E[log(R)] = E[Z] = −σ2
2
, (28)
log(1 − η(R)) = −σ2
2
⇒ η(R) = 1 − exp(−σ2
2
) > 0. (29)
9
Hence, η(w0R) = w0(1 − exp(−σ22 )).
In the case of u(x) = 1
1−γx1−γ ,
1
1 − γ (1 − η(R))1−γ = E[ 11 − γ [exp(Z)]1−γ] (30)= E[ 1
1 − γ exp((1 − γ)Z)] (31)
= 1
1 − γ exp((1 − γ)(−σ22 ) + σ22 (1 − γ)2) (32)
= 1
1 − γ exp(−σ22 (1 − γ)γ) (33)
= 1
1 − γ [exp(−σ22 γ)]1−γ (34)
⇔ (1 − η(R)) = exp(−σ2
2
γ) (35)
η(R) = 1 − exp(−σ2
2
γ) (36)
η(w0R) = w0(1 − exp(−σ2
2
γ)) (37)
We see that the solution is the same independently of the relative risk aversion coefficient.
Exercise 2.9. (i) Since W ∼ N (w,σ2w), we calculate
E[u(W )] = E[−exp(−αW )] = −exp(−αw + 1
2
α2σ2w). (38)
On the other hand,
E[u(W +D − pBID)] = E[−exp(−α(W +D − pBID))] (39)= −exp(−α(w +m − pBID) + 1
2
α2(σ2w + σ2 + 2σσwρ)) (40)
Hence, − αw + 1
2
α2σ2w = −α(w +m − pBID) + 12α2(σ2w + σ2 + 2σσwρ) (41)⇒ pBID =m − 1
2
α(σ2 + 2σσwρ) (42)
where in (40) we use the expression for the mean of the exponential of a Gaussian distributed
random variable and the fact that if the joint vector (A,B)is Gaussian,
A +B ∼ N (E[A +B], var(A +B)) (43)= N (E[A] +E[B], var(A) + var(B) + 2cov(A,B)) (44)
(ii) By similar arguments as before,
10
E[u(W −D + pask)] = −exp(−α(w −m + pask) + 1
2
α2(σ2w + σ2 − 2σσwρ)) (45)
Hence, − αw + 1
2
α2σ2w = −α(w −m + pask) + 12α2(σ2w + σ2 − 2σσwρ) (46)⇔ pask =m + 1
2
α(σ2 − 2σσwρ) (47)
Note that this simple model proposes an explanation to the fact that there is a bid-ask spread
around the average value of the asset m. This simple model captures the fact that an investor
perceives an asset to be worth more than its average price if they want to sell, and to be worth
less if they want to buy.
Exercise 2.11.
i. Following the hint, we take W such that
P[W = 2i] = 2−i; for all i = 1,2, . . . .
We easily verify
E[W ] =∞.
ii. Let us now modify the above to get the required condition. Take for example W˜ = exp(−W ),
i.e.,
P[W = e1/2i] = 2−i; for all i = 1,2, . . . .
Clearly, W˜ is always positive. Moreover,
E[log(W˜ )] = E[−W ] = −E[W ] = −∞.
11
3 Portfolio choice
One period case
Exercise 3.4. In this setting, we have only two possible assets. Let φ be the amount to be
invested in the risky asset. Note that if we set w0 − φ the amount to be invested in the risk-free
asset, the budget constraint on the initial time, is satisfied, since w0 = φ + (w0 − φ). Hence, we
can write the optimisation problem as
max
φ ∈ R ψ(θ) ∶= E[u(W (φ))]
s.t. W (φ) = (w0 − φ)R0 + φR1 (48)
In this case, we take u(x) = log(x), and we have,
E[u(W )] = 1
2
u((w0 − φ)R0 + φu) + 1
2
u((w0 − φ)R0 + φd) (49)
Using the first order conditions, we have that
d
dφ
E[u(W (φ))]∣
φ=φ∗ = 0 (50)
⇒ u′((w0 − φ∗)R0 + φ∗u)1
2
(u −R0) + u′((w0 + φ∗)R0 + φ∗d)1
2
(d −R0) = 0 (51)
⇒ u −R0(w0 − φ∗)R0 + φ∗u = R0 − d(w0 − φ∗)R0 + φ∗d (52)⇒ φ∗(u −R0)(d −R0) + (u −R0)w0R0 = φ∗(R0 − d)(u −R0) + (R0 − d)(w0R0) (53)
φ∗ = w0R0(u − 2R0 + d)
2(R0 − d)(u −R0) (54)
Compare with φ¯ = w0R0(K−1)
u−R0+K(R0−d) (Eq. 5.15 from lecture notes). Given that ρ = 1, we have
that
K = u −R0
R0 − d ⇒ φ¯ = w0R0(u −R0 −R0 − d)(u −R0)(R0 − d) + (R0 − d)(u −R0) . (55)
Thus, it follows that the expression is unchanged.
Exercise 3.5. The main difference between this problem and Exercise (3.4) is the additional
constraint on having no short positions. In terms of the variable φ (the amount invested in the
risky asset), this means that
0 ≤ φ ≤ w0. (56)
Indeed, φ cannot be negative nor be greater than the total (otherwise we would need to short
the risk-free asset). Hence, the optimisation problem now reads,
max
φ ∈ [0,w0] ψ(θ) ∶= E[u(W (φ))]
s.t. W (φ) = (w0 − φ)R0 + φR1 (57)
12
For the rest of the solution of this exercise, let us call φ¯ to the solution to the problem (48)
(i.e., the solution with allowed short positions), and let φ∗ be the solution to (57), i.e., with no
short positions. Let us remark that since [0,w0] ⊂ R, we have that
ψ(φ¯) ≥ ψ(φ¯∗).
Clearly, from this inequality we deduce that if φ¯ ∈ [0,w0], we have equality and φ¯ = φ∗. Let
us explicitly write this condition in terms of the data of the problem. From (55) we get:
0 ≤ φ¯ ≤ w0 ⇔ 0 ≤ w0R0(u − 2R0 + d)
2(R0 − d)(u −R0) ≤ w0 ⇔ 0 ≤ R0(u − 2R0 + d) ≤ 2(R0 − d)(u −R0) (58)⇔ 0 ≤ R0((u −R0) − (R0 − d)) ≤ 2(u −R0)(R0 − d) (59)
Recall that we are working under the assumption 0 < d < R0 < u. We now consider the cases
where (59) does not hold (i.e. if either the left or the right inequality fail).
• If (u −R0) < (R0 − d): we would have as optimal φ∗ = 0. To verify this, let us check the
directional derivative version of the first order conditions. We obtain
∂φE[u(W (φ))]∣φ=0 = u1(w0R0)(u −R0)
2
+ u1(w0R0)(d −R0)
2= u1(w0R0)
2
((u −R0) − (R0 − d)) < 0
Thus, ∂φE[u(W (φ))]∣φ=0(φ − 0) ≤ 0 ∀φ ∈ [0,w0]. This means that 0 is a local (and thanks
to concavity) global minimum of our problem. Note also that in this case, φ¯ < φ∗, so as a
result of not being able to short we invest more on the risky asset (or more appropriately,
we do not short it).
• If R0((u −R0) − (R0 − d)) ≥ 2(u −R0)(R0 − d) , we claim that φ∗ = w0. The verification
follows the same lines as above. In this case, φ¯∗ < φ¯, so we invest less on the risky asset (in
order not to short the risk-free one).
Exercise 3.6. The optimisation problem reads
max
φ ∈ Rn E[u(W (φ))]
s.t. w0 = φ ⋅ 1,
W (φ) = φ ⋅R1
(60)
where u(x) ∶= − exp(−αx). We can replace directly the second constraint on the utility function.
Moreover, recalling that since R1 is a Gaussian vector, (φ ⋅R1 is a Gaussian random variable
with mean φ ⋅µ and variance φ⊺Σ¯φ, it follows that
E[u(φ ⋅R1))] = E[−exp(−αφ ⋅R1)] = −exp(−αφ ⋅µ + 1
2
α2φ⊺Σ¯φ). (61)
The problem (60) then reads
max
φ ∈ Rn − exp(−αφ ⋅µ + 12α2φ⊺Σ¯φ)
s.t. w0 = φ ⋅ 1 (62)
13
The main difference of (62) with respect to section 5.5. is that the control variable is an
element of Rn, which corresponds to the fact that there is no risk-free asset. However, as before,
observing that − exp(−x) is increasing, we get that φ∗ solves (62) if and only if it also solves
max
φ ∈ Rn αφ ⋅µ − 12α2φ⊺Σ¯φ
s.t. w0 = φ ⋅ 1 (63)
We solve (63) by using the Lagrange multipliers technique1, i.e. we set
L(φ,λ) ∶= αφ ⋅µ − 1
2
α2φ⊺Σ¯φ − λ(φ ⋅ 1 −w0) (64)
and applying the first order conditions, to get
▽φL(φ∗, λ∗) = −αµ + α2Σ¯φ∗ − λ∗1 = 0 (65)
and also,
∂λL(φ∗, λ∗) = φ∗ ⋅ 1 −w0 = 0 (66)
Solving in (65) for φ∗ gives
φ∗ = 1
α2
Σ¯
−1(λ∗1 + αµ) (67)
On the other hand, multiplying (inner product) by 1 and using (66) we can write
w0 = φ∗ ⋅ 1 = ( 1
α2
Σ¯
−1(λ∗1 + αµ)) ⋅ 1 (68)
= λ∗
α2
1⊺Σ¯−11 + 1
α
1⊺Σ¯−1µ (69)
⇒ λ∗ = α2(w0 − 1α1⊺Σ¯−1µ)
1⊺Σ¯−11 = α
2w0 − α1⊺Σ¯−1µ
1⊺Σ¯−11 (70)
which gives us the value of λ∗. Replacing in (67) we conclude that
φ∗ = 1
α
Σ¯
−1
µ + ⎛⎝αw0 − 1⊺Σ¯
−1
µ
α1⊺Σ¯−11
⎞⎠ Σ¯−11 (71)
as claimed. It is interesting to compare this equation with the one in the case where a risk-free
asset is available. Note first that the optimal portfolio now does depend on the initial wealth!
The point is that because there is no risk-free investment and all the initial wealth has to be
invested, all alternatives are risky. So the small perturbations around risk-free alternatives are
not possible.
To have more insight, let us call ϕ the optimal portfolio when there is a risk-free asset. From
Section 5.5 we have that ϕ = α−1Σ¯−1µ. By replacing and reordering terms, we get
φ∗ = ϕ + (w0 − 1 ⋅ϕ) Σ¯−11
1⊺Σ¯−11
1It is possible to use the same technique as in the section 5.5., but this time is slightly more involved to write
this time
14
In other words, we are just dividing the amount that would be assigned to the risk-free asset
among the remaining ones. This is done proportionally to the sum of all covariances of each
asset.
Exercise 3.7. In this case, we find that the optimisation problem can be written
max
c0 ∈ R,φ ∈ R E[u(c0) + u(C1)]
s.t. w0 = c0 +φ ⋅ 1,
C1 = φ ⋅R
(72)
where R = [R0
Rˆ
]. Note that using the constraints, we have:
E[u(c0) + u(C1)] = u(c0) +E[u(C1)] = u(c0) +E[u(φ0R0 + φˆ ⋅ Rˆ)] (73)= u(c0) +E[u((w0 − c0 − φˆ ⋅ 1)R0 + φˆ ⋅ Rˆ)] (74)
Once again, we see that C1 = (w0 − c0 − φˆ ⋅ 1)R0 + φˆ ⋅ Rˆ is Gaussian with mean E[C1] =(w0 − c0 − φˆ ⋅ 1)R0 + φˆ ⋅µ and variance var[C1] = φˆ⊺Σ¯φ. Hence, we have that
E[u(c0) + u(C1)] = −exp(−αc0) − exp(−αE[C1] + 1
2
α2var(C1)) (75)
= −exp(−αc0) − exp(−α[(w0 − c0 − φˆ ⋅ 1)R0 + φˆ ⋅ µ] + 1
2
α2φˆ
⊺
Σ¯φˆ´udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¸udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¶
ϕ
) (76)
Let us apply the first order conditions. We get
∂c0E[u(c∗0) + u(c∗1)] = αexp(−αc∗0) − αR0exp(ϕ) = 0 (77)▽φˆE[u(c∗0) + u(c∗1)] = αexp(−αϕ)(µ − 1R0 − αΣ¯φˆ∗) = 0. (78)
Since the exponential function is always positive, from (78), we obtain
φˆ∗ = 1
α
Σ¯
−1(µ − 1R0) (79)
Let us highlight that this is exactly as in the case of the pure investment problem! The
intuition is that riskiness comes only from investment (as the consumption at initial time is
deterministic), so that this investor prioritises his decision on how to invest on the risky assets.
Also, note that if we express this problem in terms of the number of shares, we have
θi,∗ = φi,∗
Si0
= 1
αSi0
Σ¯
−1(µ − 1R0)
It remains to find c∗0 and the amount to be invested in the risk-free asset. Using (77), we
have by reordering, dividing by α and taking log that
−αc∗0 = −α[(w0 − c∗0 − φˆ∗ ⋅ 1)R0 + φˆ∗ ⋅µ] + 12α2 ˆφ∗⊺Σ¯φˆ∗] + log(R0) (80)⇒ c∗0 = 11 +R0 [(w0 − φˆ∗ ⋅ 1)R0 + φˆ∗ ⋅µ − 12α ˆφ∗⊺Σ¯φˆ∗ − 1α log(R0)] (81)
15
Noting that φˆ∗ ⋅ 1 = 1
α
1⊺Σ¯−1(µ − 1R0); φˆ∗ ⋅µ = 1
α
µ⊺Σ¯−1(µ − 1R0) and
ˆφ∗⊺Σ¯φ = 1
α2
(µ − 1R0)⊺Σ¯−1(µ − 1R0)
we finally get
c∗0 = 11 +R0 [w0R0 + 12α(µ − 1R0)⊺Σ¯−1(µ − 1R0) − 1α log(R0)].
Using that φ0,∗ = w0 − c∗0 − φˆ∗, we also get the initial amount invested in the risky-free asset.
Exercise 3.8. The problem can be written
max
θ,η
[u (ηω0) + δE [u (Ww0(1−η),φ)] = max
η
[u(ηw0) + δmax
φ
E [u (Ww0(1−η),φ)]]
By Example 3.1.3 we have that
φ∗ = w0(1 − η)R0(x − 1)
u −R0 +K (R0 − d)
solves the max problem with initial wealth w0(1 − η).
We now find the value of expected utility at this maximum. We look first at Ww0(1−eta),φ∗ .
We have that if R1 = u,
Ww0(1−η),Φ∗(ω = 1) = w0(1 − η)R0 ⎛⎝1 + (K − 1) (u −R0)u −R0 +K (R0 − d)⎞⎠
but since Kρ = u−R0
R0−d , we have
Ww0(1−η),Φ∗ = w0(1 − η)R0 (1 + K − 1
1 +K1−ρ ) .
By a similar procedure, we get that if R1 = d
Ww0(1−η),Φ∗(ω = 0) = w0(1 − η)R0 (1 + 1 −K
Kρ +K ) .
Hence,
E [u (Ww0(u−v)1φ2)] = (w0R0(1 − η))1−ρ
2(1 − ρ)
⎡⎢⎢⎢⎢⎣(K +K
1−ρ
1 +K1−ρ )1−ρ + ( Kρ + 1Kρ +K )
⎤⎥⎥⎥⎥⎦´udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¸udcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymodudcurlymod¶=C
Finally, we solve
max
η
⎡⎢⎢⎢⎢⎣(w0η)
1−ρ
1 − ρ + (w0R0(1 − η))
1−ρ
2(1 − ρ) Cδ
⎤⎥⎥⎥⎥⎦
which by a first order condition gives
16
(w0η)−ρw0 − (w0R0(1−η))−ρ2 (Cδ) (w0R0) = 0⇒ η = R0(1−η)21/ρ(Cδ)1/ρ(R0)1/ρ⇒ η = (R0) ρ−1ρ 21/ρ(Cδ)1/ρ+(R0)(ρ−1)/ρ21/ρ
Some checks are in order: note that η ∈ (0,1) as expected. Moreover, when δ → 0, η → 1.
Exercise 3.9. Since we have only one riskless asset, we have that the only uncertainty on this
case comes from the random endowment. The decision variables are c0 and φ (the amount to be
invested in the risk-free asset), and the problem reads
max
c0, φ ∈ R E[u(c0) + δu(C1)]
s.t. w0 = c0 + φ,
C1 = φR0 + Y
(82)
We can introduce both restrictions on the utility as follows: from the first restriction, we have
φ = (w0 − c0) and replacing on the end of period constraint, we get that C1 = (w0 − c0)R0 + Y .
Thus, we can solve the unconstrained problem
max
c0 ∈ R u(c0) + δE[u((ω0 − c0)R0 + Y )] (83)
which solves the first question. For the second claim, because u′′ < 0, we know that u is strictly
concave. Using first order conditions, we get
u′(c∗0) − δE[u′((ω0 − c∗0)R0 + Y )]R0 = 0. (84)
On the other hand, using u′′′ > 0, we get by Jensen’s inequality that
E[u′((ω0 − c∗0)R0 + Y )] ≥ u′(E[(ω0 − c∗0)R0 + Y ]) = u′((ω0 − c∗0)R0) (85)
where we used the fact that E[Y ] = 0. Equations (84) and (85) imply that
u′(c∗0) ≥ δu′((ω0 − c∗0)R0)R0. (86)
Assuming now that Y = 0 (denoting cˆ0 the optimal in this case), we have
u′(cˆ0) = δu′((ω0 − cˆ0)R0)R0. (87)
We claim now that (86) and (87) deduce that cˆ0 ≥ c∗0. Assume by contradiction that cˆ0 < c∗0, and
observe that the left hand of (86) is monotonously decreasing as function of c∗0 (due to concavity)
while the right-hand side is increasing with c∗0 (also due to concavity). Hence, we would get from
(87) and cˆ0 < c∗0 that u′(c∗0) < δu′((ω0 − c∗0)R0), which would contradict (86). It follows that
cˆ0 ≥ c∗0 as desired.
Multi-period case
Exercise 3.10. By definition 1.28, we have that a measure Q is risk neutral if and only if the
Radon-Nikodym density with respect to P can be written
dQ
dP
∶=MTS0T
17
for some SDF M . Hence, it follows that for any s = 0, . . . , T
EQ [Cs
S0s
] = E [MTS0T CsS0s ] = E [Es[MTS0T ]CsS0s ]
where we used the definition of the RN density and of conditional expectation, given that both
Cs, S
0
s are Fs measurable. But by the definition of SDF, Es[MTS0T ] =MsS0s . Hence,
EQ [Cs
S0s
] = E [MsCs] .
Exactly the same arguments show that
EQ [ Is
S0s
] = E [MsIs] .
Hence, the claim follows by Lemma 3.16.
Exercise 3.12. As per the question, let us call pi∗ ∈ Rn+1 the portfolio achieving the maximum
in the log-utility maximisation for one period.
Let pi ∈ Rn+1 be other non-optimal strategy (that we keep fixed), and we need to compare
St(pi) and St(pi∗) for t large enough. Let
∆ ∶= E[log(pi∗R1)] −E[log(piR1)].
By assumption, we have that 0 < ∆ <∞.
Let us now recall that for a dividend-reinvested portfolio without consumption, we have that
S0(pi) = w0; St(pi) = w0 t∏
s=1(pi⊺Rs),
Clearly the strict monotonicity of log implies that
St(pi) < St(pi∗) if and only if 1
t
log(St(pi)) < 1
t
log(St(pi∗)).
So we focus in showing a statement on the right-hand side. Indeed, note that
1
t
log(St(pi)) = log(w0)
t
+ 1
t
t∑
s=1 log(pi⊺Rs);
and, similarly for pi∗. Thus,
1
t
log(St(pi∗)) − 1
t
log(St(pi)) = 1
t
t∑
s=1{log((pi∗)⊺Rs) − log(pi⊺Rs)}. (88)
Now, the law of large numbers implies that 1
t
log(St(pi∗)) → E[log(St(pi∗))] almost surely (and
similarly for pi). Their difference converges almost surely then to ∆. Hence, there exists T > 0
such that for all t > T ,
P(∣1
t
log(St(pi∗)) − 1
t
log(St(pi)) −∆∣ ≤ ∆
2
) = 1
From where we deduce that
P(1
t
log(St(pi∗)) − 1
t
log(St(pi)) ≥ ∆
2
> 0) = 1
for all t > T , as wanted.
18

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