MTH6120 Further Topics in Mathematical Finance
Coursework 7
Exercise 1. A function is called strictly concave if in the inequalities in the definition of concavity
are strict inequalities (i.e. < rather than only ≤). So for example a twice differentiable function is
strictly concave if u00 < 0.
Are the following utility functions strictly concave?
1. u(x) = 17 + x + log(x + a)
2. u(x) = 3 − e−bx
√
3. u(x) = x + c
4. u(x) =
1−x1−q
q−1
Give reasons. Distinguish, if necessary, between positive, negative, and vanishing values of the
parameters a, b, c, q.
Also find the domain of these functions. I.e. for what x ∈ R these functions are defined.
Solution:
1. The set of x ∈ R for which the function is defined is {x ∈ R|x + a > 0}, i.e. the domain is
the interval (−a, +∞).
For x in the domain we calculate the derivatives of u:
u0 (x) = 1 +
1
x+a
and
u00 (x) = −
1
.
(x + a)2
As u00 is strictly negative we see that u(x) is strictly concave for all a and x in the domain.
2. The domain is all of R as the function is defined for all real numbers x.
The derivatives are:
u0 (x) = be−bx
and
u00 (x) = −b2 e−bx
Again as u00 (x) ≤ 0 we see that u(x) is concave for all values of b and all x. It is strictly
concave if b 6= 0.
3. The function is defined on the set {x ∈ R|x + c ≥ 0}. I.e. the domain is [−c, +∞).
Its derivatives are:
1
u0 (x) = √
2 x+c
and
−1
u00 (x) = p
4 (x + c)3
These derivatives are defined for x ∈ (−c, ∞) and here the second derivative is strictly negative
for all c. Therefore u is strictly concave for all c.
4. If q = 1 then the formula above is undefined. Otherwise u(x) is defined for x > 0 and its
derivatives are:
u0 (x) = x−q
and
u00 (x) = −qx−q−1
This will be strictly negative if q > 0.
[advanced optional add-on] We could extend the definition of u to q = 1 by taking a limit:
1 − x1−q
log(x)x1−q
= lim
= log(x)
q→1 q − 1
q→1
1
u(x) = lim
where we have used L’Hospital rule and the fact that the derivative of x1−q = exp(log(x)(1 −
q)) with respect to q is − log(x) exp(log(x)(1 − q)) = − log(x)x1−q .
So for q = 1 we obtain the known valid utility function u(x) = log(x). We can check again
that this is strictly concave:
1
1
u0 (x) =
and
u00 (x) = − 2 < 0
x
x
Exercise 2. An investor with capital x can invest any amount αx, where 0 ≤ α ≤ 1. If the amount
αx is invested he may either
√ receive 3αx with probability p or 0 with probability 1−p. The investor’s
utility function is u(x) = x. No interest is paid. How much money should be invested?
Calculate the exact value in the case p = q.
Solution: The payoff is
(
(1 − α)x + 3αx = (1 + 2α)x
Payoff =
(1 − α)x + 0αx = (1 − α)x
with probability p
with probability q = 1 − p
and the expected utility of the payoff is
f (α) = p
p
p
(1 + 2α)x + q (1 − α)x
To find the number α that maximizes this we take derivatives
√
−x
2x
1
−1
0
+q p
= x p√
f (α) = p p
+q √
=
2 1−α
1 + 2α
2 (1 + 2α)x
2 (1 − α)x
" √
#
√
√ 2p 1 − α − q 1 + 2α
p
=
x
.
2 (1 + 2α)(1 − α)
This will be zero iff
√
√
2p 1 − α = q 1 + 2α
If we square both sides of this equality we obtain
4p2 (1 − α) = q 2 (1 + 2α)
This is a linear equation in α with solution
α=
4p2 − q 2
4p2 + 2q 2
To check this is really a maximum we need to evaluate f 00 at this value. But
"
#
√
−2
−1
f 00 (α) = x p p
+q p
2 (1 + 2α)3
4 (1 − α)3
which is clearly negative. Therefore the value of α found maximizes
In the case p = q = 1/2 the value obtained is
2
2
4 12 − 21
1 − 14
3/4
α=
=
=
=
2
2
1
3/2
1+ 2
4 12 + 2 12
the expected utility.
1
.
2
Exercise 3. Let share prices Si (t), i = 1, . . . , n evolve according to independent geometric Brownian motion with parameters µi and σi . Let the period under consideration be [0, T ]. Determine
1. the rate of return Ri in this period (note this will be a random variable related to the random
variable Si (T )).
2. The expectation of the rate of return, ri .
3. The variance of the rate of return.
4. Cov(Ri , Rj ) for i 6= j.
5. Var(log(Ri + 1)).
Solution:
1. The rate of return is
Ri =
Si (T )
Si (T ) − Si (0)
=
−1
Si (0)
Si (0)
2. The expectation of the rate of return is
E (Ri ) =
E (Si (T ))
−1
Si (0)
so we just need to calculate the expectation of spot, Si (T ). But this was done in Lecture 1.
1
2
E (Si (T )) = Si (0)eµi T + 2 σi T
If we plug this in equation above we get
1
2
E (Ri ) = eµi T + 2 σi T − 1
3. The variance of the rate of return is
Var (Ri ) =
Var (Si (T ))
Si (0)2
Where we have used the fact that Var(X + a) = Var(X) and Var(aX) = a2 Var(X) for a
random variable X and a constant a.
But in exercise 1.1 we calculated the variance of GBM process to be:
2
2
Var (Si (T )) = Si (0)2 e2µi T e2σi T − eσi T
Combining these equalities yields
2
2
Var (Ri ) = e2µi T e2σi T − eσi T
4. Cov(Ri , Rj ) = 0 for i 6= j as Ri and Rj are independent random variables.
5. To calculate Var(log(Ri + 1)) we start calculating
Si (T )
Si (T )
log(Ri + 1) = log
− 1 + 1 = log
.
Si (0)
Si (0)
But now recall that a Geometric Brownian Motion with drift µ and volatility σ is by definition
S(t) = S(0) exp (µt + σW (t))
so that
log(Ri + 1) = log
Si (T )
Si (0)
= µi T + σi W (T ).
And then:
Var(log(Ri + 1)) = Var(σi W (T )) = σi2 T.