Decreasing Absolute Risk Aversion and third derivative

Question

(i) Show that if the Arrow-Pratt coefficient of absolute risk aversion ${ARA}_U(w)$ is a decreasing function, then $U^{\prime\prime\prime}>0$.

(ii) Find all the utility functions $U$ defined on $(0,\infty)$ for which ${ARA}_U (w)=\frac{2}{w}$

Answer 1

For part (b), note that $$-\frac{U''(w)}{U^\prime(w)} = -\frac{d \ln U^\prime (w)}{dw}$$ Then let $${ARA}_U (w)= -\frac{U''(w)}{U(w)} = \frac{1}{2w}$$ Integrating both sides once, we find $$\frac{d \ln U^\prime (w)}{dw} = -\frac{1}{2w} + c_0$$ where $c_0$ is an arbitrary constant. Exponentiating both sides gives $$U^\prime(w) = \frac{c_1}{w^2}$$ where $c_1$ is an arbitrary constant. Finally, integrating again yields $$U(w) = \frac{k_0}{w} + k_1w$$ where both $k_0$ and $k_1$ are arbitrary constants. (You may need to impose obvious restrictions to make sure that $U$ is increasing.)

Answer 2

First of all the utility function, $U(w)$, has to be twice continuously differentiable and monotone increasing. Thus $U'(w)>0$. You have

$ARA(w)=-\frac{U''(w)}{U'(w)}$

To get the derivative you have to apply the quotient rule.

$f(x)=\frac{u(x)}{v(x)}\Rightarrow f'(x)=\large{\frac{u'(x)\cdot v(x)-v'(x)\cdot u(x)}{[v'(x)]^2}}$

$ARA'(w)=-\frac{U'''(w)\cdot U'(w)-U''(w)\cdot U''(w)}{[U'(w)]^2}=\frac{U''(w)\cdot U''(w)}{[U'(w)]^2}-\frac{U'''(w)\cdot U'(w)}{[U'(w)]^2}$

$=ARA^2-\frac{U'''(w)\cdot U'(w)}{[U'(w)]^2}<0$

Therefore $\frac{U'''(w)\cdot U'(w)}{[U'(w)]^2}>0\Rightarrow U'''(w)\cdot U'(w)>0$

Since $U'(w)>0$, it follows $U'''(w)>0$

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Edit: For the second question I have found no solution. Maybe you have a typo somewhere.