2
$\begingroup$

I have been trying to undestand how the expected utility for a CARA negative exponential Utility function is calculated. In my particular case the variable has normally distributed returns.

Could someone help me?

$\endgroup$
3
$\begingroup$

CARA Utility function $u(c)=\frac{-e^{-ac}}{a}$ for $a>0.$

Expected utility $E(u(c))=\int_{-\infty}^{\infty} u(c) f(c) dc,$ where f is a density.

Example f(10)=0.3, f(20)=0.7, else f=0 and a=2. Then $E(u(c))=0.3\times u(10)+0.7\times u(20)=0.3\times \frac{-e^{-2*10}}{2}+0.7\times \frac{-e^{-2*20}}{2}$

Now, for normal density $f(c)=\frac{1}{\sqrt{2\pi}}\; e^{-c^2/2}$ we have $E(u(c))=\int_{-\infty}^{\infty} u(c) f(c) dc=\int_{-\infty}^{\infty} \frac{-e^{-ac}}{a} \frac{1}{\sqrt{2\pi}}\; e^{-c^2/2} dc=\frac{-1}{a\sqrt{2\pi}}\; \int_{-\infty}^{\infty} e^{-ac} e^{-c^2/2} dc=\frac{-1}{a\sqrt{2\pi}}\; \int_{-\infty}^{\infty} e^{-ac-c^2/2} dc.$

Here the integral $\int_{-\infty}^{\infty} e^{-ac-c^2/2} dc$ can be solved with help of the solution $\int_{-\infty}^{\infty} e^{-Ax^2} e^{-2Bx}\,dx=\sqrt{\frac{\pi}{A}}e^{\frac{B^2}{A}} \quad (A>0)$ from the list of integrals of exponential functions.

Comparison yields A=0.5, B=0.5a, thus $E(u(c))=\frac{-1}{a\sqrt{2\pi}}\sqrt{\frac{\pi}{0.5}}e^{\frac{(0.5a)^2}{0.5}}=\frac{-1}{a\sqrt{2\pi}}\sqrt{2\pi}e^{\frac{0.25a^2}{0.5}}=\frac{-e^{0.5a^2}}{a}$

And this can be nicely plotted with Wolfram alpha. Does your solution confirm this?

$\endgroup$

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.