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This is part of a homework problem. I think I've got it right, but it isn't fitting into the next step, so wanted to check this one.

Question: What is $E(e^{rX^2})$ when $X$ is $N(0,\sigma^2)$?

My answer:

$E(e^{rX^2})=\int_{-\infty}^{\infty}{\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{1}{2}\frac{x^2}{\sigma^2}}}e^{rx^2}dx=\frac{1}{\sqrt{2\pi\sigma^2}}\int_{-\infty}^{\infty}{e^{-x^2(\frac{1}{2\sigma^2}-r)}}dx$

The term inside the integral is just the main part of the formula for a cdf for a normal distribution with standard deviation $\gamma$ where $\frac{1}{2\gamma^2}=\frac{1}{2\sigma^2}-r=\frac{1-2r\sigma^2}{2\sigma^2}$ so it has area $\frac{\sqrt{2\pi\sigma^2}}{\sqrt{1-2r\sigma^2}}$

So, overall:

$E(e^{rX^2})=\frac{1}{\sqrt{2\pi\sigma^2}}\frac{\sqrt{2\pi\sigma^2}}{\sqrt{1-2r\sigma^2}}=\frac{1}{\sqrt{1-2r\sigma^2}}$

Is this right?

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    $\begingroup$ You are computing the moment generating function of a multiple of a chi-squared distribution. The details can be found in several threads here. $\endgroup$
    – whuber ♦
    Commented Apr 27, 2021 at 20:48
  • $\begingroup$ @whuber I'm not seeing the connection to chi-squared distributions. Here, $X$ has a normal distribution. Am I missing something? $\endgroup$
    – roundsquare
    Commented Apr 27, 2021 at 22:34
  • $\begingroup$ Yes: by definition, the sum of squares of independent standard Normal variables has a chi-squared distribution. You are computing the mgf of the square of a single variable that is a multiple of a standard Normal. $\endgroup$
    – whuber ♦
    Commented Apr 28, 2021 at 14:20
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    $\begingroup$ Ah, gotcha. I did not realize that was the definition of chi-squared. I'm not sure if it matters that the r.v. $X$ is not standard normal since the variance is not $1$. But I think I am confident that my formula is correct now. Thanks! Sadly, this means I'm stuck at the next step instead :( $\endgroup$
    – roundsquare
    Commented Apr 28, 2021 at 21:47

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