Exercises for Math in Economics

Exercises for Math in Economics

王苏生

Department of Economics

Hong Kong University of Science and Technology

June 2008

© Hong Kong University of Science & Technology, 2008

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Exercies for Chapter 1

Question 1.1. Prove similar matrices have the same characteristic polynomial.

Question 1.2. Solve

1 2 3

1 2 3

1 2 3

2 3 = 14 2 5 = 42 4 = 1

x x xx x xx x x

⎧ − +⎪⎪⎪⎪ − +⎨⎪⎪ − + −⎪⎪⎩

using elementary operations.

Question 1.3. Given

1= , = 1 ,

1 1 1 1

a b c a cA c b a B b b

c a

⎛ ⎞ ⎛ ⎞⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎝ ⎠ ⎝ ⎠

calculate AB and .AB BA−

Question 1.4. Calculate

11 12 1

12 22 2

1 2

( , ,1) .1

a a b xx y a a b y

b b c

⎛ ⎞⎛ ⎞⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎝ ⎠⎝ ⎠


1 1

.

n n

a b

a b

⎛ ⎞⎛ ⎞⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎝ ⎠⎝ ⎠

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0 01 0 .0 1

nλλ

λ

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎟⎜ ⎟⎜⎝ ⎠

Question 1.7. Suppose = , = .AB BA AC CA Prove that

( ) = ( ) , ( ) = ( ) .A B C B C A A BC BC A+ +

Question 1.8. Given

1 1 1= 2 1 0 ,

1 1 0A

⎛ ⎞− ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎟⎜ ⎟⎜ −⎝ ⎠

find 1A− using elementary operations.

Question 1.9. Given matrix

1 3

= ,4 2

A⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

find a 2 2× matrix T such that 1T AT− is diagonal.

Question 1.10. Any idempotent matrix is positive semi-definite.

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Answers for Chapter 1

Answer 1.1. For 1 = ,T AT B− we have

1 1 1

1

= = ( ) =

= = .

I B I T AT T I A T T I A T

T I A T I A

λ λ λ λ

λ λ

− − −

−

− − − −

− −

Thus, A and B have the same set of eigenvalues.

Answer 1.2. We have

2 1 3 1 2 1 3 1 2 1 3 1 2 1 0 74 2 5 4 0 0 1 2 0 0 1 2 0 0 1 2 .2 1 4 1 0 0 1 2 0 0 0 0 0 0 0 0

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞− − − −⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜⎟ ⎟ ⎟ ⎟− → − → − → −⎜ ⎜ ⎜ ⎜⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜⎟ ⎟ ⎟ ⎟⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜⎟ ⎟ ⎟ ⎟⎜ ⎜ ⎜ ⎜− − −⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠

The solution hence is

1 2

3

1= (7 )2

= 2

x x

x

⎧⎪⎪ +⎪⎨⎪⎪ −⎪⎩

where 2x is arbitrary.

Answer 1.3.

2 2 2 2

2 2 2 2

2 2 2 2 2

2 2 2

2

= 2= 2 ,

3

2 2= 2 2 .

3 2

a b c a b c ac bAB a b c ac b a b c

a b c a b c

b ac a b c b c ab ac b c aAB BA c bc ac b a b c b c ab

ac c c bc b ac

⎛ ⎞+ + + + ⎟⎜ ⎟⎜ ⎟⎜ ⎟+ + + + +⎜ ⎟⎜ ⎟⎟⎜ ⎟⎜ ⎟⎜ + + + +⎝ ⎠⎛ ⎞− + + − − − + − − ⎟⎜ ⎟⎜ ⎟⎜ ⎟− − − + + − − −⎜ ⎟⎜ ⎟⎟⎜ ⎟⎜ ⎟⎜ − − − −⎝ ⎠

Answer 1.4. The answer is: 2 211 12 22 1 22 2 2 .a x a xy a y b x b y c+ + + + +

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Answer 1.5. The answer is:

1 1

.

n n

a b

a b

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎟⎜ ⎟⎜⎝ ⎠

Answer 1.6. Denote

0 0 01 0 0 .0 1 0

A⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟≡ ⎜ ⎟⎜ ⎟⎟⎜ ⎟⎜⎝ ⎠

We have

2 3

0 0 0= 0 0 0 , = 0.

1 0 0A A

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎟⎜ ⎟⎜⎝ ⎠

Thus, = 0kA for 3.k ≥ We therefore have

2

=0 =0

1 2 2 1

2 1

1 = ( ) = =1

( 1)= = .2

( 1)2

n

nn i i n i i i n i

n ni i

n

n n n n n

n n n

I A C A C A

n nI n A A nn n n

λλ λ λ λ

λ

λλ λ λ λ λ

λ λ λ

− −

− − −

− −

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟ +⎜ ⎟⎜ ⎟⎟⎜ ⎟⎜⎝ ⎠⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟− ⎜ ⎟⎟⎜+ + ⎟⎜ ⎟⎜ ⎟⎜ ⎟−⎜ ⎟⎜ ⎟⎟⎜ ⎟⎝ ⎠

∑ ∑

Answer 1.7.

( ) = = = ( ) . ( ) = ( ) = ( ) = .A B C AB AC BA CA B C A A BC AB C BA C BCA+ + + +

Answer 1.8.

1

1 103 31 2= 0 .3 32 113 3

A−

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎟⎜ ⎟⎜ ⎟−⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎟⎜ ⎟⎜ ⎟⎜− − ⎟⎜ ⎟⎜⎝ ⎠

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Answer 1.9. Let us use eigenvalues to simplify it (hopefully to a diagonal matrix). We can

treat A as a linear operator 2 2: ,A →R R and try to find a basis of 2R consisting the eigen-

vectors of .A If such a basis exists, then A is diagonal on that basis. From the characteristic

equation:

21 3= = ( 1)( 2) 12 = 3 10 = ( 2)( 5) = 0,

4 2I A

λλ λ λ λ λ λ λ

λ− −

− − − − − − + −− −

we immediately find the eigenvalues:

1 2= 2, = 5.λ λ−

For 1 = 2,λ −

11

1

1 3 3 3( ) = = = 0

4 2 4 4I A x x x

λλ

λ

⎛ ⎞ ⎛ ⎞− − − −⎟ ⎟⎜ ⎜⎟ ⎟− ⎜ ⎜⎟ ⎟⎜ ⎜ ⎟⎟− − − −⎝ ⎠⎝ ⎠

gives us an eigenvector

1

1= .

1x

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟−⎝ ⎠

For 2 = 5,λ

22

2

1 3 4 3( ) = = = 0

4 2 4 3I A x x x

λλ

λ

⎛ ⎞ ⎛ ⎞− − −⎟ ⎟⎜ ⎜⎟ ⎟− ⎜ ⎜⎟ ⎟⎜ ⎜ ⎟⎟− − −⎝ ⎠⎝ ⎠

gives us another eigenvector

2

1= .4

3x

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎟⎜⎝ ⎠

Let

1 2

1 1( , ) = .41

3T x x

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜≡ ⎟⎜ ⎟−⎜ ⎟⎟⎜⎝ ⎠

We have

11 2 1 2 1 1 2 2 1 2

2

0( , ) = ( , ) = ( , ) = ( , ) .

0A x x Ax Ax x x x x

λλ λ

λ

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

That is,

1

2

0= .

0AT T

λλ

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠

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We therefore have

11

2

0 2 0= = .

0 0 5T AT

λλ

−⎛ ⎞ ⎛ ⎞−⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜ ⎟⎟ ⎝ ⎠⎝ ⎠

One can also verify this directly.

Answer 1.10. Suppose 1 = .T AT D− For any vector ,nx ∈ R let y T x′≡ and id be the ith

diagonal entry of ,D we have = 0id or 1 and

( ) ( ) ( ) ( )1 2

=1

= = = = = 0.n

i ii

x Ax x TDT x x TDT x T x D T x y Dy d y− ′′ ′ ′ ′ ′ ′ ′ ≥∑

Thus, A is positive semi-definite.

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Question 2.1. Prove ( ) =c c cS T S T∩ ∪ and ( ) = .c c cS T S T∪ ∩ Use a diagram to see the

intuition.

Question 2.2. Prove 1 1 1( ) = ( ) ( )f A B f A f B− − −∪ ∪ and 1 1 1( ) = ( ) ( ).f A B f A f B− − −∩ ∩

Question 2.3. Prove the Euler Law.

Question 2.4. Prove Theorem 2.9.

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Answer 2.1. We have

( ) or or

,

c c c

c c

x S T x S T x S x T x S x Tx S T

∈ ∩ ⇔ ∉ ∩ ⇔ ∉ ∉ ⇔ ∈ ∈

⇔ ∈ ∪

and

( ) and and

.

c c c

c c

x S T x S T x S x T x S x Tx S T

∈ ∪ ⇔ ∉ ∪ ⇔ ∉ ∉ ⇔ ∈ ∈

⇔ ∈ ∩

Answer 2.2. 1( ) ( ) ( )x f A B f x A B f x A−∈ ∪ ⇔ ∈ ∪ ⇔ ∈ or 1( ) ( )f x B x f A−∈ ⇔ ∈

or 1 1 1) ( ) ( ).x f B x f A f B− − −∈ ⇔ ∈ ∪

1( ) ( ) ( )x f A B f x A B f x A−∈ ∩ ⇔ ∈ ∩ ⇔ ∈ and 1( ) ( )f x B x f A−∈ ⇔ ∈ and 1 1 1( ) ( ) ( ).x f B x f A f B− − −∈ ⇔ ∈ ∩

Answer 2.3. Take differentiation w.r.t. λ on both sides of equation ( ) = ( ).f x f xλ λ We have

=1

( ) = ( ).n

x iii

f x x f xλ∑

Setting λ 1= gives us the result.

Answer 2.4. Take differentiation w.r.t. ix on both sides of equation ( ) ( )af x f xλ λ= . We

have

( ) ( )i i

ax xf x f xλ λ λ= .

We then have 1( ) ( )i ix xf x f xαλ λ −= , that is,

ixf is homogeneous of degree α 1− .

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Question 3.1. Sketch a few level sets for the following functions:

(a) 1 2=y x x

(b) 1 2=y x x+

(c) 1 2= min[ , ]y x x

Question 3.2. Any intersection of convex sets is also convex.

Question 3.3. If ,iS = 1, , ,i k… are convex sets in ,nR then their Cartesian product

1 kS S× × is also convex.

Question 3.4. Let ( )f x be a concave function, and ( )g t be an increasing concave function.

Show that the composite function ( ) [ ( )]h x g f x≡ is also concave.

Question 3.5. Prove that any strictly increasing or decreasing function :f →R R is strictly

quasi-concave.

Question 3.6. Prove the Envelope Theorem.

Question 3.7. Use the Kuhn-Tucker Theorem to derive the FOC for the Lagrange Theorem.

Question 3.8. For : , : ,n n kf ψ→ →R R R R and : ,n mϕ →R R derive the FOC for the

following problem, using the Kuhn-Tucker theorem,

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( )max

s.t. ( ) 0( ) = 0

xf x

xx

ψϕ

⎧⎪⎪⎪⎪⎪ ≥⎨⎪⎪⎪⎪⎪⎩

Question 3.9. Solve

1 2 1 2

1 1 2 2

1 1max,s.t. = ,

x x x xp x p x I

⎧⎪⎪ − −⎪⎪⎨⎪⎪ +⎪⎪⎩

using the FOCs, and verify the SOC, where 1 2,p p and I are positive constants.

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Answer 3.1. The three cases are shown in the following.

(a)

2x

1x1y=

2y=3y=

2x

(b) 1x1y=

2y=3y=

(c)

2x

1x

1y=

2y=

3y=

Figure 3.1. Level Sets

For 1 2min[ , ] = ,x x a one needs to figure out the level sets in half spaces 1 2 1 2{( , ) | < }x x x x

and 1 2 1 2{( , ) | }x x x x≥ separately, and then put them together into one level set for the level

.a

Answer 3.2. Given convex sets , ,iS i I∈ let .ii IS S

∈≡∩ For any two points , ,x y S∈ we

have , ,ix y S∈ for any .i I∈ Since iS is convex, (1 ) ,ix y Sλ λ+ − ∈ for any [0, 1].λ ∈ This is

true for any .i I∈ Therefore, (1 ) .x y Sλ λ+ − ∈ This proves that S is convex.

Answer 3.3. For simplicity, let assume that = 2.k For any two points 1 1( , ),x y 2 2( , )x y

1 2 .S S∈ × Since 1S is convex, we have 1 1 1(1 ) ,x y Sλ λ+ − ∈ for any [0, 1].λ ∈ Analogously,

Since 2S is convex, we have 2 2 2(1 ) ,x y Sλ λ+ − ∈ for any [0, 1].λ ∈ Therefore,

1 1 2 2 1 2( , ) (1 )( , ) .x y x y S Sλ λ+ − ∈ × Thus, 1 2S S× is convex.

Answer 3.4. For any 1 2, ,x x D∈ and [0, 1],λ ∈

1 2 1 2[ (1 ) ] ( ) (1 ) ( ).f x x f x f xλ λ λ λ+ − ≥ + −

Since g is increasing,

{ }1 2 1 2[ (1 ) ] [ ( ) (1 ) ( )].g f x x g f x f xλ λ λ λ+ − ≥ + − (1)

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Since g is concave,

1 2 1 2[ ( ) (1 ) ( )] [ ( )] (1 ) [ ( )].g f x f x g f x g f xλ λ λ λ+ − ≥ + − (2)

Combing (1) and (2) gives

1 2 1 2[ (1 ) ] ( ) (1 ) ( ).h x x h x h xλ λ λ λ+ − ≥ + −

That is, h is concave.

Answer 3.5. Suppose f is strictly decreasing. From ( ) ( ),f y f x≥ we know that .y x≤ If

,x y≠ then < ,y x and thus (1 ) < ,x y xλ λ+ − for all (0, 1).λ ∈ Again, by strictly decreas-

ingness, [ (1 ) ] > ( ).f x y f xλ λ+ − That is, f is strictly quasi-concave.

Answer 3.6.

[ ( ), ] = [ ( ), ] ( ) [ ( ), ].x af x a a f x a a x a f x a aa

∂ ′ +∂

Since ( )x a is an interior maximum point, by FOC, [ ( ), ] = 0.xf x a a Thus,

[ ( ), ] = [ ( ), ],af x a a f x a aa

∂∂

which is the Envelop Theorem.

Answer 3.7. The problem

max ( )s.t. ( ) = 0

f xg x

⎧⎪⎪⎨⎪⎪⎩

is equivalent to

max ( )s.t. ( ) 0

( ) 0

f xg x

g x

⎧⎪⎪⎪⎪ ≥⎨⎪⎪ − ≥⎪⎪⎩

Using Kuhn-Tucker theorem, there are , kμ δ +∈ R such that, for the optimal solution ,x∗

( ) ( ) ( )] = 0.Df x Dg x Dg xμ δ∗ ∗ ∗+ + −

Take .kλ μ δ≡ − ∈ R Then, the FOC for the Lagrange theorem holds.

Notice that λ is no longer a positive vector.

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Answer 3.8. Similar to last answer, the problem is equivalent to

max ( )s.t. ( ) 0

( ) 0( ) 0.

f xxx

x

ψϕϕ

⎧⎪⎪⎪⎪ ≥⎪⎨⎪ ≥⎪⎪⎪ − ≥⎪⎩

By the Kuhn-Tucker theorem, for solution ,x∗ there exist kλ +∈ R and , mη δ +∈ R such that

Kuhn-Tucker conditions: ( ) = 0, ( ) = 0, ( ) = 0,x x xλ ψ η ϕ δ ϕ∗ ∗ ∗⋅ ⋅ ⋅

and

FOC: ( ) ( ) ( ) ( ) = 0.Df x D x D x D xλ ψ η ϕ δ ϕ∗ ∗ ∗ ∗+ + −

Define .mμ η δ≡ − ∈ R The FOC is then

( ) ( ) ( ) = 0.Df x D x D xλ ψ μ ϕ∗ ∗ ∗+ +

Since ( ) = 0,xϕ ∗ the Kuhn-Tucker conditions is reduced to one: ( ) = 0.xλ ψ ∗⋅

Answer 3.9. Let 1 1 2 21 2

1 1 ( ).I p x p xx x

λ≡ − − + − −L The FOCs are

1 22 21 2

1 1= , =p px x

λ λ

imply 21 2

1

= .px xp

Substituting this into the budget constraint will immediately give us

*2

2 1 2

= .Ixp p p+

By symmetry, we also have

*1

1 1 2

= .Ixp p p+

For the SOC, consider

1 2

1 31

2 32

02= 0 .

20

p p

B px

px

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟−⎜ ⎟⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟− ⎟⎜ ⎟⎜ ⎟⎜⎝ ⎠

We have

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1 2 1 2 2 2

3 1 21 2 3 3

3 3 2 12 1

2 2( 1) = = < 0.2 20 0

p p p pp pB p px x

x x− − − −

− −

The SOC is thus verified.

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Question 4.1 (Money Demand). Consider an individual problem

{ } { } { } 0, , =0

1 1 1

max , ,

s t = (1 ) ,

t t t

t tt tC M B t t

t t t t t t t t t

ME u C CP

PC M B PY M i B

β ψ∞

− − −

⎡ ⎤⎛ ⎞⎟⎜⎢ ⎥⎟⎜ ⎟⎢ ⎥⎜ ⎟⎜⎝ ⎠⎢ ⎥⎣ ⎦. . + + + + +

∑ (1)

where ,t tC M and tB are respectively consumption, money stock and bond stock, { }tY is a

given stream of incomes, { }tP is a given steam of prices, and { }ti is a given stream of nominal

interest rates. Money plays the role of the medium of exchange, for which we assume that

leisure is a function = , tt t

t

ML CP

ψ⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜⎝ ⎠

of consumption and real money holdings, decreasing in

tC and increasing .t

t

MP

Since money facilitate transactions, large t

t

MP

gives the individual

more leisure time. More consumption needs more shopping time and less leisure time.

(a) Define the Lagrange function and find the FOCs for ,t tC M and .tB

(b) Show that the three FOCs equations implies a demand for money function of the form:

= ( , ).tt t

t

M f i CP

(c) Suppose 1( , ) =t t t tu C L C Lα α− and , = ,a

at tt t

t t

M MC CP P

ψ −⎛ ⎞ ⎛ ⎞⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎝ ⎠ ⎝ ⎠

with 0 < < 1,α 0 < < 1a and

1 > .aα α− Find an explicit solution of the demand function .f

Question 4.2. There are two assets: a riskless bond with gross return = 1 ,t tR r+ and a risky

asset with gross return = 1 .t tZ z+ Consider an individual problem

{ }

[ ]

1

, =

1

( ) max ( )

s t = ( ) (1 ) ,0,

s S

Ts t

t t t sC s t

s s s s s s s s

s

V A E U C

A A Y C R ZA

ωβ

ω ω

−−

+

≡

. . + − + −

≥

∑ (2)

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where tA is individual wealth, tC is consumption, tY is income, and tω is the proportion of

investment in bonds.

(a) Derive the Euler equations for the bond and for the asset, respectively.

(b) If 2( ) =U C aC bC− for , > 0a b and the real interest rate is equal to the discount rate of

preference: = ,tr θ where 1= ,

1β

θ+ show that consumption is a random walk:

1 1=t t tC C ε+ ++ with 1 = 0.t tE ε +

Question 4.3. Prove 2=0

=(1 )

t

tt ββ

β

∞

−∑ for 0 < 1.β≤

Question 4.4. The individual problem is

{ }

[ ]

1

0, =0

1

max ( )

s t = ( ) (1 ) ,0.

t t

Tt

tC t

t t t t t t t t

t

E U C

A A Y C R ZA

ωβ

ω ω

−

+. . + − + −

≥

∑ (3)

(a) Derive the following two Euler equations:

1

1

( ) = ( ) ,

( ) = ( ) .t t t t

t t t t

U C E U C R

U C E U C Z

β

β

+

+

⎡ ⎤′ ′⎣ ⎦⎡ ⎤′ ′⎣ ⎦

(b) Assume that there is no labor income and all income is derived from tradable wealth, and

( ) = ln .U C C Show that

= (1 ) .t tC Aβ−

Question 4.5 (The Capital Asset Pricing Model). Discrete-Time Version ↔ Continuous-Time

Version.

(a) Suppose that the representative consumer faces the following problem

0 0=0

1 1

0

( ) max ( )

s t = ( ),given

tt

t

t t t t t

V x E u c

x R x y cx

β∞

+ +

≡

. . + −

∑

where tc is consumption (the control), tx wealth (the state), ty exogenous endowment

random process, tR exogenous gross interest rate random process ( tR is 1 plus the net re-

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turn), and 0E the mathematical expectation based on period 0 information. Derive the fa-

mous Euler equation:

11

( )1 = .

( )

't

t t't

u cE R

u cβ +

+

⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦

(b) Assume that ty and tR are nonstochastic processes. Setup the continuous-time version of

the above problem, allowing the discount rate to depend on time, and derive the

corresponding Euler equation.

(c) Prove that when the length of period in the discrete-time model goes to zero, the Euler

equation for the discrete-time model becomes the Euler equation for the continuous-time

model.

Question 4.6. Let ( )B t be the real value of an individual's holding of bonds. The bond pays a

fixed real interest rate .r Let Y be the individual's initial endowment of wealth in real terms.

Solve the following individual's problem for optimal consumption:

0

0

= max ln( )

s t = ,= , ( ) = 0.

t

T

tcV c dt

B rB cB Y B T

∗

. . −

∫ (4)

Here, the individual puts all his wealth into bonds to earn interest: 0 = ,B Y and the individual

will consume all his wealth by the time of death: ( ) = 0.B T

Question 4.7. The individual problem is

( )( )

0log[ ( )]max

( )s.t. = [ ( ) ] ( ) ( ) ( ),

p t s

scc t e dt

dv t r t p v t y t c tdt

θ∞

− + −

≥

+ + −

∫ (5)

where ( )c t is consumption, ( )v t is wealth, and ( )y t is income. Find the Euler equation that

determines the optimal consumption path.

Question 4.8. Consider a dynamic open-economy IS-LM model:

1

= ,

= (1 ) ,

k Me Y i eP

Y Q b Y R a bT G

δδ

δ β α

−∗

⎡ ⎤⎛ ⎞⎢ ⎥⎟⎜ −⎢ ⎥⎟⎜ ⎟⎜⎝ ⎠⎢ ⎥

⎢ ⎥⎣ ⎦− − − + + − +

(6)

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where e is the nominal exchange rate, i∗ is the foreign interest rate, Y is income, T is tax,

G is government spending, M is money stock, P is price level, R is the long-term interest

rate, and = ePQP

∗

is the real exchange rate. Also, , , , , ,b aδ β α and k are constants, > 0,δ

0 < < 1,b and > 0.k

(a) Draw the phase diagram.

(b) Analyze a sudden increase in money supply.


Answer 4.1. (a) The Lagrange function is

[ ]0 1 1 1=0

= , , (1 ) .t tt t t t t t t t t t t t t

t t

ME u C C PY M i B PC M BP

β ψ λ∞

− − −

⎧ ⎫⎡ ⎤⎛ ⎞⎪ ⎪⎪ ⎪⎟⎜⎢ ⎥⎟ + + + + − − −⎜⎨ ⎬⎟⎢ ⎥⎜ ⎟⎜⎪ ⎪⎝ ⎠⎢ ⎥⎪ ⎪⎣ ⎦⎩ ⎭∑L

The FOCs for ,t tC M and tB respectively are:

1

1

== ( )

= (1 )

c l c t t

l m t t t t

t t t t

u u Pu E P

i E

ψ λψ λ β λ

λ β λ+

+

⎧ +⎪⎪⎪⎪ −⎨⎪⎪ +⎪⎪⎩

where ,cu ,lu cψ and mψ are the partial derivatives. Eliminating tP and tλ from above three

equations gives the so-called Euler equation:

( )= .1

tl m c l c

t

iu u ui

ψ ψ++

(7)

This equation and the budget constraint jointly determines the optimal ( , , ).t t tM B C

(b) (7) involves three variables , , ,tt t

t

MC iP

⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜⎝ ⎠ and under certain conditions it can be solved

for a unique :t

t

MP

= ( , ).tt t

t

M L i CP

(8)

We can show that the relation in (8) is increasing in tC and decreasing in ,ti except in cases

with extreme and unrealistic specifications for u and/or .ψ

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(c) Given the special functional forms for u and ,ψ (7) can then be solved to give

1= 1 .

1t

tt t

M a CP a i

αα α

⎛ ⎞⎟⎜ ⎟+⎜ ⎟⎜ ⎟⎜− − ⎝ ⎠

Notice that since 1( , ) = ,a

a tt t t

t

Mu C L CP

α

α α− −⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜⎝ ⎠

the condition 1 > aα α− implies an increasing

u in .tC

Answer 4.2. (a) The Bellman's equation is

[ ]1( ) = ( ) {( ) (1 ) }.max,t t t t t t t t t t t tCt t

V A U C E V A Y C R Zβ ω ωω ++ + − + − (9)

The FOCs are

[ ]{ }1 1: ( ) = (1 ) ( ) ,'t t t t t t t t tC U C E R Z V Aβ ω ω + +

′ + − (10)

1 1: ( ) ( ) = 0.'t t t t t tE R Z V Aω + +

⎡ ⎤−⎢ ⎥⎣ ⎦ (11)

Applying the envelope theorem to (9) yields

[ ]{ }1 1( ) = (1 ) ( ) .' 't t t t t t t t tV A E R Z V Aβ ω ω + ++ − (12)

By (10) and (12),

( ) = ( ),t t tV A U C′ ′ (13)

implying that the marginal value of financial wealth equals the marginal utility of consump-

tion. We can use (13) to eliminate 1 1( )'t tV A+ + in (10)-(11), which gives

[ ]{ }1( ) = ( ) (1 ) ,t t t t t t tU C E U C R Zβ ω ω+′ ′ + − (14)

1 1( ) = ( ) .t t t t t tE U C R E U C Z+ +⎡ ⎤ ⎡ ⎤′ ′⎣ ⎦ ⎣ ⎦ (15)

Substituting (15), using the fact that tω is known at time ,t yields

1( ) = ( ) ,t t t tU C E U C Rβ +⎡ ⎤′ ′⎣ ⎦ (16)

1( ) = ( ) ,t t t tU C E U C Zβ +⎡ ⎤′ ′⎣ ⎦ (17)

which are the Euler equations for the bond and the asset.

(b) Equation (16) implies

1 1 1( ) = ( ) , = 0.' 't t t t t tU C R U C Eβ ε ε+ + ++ (18)

If =tr θ and U is quadratic with 2( ) =U C aC bC− for , > 0,a b then (18) becomes

1 1 1= , = 0.t t t t tC C Eε ε+ + ++ (19)

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That is, consumption is a random walk: tC is the best predictor of 1tC + and no additional

information at time t can help the prediction.

Answer 4.3. We have

=0

1= .1

t

tβ

β

∞

−∑

Taking the derivative w.r.t. β on this equation yields

12

=0

1= .(1 )

t

t

tββ

∞−

−∑

Multiplying β on both sides of the equation then implies

2=0

= .(1 )

t

t

t ββ

β

∞

−∑

Answer 4.4. (a) The solution of (3) must satisfy the following Bellman equation:

1 1,( ) = max ( ) ( ).

t tt t t t t tC

V A U C E V Aω

β + ++ (20)

By this, the multi-period problem (5) becomes a two-period problem (20). Thus, instead of

solving for a sequence of variables 1=0{ , }T

t t tC ω − in (5), we only need to solve for two variables

( , )t tC ω at time t in (20).

When tZ follows a stationary Markov process, the value function ( )tV A can be shown to

be independent of .t In this case, we say that problem (20) is recursive, and it becomes quite

tractable. Further, for many popular forms of utility functions, ( )V A turns out to have the

same form as the underlying utility function ( ),U C in which case an explicit solution can

usually be found.

Substituting the equality constraint of (3) into (20) simplifies problem (20) to

[ ]{ }1,( ) = max ( ) ( ) (1 ) .

t tt t t t t t t t t t t tC

V A U C E V A Y C R Zω

β ω ω++ + − + − (21)

The FOCs are

[ ]{ }1 1: ( ) = (1 ) ( ) ,'t t t t t t t t tC U C E R Z V Aβ ω ω + +

′ + − (22)

1 1: ( ) ( ) = 0.'t t t t t tE R Z V Aω + +

⎡ ⎤−⎢ ⎥⎣ ⎦ (23)

Applying the envelope theorem to (21) yields

[ ]{ }1 1( ) = (1 ) ( ) .' 't t t t t t t t tV A E R Z V Aβ ω ω + ++ − (24)

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By (22) and (24),

( ) = ( ),' 't t tV A U C (25)

implying that the marginal value of financial wealth equals the marginal utility of consump-

tion. We can use (25) to eliminate 1 1( )'t tV A+ + in (22)-(23), which gives

[ ]{ }1( ) = ( ) (1 ) ,t t t t t t tU C E U C R Zβ ω ω+′ ′ + − (26)

1 1( ) = ( ) .t t t t t tE U C R E U C Z+ +⎡ ⎤ ⎡ ⎤′ ′⎣ ⎦ ⎣ ⎦ (27)

Substituting (27), using the fact that tω is known at time ,t yields

1

1

( ) = ( ) ,

( ) = ( ) .t t t t

t t t t

U C E U C R

U C E U C Z

β

β

+

+

⎡ ⎤′ ′⎣ ⎦⎡ ⎤′ ′⎣ ⎦

(28)

(b) We guess that the value function is of the same form:

( ) = ln ,V A a A b+

where a and b are two constants to be determined. Then, problem (21) becomes

[ ]

1,

1

( ) = max ln ( ln )

s.t. = ( ) (1 ) .t t

t t t tC

t t t t t t t

V A C E a A b

A A C R Zω

β

ω ω

+

+

+ +

− + −

Substituting the budget equation into the objective function reduces the problem to

[ ]{ }( ),

( ) = max ln ln ( ) (1 ) .t t

t t t t t t t t tCV A C E a A C R Z b

ωβ ω ω+ − + − + (29)

Then, by (25), we find

1 = ,

t t

aC A

implying

1= .t tC Aa

(30)

By (23), we find

1

= 0,t tt

t

R ZEA+

⎛ ⎞− ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜⎝ ⎠

implying

= 0.(1 )

t tt

t t t t

R ZER Zω ω

⎡ ⎤−⎢ ⎥⎢ ⎥+ −⎣ ⎦

(31)

Equation (30) says that consumption is a fixed proportion of wealth. Equation (31) implicitly

defines the optimal .tω If tr is fixed over time and tZ is i.i.d, this optimal tω is fixed over time.

of 29

By (29), we have

[ ]

[ ]

1ln = ln ln ln (1 )

1= (1 ) ln ln ln (1 ) .

t t t t t t t t

t t t t t t

aa A b A a a A E R Z ba

aa A a a E R Z ba

β ω ω β

β β ω ω β

⎧ ⎫⎛ ⎞−⎪ ⎪⎪ ⎪⎟⎜+ − + + + − +⎟⎨ ⎬⎜ ⎟⎜⎪ ⎪⎝ ⎠⎪ ⎪⎩ ⎭⎧ ⎫−⎪ ⎪⎪ ⎪+ − + + + − +⎨ ⎬⎪ ⎪⎪ ⎪⎩ ⎭

(32)

By (31), tω must be independent of .tA Thus, (32) implies

[ ]

= 1 ,1= ln ln (1 ) .t t t t t

a aab a a E R Z b

a

β

β ω ω β

+⎧ ⎫−⎪ ⎪⎪ ⎪− + + + − +⎨ ⎬⎪ ⎪⎪ ⎪⎩ ⎭

The first equation can determine ,a and then the second equation can determine .b We find

1= .

1a

β−

Thus, by (30),

= (1 ) .t tC Aβ− (33)

Answer 4.5.

The Bellman equation is

{ }1( ) = max ( ) [ ( )] .t

t t t t t t tcV x u c E V R x y cβ ++ + −

implying

1 1

1 1

FOC: ( ) = [ ( ) ]Envelope: ( ) = [ ( ) ]

' 't t t t

't t t t

u c E V x RV x E V' x R

ββ

+ +

+ +

⎧⎪⎪⎨⎪⎪⎩

implying

( ) = ( ),' 't tu c V x

implying

1 1( ) = [ ( ) ].' 't t t tu c E u x Rβ + +

This gives Euler equation:

11

( )1 = .

( )

't

t t't

u cE R

u cβ +

+

⎡ ⎤⎢ ⎥⎢ ⎥⎣ ⎦

(b) Let 1= 1 .

1t tt

R rr

≈ +−

Then the budget constraint for the discrete-time problem is

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1 1(1 ) = ,t t t t tx r x y c+ +− + −

implying

1 1 1= .t t t t t tx x r x y c+ + +− + −

Thus, the budget constraint for the continuous-time problem is

=x rx y c+ −

and the corresponding continuous-time version of the problem is

0( )

0

0

max ( )

s t = ,given (0) =

td

u c e dt

x rx y cx x

ρ τ τ∞

−∫

. . + −

∫

The Hamilton function is

0( )

( , , , , ) ( ) ( ),t

dH x x c t u c e rx y c x

ρ τ τλ λ

−∫≡ + + − −

implying

= .x x

c c

H Hd H Hdt

H H

λλ⎛ ⎞ ⎛ ⎞⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎟ ⎟⎟ ⎟⎜ ⎜⎟ ⎟⎜ ⎜⎝ ⎠ ⎝ ⎠

is

0

( )= .

0 ( )t

d'

rddt u c e

ρ τ τ

λλ

λ−

⎛ ⎞⎛ ⎞ ⎟− ⎜ ⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎟⎟ ⎜⎜ ⎟ ⎟∫⎝ ⎠ ⎜ ⎟−⎝ ⎠

implying

0

( )

=

= ( )t

d

r

u c eρ τ τ

λ λ

λ−

⎧⎪ −⎪⎪⎨⎪ ∫⎪ ′⎪⎩

By substituting the 2nd equation into the 1st one, we have the Euler equation:

( )= ( ) .( )

u cc ru c

ρ′

−′′

(c) Let 1=

1β

ρ+ and

1=1t

t

Rr−

. Then the discrete-time nonstochastic version of Euler

equation becomes

1 1( )(1 )(1 ) = ( ).' 't t tu c r u cρ + ++ −

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Let the length of period be .tΔ Then, the above equation becomes

( )(1 )(1 ) = ( ),' 't t t t tu c t r t u cρ +Δ +Δ+ Δ − Δ

implying

2( )[1 ] = ( ).' 't t t t t t tu c r t t r t u cρ ρ+Δ +Δ +Δ− Δ + Δ − Δ

implying

( ) ( )

( )( ) = .' '

' t t tt t t t t

u c u cu c r r t

tρ ρ +Δ

+Δ +Δ

−− − Δ

Δ

Letting 0tΔ → implies that

( )( ) = ( )' '' tt t t

dcu c r u cdt

ρ−

which is the continuous-time version of the Euler equation.

Answer 4.6. The Hamiltonian for (4) is

= ln ( ).c rB cλ+ −H

The Euler equations are

1= 0 : = ,

= : = .

c

B

cr

λ

λ λ λ− −

H

H

Then,

( ) = 0,

rtd edtλ

implying

= ,rtAeλ −

where A is a constant. Then,

1= .rt

tc eA

(34)

To determine ,A use the two conditions in (4), implying

( ) = ,

rtrtd Be ce

dt

−−−

implying

0 0( ) = ,

TrT rt

tB T e B c e dt− −− −∫

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implying

0

= .T

rttc e dt Y−∫ (35)

Substituting (34) into (35) gives

= ,T YA

implying = / .A T Y Thus,

= .rtt

Yc eT

Answer 4.7. The Hamiltonian is

[ ]= log ( ) .c r p v y cλ+ + + −H

The Euler equations are

1= 0 : = ,c c

λH (36)

= ( ) : = ( ) ( ),vp p r pλ θ λ λ θ λ λ+ − + − +H (37)

( )transversality: = 0.lim p t

tve θλ − +

→∞ (38)

Solving for c from (36) and substituting it into (37) yields

2 = ,c rc c

θ−−

i.e.,

= [ ( ) ] ,c r t cθ−

which is the Euler equation determining ( ).c t

Answer 4.8. (a) The IS and LM curves are

1

: = 0 or = ,

: = 0 or (1 ) = 0.

k MLM e Y iP

IS Y Q b Y R a bT G

δδ

δ β α

−∗⎛ ⎞⎟⎜ ⎟⎜ ⎟⎜⎝ ⎠

− − − + + − +

of 29

. E

IS

LM

Y

e

Y

S

Figure 2-4. Phase Diagram for an Open Economy

(c) With an increase in money stock, the LM curve shifts out, resulting a new equilibrium .'E

The converging path S shifts to .'S The dynamics is: as soon as the money stock is

increased, the economy jumps from E to point ,A and then continuously moves towards 'E along the new converging path .'S

. E

IS

LM

Y

e

Y

E’

A..

S

'S

Figure 2-5. A Sudden Money Increase

The nominal exchange rate initially jumps up and then continuously comes down to a new

level. That is, there is an exchange rate overshooting in the short-run. This is the well-known

exchange rate overshooting phenomenon first analyzed by Dornbusch.

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Question 5.1. Find the implicit solution for the following differential equation:

3 4 24 5 2 = 0.x dx y dy xydx x dy+ + +

Question 5.2. Solve equation

5 4

5 4

1 = 0.d x d xdt t dt

−

of 29


Answer 5.1. We have

3 4 2 4 5 2 2

4 5 2

4 5 2

4 5 2 == ( )= ( ).

x dx y dy xydx x dy dx dy ydx x dydx dy d yxd x y yx

+ + + + + ++ +

+ +

The original equation is thus equivalent to

4 5 2( ) = 0.d x y yx+ +

This equation obviously has the general solution of the form

4 5 2 = ,x y yx c+ +

where c is an arbitrary constant.

Answer 5.2. Let

4

4= .d xydt

Then the equation becomes a 1st-order equation:

1 = 0.dy y

dt t−

This can be easily solved to get = ,y ct where c is an arbitrary constant. Then,

4

4 = .d x ctdt

(3)

Laplace method gives the general solution for (1):

5 3 21 2 3 4 5= ,x c t c t c t c t c+ + + + (4)

for arbitrary 2 3 4, ,c c c and 5c and some 1.c (2) is thus the general solution for the original

equation, for arbitrary 1 2 3 4, , ,c c c c and 5.c

Exercises for Math in Economics

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