# Where does the “2” come from in deriving Normal PDF from its kernel?

I'm trying to train myself to recognize probability densities by deriving PDFs from bare kernel functions. In other words, I find a constant expression by integrating a kernel function over its support range, and divide the original kernel function by that expression, so that it equals 1 and thus is a PDF.

For the Normal distribution the kernel is: $e^{-{{(x-a)^2} \over b}}$. So, I let $u = {{x-a} \over \sqrt b}$ then $du = {1 \over \sqrt b}$, so...

$$\int e^{-{{(x-a)^2} \over b}}dx = \sqrt b \int e^{-{{(x-a)^2} \over b}}{1 \over \sqrt b}dx = \sqrt b \int e^{-u^2}du = \sqrt b \sqrt \pi$$

Let $b = \sigma^2 , a = \mu$ and we have... $$\sqrt{\sigma^2 \pi} = \int e^{-{{(x-\mu)^2} \over \sigma^2}}dx \Rightarrow 1 = \int {{e^{-{{(x-\mu)^2} \over \sigma^2}}} \over {\sqrt{\sigma^2 \pi}}} dx$$

...but the problem is that ${{e^{-{{(x-\mu)^2} \over \sigma^2}}} \over {\sqrt{\sigma^2 \pi}}}$ is not the Normal PDF! This is the Normal PDF:

$${{e^{-{{(x-\mu)^2} \over 2\sigma^2}}} \over {\sqrt{2\sigma^2 \pi}}}$$

It seems to me that I could have defined $b$ in the original expression to have any coefficient, and would have still satisfied the conditions for a PDF. So, how would I have arrived at the $2\sigma^2$ parameterization if I didn't expect it ahead of time?

Thanks.

The problem is that you cannot do $b=\sigma^2$, because the kernal is actually $$e^{\frac{-(x-\mu)^2}{2\sigma^2}}$$
So you should set $b=2\sigma^2$, and you will be fine.