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

A: 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.
