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?


1 Answer 1


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?


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