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 some values of $x$. 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 $u((1-\pi)w)=E[u(1+\epsilon \tilde)w)]$. But I'm not sure how to apply this information to solve the question.

  • $\begingroup$ Welcome to Econ.SE. I formatted your question, notice that you can use dollar signs to add latex-like mathematic formulas. I was unsure about the last part, where you had tilde. $\endgroup$
    – FooBar
    Feb 4, 2015 at 16:52

1 Answer 1


Whenever you need to make someone indifferent between $x$ and $y$, it means that

$$U(x) = U(y)$$


Denote amount he would pay by $z$. Paying $z$ to avoid the lottery gives him "certain utility" $u(100.000 - z)$. With Von Neumann-Morgenstern utility as given,
we can denote the lottery utilities as $$prob*U(10.000 + x) + (1-Prob)*U(10.000-x)$$ (Understand why we can do that!)


Without uncertainty, he will have $u(w)$. Then, you need to compute the utility of the lottery as the function of $p, x,y$.

That is, you need to solve

$$U(100000) = prob*U(w+y) + (1-prob)*U(w-x)$$

Now, I'm not sure whether you were meant to say loses x or wins y, I suppose it is loses x or wins y. Anyhow, the value of the lottery is on the right hand side. Can it ever dominate the left hand side, given the additional information for $p$ that you were told in that subquestion?

  • $\begingroup$ Thanks this is great! I have one question though..how come in part a) if I take U=2w^(1/2) or U= ln(2w) which are both valid for p=0.5, I get that the answer for U=ln(2w) is twice as much? $\endgroup$
    – User38
    Feb 4, 2015 at 19:07

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