slide-co minh nt

8/11/2019 Slide-co Minh NT

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MATHEMATICS FOR ECONOMICS

ASS.PROF. NGUYEN THI MINH

[email protected]


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COURSE CONTENTS

Review on probability and statistics

Review on static optimization

Difference equation

Differential equation

Dynamic optimization


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PART 1: PROBABILITY ANDSTATISTICS

LECTURE 1: CONCEPTS IN PROBABILITY


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KEY CONCEPTS IN PROBABILITY

Random variable

Mean

Variance Standard deviation

Covariance / Correlation coefficient

Distribution function

Distribution function for a discrete variable

Distribution function for a continuousvariable


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DISTRIBUTION OF A R.V

6

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

X

P(X>20)


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SOME COMMON DISTRIBUTIONS

N(, 2): normal distribution

X~> N(, 2)

=> E(X) = , var(X) = 2

The rule of large number:

If X1,.., Xnare i.i.d with mean , var = 2

then:

7

21 ... ( , )nX X

Nn


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N(0,1): standardized normal distribution

Z~> N(0,1) => E(Z) = 0, var(Z) =1

Critical values:

8

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

Critical value


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We have the following:

Relationship between N(0,1) and N(, 2):

9

~ (0,1)X N


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EXAMPLE

P(-1.96 want to

find a range in which X will probably takevalue?

With probably = 0.95, we have:

P(a< X


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KEY CONCEPTS IN PROBABILITY

Principle of small probability

Principle of large probability


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Some other distributions:

Student distribution T(n)

Fisher distribution (F(n1,n2)

Chi-squared distribution

Uniform distribution U(a,b)

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SUMMARY

With a r.v we want to know:

Mean (population mean)

Variance (population variance)

V.v

With many r.v(s) we want to know:

Correlation (covariance)

Independency between them Q: How could we get those?

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LECTURE 2: STATISTICS:

Key concepts

Descriptive statistics

Inferential statistics


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KEY CONCEPTS IN STATISTICS

Want to know about some R.V X (population)

(?, ?, ?)

We normally do not have the whole pop.

=> sample

Statistics:

Describe samples (descriptivestatistics)

Make inference about the population(inferentialstatistics)


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Key statistic:

Sample mean

Sample standard deviation/ variance

Sample correlation

Standard error

Etc.


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Example

A sample:

(10,12, 28, 10, 30)

=> sample mean?

(20, 12, 20, 18, 18) => sample mean may take different values

depending on a particular sample

=> sample mean is a r.v, calledestimator of pop. mean

17

X


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Estimators

Unbiased estimator

Efficient estimator Consistent estimator

Sample mean is unbiased, efficient

estimator for pop. mean


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DESCRIPTIVE STATISTICS

A random sample from a population

Summarize the sample to get a clear picture

about essential characteristics of the pop.

Sample mean Sample variance

Sample proportion

Sample correlation, etc. They are point estimators of the pop.

parameters


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INFERENTIAL STATISTICS

Imagine:

Ask 3 students, one likes maths =>?

Ask 300 students, 100 like maths =>

Point estimators do not reflect the precisenessof the estimates

=> need more => inferential statistics


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Information from a sample =>

Point estimator =>

Confidence interval for a population

parameter Hypothesis testing about pop. Parameters

Hypothesis testing about pop. distribution


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Just work with the mean

We use the following result

If X ~>N or n is large

1. ( )

n

XT

s e X


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C.I FOR A MEAN

We have

1~( )

n

Xt T

se X

0.025, 1 1 0.025, 1( ) 0.95 n n nP t t t

0.025, 1 0.025, 1( ) 0.95

( )

n n

XP t t

se X

0.025, 1 0.025, 1( ( ) ( )) 0.95 n nP X t se X X t se X


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C.I FOR A MEAN

The 95% CI for is

The (1- )100% CI of is:

0.025 1 0.025, 1( , ( ), ( )) n nX t se X X t se X

/2, 1 /2, 1( ( ), ( ))n nX t se X X t se X


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GENERAL FORMULA FOR A C.I

A (symmetric) 95% C.I for some

A (symmetric) (1-)100% C.I for some

25

0.025, 0.025, ( ( ), ( ))df df t se t se

/2, /2,

( ( ), ( ))df df t se t se


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HYPOTHESIS TESTING - EXAMPLE

Weight of milk boxes

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w 45 46.5 47 48 49 50.5 51 52 54

N0 5 3 5 30 40 20 8 7 1


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HYPOTHESIS TESTING

Example: Each milk powder box weighted: 500 grams

You bought 1 box, and it weighs 470grams

only, so does your neighbors. Want to know if the producer cheated.

Given that the weight follows a Normal

distribution


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HYPOTHESIS TESTING

Example: => Hypothesis H0: = 500

=> Alternative H1:

If H0is correct then:

500

1

500

( ) n

Xt

se X


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HYPOTHESIS TESTING

=> if H0correct then:

i.e: this event probably does not take place if

H0is true

From a sample: if it takes place => evidence against H0

if it does not take place => no evidence

against H0

0.05, 1

500( ) 0.05

( )

n

XP t

se X


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EXAMPLE

Some people presume that the average salaryof MDEans is 12m per month. A (random)

sample of 100 MDEans is collected, the data

are as follow:

Test if the presumption is correct with 5%

significance level?31

N0 10 30 40 15 5

wage 6 8 12 15 50


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REVIEW EXAMPLE

Data from a (randomly selected) sample of 100newborn babies in the rural area:

Construct a 95% CI for the mean weight

Test a claim that newborn babies are heavier inurban than in rural area, given that the

average weight of newborn babies in r.r area is

3.132

N0 10 20 40 15 5

weight 2.8 2.9 3 3.1 3.4


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DO IT IN STATA

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PART 2: OPTIMIZATION


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LECT 3: STATIC OPTIMIZATION -REVIEW


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OPTIMIZATION PROBLEM

Want to maximize:

Profit/Happiness/Utility

Minimize: Total cost/Risk/Etc.

=> just focus on maximization problems

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OPTIMIZATION PROBLEMS WITH CONSTRAINTS

Lagrange method

Interpretation of the Lagrange multiplier

Second order conditions Applications


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EXAMPLE 4

Example 4: maximize utility: Y = 2X1

0.3X20.7;

s.t: 2X1+3X2= 100

maximize utility function U = lnC + ln(24-L),s.t: C = 3L

where C is consumption and L is labor supply


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THE GENERAL PROBLEM WITH EQUALITYCONSTRAINTS

Maximize the objective fn: f(x1,..xn) s.tg1(x1,..,xn) = b1

g2(x1,..,xn) = b2

gm(x1,..,xn) = bm

Vector x that satisfies the m constraints:

feasible vector (feasible solution) These constraints are independent

Condition: m < n to make sure that the system

has more than one feasible solution

(2.1)


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THE METHOD OF LAGRANGE

Form the Lagrangnian:L = f(x) + 1 (b1- g

1(x))+..+ m (bm- gm(x))

Then solve the following (F.O.C):

f1(x) + 1g11(x)+..+ mg1m(x)= 0--

fm(x) + 1gm1(x)+..+ mgm

m(x)= 0

g1(x) =b1;..;gm(x) = bm1,.., m: Lagrange multipliers


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THE METHOD OF LAGRANGE (CONT.)

Check for sufficient condition: f concave and g linear, or

Bordered Hessian matrix is negative definite


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THE METHOD OF LAGRANGE (CONT.)

Check for sufficient condition: f concave, linear g

Bordered Hessian matrix is negative definite


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THE SECOND ORDER CONDITIONS

In the sense that:


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EXAMPLE 5

Y = 2K0.3L0.7max; 3K +2L = 100

S1: The Lagrange: L = 2K0.3L0.7+ (100-3K-2L)

S2: The F.O.C:

0.6K-0.7L0.7- 3= 0

1.4K0.3L-0.3- 2= 0

3K + 2L -100 = 0

=> L=35, K = 10

S3: Check the sufficient condition

In class exercise: 5,6, 7/page 521


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EXERCISES

Derive a consumers demand for good, givenher utility function and budget constraint as:

what is her own-price elasticity?

What if the utility function is (homework):

1

1 2 1 2 1 1 2 2( , ) ;u x x x x p x p x m

11 2 1 1 2 2( , ) ( ) ( )u x x x c x c

INTERPRETATION OF LAGRANGE


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INTERPRETATION OF LAGRANGEMULTIPLIERs

Q: what happens to f* when b increases

Lets get back to the simple problem in slide 16

L*=f(x1*,x2*)+ *(b-g(x1*,x2*)) {= f(x1*,x2*)}

INTERPRETATION OF LAGRANGE


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O O G GMultipliers

When g(x) = b can be thought of as a budget

constraint or resource constraint.

* marginal contribution of budget or resource

to the objectives maximized value

* can be explained as:


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CENTRALIZATION OR INVISIBLE HAND?

Consider an economy which consists of 2

industries; production fns: F(x1), H(x2) where xi

and piare the amount of a resource used and

output price in industry i. The total resource is 9

F(x1) = 3x11/3; p1= 1; H(x2) = x21/3; p2=12


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Case a: under centralized economy


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Case b:


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LECTURE 3+ (Read yourself)

CONCAVE PROGRAMMING

THE KUHN-TUCKER CONDITIONS


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INTRODUCTION

The previous problem is also named as theclassical optimization problem: constraints are

equalities, always binding

However, in many cases, it requires inequality

constraints:

K 0, L 0

b 2x13x2 0

Problems with inequality constraints


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INTRODUCTION

We are confined with concave functions (forthe objective) =>concave Programming

Problems in which the objective fn and the

constraint fn are convex can be handled by

noting that (- concave fn = convex fn)


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THE KUHN-TURKER CONDITION

Problem:f(x1,..,xn) -> max

g(x1,..xn) 0; xj 0

=> L = f(x1,..xn)+ g(x1,..xn) The Kuhn-Tucker condition: (x*, *):

L/xj 0 , xj 0, xjL/xi=0 (*)

L/ 0 , 0, L/ =0 (**)


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INTERPRETATION

Condition (*) implies that L is maximized onxs direction

Condition (**) implies that L is minimized on

s direction

=> Kuhn-Tucker conditions mean that we are

searching for a saddle point.

In other words, we are solving the following:

Find (x1,..xn, ) such that:

MinMaxxL = f(x1,..xn)+ g(x1,..xn)


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SADDLE POINT
http://upload.wikimedia.org/wikipedia/commons/4/40/Saddle_point.pnghttp://upload.wikimedia.org/wikipedia/commons/4/40/Saddle_point.png


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THE KUHN-TURKER THEOREM

The Kuhn-Tucker theorem:

if f and g are concave and differentiable

there exists a point (x10

,..,xn0

):g(x1

0,..,xn0)>0, then

=> x* is a solution IFF there exists *such

that (x*, *) satisfies K-T condition


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EXAMPLE 6

f(x1,x2) = ln(x1) + ln(x2+2) => Max s.t 4-x1x2 0; x1 0, x2 0

The K-T condition:

Lx1= x1-1

0, x1 0, (x1-1

) x1= 0 Lx2= (x2+2)

-10, x2 0, (x2 +2)-1)

x2= 0

L = 4-x1x2 0, 0, (4-x1x2) =0

EXAMPLE 7


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EXAMPLE 7

The consumer problem:

Max u(x1,x2) s.t m-p1x1p2x2 0, x1, x2 0

L = u(x1,x2) + (m-p1x1p2x2)

The K-T conditions:

ui- *pi 0, xi* 0, (ui- *pi ) xi* =0 (1)

m-p1x*1p2x*2 0,*0,(m-p1x*1

p2x*2)*=0(2) Solve??

EXERCISE


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EXERCISE

THE MIXED CASE


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THE MIXED CASE

The problem can be generalized as:f(x1,..,xn) -> max, s.t

g1(x1,..xn) 0; ..;gm1 (x1,..xn) 0;

g(m1+1)

(x1,..xn) =bm1+1, .., g(m)

(x1,..xn) =bmxj 0

=> K-T condition will be:

THE MIXED CASE ( t )


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THE MIXED CASE (cont.)

f(x1,x2) -> max, s.tg1(x1,x2) 0;g

2 (x1,x2) = 0; x2 0

K-T condition will be:

Lx1 = 0, Lx2 0, x2 0, x2Lx2 = 0

L 0, 0, L = 0

SUMMARY


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SUMMARY

Static optimization Deal with one point- in- time- problem

Appropriate for studying the system in an

equilibrium state Can be able to estimate the total effect of a

change in exogenous variables on

endogenous variables

SUMMARY


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SUMMARY

What we have learn so far: static optimization problem without constraint

f.o.c and second order condition

problem with equality constraint: The method of Lagrange

The Lagrange multiplier

know how to derive demand functionfrom utility fn & budget constraint (for a

consumer), demand fn from cost function

and production function (for a firm),..

with mixed constraint: The Kuhn-Tucker


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PART II: DYNAMIC OPTIMIZATION


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INTRODUCTION


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INTRODUCTION

Dynamic: decisions made today affecttomorrows outcome

Variables in the system are a function of time

=> if time is considered as continuous =>differential equation (s)

=> if time is discrete => difference equations

We just focus on differential equations in this

course


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LECTURE 4: ORDINARY LINEAR FIRSTORDER DIFFERENTIAL EQUATION


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NOTATION AND DEFINITION


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NOTATION AND DEFINITION

Example:

we want to:

solve for K as a function of time

see what happens in the infinity

( ) 10 0.02 K t K

AUTONOMOUS EQUATIONS


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AUTONOMOUS EQUATIONS

Equation:

Linear: linear in x Autonomous: does not depend explicitly on t

First order: contains only the first order

differential term

Example

( ) ( )x t ax t b

( ) 2 ( ) 5x t x t

(4.1)

SOLUTION TO A FODE


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SOLUTION TO A FODE

Solution to FODE (4.1): function of time thatmakes the equation true. It can be written as:

x = xh+ xp

where xh: homogenous solution

xp: a particular solution

HOMOGENOUS F O D E


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HOMOGENOUS F.O.D.E

Equation:

Homogenous solution:

( ) ( ) 0x t ax t

xa

x xdt adt

x

1 2lnx c at c atx ce


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SOLUTION TO FODE


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SOLUTION TO FODE

Example:

=> solution is:

( ) 2 ( ) 5x t x t

2( ) 2.5tx t ce

THE INITIAL VALUE PROBLEM


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THE INITIAL VALUE PROBLEM

An initial value of x is given: x(t0) = x0 =>

=> solution will be:

0

0

at bx ce

a

00( )

atbc x e

a

0( )

0( ) ( ) a t tb b

x t x ea a


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EXERCISE- NATIONAL DEBT


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Q: persistent budget deficits causebankruptcy?

Ignore inflation

Persistent budget deficits:

Y grows at rate g:

Interest rate obligation: rD

Q: if rD>Y in some day?

; 0D bY b

; 0Y gY g



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=>

If rb/g >1? let r = 2%, b=1.6, g=3%

Excercis: 3-4-5/869

0 gtY Y e

00 0

gtgt bY eD bY bY e D c

g

0( )rcrbrD Yg Y

S S AND CONVERGENCE


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S.S AND CONVERGENCE

Q: does the system converge to its s.s?or:

The solution to the initial value problem:

y(t) = (y0-yss)e-at + (yss)

yss= b/a: the only steady state value.

Theorem: For a linear, autonomous, FODE, the

solutions converge to its ss iff a>0

( ) ?ty t y


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LECTURE 5-1: NON-LINEAR FIRSTORDER DIFFERENTIAL EQUATION


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NON-LINEAR FODE

Example

k: capital per labor, and no population growth=> the accumulation of capita per labor

Or:

( )k t ak k (4.2)

22 3 1x x x


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NON-LINEAR FODE

In general:

Nonlinear In some cases, the equation can be solved

analytically (try with this one !!!)

In the others => it cannot => qualitative

analysis => phase diagram

( )x g x (5.3)

PHASE DIAGRAM


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PHASE DIAGRAM

k

y=0.02k

1/22y kinvestment >depreciation =>

K increases investment kdecreases

k*

PHASE DIAGRAM


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PHASE DIAGRAM

Consider:

K

K

10000

stableequilibrium

unstableequilibrium

?

STABILITY ANALYSIS


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STABILITY ANALYSIS

For a non-linear autonomous F.O.D.E, there may

be more than one equilibrium.

=> which one the system converges to?

=> which one the system will stay/ move away?

STABILITY ANALYSIS


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STABILITY ANALYSIS

Theorem: a steady state of a FODE is stable if

g(x) 0 at

that point

If g(x)|x*>0 =>x* unstable

If g(y)|y* x* stable

Meaning:

( )x g x

EXERCISE


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EXERCISE

A fishery model: the change in the stock of fish

Where the first term indicates the naturalgrowth rate of the fish population, and the

second term represents the amount of fish

harvested by a fishing industry.

Q: if the fish is going to be extinct? if any

positive equilibrium exists?

(5 ) 6y y y

EXERCISE


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EXERCISE

A neoclassical growth model:

y = f(k);

Population grows at rate n

=>

Draw a phase diagram and make comments

K sY

2

( )KdLK KL K L sYLk k kn sy kn

dt L L L L

( )k sf k kn

APPENDIX - BERNOULLIS EQUATION


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APPENDIX BERNOULLI S EQUATION

nx ax bx

1n n

x x ax b

1where1

ny ay b y xn


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LECTURE 5-2: SYSTEM OF LINEAR1-ST ORDER EQUATION

EXAMPLE


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EXAMPLE

Prey and predator: Consider a rabbit and tigerpopulation

Q: How does the population of rabbits and

tigers evolve over time?

2 0.1

0.3 0.2

R R T

T R T

2 0.1

0.3 0.2

R R

TT

A LINEAR SYSTEM OF TWO EQUATIONS


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A LINEAR SYSTEM OF TWO EQUATIONS

Consider:

=> the homogenous system will be:

As before, the complete solution to (6.1) can be

expressed as:

1 11 1 12 2

2 21 1 22 2

x a x a x

x a x a x

1 11 1 12 2 1

2 21 1 22 2 2

x a x a x bx a x a x b

(6.1)

(6.2)

1 1 1 2 2 2;h p h px x x x x x

HOMOGENOUS SOLUTION


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Equation system (6.2) can be rewritten as:

As in the single equation case the solution will

involve exponential of the type et.

Suppose a solution is aet, a is a non-zero vector

From (6.3) we then have:

1 11 12 1

21 22 22

;x a a x

or x Axa a xx

(6.3)

( ) ( ) 0t tx A ae ae Aa a A I a

HOMOGENOUS SOLUTION


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For the system to have solutions other thana=0, it must be the case that: det(A-I) = 0

is the solution to:

2(a11+a22) +(a11a22-a21a12)=0

The roots may be real & distinct, or real &

equal, or complex, depending the sign of

11 22 11 221 2

( ) ( ); ;

2 2

a a a a

(6.4)

HOMOGENOUS SOLUTION


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Real and distinct roots =>

Real and equal roots:

1 2

1 2

1 1 2

1 11 2 112 1 2

12 12

t t

t t

x C e C e

a ax C e C e

a a

1 1 2

11 22 1 2

12 12

( ) )

t t

t

x C e C tea C

x C C t ea a

(6.5)

(6.6)

HOMOGENOUS SOLUTION


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Complex roots

Where h=(a11+a22)/2; v=0.5sqrt(-)

Example: Solve the following

1 1 2

11 1 2 11 2 12

12 12

( cos( ) sin( ))

( ) ( )( cos( ) sin( ))

ht

ht

x A vt A vt e

h a A vA h a A vAx vt vt e

a a

1 1 2

2 1 2

2

5 4

x x x

x x x

(6.7)

THE COMPLETE SOLUTION


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The particular solution:

Hence the complete solution is the sum of(6.8) with one of (6.5), (6.7) and (6.8),

respectively for the case of 2 different real

roots, equal real roots, and complex roots

1( ) 0x t Ax b x A b (6.8)

STABITILITY OF THE SYSTEM


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From (6.5)-(6.8) => the stability of the system

depends on the sign of the roots of its characteristic

equation:

2 distinct real roots:

both negative: the steady state is asym. stable one (+), one (-): saddle point

both (+): unstable

equal real roots: negative: stable

positive: unstable

STABITILITY OF THE SYSTEM


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Complex: The real part (-): stable

The real part (+): unstable

More notes on saddle point: there is atrajectory that converges to the s.s solution

if the initial conditions satisfy:

1 112 2 1 1

12

( ) ( )ax x x xa


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PHASE DIAGRAM


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Example 1:

82

42

22

11

xx

xx

x1

x2 01 x

02 x4

2 x1

x20

1 x

02 x4

2

stable node

PHASE DIAGRAM


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example 2:

12

52

22

11

xx

xx

x1

x20

1 x

02 x0.5

2.5

unstable node

PHASE DIAGRAM


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Example 3:

1

2

12

21

xx

xx

x1

x20

1 x

02 x4

2 x1

x20

1 x

02 x2

1saddle

point

PHASE DIAGRAM


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Example 4: complex roots (-ve real part)

212

21

5.2

2

xxx

xx

x*=(x1*= 0,8; x2*=2 )

x1

x2

01 x

02 x

2

stable focus

= spiral + converge


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LECTURE 7

SYSTEM OF NONLINEAR F.O.D.E

SYSTEM OF NON-LINEAR EQUATIONS


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Consider the non-linear system:

f and g are non-linear

Denote (x1*, x2*) an equilibrium

Q: how the system behaves when being near the

(x1*, x2*)?

1 1 2

2 1 2

( , )

( , )

x f x x

x g x x



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Around this point, the system can be

approximately linearized as:

* * * * * * * *

1 2 1 2 1 1 2 1 1 2 1 2 2 2

* * * * * * * *

1 2 1 2 1 1 2 1 1 2 1 2 2 2

( , ) ( , ) ( , )( ) ( , )( )

( , ) ( , ) ( , )( ) ( , )( )

f x x f x x f x x x x f x x x x

g x x g x x g x x x x g x x x x



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Hence:

Theorem: If det(A) 0 then the qualitative

behavior of the trajectories of the nonlinear

system in the neighbor of the ss point is thesame that of the trajectories of the

corresponding linear system

*1 111 12 11 121 1 1

*21 22 21 222 22 2 2

; : ;x za a a ax x z

ora a a ax zx x z

EXAMPLE 7


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Pollution and growth: K and P: the stock of

capital and pollution. K and P depreciate at rate

1 and 2, respectively.

Y1= K, 0 Come up with the following system

Use phase diagram to analyze the system?

(practice at the class)

1

2

K sK K

P K P

EXAMPLE 8


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Competing species:

Two populations compete for foods in the same

area =>

Use phase diagram to qualitatively analyze the

stability of the system

Using linear approximation to analyze the

stability of the system

/ 1 2

/ 1 2

X X X Y

Y Y Y X

EXAMPLE 9- OVERSHOOTING MODEL


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113

DETERMINING STABILITY


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114

Recall the characteristic equation of the system:

2(a11+a22) +(a11a22-a21a12)=0

Theorem: The stability properties of the ss

equilibrium of the system is also determined as:If |A| SS is a saddle point

If |A|>0, tr(A) 0, tr(A) >0: the real parts ofroots are positive; unstable

where tr(A) := (a11+a22)


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115

LECTURE 8: DYNAMICOPTIMIZATION

INTRODUCTION


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116

Example 8.1: At t=0, a person has s0units of

money in a bank account. The money earns

interest at the rate r. He is choosing a time

path of consumption c(t) that maximizes

0

0

ln

. : ; (0) ; ( ) 0

T

cdt

s t s rs c s s s T b

INTRODUCTION


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117

Ex 8.2: A profit maximizing farmerpurchases machines to tend his farm

Can be applied in many applications: optimal

investment, optimal borrowing, resource

exploiting, etc.

A SIMPLE PROBLEM


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Find c(t) that maximizes

s(t): the state variable, c(t): the control

variable

(8.3): boundary condition

(x(t),c(t)) that satisfy (8.2),(8.3): feasible

solutions

MaxdtttctsvVT

0

)),(),((

)),(),(( ttctsfs

s(0)=s0,

(8.1)

(8.2)

(8.3)

NOTES ON THE SIMPLE PROBLEM


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In this simple problem:

No restrictions on c(t)

The planning horizon [0,T] is given

s(0), s(T), T are given

These conditions will be relaxed later

Also assume that the optimal control c*(t)

exists, unique and differentiable w.r.t t


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THE MAXIMUM PRINCIPLE


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121

Theorem 8.1: An optimal solution to problem

(8.1)-(8.3) must satisfy the boundary condition

and the following :

1. c(t) maximizes H(t):

1. the state and costate variables:

0 0( ) ( ) ( )

H v f

c t c t c t

; ( )( ) ( ) ( ) ( )

H H v fs t

t s t s t s t

(8.4)

(8.5)

EXAMPLE 9


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122

At t=0, a person has s0units of money in a bankaccount. The money earns interest at the rate r.

He is choosing a time path of consumption c(t)

that maximizes:

The Hamiltonian:

0)(;)0(;:.

ln

0

0

bTssscrssts

cdtT

)(ln crscH

EXAMPLE 9


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123

The necessary conditions imply:

EXAMPLE (CONT.)


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124

(s0-s

Texp(-rT)) must be positive; i.e the final

stock does not exceed the level to which the

initial stock would have naturally grown if

consumption was null

consumption grows exponentially depending

on r.

PHASE DIAGRAM


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125

s

c

0s

0c

sT s0

1

2

3

I

II

sT

4

5


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REMARKS


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127

Among three trajectories: (1) takes least time

and (3) longest time among the three.

As T is given, the smaller T, the higher the

curve is.

What about the boundary conditions? What if,

say, sTis larger than s0?

What if sT is large and T is not?

=> we may end up with no feasible solution at

all, let alone an optimal one

ECONOMIC INTERPRETATION OF THE MP


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128

Let state the general problem

s: stock of capital, c; flow of consumption, v:

instantaneous value function, V* the maximum

value fn, f: the growth fn of capital stock.

The Hamiltonian: H= v(c,s,t)+ f(s,c,t)

Then the MP yields:

)(;)0();),(),((:.

))),()((max),0,,(*

0

0

0

Tsssttctsfsts

dtttctsvTssVT

T



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Can be shown that:

V*/ s0= *(0); V*/ sT= -*(T);

the contribution of one extra

unit of initial capital stockto the total value

the deduction from the

total value if requiring one

more unit of final capital stock

*(t): the imputed value of stock atany instant t along the optimal path

(shadow price)



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(1) v/ c(t) +(t) f/ c(t) = 0 f/ c(t): the marginal effect of

consumption on the rate of growth of stock

at time t => (t)f/ c(t): the marginal

contribution of current consumption to

future value

v/ c(t): the marginal contribution of

consumption to the instantaneous value Thus (1) implies: the total contribution of

one more unit of consumption at time t is

null along the optimal part



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ss fv **

the rate of change in

the value of stock

M.E on instantaneous

value function

the M.E on

future value

* * 0s sv f

(2)

(2)

(2) and (2): the depreciation rate = net marginalcontribution of capital (current + future)

SUFFICIENCY


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132

Theorem 8.2: The necessary conditions in

theorem 8.1 are also sufficient for the opt

solution of the problem (8.1)-(8.3) if

v(s,c,t) is concave in (s,c) jointly and

differentiable

One of the following satisfies:

f(s,c,t) if linear in (s,c)

f(s,c,t) is concave in (s,c) and (t) 0

f(s,c,t) is convex in (s,c) and (t) 0

PROBLEM WITH DISCOUTING FACTOR


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Find c(t) that maximizes:

u(.): utility function, F(.): production function

u>0; u0; F


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The Hamiltonian: H = exp(-t)u(c)+f(s,c)

The MP yields:

The problem is: t in the system as an

independent argument =>

1/ ( , ) 0

( , ),

/ /

t

t

H c e u c s

s f s c

e u s f s

(7.6)

THE CURRENT VALUED HAMILTONIAN


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135

The Hamiltonian: H = exp(-t)u(c)+f(s,c)

t as an independent argument in the H=>

want to move it away

Current value Hamiltonian:

Hc= u(c)+ (t)f(s,s)

where: (t) = exp(t) (t)=>

t t t u fe e es s

(7.7)

THE CURRENT VALUED HAMILTONIAN


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136

Then the MP for the current H can be written

as:

Example with: u(c,s) = u(c); f(c,s)=f(s)-c=>

( , )s f s c (7.8)

(7.7)/ 0 0C

u fH c

c c

u f

s s

(7.9)

THE PHASE DIAGRAM IN STATE-COSTATEPLANE


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0

0 s

0sE


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EXAMPLE 10


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139

Consider again:

We consider corresponding conditionstranservality conditions

0

0

ln

. : ; (0)

T

cdt

s t s rs c s s

( )?s T

TRANSERVALITY CONDITIONS


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s(T) = b

s(T) b =>

s(T) free

None

( ( ) ) 0, ( ) 0

,( ( ) ) ( ) 0

s T b T

s T b T

( ) 0T

EXAMPLE 10.1


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141

Consider again:

How to solve?

0

0

ln

. : ; (0)

T

cdt

s t s rs c s s

( ) 0s T

EXAMPLE 10.1


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1. c(t) maximizes H(t):

2. the state and costate variables:

3. Transerversality:

0 0( ) ( ) ( )

H v f

c t c t c t

; ( )( ) ( ) ( ) ( )

H H v fs t

t s t s t s t

(8.4)

(8.5)

EXAMPLE 10.2


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143

Consider again:

How to solve?

0

0

ln

. : ; (0)

T

cdt

s t s rs c s s


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APPENDIX FOR TRANSERVALITY


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145

Read the text


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146

LECTURE 9

PROBLEM WITH CONSTRAINTS

EXAMPLE 11


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147

You own a mineral spring, from which spring

flows at a constant rate. Outflow may be soldimmediately or stocked at a reservoir at no cost,

however the reservoir leaks and a constant

proportion of current stocks is lost per unit of

time. You want to maximize profit over [0,T]

c(t): quantity solved at time t, U(c(t)):profit

R = F(s) rate of outflow of the spring

s(t): level of stock

m: proportion of stocks that leaks

EXAMPLE


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148

Extraction problem :

Constraints:

c(t) c0; s(t) 0

F(s(t)) c(t): consumption constraint

)()())(()( tmstctsFts

T

dttcuV0

))((

EQUALITY CONSTRAINT PROBLEM


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The problem:

s(0) = s0; s(T) = sT;

c(t) = F(s(t))(cant be stored)

The Hamiltonian:

H(s,c,t,) = u(c)+ (F(s) c - ms)


T

dttcuV0

))((

EQUALITY CONSTRAINT PROBLEM


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The M.P yields:

H/ c = 0=>u(c) = 0

Boundary condition

( '( ) )H

F s ms

( ) ( )H

s t F s c

INEQUALITY CONSTRAINT PROBLEM


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s(0) = s0; s(T) = sT;

F(s(t)) c(t)(can be stored)


T dttcuV0

))((



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The M.P yields:L/ c = 0 => u(c) = 0

F(s)-c0, 0, (F(s)-c). =0

mscsFts )()(

)('))('(/ sFmsFsL

(8.1)

(8.2)

(8.3)



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153

Interpretation of (8.1):

imputed value of the final good at t0equals

the marginal contribution of its consumption

at t0to total value (the same as before)

instantaneous marginal contribution of cons.

is greater or equal then the contribution to

future value (agent should have consumed

more but was restricted by the constraint=> consumed less than he would like to)

Interpretation of (8.2):

If >0 => F(s) c = 0: consume at most as

Phase diagram


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Case 1: c

(8.3)and (8.4) make up a system for this case

Case 2: >0, c = F(s):

(8.3) =>

))(()("

)('sFm

cu

cuc (8.4)

0 mss 0)(' ssFc

PHASE DIAGRAM


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0c

0s

c=F(s)

s

c

c > F(s)


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THE GENERAL PROBLEM


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The general problem

T

dtttctsvV0

)),(),((

nittctsfts i ,..,1);),(),(()(

1,..1,0)),(),(( mjttctsg

j

mmjttctsg j ,..,1,0)),(),(( 1

si(0) = si0; si(T) = siT

THE GENERAL PROBLEM


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The Hamiltonian

Theorem 8.3: Let c*(t) be an opt. solution to

the problem and s*(t) be the corresponding

time path of the state variables. Then there

exist -variable (t) and multipliers (t) so that:

THE GENERAL PROBLEM


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1

1

/ 0( , , ) 0, ( ) 0, ( , , ) ( ) 0( 1, )

( , , ) 0

/

/

( ) ( ( ), ( ), )'( ) /

ij j

j j

j

L c for all i

g s c t t g s c t t j m

g s c t for j m

L s

s L

t L s t c t t t L t

continuous

SUFFICIENT CONDITION


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Theorem 8.4 (The sufficiency theorem): Let(s*(t), c*(t)) satisfies conditions in the

necessary theorem, and assume that

L*= L(s(t), c(t), *(t), *(t),t) is concave in

(s,c) then (s*(t), c*(t)) is an opt. solution. If

The L*is strictly concave then the solution is

unique opt. solution

EXERCISE


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161

Set up the control problem and solve it. A pondcontains a fish population of size s(t); it natural

growth rate is f(s(t)), but this can be reduced

by c(t)- the catch of fish at time t. This catch

can be sold and provides a revenue R(c(t)) at

time t. The exponential rate of discounting for

revenue is

SUMMARY


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No constraint on end-

point

s(T) = a

s(T) a

g(s,c) 0

(T) = 0

s(T) = a

(s(T) a) 0, (T) 0,

(s(T) a) (T) =0

g(s,c) 0,0,

g(s,c)=0

slide-co minh nt

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