Question

4 Problem 4 (2.5 pts) An investor has utility function: u(C++1) = 1- Suppose the investor will consume $50,000 next year. How much would he be willing to give up at t +1 to avoid an even bet of gaining or losing $5, $50, $500, $5000 if y = 0, 1, 2, 10, 50? Hints: To answer this question, equate utility of $50,000 minus x for sure to the expected utility of winning and losing the bet. It's prettier if you set the problem up as, what fraction of consumption x would you give up for sure to avoid a bet on a fraction of consumption y, i.e. how much would you give up for sure xc to avoid the 50/50 chance of gaining or losing yc: u(c – xc) = fu(c – ye) + fu(c + yc)

Answer 1

Using the hint and the equation:
u(c-xc) = 1/2u(c-yc) + 1/2u(c+uc)

equates the utility of a certain consumption of c-xc(LHS) or an even bet of yc on the initial amount c (RHS)

It means we have to find x in each of the cases where y= 0.0001 (=5/50,000) , 0.001, 0.01 , 0.1 and gamma γ(Which I will denote with g for ease) 0,1,2,10,50

Now writing it down

(c-xc)^1-g/1-g = 1/2(c-yc)^1-g/1-g + 1/2(c+yc)^1-g/1-g

Multiplying by 1-g on both LHS and RHS we get

(c-xc)^1-g = 1/2(c-yc)^1-g + 1/2(c+yc)^1-g

For g=0, we get c-xc= 0.5c+0.5c

Therefore x=0 (The investor won't pay any amount for a certain loss and is thus indifferent to facing an even bet on the 50,000$. We can also see it from the utility function u(c)=c so his utility for the even bet will be the average expected outcome of the bet= 50,000$)

For g=1 our current equation and utility function breaks down so we cannot calculate it.

For g=2,

1/(c-xc)= 0.5/(c-yc)+0.5/(c+yc)

1) Calculating it for y=0.0001 or yc=5$

1/(c-xc)= 0.5/49,995+0.5/50,005

x=10^-8

2) For yc= 50$

1/50,000(1-x))= 0.5/49,950+0.5/50,050

x= 10^-6

3) For yc= 500$

1/(c-xc)= 0.5/49,500+0.5/50,500

x= 10^-4

⁴⁾

For yc= 5,000$

1/(c-xc)= 0.5/45,000+0.5/55,000

x=0.01

Now to solve for for g= 10 and 50 we get

c-xc= (1/2(c-yc)^1-g + 1/2(c+yc)^1-g)^1/(1-g)

1-x= (1/50,000)(1/2(c-yc)^1-g + 1/2(c+yc)^1-g)^1/(1-g)

g=10

x=1-(1/50,000)(1/2(c-yc)^1-g + 1/2(c+yc)^1-g)^1/(1-g)

For g=10

yc=5$

x= 1-(1/50000)((0.5 × 49995)^^-9 + (0.5 × 50005)^^-9)^^{-(1 ÷ 9)}=0.537062667

yc=50$

x= 0.5370649585

yc=500$, x= 0.5372937892

yc=5000$,x= 0.5575422572

g=50

yc=5$, x= 0.5070232624

yc=50$,x= 0.5070354586

yc=500$,x= 0.5082087824

yc=5000$ x= 0.5500004928

Here's a table of all the compiled answers. Apologies if I might have made a mistake.

g/y	0.1	0.01	0.001	0.0001
0	0	0	0	0
1	-	-	-	-
2	0.01	0.0001	0.000001	0.00000001
10	0.5575422572	0.5372937892	0.5370649585	0.537062667
50	0.5500004928	0.5082087824	0.5070354586	0.5070232624

Hope it helps