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Say I have the following utility function:

$$u(x,w)=f(w+x)$$

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!

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2 Answers 2

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We have $$-\frac{f''}{f'}-\frac 1 x - a = 0 \implies f'' = (-\frac 1x -a ) f'.$$

Set $y\equiv f'$ to arrive at $$y' = (-\frac 1x -a ) y.$$

This has solution $$f' = y : \int \frac 1y dy = \int (-\frac 1x -a ) dx$$ $$\implies \ln y = -\ln x -ax \implies y = \frac 1x e^{-ax}.$$

So $$f' = \frac 1x e^{-ax} \implies f(x) = \int^{ax}_{-\infty}\frac 1t e^{-t}dt.$$

This is related to the Exponential Integral. It satisfies the required inequality for any $a>0$. The lower bound need not be minus infinity, for your purposes. Explore how you can insert the exogenous parameter $w$ in there.

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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 $x\in R_+$.

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