Mock Exam 4

QUANTITATIVE METHODS FOR FINANCE

Mock Exam 4 (Academic Year 2013-14)

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

1 Consider a stock that pays out the dividend (3X + 2m) dt every �second� (with dX =

�k (X �m) dt+X�dz).

[8 points] Show that the equilibrium price S (X) of the stock has the following dynamics (ex-

pressed in total-gain form):

dS + (3X + 2m) dt =

�Sr +

3X

r + ��+ k��

�dt +

3X

r + ��+ k�dz .

[7 points] Show that, if �� > 0, the following statement holds true:

E [ S (X) ] <5m

r.

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

maxw

Ehlog� fW � i

sub w � 150% , fW = 100 ( (1 + r) + w (er � r) ) ;

where

r = 2% ; er =

(+60% with probability p 2 (0; 1)�20% with probability 1� p

.

The risk premium is E [ er ]� r = 26%. The shadow price l� of the portfolio constraint is:

a) 5: 659 075 22� 10�2;b) 8: 659 075 22� 10�2;c) 3: 659 075 22� 10�2;d) 1: 659 075 22� 10�2.

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

y

#=

"400� 0:8Y � 1:2X400� 1:2Y � 0:8X

#.

The production costs are C (x; y) = 2x + 3y + xy + 150. The government imposes the constraint

Y � 60 (y�s sale price must not exceed 60). The shadow price l� associated to the constraint is:

a) l� = 101: 777 778;

b) l� = 107: 777 778;

c) l� = 120: 777 778;

d) l� = 89: 777 778.

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

rate (r = 0):

M =

26664�1:0 �0:9 �1:01 3 2

1 0 0

1 0 1

37775 .

The no-arbitrage price of a European put option written on the risky security 2 (the strike price is 2)

is:

a) 1:00;

b) 2:00;

c) 1:25;

d) 0:75.

5 [4 points] Given the one-period market

M =

26664�1 �0:351:02 1

1:02 0

1:02 0

37775and the payo¤

hX (1) (!1) X (1) (!2) X (1) (!3)

iT=h2 0 1

iT, the maximum-in�ow

strategy #l that super-replicates � eX (1) is such that:a) #l0 = �2=1:02;b) #l0 = �1=1:02;c) #l1 = �1;d) #l1 = �2.


SOLUTIONS

1 The equilibrium-valuation problem is

1

dtEt [dS] + 3X + 2m = Sr + SXX�� , where

1

dtEt [dS] = SX (�k (X �m)) +

1

2SXXX

2�2 .

Let us formulate the educated guess

S (X) = BX + C ,

where B and C are constants to be determined. Given

SX = B ,

SXX = 0 ,

the dynamic equilibrium restriction becomes

B (�k (X �m)) + 3X + 2m = (BX + C) r +BX��

m

Bkm+ 2m� Cr| {z }= 0

= (B (r + ��+ k)� 3)| {z }= 0

X

m

B =3

r + ��+ k,

C =3m

r

k

r + ��+ k+2m

r.


The total gain on the stock is

dS + (3X + 2m) dt =

�SX (�k (X �m)) +

1

2SXXX

2�2 + 3X + 2m

�dt + SXX�dz

= ( Sr + SXX�� ) dt + SXX�dz .

=

�Sr +

3X

r + ��+ k��

�dt +

3X

r + ��+ k�dz .

If �� > 0, we have

E [S (X)] =r

r

3E [X]

r + ��+ k+

3m

r

k

r + ��+ k+2m

r

=3m

r

r

r + ��+ k+

3m

r

k

r + ��+ k+2m

r

=3m

r

r + k

r + ��+ k+2m

r

<5m

r.


SOLUTIONS

2 The correct answer is a).

The probability mass p implied by the risk premium is 60%:

0:60p� 0:20 (1� p)� 0:02 = 0:26 =) p = 60% .

The investor�s expected utility is

Ehlog� fW � i

= 0:6 ln (58w + 102) + 0:4 ln (102� 22w)

and the Lagrangian function is

L (w; l) = Ehlog� fW � i

� l ( w � 1:5 ) .

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

= 0:458

58w + 102+ 0:6

�22102� 22w

=�319w + 663

2601 + 918w � 319w2 = 0 () w =663

319� 1:5 (unfeasible) .

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

8><>:Lw =

�319w+6632601+918w�319w2 � l = 0

Ll = �w + 1:5 = 0 (the constraint is binding)

()

8><>:l� = 5: 659 075 22� 10�2 > 0

w� = 1:5 .


SOLUTIONS

3 The correct answer is b).

The inverse demand functions are "X = 200 + y � 1:5xY = 200 + x� 1:5y

#so that the monopolist�s problem is

maxx;yP (x; y) s.t. 200 + x� 1:5y| {z }

y�s sale price as a function of outputs

� 60 � 0 ,

with

P (x; y) = x (200 + y � 1:5x) + y (200� 1:5y + x)� (2x+ 3y + xy + 150) :

The First Order Conditions for constrained optimality will be su¢ cient because the constraint function

is linear (the feasible set f(x; y) 2 R2 : 200 + x� 1:5y � 60g is convex) and the pro�t function P (x; y)is strictly concave:

H =

264 Pxx Pxy

Pyx Pyy

375 =264 �3 1

1 �3

375 with Pxx = �3 < 0 and det (H) = 8 > 0 :

The corresponding Lagrangian function is

L (x; y; l) = P (x; y)� l (200� 1:5y + x� 60)

and the Kuhn-Tucker First Order Conditions are are:8>>>>>>>>>>><>>>>>>>>>>>:

Lx = 0

Ly = 0

l � 0Ll � 0

l � Ll = 0

,

8>>>>>>>>>>><>>>>>>>>>>>:

y � 3x� l + 198 = 01: 5l + x� 3y + 197 = 0

l � 0� (200� 1:5y + x� 60) � 0

�l (200� 1:5y + x� 60) = 0 .


For l = 0 (we assume a painless constraint), we have:

8><>:Lx (x; y; 0) = Px = 0

Ly (x; y; 0) = Py = 0

,

8><>:x = 98: 875

y = 98: 625

.

Such a solution corresponds to the unconstrained maximum-pro�t point (P (98: 875; 98: 625) = 19353:

187 5). The constraint is violated:

Y (98: 875; 98: 625) = 150: 937 5 � 60 :

For l > 0 (we assume a painful constraint), we have8>>>>>><>>>>>>:

Lx = 0

Ly = 0

Ll = 0 (the constraint is binding)

,

8>>>>>><>>>>>>:

x� = 78: 666 666 7

y� = 145: 777 778

l� = 107: 777 778

.

The constrained maximum pro�t is

P (x�; y�) = 14452: 666 6 .


SOLUTIONS

4 The correct answer is a).

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 1:00:91:0

375 =1

1 + 0

264 1 + 0 3 2

1 + 0 0 0

1 + 0 0 1

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

Q (!3)

375 .

Since

det

0B@264 1 3 2

1 0 0

1 0 1

3751CA = � 3 ,

the unique measure Q is:

264 Q (!1)Q (!2)

Q (!3)

375 =

0BB@264 1 3 2

1 0 0

1 0 1

375T1CCA�10B@(1 + 0)

264 1:00:91:0

3751CA =

264 0:30:30:4

375

with

0BB@264 1 3 2

1 0 0

1 0 1

375T1CCA�1

=

0B@264 1 1 1

3 0 0

2 0 1

3751CA�1

=1

�3

264 0 �3 0

�1 �1 2

0 3 �3

375| {z }matrix of cofactors

T

.


The payo¤ to be priced is

eX (1) = max�2� eS2 (1) ; 0 �

m

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

X (1) (!3)

375 =

264 max ( 2� 2 ; 0 )max ( 2� 0 ; 0 )max ( 2� 1 ; 0 )

375 =

264 021

375 .

Its no-arbitrage price is

X (0) =1

1 + 0

264 021

375T 264 0:30:3

0:4

375 = 1 .

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

264 #X0

#X1#X2

375 =

264 1 3 2

1 0 0

1 0 1

375�1 264 02

1

375 =1

�3

264 0 �1 0

�3 �1 3

0 2 �3

375| {z }matrix of cofactors

T 264 021

375 =

264 2

0

�1

375

and

V#X (0) =

264 2

0

�1

375T 264 1:00:9

1:0

375 = 1:0 .


SOLUTIONS

5 The correct answer is d).

The market is obviously incomplete (2 securities and 3 states of the world). The payo¤ is not

replicable:

det

0B@264 1:02 1 2

1:02 0 0

1:02 0 1

3751CA = � 1: 02 .

The candidate super-replicating strategies # = [#0; #1]T solve the system

264 1:02 1

1:02 0

1:02 0

375# � �264 201

375 ()

8><>:1:02 � #0 + 1 � #1 � �2 (not below the green line)

1:02 � #0 + 0 � #1 � 0 (to the right of the yellow line)

1:02 � #0 + 0 � #1 � �1 (to the right of the blue line)

:

3 2 1 1 2 3

5

4

3

2

1

1

2

theta_0

theta_1

super

replic

ating

X(1)

We determine the maximum-in�ow strategy among all those that super-replicate� eX (1) by studyingthe initial-in�ow function

f#(0) = �#0 � 1� #1 � 0:35 ( = � V#(0) ).

The level curve f of such a function is identi�ed by the straight line of equation

�#0 � #1 � 0:35 = f () #1 = �f

0:35� #0 �

1

0:35(see the red solid lines below).


3 2 1 1 2 3

5

4

3

2

1

1

2

theta_0

theta_1

super

replic

ating

X(1)

f = 0.7 f = 0

The maximum-in�ow strategy is given by the couple [#l0; #l1]T represented by the intersection between

the green line and the yellow line. Hence, we must solve the system

(1:02 � #l0 + 1 � #l1 = �21:02 � #l0 + 0 � #l1 = 0

to obtain the seeked strategy

"#l0#l1

#=

"0

�2

#.


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Mock Exam 4