QUANTITATIVE METHODS FOR FINANCE

January-2014 Exam

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

1 [8 points] Assume dX = �k (X �m) dt +X�dz with k > 0 and m > 0. Given m < 5

and � = 0, show that the equilibrium price S (X) of the stock that pays out 3 (X + 5) dt every �second�

is such that

E [ S (X) ] >6m

r.

[7 points] Your initial capital is H = 100 Euro. Assume dX = X�dt + X�dz (the associated

parabola is y ( ) = 12�2 2 +

��� ���� 1

2�2� � r ). By borrowing 500 Euro and by cashing in 400

Euro via the short sale of the stocks that pay out X12dt, you invest 1000 Euro in the stock that pays

out 3X15dt every �second�(assume y

�12

�< 0, which implies y

�15

�< 0). Work out the total return

1HdH on your portfolio.

2 [4 points] Consider the problem of maximizing the risk-adjusted expected return on

a portfolio with constrained exposure to the second risky security (assume A > 0, r2 > r1 > r,

�2 > �1 > 0, and �1 < � < 0):

maxw1; w2

r + w1 ( r1 � r ) + w2 ( r2 � r ) �A

2

��21w

21 + 2��1�2w1w2 + �22w

22

�sub w2 =

1

2.

The shadow price l� of the constraint is:

a) l� = (r1 � r)� ��2�1 (r2 � r)�A2�22 (1� �2);

b) l� = (r2 � r)� ��2 (r1 � r)� A2�21 (1� �2);

c) l� = (r2 � r) + ��2�1 (r1 � r)�A2�22 (�

2 � 1);

d) l� = (r2 � r)� ��2�1 (r1 � r)�A2�22 (1� �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 = 1000� 23Y � 4

3X ; y = 1000� 4

3Y � 2

3X :

Given that the production costs are C (x; y) = 15x+ 10y + xy + 5000 and that the government sets

the production-target constraint (y � 300)2 � 100, the shadow price l� of the constraint is:

a) l� = 911;

b) l� = 914;

c) l� = 98;

d) l� = 92.

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

rate (r = 0):

M =

26664�1:0 �3: 1 �3: 7 �1: 9 �1: 11 + 0 2 7 3 1

1 + 0 3 4 2 0

1 + 0 4 1 1 2

37775 .

The no-arbitrage price of the payo¤ eX (1) = 7B(1) + 3 eS3 (1)2 is:a) 21: 9;

b) 19: 9;

c) 15: 9;

d) 17: 9.

5 [4 points] Given the one-period market (with riskfree rate r = 0)

M =

26664�1 �1:51 + 0 2

1 + 0 0

1 + 0 4

37775and the payo¤

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

iT=h0 1 2

iT, the maximum-in�ow

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

c) #l1 = �1=4;d) #l1 = 1=4.

Alessandro Sbuelz - SBFA, Catholic University of Milan 2

SOLUTIONS

1 Given � = 0, the equilibrium-valuation problem is

1

dtEt [dS] + 3X + 15 = Sr , 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 + 15 = (BX + C) r

m

Bkm+ 15� Cr| {z }= 0

= (B (r + k)� 3)| {z }= 0

X

m

B =3

r + k,

C =3m

r

k

r + k+15

r.

Alessandro Sbuelz - SBFA, Catholic University of Milan 3

Then, we have

E [S (X)] =r

r

3E [X]

r + k+

3m

r

k

r + k+15

r

=3m

r

r

r + k+

3m

r

k

r + k+15

r

=3m

r

r + k

r + k+

15

r|{z}> 3m

r

>3m

r+

3m

r.

The equilibrium value of the stock that pays out X12dt every �second�is

G (X) =X

12

r + 12����

�12�� 1

8�2�

( it solves the problem1

dtEt [dG] +X

12 = Gr +GXX��� with G (0) = 0 ) :

The equilibrium value of the stock that pays out 3X15dt every �second�is

Alessandro Sbuelz - SBFA, Catholic University of Milan 4

F (X) =3X

15

r + 15����

�15�� 2

25�2�

( it solves the problem1

dtEt [dF ] + 3X

15 = Fr + FXX��� with F (0) = 0 ) :

The total gain on your portfolio is

dH =1000

F

�dF + 3X

15dt

�� 400

G

�dG + X

12dt

�� 500rdt

= 1000

dF + 3X

15dt

F

!� 400

dG + X

12dt

G

!� 500rdt

= 1000

��r +

1

5���

�dt +

1

5�dz

�� 400

��r +

1

2���

�dt +

1

2�dz

�� 500rdt

= ( Hr + 0 ��� ) dt + 0 �dz .

The total return on your portfolio is

1

HdH = r dt .

Alessandro Sbuelz - SBFA, Catholic University of Milan 5

SOLUTIONS

2 The correct answer is d).

For this equality-constrained problem, the Lagrangian function is

L (w1; w2; l) = r + w1 (r1 � r) + w2 (r2 � r) � A

2

�w21�

21 + w

22�

22 + 2��1�2w1w2

�� l

�w2 �

1

2

�

and the First Order Conditions (FOCs) are su¢ cient. The objective function is strictly concave in w1and w2. The constraint function is convex as it is linear in w2. The FOCs are:

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

L w1 = 0

L w2 = 0

L l = 0

,

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

�Aw1�21 � Aw2��2�1 + (r1 � r) = 0

�Aw2�22 � Aw1��1�2 � l + (r2 � r) = 0

��w2 � 1

2

�= 0

Given the explicit form of the third equation (w�2 =12), the �rst two equations can be rewritten as

�"

A�21 0

A��1�2 1

#"w1

l

#=

"� (r1 � r) + 1

2A��1�2

� (r2 � r) + 12A�22

#with det

"A�21 0

A��1�2 1

#!= A�21 > 0 .

Hence,

"w�1l�

#=

1

A�21

"1 0

�A��1�2 A�21

# "(r1 � r)� 1

2A��1�2

(r2 � r)� 12A�22

#,

that is,

w�1 =r1 � rA�21

� �2

�2�1

and l� = (r2 � r)� ��2�1(r1 � r)�

A

2�22�1� �2

�.

Alessandro Sbuelz - SBFA, Catholic University of Milan 6

SOLUTIONS

3 The correct answer is d).

The inverse demand functions are "X = 1

2y � x+ 500

Y = 12x� y + 500

#so that the monopolist�s problem is

maxx;yP (x; y) s.t. (x� 300)2 � 100 � 0

with

P (x; y) = x

�1

2y � x+ 500

�+ y

�1

2x� y + 500

�� (15x+ 10y + xy + 5000) :

The First Order Conditions for constrained optimality will be su¢ cient because the feasible set

�(x; y) 2 R2 : (x� 300)2 � 100

=

�(x; y) 2 R2 : 290 � x � 310

is convex and the pro�t function P (x; y) is strictly concave:

H =

264 Pxx Pxy

Pyx Pyy

375 =264 �2 0

0 �2

375 with Pxx = �2 < 0 and det (H) = 4 > 0 :

Given the Lagrangian function

L (x; y; l) = P (x; y)� l�(y � 300)2 � 100

�,

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

Lx = 0

Ly = 0

l � 0Ll � 0

l � Ll = 0

,

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

485� 2x = 0600l � 2y � 2ly + 490 = 0

l � 0��(y � 300)2 � 100

�� 0

�l�(y � 300)2 � 100

�= 0

.

Alessandro Sbuelz - SBFA, Catholic University of Milan 7

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

8><>:485� 2x = 0

�2y + 490 = 0,

8><>:x = 485

2

y = 245

.

The unconstrained maximum-pro�t point is such that P�881063; 874063

�= 113831: 25. It turns out to be

unfeasible as the constraint is violated:

(245� 300)2 � 100 :

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

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

485� 2x = 0

600l � 2y � 2ly + 490 = 0

(y � 300)2 � 100 = 0 (the constr. is binding)

,

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

x = 4852

l = �y�245y�300

x =

(290

310with l > 0

,

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

x = 4852

l = 92> 0

y = 290 .

The constrained maximum pro�t is

P

�485

2; 290

�= 111806: 25 .

Alessandro Sbuelz - SBFA, Catholic University of Milan 8

SOLUTIONS

4 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):

266666641:0

3: 1

3: 7

1: 9

1: 1

37777775 =1

1 + 0

264 1 + 0 2 7 3 1

1 + 0 3 4 2 0

1 + 0 4 1 1 2

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

Q (!3)

375 .

Since

det

0B@264 1 + 0 3 1

1 + 0 2 0

1 + 0 1 2

3751CA = �3 ,

we can focus on the riskless security and on the two last risky securities to work out the unique measure

Q:

264 Q (!1)Q (!2)

Q (!3)

375 =

0BB@264 1 + 0 3 1

1 + 0 2 0

1 + 0 1 2

375T1CCA�10B@(1 + 0)

264 1:0

1: 9

1: 1

3751CA =

264 0:30:30:4

375 .

The other two risky securities are also properly priced:

3: 1 =1

1 + 0

264 234

375T 264 0:30:3

0:4

375 ,

3: 7 =1

1 + 0

264 741

375T 264 0:30:3

0:4

375 ,

Alessandro Sbuelz - SBFA, Catholic University of Milan 9

The payo¤ to be priced is

eX (1) = 7B(1) + 3 eS3 (1)2m

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

X (1) (!3)

375 = 7

264 111

375+ 3264 322212

375 =

264 341910

375 .

Its no-arbitrage price is

X (0) =1

1 + 0

264 341910

375T 264 0:30:3

0:4

375 = 19: 9 .

An alternative would be the calculation of the intial cost of the replicating strategy #X that involves

only the three mentioned securities (#X1 = #X2 = 0):

264 #X0

#X3#X4

375 =

264 1 + 0 3 1

1 + 0 2 0

1 + 0 1 2

375�1 264 3419

10

375 =

264 �7132

375and

V#X (0) =

26666664�70

0

13

2

37777775

T 266666641:0

3: 1

3: 7

1: 9

1: 1

37777775 = 19: 9 .

Alessandro Sbuelz - SBFA, Catholic University of Milan 10

SOLUTIONS

5 The correct answer is b).

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

replicable:

det

0B@264 1 + 0 2 0

1 + 0 0 1

1 + 0 4 2

3751CA = � 5 .

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

264 1 + 0 2

1 + 0 0

1 + 0 4

375# � �264 012

375 ()

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

1 � #0 + 0 � #1 � �1 (not to the left of the yellow line)

1 � #0 + 4 � #1 � �2 (not below the blue line)

:

­3 ­2 ­1 1 2 3

­2

­1

1

2

theta_0

theta_1

super­

replic

ating

 ­X(1)

f = 0.7 f = 0

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 � 1:5 ( = � V#(0) ).

Alessandro Sbuelz - SBFA, Catholic University of Milan 11

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

�#0 � 1� #1 � 1:5 = f () #1 = �f

1:5� #0 �

1

1:5(see the red solid lines below).

­2 ­1 1

­1

1

theta_0

theta_1

super­

replic

ating

 ­X(1)

f = 0.25 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 � #l0 + 2 � #l1 = 01 � #l0 + 0 � #l1 = �1

to obtain the seeked strategy

"#l0#l1

#=

"�112

#

and the associated maximum in�ow

�V#l(0) = � (�1) � 1��1

2

�� 1:5 = 0:25 .

Alessandro Sbuelz - SBFA, Catholic University of Milan 12