Find the standard deviation of $ \frac{\gamma}{\sqrt{2\pi\sigma}}\exp\left(-\frac{\gamma^2}{\sigma}\frac{(x-\mu)^2}{2}\right)$

Question

Given $\frac{\gamma}{\sqrt{2\pi\sigma}}\exp\left(-\frac{\gamma^2}{\sigma}\frac{(x-\mu)^2}{2}\right)$ as a normal distribution PDF with mean $\mu$, I'd like to solve for the std deviation in terms of $\gamma$ and $\sigma$. Any hints or tips on how to think about this and solve it would be appreciated. I am familiar with the normal distribution PDF: $\frac{1}{\sigma\sqrt{2\pi}}\exp\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)$ and I can see that the former and latter PDF are slightly different in terms of $\sigma $ and $\gamma$ and I know that the integral of the PDF over the real line equals 1. I tried doing substitutions with different quotients of $\sigma$ and $\gamma $, like $\frac{\gamma}{\sqrt{\sigma}} $ and then simplifying, etc. Still, as of now I haven't figured it out. Thanks.

Answer 1

The expression in your title does not define a normal density unless you assume that $\sigma > 0$, and indeed most people would use $\sigma^2$ in the normal density as you have indeed done in the text of your question. Provided that $\sigma > 0$, note that $$\frac{\gamma}{\sqrt{2\pi\sigma}}\exp\left(-\frac{\gamma^2}{\sigma}\frac{(x-\mu)^2}{2}\right) = \frac{1}{\sqrt{2\pi(\sigma/\gamma^2)}}\exp\left(-\frac{(x-\mu)^2}{2(\sqrt{\sigma}/\gamma)^2}\right)$$ is a normal pdf with mean $\mu$ and variance $\left(\frac{\sqrt{\sigma}}{\gamma}\right)^2 = \frac{\sigma}{\gamma^2}$ and so you can read off the standard deviation as $\frac{\sqrt{\sigma}}{\gamma}$ without resorting to integration.