3. ejer 2.1_Gibbons

Problem Set 4 Solution 17.881/882 October 26, 2004

1

Gibbons 2.1 (p.130) (p.130)

This is a dynamic game of perfect information, we will use backward induction to solve. We start at the final stage. The parent’s objective is max V  (I  p (A) − B ) + kU (I c (A) + B ) B

The first-order condition is: 0

0

−V   (I  p (A) − B ) + kU  (I c (A) + B ) = 0

(1)

(I’ll omit discussion of the second-order condition). This equation is defining an implicit relation B (A). In the first stage, the child anticipates his choice of A to a ff ect ect B according to 1. The child’s problem is max U (I c (A) + B (A)) A

The first-order condition is: 0

0

0

U  (I c (A) + B )[I c (A) + B (A)]

=0

(2)

(I’ll omit discussion of the second-order condition). Since U  >  0,  0 , the only way for 2 to hold is to have 0

0

0

I c (A) =

To find

B 0 (A),

−B (A)

(3)

let us use the implicit function theorem on 1 . dB dA

00

=−

−V  

I  p00

V  00

00

+ kU  + kU  00

I c0

Using 3, and solving for

I c0 ,

we find

00

0

0

V   [I c (A) + I  p (A)]

=0 0

0

Since V    is strictly concave, this can only hold if  I c (A) + I  p (A) = 0, which is exactly the first-order condition of the joint-income maximization problem: max I c (A) + I  p (A) A