Calculating Expected CARA Utility for Normally Distributed Returns

Summary

The document shows how to calculate expected utility under a constant absolute risk aversion (CARA) negative exponential utility function. It first expresses expected utility as the integral of utility over the distribution of the outcome, then illustrates the idea with a discrete distribution by weighting utility values by their probabilities.

For a standard normal variable, it substitutes the density into the expectation and evaluates the resulting exponential integral by completing the square or using a Gaussian integral identity. This yields an expression for expected utility in that specific standard-normal case. The derivation is useful for understanding the calculation, but the stated normal density is standardized; applying the result to returns with a different mean or variance requires adjusting the distribution parameters. The document does not discuss how to estimate risk aversion or use the result to select a portfolio.

Key ideas

• Expected utility is the probability-weighted average of utility across possible outcomes.
• For continuous outcomes, expected utility is computed by integrating utility against the outcome density.
• A discrete example illustrates the same calculation as a sum of probability-weighted utility values.
• For the standard normal case shown, the exponential integral can be evaluated with a Gaussian integral identity.
• The displayed normal density is standardized, so other means or variances require adapting the calculation.

Tags

# CARA Utility function expected utility


# CARA Utility function expected utility












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?

## Answer by Konsta (score 3)

https://quant.stackexchange.com/a/4308

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?

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

This summary was written by Stratmill's research agent from the original; it is not a copy of the source.