Say I have the following utility function:


This utility exhibits decreasing absolute risk aversion ($f'>0$, $f''<0$ and $f'''>0$). $x\in R_+$ is the control variable and $w\in R_+$ is an exogenous parameter.

Is there a parametric form of $f$ for which $A(x)>\frac{1}{x}$?

Where $A(x)=\frac{-f''}{f'}$ is the Arrow-Pratt measure of absolute risk-aversion. I cannot manage to find such a function, which makes me believe that maybe the Arrow-Pratt measure has a lower bound when the utility is DARA.

Many thanks!


1 Answer 1


Ok, I think the answer is no. There does not exist a utility function satisfying DARA such that $A(x)>\frac{1}{x}$ for all $x\in R_+$.

To prove that, let's assume such a function exists. Then, there must exists $a\in R_+$ such that: $$\frac{-f''}{f'}>\frac{1}{x}\Leftrightarrow\frac{-f''}{f'}-\frac{1}{x}-a=0$$ If we rearrange this function a bit by taking $y=x+w$, we have: $$-\frac{f''(y)}{f'(y)}=\frac{1+a(y-w)}{y-w}\qquad\text{(1)}$$ Expression (1) is a second-order nonlinear ordinary differential equation. Solving it (here pardon my laziness, I used Wolfram) gives: $$f(y)=\int_1^{y}exp\bigg(\int_{1}^{\phi}\frac{a(\rho-w)+1}{w-\rho}d\rho\bigg)c_1d\phi+c_2$$ Taking the derivative of $f$ we get: $$f'(y)=exp\bigg(\int_1^{y}\frac{a(\rho-w)+1}{w-\rho}d\rho\bigg)c_1$$ So we do have the condition $f'>0$ fulfilled as long as $c_1>0$. Taking the second-order derivative of $f$, we get: $$f''(y)=\frac{a(y-w)+1}{w-y}exp\bigg(\int_1^{y}\frac{a(\rho-w)+1}{w-\rho}d\rho\bigg)c_1$$ This expression is negative if and only if: $$\frac{a(y-w)+1}{w-y}>0$$ Using the fact that $x=y-w$: $$-\frac{1+ax}{x}>0\Leftrightarrow x<-\frac{1}{a}$$ We have a contradiction -- if I did not make any mistake in the computations -- there does not exist a DARA utility function where $A(x)>\frac{1}{x}$, for all $x\in R_+$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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