# Constant Relative Risk Aversion

The question:

Consider a person with constant relative risk aversion p.

(a) Suppose the person has wealth of 100,000 and faces a gamble in which he wins or loses x with equal probabilities. Calculate the amount he would pay to avoid the gamble, for various values of p (say, between 0.5 and 40), and for x=100, x=1000, x=10,000 and x= 25000. For large gambles, do large values of p seem reasonable? What about small gambles?

(b) Suppose p > 1 and the person has wealth w. Suppose he is offered a gamble in which he loses x or wins y with equal probabilities. Show that he will reject the gamble no matter how large y is if p >= (log(0.5)+log(1-x/w))/log(1-x/w).

I'm not sure where to start with this. Am I solving for the risk premium and multiplying by w?

I know that for someone with CRRA utility u(w)= (1/(1-p))w^(1-p) and that an individual will pay pi(w) to avoid the gamble if u((1-pi)w)=E[u(1+epsilon tilda)w)]. But I'm not sure how to apply this information to solve the question. Any help is appreciated.

if you have $p=0.5$ For example: $U(w)=ln(2w)$

why is that? relative risk aversion is given by

$$RRA=\frac{-wU''(w)}{U('w)}=\frac{-w*(-1/4w^2)}{1/2w}=0.5$$

Now you can apply your formula.

take for example: $x= 10000$ and $\pi=0.5=1-\pi.$ then expected utility is equal to

$EU(x,w)=0.5*ln(2*(w+x))+0.5*ln(2*(w-x))=0.5ln(220000)+0.5ln(180000)$

you want to know $RP:Risk~premium$

so you need to solve $U(w-RP)=EU(x,w)=0.5ln(220000)+0.5ln(180000)$

or $ln(200000-2RP)=12.20$

from which follows that $$RP=\frac{200000 -e^{12,2}}2=501.25$$

which is the amount the investor is ready to pay in order to avoid the gamble.

• Thanks @Dachser this is great. How come when I take u= 2w^(1/2) for p=0.5 I come up with an answer half that of yours? – User38 Feb 4 '15 at 18:37
• Your assumuption is right, too. I am sorry - I thought mine was the only one. It is clear that if you take another utility function as a basis you get a different risk premium. It's just the nature of mathematics :) The comparison of different values for $x$ and $p$ should make no difference though. – Dachser Feb 5 '15 at 9:33