for CRRA, does increasing gamma leads to increase in risk-aversion?

Looking at the curve, I think increasing gamma leads to less in risk-aversion (since the risk preimum is less). But in terms of absolute risk aversion, CRRA = $\gamma /X$. Looks like increasing $\gamma $ leads to high risk-aversion. Which is right?


If the utility function $W \mapsto U(W)$ (where $W$ is wealth) is concave, then the individual is risk-averse and unwilling to accept any actuariually fair gamble.

We can distinguish between absolute risk aversion ($ARA$) and relative risk aversion ($RRA$)

$$ARA(W) = - \frac{U''(W)}{U'(W)},\quad RRA(W) = - W\frac{U''(W)}{U'(W)}$$

Here, $ARA(W)$ determines the absolute amount the individual is willing to pay to avoid a gamble of a given absolute size. Similarly, $RRA(W)$ determines the relative amount, i.e., fraction of wealth, the individual is willing to pay to avoid a gamble of a given size relative to wealth. A derivation for $ARA$ is given here and is easily modified for $RRA$ by replacing $\epsilon$ and $\delta$ with $\epsilon/W$ and $\delta/W$, respectively.

As you would expect, a $CRRA$ utility function has constant relative risk aversion $\gamma$,

$$RRA(W) = -W \frac{U''(W)}{U'(W)} = - W \frac{d}{dW} \log U'(W) = \gamma$$

Without loss of generality in terms of constants, we can solve for $U$ as

$$U(W) = \frac{W^{1-\gamma}-1}{1-\gamma}$$

To ensure concavity (risk aversion) we must have $\gamma > 0$. The case where $\gamma = 0$ corresponds to a linear utility function (risk neutrality) and in the limit as $\gamma \to 1$ we have , by L'Hopital's rule,

$$\lim_{\gamma \to 1}U(W) = \log W$$

With $\gamma$ fixed the fraction of wealth the individual pays to avoid a gamble is, of course, independent of wealth since this is $CRRA$. Nevertheless the fraction of wealth paid would increase as $\gamma$ increases.

However, since $ARA(W) = RRA(W)/W$, the absolute amount paid decreases with increasing wealth.

| improve this answer | |

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.