Easiest way to see this expectation?

Question

Say you have some normal random variable $X \sim N\left( {0 ,{\sigma ^2}} \right)$ and need the expectation ${\rm{E}}\left[ {{\rm{exp}}\left( {\gamma {{\rm{X}}^2}} \right)} \right]$. Any quick ways to calculate it out? I have been finding the equivalent gamma distribution and then using the moment generating function, but that's easy to mess up under exam conditions. Thanks for the help.

Answer 1

If $X \sim \operatorname{Normal}(0, \sigma^2)$, then $X/\sigma \sim \operatorname{Normal}(0,1)$, and $$Y = X^2/\sigma^2 \sim \chi^2(1);$$ that is to say, $X^2/\sigma^2$ is chi-square distributed with one degree of freedom. Thus $$\operatorname{E}[e^{\gamma X^2}] = M_Y(\gamma \sigma^2)$$ is the moment-generating function of a chi-square distribution evaluated at $\gamma\sigma^2$.

If you do not recall the MGF, the integration itself is not that hard: $$\operatorname{E}[e^{\gamma X^2}] = \int_{x=-\infty}^\infty e^{\gamma x^2} \frac{1}{\sqrt{2\pi} \sigma} e^{-x^2/(2\sigma^2)} \, dx = \frac{K}{\sigma} \int_{x=-\infty}^\infty \frac{1}{\sqrt{2\pi} K} e^{-x^2/(2K^2)} \, dx = \frac{K}{\sigma},$$ where we require $K > 0$ to be a constant satisfying $$\frac{1}{2K^2} = \frac{1}{2\sigma^2} - \gamma,$$ or $$K = \frac{\sigma}{\sqrt{1 - 2\gamma\sigma^2}}.$$ Thus $$\operatorname{E}[e^{\gamma X^2}] = (1 - 2\gamma\sigma^2)^{-1/2}, \quad \gamma < \frac{1}{2\sigma^2}.$$