3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – User38 Feb 4 '15 at 18:37
  • $\begingroup$ 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. $\endgroup$ – Dachser Feb 5 '15 at 9:33

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.