QUANTITATIVE METHODS FOR FINANCE

Mock Exam 6 (Academic Year 2013-14)

[5 exercises; 31 points available; 90 minutes available]

1 Consider a stock that pays out the dividendX3dt every �second�(with dX = X�dt+X�dz).

[8 points] Show that, under the condition

y (3) < 0 with y ( ) =1

2�2 2 +

��� ���� 1

2�2� � r ,

the equilibrium price F (X) of the stock has the following dynamics (expressed in total-return form):

dF + X3dt

F= (r + 3���) dt + 3�dz .

[7 points] Your initial capital is H = 100 Euro. You invest 50 Euro in the stock and 50 Euro in

the riskfree asset. Work out the per-annum expected total return 1dt1HEt [dH] on your portfolio.

2 [4 points] Consider a constrained log-utility investor whose problem is

maxw

Ehlog� fW � i

sub w � 100% , fW = 100 ( (1 + r) + w (er � r) ) ;

where

r = 1% ; er =

(+40% with probability 1

2

�20% with probability 12

.

The shadow price l� of the portfolio constraint is:

a) 4: 035 714 29� 10�3;b) 8: 035 714 29� 10�3;c) 12: 035 714 29� 10�3;d) 1: 035 714 29� 10�3.

Alessandro Sbuelz - SBFA, Catholic University of Milan 1

3 [4 points] A �rm produces two outputs x and y, whose sale prices are X and Y , respec-

tively. The �rm is monopolist in both markets and faces the following demand functions (x and y are

complementary goods):

x = 150� 13Y � 2

3X ; y = 150� 2

3Y � 1

3X :

Given that the production costs are C (x; y) = x+ 2y + 2xy + 300, the �rm�s maximum pro�t is:

a) 6218: 25;

b) 4255: 125;

c) 5775: 25;

d) 5213: 125.

4 [4 points] Consider the following one-period arbitrage-free market with a zero riskfree

rate (r = 0):

M =

26664�1:0 �0:5 �0:751 0 0

1 1 1

1 0 1

37775 .

The no-arbitrage price of the payo¤ eX (1) = 2B(1) + eS2 (1)2 + 5 eS1 (1)2 is:

a) 8: 25;

b) 7: 50;

c) 5: 25;

d) 9: 15.

5 [4 points] Consider the following one-period market with a zero riskfree rate (r = 0):

M =

26666664�1:0 �5

8�38

1 0 0

1 1 1

1 0 1

1 2 0

37777775 .

The following holds true:

a) the market is not arbitrage-free;

b) Q (!2) =58� 2q with q 2

�18; 516

�;

c) Q (!3) =58� q with q 2

�18; 516

�;

d) Q (!1) =58.

Alessandro Sbuelz - SBFA, Catholic University of Milan 2

SOLUTIONS

1 We are dealing with a �power�stock that pays out X3dt every �second�(with dX = X�dt+

X�dz). The equilibrium-valuation problem is

1

dtEt [dF ] + X3 = Fr + FXX��� and F (0) = 0 , where

1

dtEt [dF ] = FXX�+

1

2FXXX

2�2 .

Let us formulate the educated guess

F (X) = AX3 (the boundary condition F (0) = 0 is met by construction) ,

where A is a constant to be determined (we want it positive to support a non-negative stock price).

Given

FX = 3AX2 ,

FXX = 6AX ,

the dynamic equilibrium restriction becomes

3AX3�+1

26AX3�2 +X3 = AX3r + 3AX3���

m

3A�+ 3A�2 + 1 = Ar + 3A���

m

A = 1 =�r + 3����

�3�+ 3�2

� �.

Alessandro Sbuelz - SBFA, Catholic University of Milan 3

Under the condition

y (3) = ��r + 3����

�3�+ 3�2

� �< 0 with y ( ) =

1

2�2 2 +

��� ���� 1

2�2� � r ,

the non-negativity of the equilibrium stock price is granted.

Since

dF + X3dt =

�FXX�+

1

2FXXX

2�2 + X3

�dt + FXX�dz

m

dF + X3dt = ( Fr + FXX��� ) dt + FXX�dz ,

the total return on the �power�stock is

dF +X3dt

F=

�r +

FXFX���

�dt+

FXFX�dz ,

with the elasticity being

FXFX = 3 .

The per-annum expected total gain is

1

dtEt [dH] =

50

F

�1

dtEt [dF ] + X3

�+ 50r

= 50 (r + 3���) + 50r .

Hence, the per-annum expected total return is

1

dt

1

HEt [dH] = r +

150

100��� .

Alessandro Sbuelz - SBFA, Catholic University of Milan 4

SOLUTIONS

2 The correct answer is b).

The investor�s expected utility is

Ehlog� fW � i

= 0:5 ln (39w + 101) + 0:5 ln (101� 21w)

and the Lagrangian function is

L (w; l) = Ehlog� fW � i

� l ( w � 1 ) .

The Kuhn-Tucker First Order Conditions are:8>>>>>>>><>>>>>>>>:

L w = 0

l � 0

L l � 0

l � L l = 0 .

If l = 0 (we assume a painless constraint), the F.O.C.s become

d

dwEhlog� fW � i

= 0:539

39w + 101+ 0:5

�21101� 21w

=819w � 909

(39w + 101) (21w � 101) = 0 () w =909

819� 1 (unfeasible) .

If l > 0 (we assume a painful constraint), the F.O.C.s become

8><>:Lw =

819w�909(39w+101)(21w�101) � l = 0

Ll = �w + 1 = 0 (the constraint is binding)

()

8><>:l� = 8: 035 714 29� 10�3 > 0

w� = 1 .

Alessandro Sbuelz - SBFA, Catholic University of Milan 5

SOLUTIONS

3 The correct answer is d).

The inverse demand functions are "X = 150 + y � 2xY = 150� 2y + x

#

so that the monopolist�s problem is

maxx;yP (x; y)

with

P (x; y) = x (150 + y � 2x) + y (150� 2y + x)� (x+ 2y + 2xy + 300) :

The First Order Conditions are:

8><>:Px = �4x+ 149 = 0

Py = �4y + 148 = 0,

8><>:x = 37:25

y = 37

.

The pro�t function P (x; y) is strictly concave, as the Hessian matrix is negative de�nite everywhere:

H =

264 Pxx Pxy

Pyx Pyy

375 =264 �4 0

0 �4

375 with Pxx = �4 < 0 and det (H) = 16 > 0 :

Hence, the maximum pro�t is

P (37:25; 37) = 5213: 125 .

Alessandro Sbuelz - SBFA, Catholic University of Milan 6

SOLUTIONS

4 The correct answer is c).

By the First Fundamental Theorem of Asset Pricing, any arbitrage opportunity is ruled out if the

market M supports a risk-neutral probability measure Q (recall that the riskfree rate is r = 0):

264 1:0

0:5

0:75

375 =1

1 + 0

264 1 + 0 0 0

1 + 0 1 1

1 + 0 0 1

375T 264 Q (!1)Q (!2)

Q (!3)

375 .

Since

det

0B@264 1 0 0

1 1 1

1 0 1

3751CA = 1 ,

the unique measure Q is:

264 Q (!1)Q (!2)

Q (!3)

375 =

0BB@264 1 0 0

1 1 1

1 0 1

375T1CCA�10B@(1 + 0)

264 1:0

0:5

0:75

3751CA =

264 0:250:50:25

375

with

0BB@264 1 0 0

1 1 1

1 0 1

375T1CCA�1

=

0B@264 1 1 1

0 1 0

0 1 1

3751CA�1

=1

1

264 1 0 0

0 1 �1�1 0 1

375| {z }matrix of cofactors

T

.

Alessandro Sbuelz - SBFA, Catholic University of Milan 7

The payo¤ to be priced is

eX (1) = 2B(1) + eS2 (1)2 + 5 eS1 (1)2m

264 X (1) (!1)X (1) (!2)

X (1) (!3)

375 =

264 222

375 +

264 021212

375 +

264 5 � 025 � 12

5 � 02

375 =

264 283

375 .

Its no-arbitrage price is

X (0) =1

1 + 0

264 283

375T 264 0:250:5

0:25

375 = 5: 25 .

An alternative would be the calculation of the intial cost of the unique replicating strategy #X :

264 #X0

#X1#X2

375 =

264 1 0 0

1 1 1

1 0 1

375�1 264 28

3

375 =1

1

264 1 0 �10 1 0

0 �1 1

375| {z }matrix of cofactors

T 264 283

375 =

264 251

375

and

V#X (0) =

264 251

375T 264 1:0

0:5

0:75

375 = 5: 25 .

Alessandro Sbuelz - SBFA, Catholic University of Milan 8

SOLUTIONS

5 The correct answer is b).

By the First Fundamental Theorem of Asset Pricing, any arbitrage opportunity is ruled out if the

market M supports a risk-neutral probability measure Q (recall that the riskfree rate is r = 0):

264 1:05838

375 =1

1 + 0

266641 + 0 0 0

1 + 0 1 1

1 + 0 0 1

1 + 0 2 0

37775T 26664

Q (!1)

Q (!2)

Q (!3)

Q (!4)

37775 .

Let�s �x Q (!4) = q. Since

det

0B@264 1 0 0

1 1 1

1 0 1

3751CA = 1 ,

the candidate measure Q must be such that:

264 Q (!1)Q (!2)

Q (!3)

375 =

0BB@264 1 0 0

1 1 1

1 0 1

375T1CCA�1 0B@(1 + 0)

264 1:05838

375� q264 120

3751CA .

=

264 1 0 �10 1 0

0 �1 1

375264 1� q

58� 2q38

375

=

26458� q

58� 2q

2q � 14

375 .

Alessandro Sbuelz - SBFA, Catholic University of Milan 9

By imposing the positivity of the probability masses, we have

8>>><>>>:58� q > 0

58� 2q > 0

2q � 14> 0

q > 0

() q 2�1

8;5

16

�.

Hence, the market M is arbitrage-free and, by the Second Fundamental Theorem of Asset Pricing,

Q�s multiplicity implies M�s incompleteness.

Alessandro Sbuelz - SBFA, Catholic University of Milan 10