QMF Lectures 2013-14

QUANTITATIVE METHODS FOR FINANCE

[email protected]

Room 221, 2nd oor, Via Necchi 9

Phone: +39-02-72342345

O�ce hours: 14:30-16:30 on Tuesday

Alessandro Sbuelz - SBFA, Catholic University of Milan - A.Y. 2013-14 1

TEXTBOOK

I Lecture notes.

I A. BATTAUZ - F. ORTU, Arbitrage Theory in Discrete and Continuous

Time (Cod. 8444), Edizioni EGEA, last edition (available at Libreria EGEA, Via

Bocconi, 8).

I K. SYDSAETER - P. J. HAMMOND, Mathematics for economic analysis,

Prentice-Hall, 1995.


THE COURSE MATERIAL IS ON BLACKBOARD (BB)

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ation on how to enroll in BB courses.

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I For any problem, contact [email protected].


COURSE OUTLINE

I No-arbitrage pricing of assets replicated by existing traded securities.

I No-arbitrage pricing of unreplicable assets.

I Constrained optimization with applications to �nance/economics.

I Equilibrium valuation of assets.


LINEAR ALGEBRA (REVIEW)


I IS THE NEUTRAL ELEMENT OF THE MULTIPLICATION

I The n�n identity matrix I has 1s down the main diagonal and 0s elsewhere.The 2� 2 identity matrix, for example, is

I =

"1 00 1

#.

"1 00 1

# "3 4 05 2 2

#=

"3 4 05 2 2

#.

264 5 18 31 0

375 "1 00 1

#=

264 5 18 31 0

375 .


TRANSPOSITION AND MULTIPLICATION

I If you transpose a matrix product you multiply the transposed matrices inreverse order:

0BBBBBBBB@

Az }| {"1 2 01 1 1

#bz }| {264 581

375

1CCCCCCCCA

T

=

"2114

#T

m

bTz }| {h5 8 1

iATz }| {264 1 12 10 1

375 =h21 14

i


INVERSE AND INVERTIBLE MATRICES

I If it exists, the inverse of an n�n matrix C is an n�n matrix C�1 satisfying

CC�1 = I and C�1C = I :

I For example,

Cz }| {"1 21 3

#C�1z }| {"3 �2�1 1

#=

"1 00 1

#

and

C�1z }| {"3 �2�1 1

#Cz }| {"1 21 3

#=

"1 00 1

#:

I An n� n matrix C is invertible if its inverse matrix C�1 exists.


DETERMINANTS OF SQUARE MATRICES 1

I C is invertible if and only if (i�) it is non-singular, that is, its determinant

is non-zero.

det

"1 21 3

#!= 1 � (�1)1+1 � det (3)| {z }

the 1-1 cofactor

+

1 � (�1)2+1 � det (2)| {z }the 2-1 cofactor

= 1 � 3 + 1 � (�2) = 1 .


DETERMINANTS OF SQUARE MATRICES 2

det

0B@264 1 2 32 3 23 3 2

3751CA = 1 � (�1)1+1 � det

"3 23 2

#!| {z }

the 1-1 cofactor

+

2 � (�1)2+1 � det "

2 33 2

#!| {z }

the 2-1 cofactor

+

3 � (�1)3+1 � det "

2 33 2

#!| {z }

the 3-1 cofactor

= 1 � (0) + 2 � 5 + 3 � (�5) = �5


CONSTRUCTING THE INVERSE MATRIX 1

"1 21 3

#�1=

1

det

"1 21 3

#!| {z }

=1

matrix of cofactors of

"1 21 3

# !T



matrix of cofactors of

"1 21 3

#=

266666666664

(�1)1+1 � det (3)| {z }the 1-1 cofactor




377777777775

=

264 3 �1

�2 1

375



"1 21 3

#�1=

1

1

"3 �1�2 1

#T

=

"3 �2�1 1

#


A FORMULA FOR INVERTING 2� 2 MATRICES

I Given ad� cb 6= 0, consider the square matrix

"a bc d

#.

I Its inverse is

1

ad� cb

"d �b�c a

#.


INVERTING A 3� 3 MATRIX

I Show that

266666641 2 3

2 3 2

3 3 2

37777775

�1

=1

�5

266666640 5 �5

2 �7 4

�3 3 �1

37777775 .


SYSTEMS OF LINEAR EQUATIONS (REVIEW)


A FIRST EXAMPLE 1

I Consider the following linear system in the unknowns x1, x2, and x3 :

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

x1 + 2x2 + 3x3

2x1 + 3x2 + 2x3

3x1 + 3x2 + 2x3

=

=

=

6

�1

0

I We can rewrite it as follows:

264 1 2 32 3 23 3 2

375264 x1x2x3

375 =

264 6�10

375


A FIRST EXAMPLE 2

I We can also rewrite the system as a `constrained' linear combination:

x1

264 123

375+ x2264 233

375+ x3264 322

375 =

264 6�10

375


A FIRST EXAMPLE 3

I The system admits a solution i� the vector

264 6�10

375can be seen as a linear combination of the vectors

264 123

375 ,

264 233

375 , and

264 322

375

with proper weights x�1, x�2, and x

�3.


A FIRST EXAMPLE 4

I This can happen i� (Rouch�e-Capelli Theorem)

the number of independent columns of

264 1 2 32 3 23 3 2

375

equals

the number of independent columns of

264 1 2 3 62 3 2 �13 3 2 0

375| {z }complete matrix

.


A FIRST EXAMPLE 5

I The rank of a matrix measures how many indipendent columns the matrix

has.

I The rank of a matrix is the highest order of non-singular square submatrices.

264 1 2 32 3 23 3 2

375 is non-singular so that

264 1 2 3 62 3 2 �13 3 2 0

375| {z }complete matrix

has maximum rank (3) .


A FIRST EXAMPLE 6

I We can work out the solution x�1, x�2, and x

�3 by means of the Cramer's

Theorem:

x�1 = det

0B@264 6 2 3�1 3 20 3 2

3751CA = (�5) =

�5�5

= 1 ;

x�2 = det

0B@264 1 6 32 �1 23 0 2

3751CA = (�5) =

19

�5= �3:8 ;

x�3 = det

0B@264 1 2 62 3 �13 3 0

3751CA = (�5) =

�21�5

= 4:2 :


A FIRST EXAMPLE 7

I We can work out x�1, x�2, and x

�3 also by means of inversion (given non-

singularity):

264 1 2 32 3 23 3 2

375�1

�

264 1 2 32 3 23 3 2

375 �264 x�1x�2x�3

375 =

264 1 2 32 3 23 3 2

375�1

�

264 6�10

375

m

264 1 0 00 1 00 0 1

375 �264 x�1x�2x�3

375 =

264 1 2 32 3 23 3 2

375�1

�

264 6�10

375 .


A FIRST EXAMPLE 8

26666664x�1

x�2

x�3

37777775 =1

�5

266666640 5 �5

2 �7 4

�3 3 �1

37777775 �266666646

�1

0

37777775 =

266666641

�195215

37777775


A SECOND EXAMPLE 1

I Consider the following linear system in the unknowns x1, x2, x3 and x4 :

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

x1 � x2 + 2x3 � x4

2x1 � x2 � x3 + 2x4

�x1 + 2x2 + 2x3 + x4

=

=

=

1

0

1

I Let's express it as follows:

x1

264 12�1

375+ x2264 �1�12

375+ x3264 2�12

375+ x4264 �121

375 =

264 101

375 .


A SECOND EXAMPLE 2

I The �rst three vectors are independent:

det

0B@264 1 �1 22 �1 �1�1 2 2

3751CA = 9


A SECOND EXAMPLE 3

I If the fourth vector is brought on the right-hand side,

x1

264 12�1

375+ x2264 �1�12

375+ x3264 2�12

375 =

264 101

375� x4264 �121

375 ,

the solution (with one degree of freedom) can be written as

26666664x�1

x�2

x�3

37777775 =266666641 �1 2

2 �1 �1

�1 2 2

37777775

�10BBBBBB@

266666641

0

1

37777775� x426666664�1

2

1

37777775

1CCCCCCA =

26666664

13 �

53x4

29 �

169 x4

49x4 +

49

37777775 .


A THIRD EXAMPLE 1

I Consider the following linear system in the unknowns x1, x2, and x3 :

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

2x1 + 7x2 + x3

5x1 + 6x2 + 8x3

9x2

7x1 + 6x2 + 7x3

=

=

=

=

15

3

27

�3

I Let's state it as follows:

x1

266642507

37775+ x2266647696

37775+ x3266641807

37775 =

2666415327�3

37775 .


A THIRD EXAMPLE 2

I The �rst three vectors are independent because

det

0B@264 2 7 15 6 80 9 0

3751CA = �99 .

I The last vector linearly depends on the �rst three vectors as

det

0BBB@266642 7 1 155 6 8 30 9 0 277 6 7 �3

377751CCCA = 0 .

I Hence, the system admits solution.


A THIRD EXAMPLE 3

I We can work out x�1, x�2, and x

�3 by focusing on the �rst three equations

(given non-singularity):

x�1

264 250

375+ x�2264 769

375+ x�3264 180

375 =

264 15327

375m

264 x�1x�2x�3

375 =

264 2 7 15 6 80 9 0

375�1 264 153

27

375 =

264 �330

375 .

I The last equation is met by construction:

x�1 � 7 + x�2 � 6 + x�3 � 7 = �21 + 18 + 0 = � 3 .


ONE-PERIOD FINANCIAL MARKETS


TIMING

I Investors face two trading dates only, namely t = 0 and t = 1.

I At time t = 0, investors choose their investment strategy.

I At time t = 1 they receive the liquidation value of their strategy.

I At time 0, investors can choose among N + 1 securities, for which we use

the index j, with j = 0; : : : ; N .


THE RISKLESS SECURITY

I The security with j = 0 represents a riskless security.

I B(0) � 1 denotes the time-0 price of the riskless security.

I B(1) � 1 + r denotes its time-1 price.

I The quantity r is the risk-free rate, with r > 0.


THE RISKY SECURITIES

I For the N securities with j > 0, Sj(0) denotes their time-0 price.

I eSj(1) is a random variable that denotes their time-1 price, with j = 1; : : : ; N .


UNCERTAINTY

I By time 1, the market uncertainty will resolve in one of K possible states

of the world.

I !k indicates the generic k-th state of the world at time 1.

I The !k's are relevant economic/�nancial scenarios.

I indicates the set of all states of the world, i.e. = f!1; : : : ; !Kg.

I Sj(1)(!k) indicates the time-1 price of the j-th security in scenario !k.


THE PAYOFF MATRIX M

I The payo� matrix M has K + 1 rows and N + 1 columns.

I Each column j of M represents the out ows/in ows from buying 1 unit of

the j-th security (j = 0; 1; :::; N):

M �

26666666666666664

�1 �S1(0) �S2(0) � � � �SN(0)

1 + r S1(1)(!1) S2(1)(!1) � � � SN(1)(!1)

1 + r S1(1)(!2) S2(1)(!2) � � � SN(1)(!2)

... ... ... ...

1 + r S1(1)(!K) S2(1)(!K) � � � SN(1)(!K)

37777777777777775


EXEMPLIFYING A MARKET WITH 3 SECURITIES AND 3 STATES 1

I Consider a one-period market with N = 2 and K = 3.

I Assume that B(1) = 1:05, so that r = 5%.

I The time-0 prices of the risky securities are S1(0) = 0:5 and S2(0) = 2:5.

I The time-1 prices of the risky securities are

S1(1)(!1) = 0

S1(1)(!2) = 2

S1(1)(!3) = 0

S2(1)(!1) = 0

S2(1)(!2) = 0

S2(1)(!3) = 5


EXEMPLIFYING A MARKET WITH 3 SECURITIES AND 3 STATES 2

I The payo� matrix is then the following 4� 3 matrix:

M =

266666666664

�1 �0:5 �2:5

1:05 0 0

1:05 2 0

1:05 0 5

377777777775


INVESTMENT STRATEGIES 1

I #0 is the position in the risk-free security.

I #1; #2 are the positions in the two risky securities.

I The positions represent the units of each security bought, or sold short, at

time 0.



#1 = 3 (buying 3 units of security 1)

initial out ow: 3 (�0:5) = �1:5

�nal in ows: 3

266666640

2

0

37777775 =

266666640

6

0

37777775


INVESTMENT STRATEGIES 3 (LENDING)

#0 = 10 (buying 10 units of the riskless security)

initial out ow: 10 (�1) = �10

�nal in ows: 10

266666641:05

1:05

1:05

37777775 =

2666666410:5

10:5

10:5

37777775



#2 = �2 (selling short 2 units of security 2)

initial in ow: �2 (�2:5) = 5

�nal out ows: �2

266666640

0

5

37777775 =

266666640

0

�10

37777775


INVESTMENT STRATEGIES 5 (BORROWING)

#0 = �7 (selling short 7 units of the riskless security)

initial in ow: �7 (�1) = 7

�nal out ows: �7

266666641:05

1:05

1:05

37777775 =

26666664�7: 35

�7: 35

�7: 35

37777775



I The generic investment strategy is a (column) vector # 2 R3:

# =

0B@ #0#1#2

1CA :

I Let's look at the following strategy #�:

#� =

0BBBBBB@4

5

2

1CCCCCCA =0BBBBBB@buying 4 units of the riskless security

buying 5 units of security 1

buying 2 units of security 2

1CCCCCCA :



I The market is competitive: no one has price impact.

I The market is frictionless:

B no indivisibilities (#is are not only integers);

B no short-selling constraint (#is can be negative);

B no trading limits (#is are unbounded);

B no margin requirements;

B no bid-ask spreads;

B no taxation on capital gains.


THE INITIAL COST OF AN INVESTMENT STRATEGY

I At time 0, the quantity V# (0) is the cost an investor must face to set up

the investment strategy #.

I Let's assess the initial cost V#� (0) for the strategy #�:

#� =

0BBBBBB@4

5

2

1CCCCCCA :

I The corresponding initial out ow is �V#� (0) :


THE FINAL PROCEEDS OF AN INVESTMENT STRATEGY

I The �nal proceed V# (1) (!k) is the time-1 cash ow in the state !k from

liquidating the investment strategy #.

I Let's assess the �nal proceeds of the strategy #�.


THE STRATEGY FOR GIVEN FINAL PROCEEDS

I Let's �nd the strategy # whose �nal proceeds are:

26666664V# (1) (!1)

V# (1) (!2)

V# (1) (!3)

37777775 =

266666640:5

0:5

0:5

37777775 :


BACK-ENGINEERING STRATEGIES

I Find the strategy # whose initial cost and �nal proceeds are:

V# (0) = 0;

V# (1) (!2) = 4;

V# (1) (!3) = 2.


`FREE LUNCHES'?

I They are possible if there are:

I violations of the law of one price;

I arbitrage opportunities of the �rst type;

I arbitrage opportunities of the second type.


VIOLATIONS OF THE LAW OF ONE PRICE

Two strategies have di�erent costs at time 0

despite

providing the same �nal proceeds in any state at time 1.

I In our market M there are no such violations.


ARBITRAGE OPPORTUNITIES OF THE FIRST TYPE

There is a strategy with non-positive initial costthat

never yields negative �nal proceeds

while

granting strictly-positive �nal proceedsin

at least one of the states at time 1:

I In our market M there are no such opportunities.

I For example, let's work out the initial cost of a strategy that pays o� nothing

but 1 cent in the state !3:


ARBITRAGE OPPORTUNITIES OF THE SECOND TYPE

There is a strategy with strictly-negative initial cost

that

never yields negative �nal proceeds.

I In our market M there are no such opportunities.

I For example, let's calculate the �nal proceed prevailing in the state !3 of a

strategy that costs �10 cents and pays o� nothing in the states !1 and !2.


`FREE LUNCHES' ARE IMPLAUSIBLE

I Any non-satiated investor (in the sense that she always prefers more wealth

to less) would exploit these situations.

I There would be no equilibrium.


THE NO-ARBITRAGE ASSUMPTION

I A market satis�es the no-arbitrage condition if it does not permit

I violations of the law of one price;

I arbitrage opportunities of the �rst type;

I arbitrage opportunities of the second type.

I Our market

M =

26664�1 �0:5 �2:51:05 0 01:05 2 01:05 0 5

37775is arbitrage-free.


THE RISK-NEUTRAL PROBABILITY MEASURE Q 1

I A risk-neutral probability measure Q is made ofK probability masses Q (!k)

such that:

Q (!k) > 0 for any k = 1; ::::;K .

1 The masses are strictly-positive .


THE RISK-NEUTRAL PROBABILITY MEASURE Q 2A

Q (!1) + :::+Q (!K) = 1 .

2 The masses sum to 1 .


THE RISK-NEUTRAL PROBABILITY MEASURE Q 2B

B (0) =1

1 + rEQ [B (1)]

=1

1 + r[Q (!1)B (1) + :::+Q (!K)B (1)]

= 1 .

2 The masses sum to 1 .

I For the riskless security, the price is the discounted Q-expectation of the

payo�.


THE RISK-NEUTRAL PROBABILITY MEASURE Q 3

Sj (0) =1

1 + rEQ

h eSj (1)i

=1

1 + r

hQ (!1)Sj (1) (!1) + :::+Q (!K)Sj (1) (!K)

i, j = 1; :::; N .

3 Prices are discounted Q-expectations of payo�s.


WHY RISK-NEUTRAL? 1

I The statement

Sj (0) =1

1 + rEQ

h eSj (1)i , j = 1; :::; N ,

is equivalent to the statement

EQ" eSj (1)� Sj (0)

Sj (0)

#= r, j = 1; :::; N .

I Q-expected returns equal the risk-free rate.


WHY RISK-NEUTRAL? 2

I UnderQ, the initial cost of a strategy # equals the discountedQ-expectationof its �nal proceed:

V# (0) =1

1 + rEQ

h eV# (1)i .

I The Q-expected return from # equals the risk-free rate.


THE FIRST FUNDAMENTAL THEOREM OF ASSET PRICING

I The following statements are equivalent:

I the no-arbitrage condition holds;

I a risk-neutral probability measure Q exists.


OUR ARBITRAGE-FREE MARKET ADMITS A Q

I The risk-neutral probability masses are:

26666664Q (!1)

Q (!2)

Q (!3)

37777775 =

266666640:212 5

0:262 5

0:525

37777775 .


A LONG FORWARD CONTRACT ON SECURITY 2

I A long forward contract is an agreement to buy the risky security at date 1

at a pre-speci�ed price (the delivery price E = 1).

I The payo� of the contract is

26666664f (1) (!1)

f (1) (!2)

f (1) (!3)

37777775 =

26666664S2 (1) (!1)� E

S2 (1) (!2)� E

S2 (1) (!3)� E

37777775 =

26666664�1

�1

4

37777775 .


NO-ARBITRAGE PRICING OF THE LONG FORWARD 1

I The unique strategy that replicates the long forward payo� ef (1) is

#f =

264 �0:952 380 9501

375 .

I The no-arbitrage price of the long forward equals the initial cost of the

replicating strategy #f :

V#f (0) = 1:547 619 1 .


NO-ARBITRAGE PRICING OF THE LONG FORWARD 2

I The no-arbitrage price of the long forward is

f (0) =1

1 + rEQ

h ef (1)i

=1

1 + rEQ

h eS2 (1)� Ei

= 1:547 619 1 .


NO-ARBITRAGE PRICING OF A EUROPEAN CALL

I A European call option on the risky security 2 with strike price E = 1 is the

right to buy the risky security at date 1 at the pre-speci�ed strike price.

I Find its no-arbitrage price c (0).


NO-ARBITRAGE PRICING OF A EUROPEAN PUT

I A European put option on the risky security 2 with strike price E = 1 is the

right to sell the risky security at date 1 at the pre-speci�ed strike price.

I Find its no-arbitrage price p (0).


THE PUT-CALL PARITY

c (0) � p (0) = f (0)

= S2 (0)�E

1 + r.


MARKET COMPLETENESS


COMPLETE MARKETS 1

I A payo� fX (1) is

replicable

if there exists a strategy # such that

V# (1) (!k) = X (1) (!k) for all k = 1; : : : ;K:


COMPLETE MARKETS 2

I Our market M is

complete

if every possible payo� is replicable, that is, if there is a # such that

266666641:05 0 0

1:05 2 0

1:05 0 5

37777775

26666664#0

#1

#2

37777775 =

26666664X (1) (!1)

X (1) (!2)

X (1) (!3)

37777775

for any X (1) (!1), X (1) (!2), and X (1) (!3):


COMPLETE MARKETS 3

I M is complete:

26666664#0

#1

#2

37777775 =

266666641:05 0 0

1:05 2 0

1:05 0 5

37777775

�1 26666664X (1) (!1)

X (1) (!2)

X (1) (!3)

37777775 .


THE SECOND FUNDAMENTAL THEOREM OF ASSET PRICING

I The following statements are equivalent:

I Both no-arbitrage and market completeness hold;

I There exists one, and only one, risk-neutral probability Q.


A UNIQUE Q FOR OUR COMPLETE M 1

I There is a unique triplet of strictly-positive masses Q (!1), Q (!2), andQ (!3) that meets the following constraints:

discounted Q�expected �nal pricesz }| {

11+0:05

266666641:05 0 0

1:05 2 0

1:05 0 5

37777775

T 26666664Q (!1)

Q (!2)

Q (!3)

37777775=

initial pricesz }| {266666641

0:5

2: 5

37777775.


A UNIQUE Q FOR OUR COMPLETE M 2

I The unique triplet is

26666664Q (!1)

Q (!2)

Q (!3)

37777775 = 1:05

0BBBBBBB@

266666641:05 0 0

1:05 2 0

1:05 0 5

37777775

T1CCCCCCCA

�1 266666641

0:5

2: 5

37777775

=

266666640:212 5

0:262 5

0:525

37777775 .


AN INCOMPLETE MARKET 1

I The market

M 0 =

266666666664

�1 �0:5

1:02 0

1:02 2

1:02 0:10

377777777775, N + 1 = 2, K = 3,

is incomplete.


AN INCOMPLETE MARKET 2

I Indeed, the payo�

26666664X (1) (!1)

X (1) (!2)

X (1) (!3)

37777775 =

266666641

1:5

2

37777775

cannot be replicated:

det

0BBBBBB@

266666641:02 0 1

1:02 2 1:5

1:02 0:10 2

37777775

1CCCCCCA = 1: 989 .


OUR INCOMPLETE M 0 IS ARBITRAGE-FREE 1

I M 0 supports many risk-neutral probability measures:

discounted Q-expected �nal pricesz }| {

11+0:02

264 1: 02 1: 02 1: 02

0 2 0:1

37526666664Q (!1)

Q (!2)

Q (!3)

37777775=

intial pricesz }| {264 1

0:5

375

m

264 1: 020

375Q (!1) +264 1: 02

2

375Q (!2) +264 1: 020:10

375Q (!3) = 1:02

264 1

0:5

375



264 1: 020

375Q (!1) +264 1: 02

2

375Q (!2) = 1:02

264 1

0:5

375 �

264 1: 020:10

375 qz }| {Q (!3)

m

264 1: 02 1: 02

0 2

375264 Q (!1)Q (!2)

375 =

264 1: 02� 1:02q0:51� 0:10q

375m

264 Q (!1)Q (!2)

375 =

264 1: 02 1: 02

0 2

375�1 264 1: 02� 1:02q

0:51� 0:10q

375 .



I We have

264 Q (!1)Q (!2)

375 =

266411:02 �12

0 12

3775264 1: 02� 1:02q0:51� 0:10q

375 =

264 0:745 � 0:95q0:255 � 0:05q

375

with the strict-positivity constraints

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

Q (!1) = 0:745 � 0:95q > 0

Q (!2) = 0:255 � 0:05q > 0

Q (!3) = q > 0

() q 2 (0; 0:784 210 53) .

.


NO-ARBITRAGE PRICING OF AN UNREPLICABLE PAYOFF 1

I The no-arbitrage prices of the unreplicable payo�

26666664X (1) (!1)

X (1) (!2)

X (1) (!3)

37777775 =

266666641

1:5

2

37777775are

1

1 + rEQ

hfX (1)i=

1

1:02

266666641 � (0:745 � 0:95q) +

1:5 � (0:255 � 0:05q) +

2 � q

37777775= 0:95588235q + 1:1053922 .


NO-ARBITRAGE PRICING OF AN UNREPLICABLE PAYOFF 2

I The no-arbitrage prices range from

infQ

1

1 + rEQ

hfX (1)i= inf

q2(0;0:784 210 53)[0:955 882 35q + 1:1053922] = 1:1053922

to

supQ

1

1 + rEQ

hfX (1)i= sup

q2(0;0:784 210 53)[0:955 882 35q + 1:1053922] = 1:8550052 .


WHERE DOES THE UPPER BOUND COME FROM?

1: 855 005 2 = supQ

1

1 + rEQ

hfX (1)i

= V#u (0)

where

#u = argmin#

V# (0) subject to V# (1) (!k) � X (1) (!k) in each state k:

I #u denotes a minimum-cost super-replicating strategy.

I V#u (0) is the minimum cost of super-replicating fX (1) .


MINIMUM-COST SUPER-REPLICATION OF fX (1) 1

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

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

1:02 � #0 + 0 � #1 � 1

1:02 � #0 + 2 � #1 � 1:5

1:02 � #0 + 0:10 � #1 � 2

;

the solution of which can be found graphically.



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

r1 : 1:02 � #0 + 0 � #1 = 1

r2 : 1:02 � #0 + 2 � #1 = 1:5

r3 : 1:02 � #0 + 0:10 � #1 = 2

;

5 4 3 2 1 1 2 3 4 5 6 7 8 9 10

4

2

2

4

6

8

10

12

14

theta_0

theta_1

r_1 r_3

r_2



5 4 3 2 1 1 2 3 4 5 6 7 8 9 10

4

2

2

4

6

8

10

12

14

theta_0

theta_1

r_1 r_3

(1.5 , 8)

r_2sup

errep

licati

ng X

(1)

I The strategy [#0; #1]T = [1:5; 8]T super-replicates fX (1) .



I The initial-cost function of any strategy # is

V#(0) = #0 + #1 � 0:5 .

I Its level curve c is the straight line #1 =c0:5 �

#00:5 :

5 4 3 2 1 1 2 3 4 5 6

5

4

3

2

1

1

2

3

4

5

6

7

8

theta_0

theta_1

r_1 r_3

r_2

super

replic

ating

X(1)

c = 1.855

c = 0



I The minimum-cost super-replicating strategy [#u0 ; #u1 ]T is the point of in-

tersection between r2 and r3:

I Hence, we must solve the system

8><>:1:02 � #u0 + 2 � #u1 = 1:5

1:02 � #u0 + 0:10 � #u1 = 2

to obtain

264 #u0#u1

375 =

264 1: 986 58

�0:263 158

375 .


WHERE DOES THE LOWER BOUND COME FROM?

1: 105 392 2 = infQ

1

1 + rEQ

hfX (1)i

= � V#l(0)

where

#l = argmax#

� V# (0) subject to V# (1) (!k) � �X (1) (!k) in each state k:

I #l denotes a maximum-in ow strategy that generates a liability lighter than�fX (1).

I �V#l(0) is the maximum in ow from super-replicating �fX (1) .


MAXIMUM-INFLOW SUPER-REPLICATION OF �fX (1) 18>>>>>><>>>>>>:

h1 : 1:02 � #0 + 0 � #1 = �1

h2 : 1:02 � #0 + 2 � #1 = �1:5

h3 : 1:02 � #0 + 0:10 � #1 = �2

.

3.0 2.5 2.0 1.5 1.0 0.5 0.5 1.0 1.5 2.0 2.5 3.0

5

4

3

2

1

1

2

3

4

5

theta_0

theta_1

h_1

h_2

h_3

super

replic

ating

X(1)

I The strategy [#0; #1]T = [�0:25;�0:5]T super-replicates �fX (1) .


MAXIMUM-INFLOW SUPER-REPLICATION OF �fX (1) 2

I The initial-in ow function of any strategy # is

f#(0) = �#0 � #1 � 0:5 ( = � V#(0) ) .

I Its level curve f is the straight line #1 = � f0:5 �

#00:5 :

3.0 2.5 2.0 1.5 1.0 0.5 0.5 1.0 1.5 2.0 2.5 3.0

5

4

3

2

1

1

2

3

4

5

theta_0

theta_1

h_1

h_2

h_3

super

replic

ating

X(1)

f = 1.10539

f = 0


MAXIMUM-INFLOW SUPER-REPLICATION OF �fX (1) 3

I The liability-type strategy [#l0; #l1]T is the point of intersection between h1

and h2:

I Hence, we must solve the system

8><>:1:02 � #l0 + 0 � #l1 = �1

1:02 � #l0 + 2 � #l1 = �1:5

to obtain

264 #l0#l1

375 =

264 �0:980 392�0:25

375 .


REPLICABLE PAYOFFS ADMIT A UNIQUE NO-ARBITRAGE PRICE

I Consider the following replicable payo�:26666664X (1) (!1)

X (1) (!2)

X (1) (!3)

37777775 =

266666642: 04

4: 04

2: 14

37777775 .

I Its no-arbitrage price does not depend on q :

1

1 + rEQ

hfX (1)i

=1

1:02

266666640:745 � 0:95q

0:255 � 0:05q

q

37777775

T 266666642: 04

4: 04

2: 14

37777775 = 2: 5 .


OPTIMIZATION: AN INTRODUCTION


WEALTH ALLOCATION AND FINAL WEALTH

I An investor allocates her initial wealth of 100 between a riskless security

( its net return is r ) and a risky security ( its net return is er ).

I The proportion of initial wealth allocated to the risky security is w .

I The �nal wealth achieved with the allocation w is

fW = 100 ( 1� w ) (1 + r) + 100 w (1 + er)

= 100 ( (1 + r) + w (er � r) ) .


PORTFOLIO RETURNS AND BORROWING

I The portfolio based on the allocation w yields a net return of

fW � 100100

= r + w (er � r) .

I If the investor invests more than her initial wealth in the risky security

(w > 1), she needs to borrow:

1� w| {z }proportion allocated to the riskless security

< 0 .


THE RETURN ON THE RISKLESS SECURITY

I Assume for now that

r = 0% .

I Hence, a purely riskless allocation (w = 0) implies

fW = 100 with probability 1 .


THE RETURN ON THE RISKY SECURITY

er =

8><>:g = 30% with probability mass p

b = �10% with probability mass 1� p

I If p = 0:25 , the expectation is

E [ er ] = 0% .

I Hence, a purely risky allocation (w = 1) implies

Eh fW i

= 100 .


THE INVESTOR IS RISK AVERSE 1

I If the investor were to pursue a purely riskless allocation (w = 0), she would

get the following utility out of the �nal wealth fW = 100 (with probability 1)

U ( 100 ) = log ( 100 ) .



I If the investor were to pursue a purely risky allocation (w = 1), she would

get the following expected utility out of the �nal wealth fW = 100 ( 1 + er ):

E [ log ( 100 (1 + er) ) ] = 0:25 � log ( 130 ) + 0:75 � log ( 90 )

< log ( 0:25 � 130 + 0:75 � 90 )

= log ( 100 ) .



70 75 80 85 90 95 100 105 110 115 120 125 130 135 140 145 1504.50

4.52

4.54

4.56

4.58

4.60

4.62

4.64

4.66

4.68

4.70

4.72

4.74

4.76

4.78

4.80

4.82

4.84

4.86

wealth W

(expected) utility

log ( W

)E [ l

og ( W

) ]

log ( 100 )

p = 0

p = 1

p = 0.25


THE RETURN ON THE RISKLESS SECURITY

I Assume

r = 2% .


A POSITIVE RISK PREMIUM ON THE RISKY SECURITY

I Consider

p = 0:5

so that the risk premium is

E [ er ] � r = 10% � 2% = 8% .

I The standard deviation of the risky return is

std [ er ] = 20% .


OPTIMAL ALLOCATION WITHOUT PORTFOLIO CONSTRAINTS 1

I Recall that the �nal wealth is

fW = 100 ( ( 1 + r ) + w ( er � r ) ) .

I The optimal allocation is w� = argmaxw

Ehlog

� fW � i.



w� = 242:86% ()

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

ddw E

hlog

� fW � i> 0 for w < w�

ddw E

hlog

� fW � i��w = w�

= 0 (stationarity)

ddw E

hlog

� fW � i< 0 for w > w�



I Recall that log x is a real number only if x > 0 .

I w must be such that the �nal wealth fW is always strictly positive:

8>>><>>>:100 ( (1 + r) + w (g � r) ) = 102 + w28 > 0

100 ( (1 + r) + w (b� r) ) = 102� w12 > 0

() w 2�� 5114;17

2

�.



d

dwEhlog

� fW � i� 0

m

1

2� 1

102 + w28� 28 +

1

2� 1

102� w12� ( � 12 ) � 0



1

2� 1

102 + w28� 28 +

1

2� 1

102� w12� ( � 12 ) � 0

m

14 ( 102 � w 12 ) � 6 ( 102 + w 28)

( 102 + w 28) ( 102 � w 12 )� 0

m

816 � 336 w

( 102 + w 28) ( 102 � w 12 )� 0



816 � 336w

( 102 + w28 ) ( 102� w12 )� 0

I The denominator is striclty positive for w 2�� 5114;

172

�.

I The numerator is non-negative for w 2 ( �1; 177 ] with

17

7= 242:86% :



5 4 3 2 1 1 2 3 4 5 6 7 8 9 10

1.0

0.5

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

allocation w

expected utility

w *


w� WITH A PORTFOLIO CONSTRAINT ( w � 1 ) 1

I The investor cannot borrow to allocate more than 100 into the risky security:

w� = 1

5 4 3 2 1 1 2 3 4 5 6 7 8 9 10

1

1

2

3

4

5

allocation w

expected utility



I The portfolio constraint is binding at the optimum:

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

ddw E

hlog

� fW � i��w = w�

> 0

w� � 1 = 0



I Let's introduce the Lagrangian function

L (w; l) = Ehlog

� fW � i� l ( w � 1 ) .

I l is the Lagrange multiplier with

l � 0 .



I Let's use the Lagrangian function to re-express the optimality conditions:

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

@@w L

��w = w�; l = l�

= 0

l� � 0

@@l L

��w = w�

� 0

l� � @@l L

��w = w�

= 0

=)

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

ddw E

hlog

�fW�i��w = w�

� l� = 0

l� = 0:041026 > 0

� ( w� � 1 ) = 0

l� � ( � ( w� � 1 ) ) = 0



I If the constraint is relaxed ( say to w � 1:01 ), the optimized expected

utility goes up by

d

dwEhlog

� fW � i��w = w�

� 0:01 = l� � 0:01 .



0.960 0.965 0.970 0.975 0.980 0.985 0.990 0.995 1.000 1.005 1.010 1.015 1.020 1.025 1.030 1.035 1.0404.6820

4.6825

4.6830

4.6835

4.6840

4.6845

4.6850

allocation w

expected utility

about 0.00041



I The Lagrange multiplier l� is a shadow price :

l� = 0:041026 .

( optimized-expected-utility gain per unit of constraint relaxation )


UNCONSTRAINED OPTIMIZATION


A PROFIT FUNCTION

I A �rm produces two outputs x and y (they can be sold at the �xed prices

20 and 30, respectively).

I The production costs are quadratic:

C (x; y) = x2 + 2y2 � 2xy + 20 .

I The pro�t function remains quadratic:

P (x; y) = 20x+ 30y ��x2 + 2y2 � 2xy + 20

�.


P 'S LEVEL CURVES 1

I The level curve p = 300 is the set of output pairs (x; y) such that the

pro�t P (x; y) equals 300.

0 10 20 30 40 50 60 700

10

20

30

40

50

60

70

x

y


P 'S LEVEL CURVES 2

I The level curve p = 600 is in black

0 10 20 30 40 50 60 700

10

20

30

40

50

60

70

x

y


P 'S LEVEL CURVES 3

I The level curve p = 705 is the singleton (x� = 35; y� = 25), which isthe maximum-pro�t production in the absence of constrains.

0 10 20 30 40 50 60 700

10

20

30

40

50

60

70

x

y


FIRST PARTIAL DERIVATIVE WITH RESPECT TO x 1

Px (35; 25) = limh!0

P (35 + h; 25)� P (35; 25)h

= 0

I It is the slope of P 's graph in (x = 35; y = 25) along the direction (x; y = 25)

423040

3836

output x26

3424

output y

32

28

302228

20

550

600

700

650

profit level z


FIRST PARTIAL DERIVATIVE WITH RESPECT TO x 2

Px (35; 25) =@

@x

�20x+ 30y � x2 � 2y2 + 2xy � 20

��( x=35; y=25 )

= (20� 2x+ 2y)j ( x=35; y=25 )

= 0


FIRST PARTIAL DERIVATIVE WITH RESPECT TO y 1

Py (35; 25) = limk!0

P (35; 25 + k)� P (35; 25)k

= 0

I It is the slope of P 's graph in (x = 35; y = 25) along the direction (x = 35; y)

423040

26 363828

output x

3424

output y

323022

2820

550

600

profit level z

650

700


FIRST PARTIAL DERIVATIVE WITH RESPECT TO y 2

Py (35; 25) =@

@y

�20x+ 30y � x2 � 2y2 + 2xy � 20

��( x=35; y=25 )

= (30� 4y + 2x)j ( x=35; y=25 )

= 0


THE (NECESSARY) FIRST-ORDER CONDITIONS FOR OPTIMALITY

I P : X � R2 �! R admits �rst partial derivatives at the interior point

(x�; y�) 2 X.

I If P has a local/global maximum at the interior point (x�; y�), then (x�; y�)is a stationary point for P :

Px ( x�; y� ) = Py ( x

�; y� ) = 0 .


THE HESSIAN MATRIX

I If P : X � R2 �! R admits second partial derivatives at the interior

point (x0; y0) 2 X, then P 's Hessian matrix at (x0; y0) is

264 Pxx (x0; y0) Pxy (x0; y0)

Pyx (x0; y0) Pyy (x0; y0)

375 .

I Schwartz's Theorem: Pxy (x0; y0) = Pyx (x0; y0) .

I In our case, the Hessian is264 �2 2

2 �4

375 .


TAYLOR'S FORMULA STOPPED AT THE SECOND ORDER

I Given any point (x0; y0), our quadratic pro�t function can be re-expressed

as follows:

P (x; y) = P (x0; y0) +

Px (x0; y0) (x� x0) + Py (x0; y0) (y � y0) +

1

2

264 x� x0y � y0

375T 264 Pxx (x0; y0) Pxy (x0; y0)

Pyx (x0; y0) Pyy (x0; y0)

375264 x� x0y � y0

375 .


P 'S HESSIAN AT THE MAX. POINT (x� = 35; y� = 25) 1

P (x; y) = 705 +

0 (x� 35) + 0 (y � 25) + ( stationarity )

1

2

264 x� 35y � 25

375T 264 �2 2

2 �4

375264 x� 35y � 25

375| {z }

< 0 for any x 6= 35 or y 6= 25 (negative de�nite)

.


P 'S HESSIAN AT THE MAX. POINT (x� = 35; y� = 25) 2

Pxx (x�; y�) = � 2 < 0

det

0B@264 Pxx (x�; y�) Pxy (x�; y�)

Pyx (x�; y�) Pyy (x�; y�)

3751CA = det

0B@264 �2 2

2 �4

3751CA = 4 > 0


THE (SUFFICIENT) SECOND-ORDER CONDITIONS FOR A LOCAL

MAXIMUM

I P : X � R2 �! R admits continuous �rst and second partial derivatives

in an open ball of the interior point (x�; y�) 2 X.

I If (x�; y�) is a stationary point for P and if

I P 's Hessian matrix at (x�; y�) is negative de�nite,

I equivalently, Pxx (x�; y�) < 0 and

det

0B@264 Pxx (x�; y�) Pxy (x�; y�)

Pyx (x�; y�) Pyy (x�; y�)

3751CA > 0

then (x�; y�) is a local maximum point for P .


CONVEX SETS

I X � R2 is convex when the segment that connects two arbitrary points ofthe set does not lie outside the set.

I R2 is strictly convex (the connecting segment lies strictly inside R2).

In( x; y ) 2 R 2 : x � 0 ^ y � 0

ois convex:

2 1 1 2 3 4

2

1

1

2

3

4

x

y

here


P 'S STRICT CONCAVITY 1

I P : X � R2 �! R is strictly concave in its strictly convex domain X when

the segment that connects two arbitrary points of P 's graph lies strictly below the

graph.

profit level z

600

400

200

00

10

output y

20

0

20

output x40

60

806050

4030

I 20x+ 30y ��x2 + 2y2 � 2xy + 20

�is strictly concave.



I P : X � R2 �! R is strictly concave in its strictly convex domain X

if and only if the non-empty sets f (x; y) 2 X : P (x; y) � a g are strictly

convex .

0 10 20 30 40 50 60 700

10

20

30

40

50

x

ylevel 300

level 600

level 700

I 20x+ 30y ��x2 + 2y2 � 2xy + 20




I Assume that P : X � R2 �! R is twice derivable in all the points of its

open and strictly convex domain X.

I P is strictly concave in X if and only if the Hessian matrix is negative

de�nite in any point of X.

I 20x+ 30y ��x2 + 2y2 � 2xy + 20

�is strictly concave:

its Hessian

264 �2 2

2 �4

375 is always negative de�nite.


STRICT CONCAVITY AND THE UNIQUE GLOBAL MAXIMUM

I P : X � R2 �! R is strictly concave in its strictly convex domain X.

I If a local maximum point exists for P , it is also of global maximum andunique.

profit level z

600

400

200

00

10

output y

20

0

20

output x40

60

806050

4030

I (35; 25) is the unique global maximum point.


C'S STRICT CONVEXITY 1

I C : X � R2 �! R is strictly convex in its strictly convex domain X whenthe segment that connects two arbitrary points of C's graph lies strictly above thegraph.

0

2

0

54

output x

6

output y

10

158

20

010

40cost level z60

80

100

I The cost function C (x; y) = x2+2y2�2xy+20 is strictly convex.



I C : X � R2 �! R is strictly convex in its strictly convex domain X if

and only if �C is strictly concave there.

106

25y x

4

0 0

810

100

80

60z40

20

15

0

I ��x2 + 2y2 � 2xy + 20




I C : X � R2 �! R is strictly convex in its strictly convex domain X if and

only if the non-empty sets f (x; y) 2 X : C (x; y) � a g are strictly convex .

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 150

1

2

3

4

5

6

7

8

9

10

x

y

level 40

level 80

level 120

I x2 + 2y2 � 2xy + 20 is strictly convex.


MAXIMIZATION WITH EQUALITY CONSTRAINTS


RECALLING THE COST FUNCTION

I A �rm produces two outputs x and y (they can be sold at the �xed prices

20 and 30, respectively) and its cost function is:

C (x; y) = x2 + 2y2 � 2xy + 20 .


A COST CONSTRAINT THAT DEFINES AN IMPLICIT FUNCTION 1

I The output pair (x = 10; y = 10) commands a cost of 120. In a neighbor-hood of x = 10, the cost constraint

C (x; y) = 120

implies the existence of a unique function

y = � (x)

such that

10 = � (10) and ddx� (x) = �

C x ( x; �(x) )

C y ( x; �(x) ).


A COST CONSTRAINT THAT DEFINES AN IMPLICIT FUNCTION 2

5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

4

2

2

4

6

8

10

12

14

output x

output y

( level curve ) cost = 120

( 10 , 10 )

10 = � (10) and ddx� (x)

��x = 10

= �C x ( 10; 10 )

C y ( 10; 10 )= 0


PROFIT MAXIMIZATION WITH A COST CONSTRAINT 1

I The �rm's pro�t is

P (x; y) = 20x + 30y � C (x; y)

I The optimization problem is

maxx; y

P (x; y) subject to C (x; y) = a



I From the constraint equation

C (x; y) = a

we can work out y from x via the implicit function

y = � (x) :

I By plugging y = � (x) into P , the problem becomes:

maxx

P ( x; � (x) ) .



I If P ( x; � (x) ) exhibits an interior extremum point, then such a point

must be stationary:

dP

dx= Px + Py �

d

dx� = 0

m

Px

Py= � d

dx�

m

Px

Py= �

�CxCy

!.



I The necessary condition for optimality

Px = Py = Cx = Cy

holds if and only if there exists a real number l (the Lagrange multiplier )

such that

Px = l Cx ;

Py = l Cy :



I By introducing the Lagrangian function

L (x; y; l; a) = P (x; y) � l (C (x; y)� a) ;

the necessary condition for optimality and the constraint can be expressed as

follows (stationarity for L):

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

Lx = Px � lCx = 0

Ly = Py � lCy = 0

Ll = � (C � a) = 0 (constraint)



8>>><>>>:Lx = 0 , 20� 2x+ 2y � l (2x� 2y) = 0

Ly = 0 , 30� 4y + 2x� l (4y � 2x) = 0

,

8>>>><>>>>:x� = 35

l+1

y� = 25l+1

��x�2 + 2y�2 � 2x�y� + 20 � a

�= 0| {z }

Ll = 0

,

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

l = +�725a�20

�12 � 1 (select)

l = ��725a�20

�12 � 1 (discard)



l� =�725

a� 20

�12 � 1

x� =35

l� + 1= 35

�a� 20725

�12

y� =25

l� + 1= 25

�a� 20725

�12



L (x�; y�; l�; a) = P (x�; y�) � l� (C (x�; y�)� a)| {z }= 0

= P (x�; y�)

= 2 (725)12 (a� 20)

12 � a .


THE LAGRANGE MULTIPLIER l� IS A SHADOW PRICE 1

dda P ( x

�; y� ) =d

da

�2 (725)

12 (a� 20)

12 � a

�

=�725

a� 20

�12 � 1 = l�


THE LAGRANGE MULTIPLIER l� IS A SHADOW PRICE 2

I If we �x the cost constraint at a = 120 , then the shadow price is

l� = 1:6926 .

I The �rm is willing to surrender up to 1:6926 cents of pro�ts to obtain a

1-cent relaxation of the cost constraint ( da = 0:01 ):

d

daP ( x�; y� ) � 0:01 = l� � 0:01 ' 1:6926 cents .


GRAPHICAL ANALYSIS WITH a = 120 1

I The optimal production is x� = 12:9987 and y� = 9:2848. The constrained-maximum pro�t is P (x�; y�) = 418: 5165:

5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

4

2

2

4

6

8

10

12

14

output x

output y

( constraint ) cost = 120

( level curve ) profit = 418.52


GRAPHICAL ANALYSIS WITH a = 120 2

I At the point (x�; y�) there is tangency between the constrained-cost

level curve and the maximum-pro�t level curve:

� P x (x�; y�) = P y (x�; y�) = � C x (x�; y�) = C y (x�; y�)

5 4 3 2 1 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

4

2

2

4

6

8

10

12

14

output x

output y

( constraint ) cost = 120

( level curve ) profit = 418.52


OPTIMIZATION WITH INEQUALITY CONSTRAINTS


THE PORTFOLIO RETURN

I The portfolio return is

erp = w1 er1 + w2 er2 + (1� w1 � w2) r .

wj = fraction of initial wealth in the risky asset j ( j = 1; 2 ) ,

erj = net return on the the risky asset j ( j = 1; 2 ) ,

r = net return on the riskless asset .


THE EXPECTED RETURN ON A PORTFOLIO

I The expectation E [ erp ] is

T (w1; w2) = r + w1 ( r1 � r ) + w2 ( r2 � r ) ,

Eh erj i = rj ( j = 1; 2 ) .


T 'S LEVEL CURVES 1

I The allocation pairs (w1; w2) such that T (w1; w2) = 8% are in black.

1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 1.0 1.2 1.4

1.4

1.2

1.0

0.8

0.6

0.4

0.2

0.2

0.4

0.6

0.8

1.0

1.2

1.4

w_1

w_2

r1 = 12%; r2 = 8%; r = 2%:


T 'S LEVEL CURVES 2

I The allocation pairs (w1; w2) such that T (w1; w2) = 10% are in black.

1.4 1.2 1.0 0.8 0.6 0.4 0.2 0.2 0.4 0.6 0.8 1.0 1.2 1.4

1.4

1.2

1.0

0.8

0.6

0.4

0.2

0.2

0.4

0.6

0.8

1.0

1.2

1.4

w_1

w_2

r1 = 12%; r2 = 8%; r = 2%:


VARIANCES AND CORRELATION

V arh erj i = �2j ( j = 1; 2 )

Cov [ er1 , er2 ] = ( �1 �2 ) = �


THE VARIANCE OF A PORTFOLIO RETURN

I The variance V ar (erp) is

V (w1; w2) =

264 w1w2

375T 264 �21 �1�2�

�1�2� �22

375264 w1w2

375

= �21w21 + 2��1�2w1w2 + �22w

22

� 0 for any portfolio choice (w1; w2) .


V 'S LEVEL CURVES 1

I The allocation pairs (w1; w2) such that V (w1; w2) = (20%)2 are inblack.

1 1

1

1

w_1

w_2

�1 = 30%, �2 = 20%,

� = �0:5


V 'S LEVEL CURVES 2

I The allocation pairs (w1; w2) such that V (w1; w2) = (10%)2 are inblack.

1 1

1

1

w_1

w_2

�1 = 30%, �2 = 20%,

� = �0:5


V 'S LEVEL CURVES 3

I The allocation pair (w1; w2) such that V (w1; w2) = 0 is the point�w�1 = 0; w

�2 = 0

�.

1 1

1

1

w_1

w_2

�1 = 30%, �2 = 20%,

� = �0:5


MAXIMIZING THE RISK-ADJUSTED EXPECTED RETURN 1

I The investor's risk aversion is measured by the parameter A > 0 .

I The risk-adjusted expected return on the portfolio is

J (w1; w2) = T (w1; w2) � A

2V (w1; w2) .

I The problem is

maxw1; w2

J (w1; w2) sub

( total risk exposure must not exceed q )z }| {w1 + w2 � q � 0 .



I If q = 3 , the constraint is painless and not binding with w�1 = 107:4%and w�2 = 155:6% (the shadow price of the constraint is l� = 0).

1.0 0.5 0.5 1.0 1.5 2.0 2.5 3.0

1

1

2

3

w_1

w_2

h e r ew_1 + w_2 = 3

J's level curves

A = 2 ; �1 = 30%; �2 = 20%; � = �0:5; r1 = 12%; r2 = 8%; r = 2%:



I Given L (w1; w2; l) = J (w1; w2) � l (w1 + w2 � q) , the �rst-order

conditions for constrained optimality can be written �a la Kuhn-Tucker :8>>>>>>>>>>>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>>>>>>>>>>>:

L w1

�w�1; w

�2; l

��= 0

L w2

�w�1; w

�2; l

��= 0

l� � 0

L l

�w�1; w

�2; l

�� 0

l� � L l

�w�1; w

�2; l

��= 0

q = 3

=)

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

J w1

�w�1; w

�2;�= l�

J w2

�w�1; w

�2;�= l�

l� = 0w�1 + w

�2 � 3 < 0



I The feasibility conditions are:

8>><>>:l� � 0

L l

�w�1; w

�2; l

�� 0 ( w�1 + w

�2 � q � 0 )

I The complementary slackness condition is:

l� � L l

�w�1; w

�2; l

��

= 0



I If q = 1 , the constraint is painful and binding with w�1 = 47:4% and

w�2 = 52:6% (the shadow price of the constraint is l� = 4:63%).

1.0 0.5 0.5 1.0 1.5 2.0 2.5 3.0

1

1

2

3

w_1

w_2

[ 1.074 , 1.556 ]

h e r e

w_1 + w_2 = 1

J's level curves

A = 2 ; �1 = 30%; �2 = 20%; � = �0:5; r1 = 12%; r2 = 8%; r = 2%:



I Given L (w1; w2; l) = J (w1; w2) � l (w1 + w2 � q) , the �rst-order

conditions for constrained optimality can be written �a la Kuhn-Tucker :8>>>>>>>>>>>>>>>>>>>>>>>>>>><>>>>>>>>>>>>>>>>>>>>>>>>>>>:

L w1

�w�1; w

�2; l

��= 0

L w2

�w�1; w

�2; l

��= 0

l� � 0

L l

�w�1; w

�2; l

�� 0

l� � L l

�w�1; w

�2; l

��= 0

q = 1

=)

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

J w1

�w�1; w

�2

�= l�

J w2

�w�1; w

�2

�= l�

l� > 0

w�1 + w�2 � 1 = 0


STOCK PRICING


THE DIVIDEND PROCESS

I The current date is t

I The stock pays out the dividend Xdt every `second' (a period of length dt)

I The dynamics of the dividend process is

dX = a (X; t) � dt| {z }expected change Et [dX]

+ b (X; t) � dz| {z }unexpected change

,

where dz is Gaussian with Et [dz] = 0 and Ethdz2

i= dt


A FUNCTION OF THE DIVIDEND X AND ITS DYNAMICS

I S (X) is a function of X

I Given the notation

SX =d

dXS and SXX =

d2

dX2S ,

we have

dS = SXdX +1

2SXXb

2dt

=�SXa+

1

2SXXb

2�� dt| {z }

expected change Et [dS]

+ SXb � dz| {z }unexpected change


EQUILIBRIUM STOCK PRICING 1

I The stock price is S (X)

I The per-annum expected total gain on the stock is

1

dtEt [dS] + X = expected capital gain + income gain

I In equilibrium, the following restriction holds true:

1dtEt [dS] + X = Sr + SX b � � � �

I Sr is the cost of borrowing expressed in Euro (r is the riskfree rate)

I SX b � � � � is the risk premium expressed in Euro


EQUILIBRIUM STOCK PRICING 2

I SX b is the equity risk expressed in Euro

I � is the premium per unit of systematic risk

I � is the fraction of equity risk that is systematic.


TRENDING DIVIDENDS

I X's dynamics is

dX = X� � dt+X� � dz

I X is expected to grow at the rate �

I X = 0 is an absorbing boundary


STOCK PRICING WITH TRENDING DIVIDENDS 1

I The equilibrium restriction is

SXX�+1

2SXXX

2�2| {z }= 1

dtEt [dS]

+X = Sr + SXX��

I If X = 0, the stock is worthless (it is unable to generate dividends):

S (0) = 0 (boundary condition)



I The solution is

S (X) =X

r + �� (non-negative and meaningful i� r + �� > 0)

I Given the parabola

y ( ) =1

2�2 2 +

�� 1

2�2� � r

we impose

y (1) = � (r + �� ) < 0



I The elasticity of S (X) with respect to X is

SXSX = 1

0@ =dSSdXX

1A

I A 1% change in the dividend implies a 1% change in the stock price



I The total return on the stock is

dS + Xdt

S=

�r +

SXX

S��

�dt

| {z }expected total return

+SXX

S� dz| {z }

unexpected total return



I The per-annum expected total return on the stock is

1

dt

1

SEt [dS] +

X

S|{z}per-annum dividend yield

= r +SXSX��

= r + ��


THE EXPECTED TOTAL RETURN ON A PORTFOLIO

I Our initial capital is H = 100 Euro. We invest 60 Euro in the stock and

40 Euro in the riskfree asset

I The per-annum expected total gain is

1

dtEt [dH] =

60

S

�1

dtEt [dS] +X

�+ 40r

= 60 (r + ��) + 40r

I The per-annum expected total return is

1

dt

1

HEt [dH] = r +

60

100��


RATIONAL BUBBLES 1

I Another solution to

SXX�+1

2SXXX

2�2 +X = Sr + SXX�� with S (0) = 0

is given by

S (X) =X

r + �� | {z }fundamental value

+ c �X 1| {z }bubble

c = positive constant

1 = �� 2

� 12

�+

"�� 2

� 12

�2+ 2

r

�2

#12

> 1


RATIONAL BUBBLES 2

I The elasticity of S (X) with respect to X is `excessive':

SXSX =

S + ( 1 � 1) cX 1S

> 1


THE COEFFICIENT 1

I Given the parabola y ( ) = 12�2 2 +

�� 1

2�2� � r we have

y (1) = � (r + �� ) < 0 and y ( 1) = 0

gamma

y

gamm

a_1

( r + sigma*lambda*rho mu ) < 0

1


`POWER' STOCK PRICING 1

I The stock that pays out X2dt every `second' has the following equilibrium price:

P (X) =X2

r + 2��2�+ �2

�

I Given the parabola

y ( ) =1

2�2 2 +

�� 1

2�2� � r

we impose

y (2) = ��r + 2��

�2�+ �2

��< 0


`POWER' STOCK PRICING 2

I The total return on the stock is

dP + X2dt

P=

�r +

PXX

P��

�dt

| {z }expected total return

+PXX

P� dz| {z }

unexpected total return

= ( r + 2�� ) dt + 2� dz


MEAN-REVERTING DIVIDENDS

I X's dynamics is

dX = �k (X �m) � dt+X� � dz

I The long-run mean is E [X] = m > 0

I k > 0 is the speed of mean reversion

I If k �! +1, X remains �xed at the level m

I X = 0 is a re ecting boundary


STOCK PRICING WITH MEAN-REVERTING DIVIDENDS 1

I The per-annum expected capital gain is

1

dtEt [dS] = SX (�k (X �m)) + 1

2SXXX

2�2

I The equilibrium stock price solves

1

dtEt [dS] + X = Sr + SXX��

and is

S (X) =m

r� k

r + ��+ k+

X

r + ��+ k


STOCK PRICING WITH MEAN-REVERTING DIVIDENDS 2

I The absence of mean reversion brings us back to the trending-dividends case

(without growth):

limk!0

S (X) =X

r + ��

I A massive mean reversion brings us back to the riskless case (without growth):

limk!+1

S (X) =m

r

I Systematic risk and risk aversion (�� > 0) do have a long-run impact:

E [ S (X) ] =m

r� k + r

r + ��+ k<

m

r


APPENDIX


THE CAPITAL ASSET PRICING MODEL 1

I The stock market portfolio is worth M with dynamics

dM +XMdt = M ( r + �M � ) dt + M �M dzM

I The premium for systematic risk (expressed in Euro) is M�M � �

I The CAPM states

1

dtEt [dM ] +XM = Mr +

covt [dM; dM ]

vart [dM ]| {z }`beta' equal to 1

� M �M � �


THE CAPITAL ASSET PRICING MODEL 2

I For the single stock, the CAPM states

1

dtEt [dS] +X = Sr +

covt [dS; dM ]

vart [dM ]| {z }`beta' equal to SX b � �

M �M

� M �M � �

= Sr + SX b � � � �


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QMF Lectures 2013-14