The sum of Two Identical Normals is Normal

Normal Density function

Normally distributed variable x with mean $\left.\mu\right.$ and standard deviation $\left.\sigma\right.$ has density function

$d(x)=\frac{1}{\sqrt{2\pi}\sigma}e^{\frac{-(x-\mu)^2}{2\sigma}}$

Consider the sum s of two of these random variables x. The density of s is given by the convolution of the densities of the two:

$\left.d(s)\right.$	$=\int_{-\infty}^{\infty}d_1(x)d_2(s-x)dx$
	$=\frac{1}{\left(\sqrt{2\pi}\sigma\right)^2}\int_{-\infty}^{\infty}e^{\frac{-(x-\mu)^2}{2\sigma^2}}e^{\frac{-(s-x-\mu)^2}{2\sigma^2}}dx$
	$=\frac{1}{\left(\sqrt{2\pi}\sigma\right)^2}\int_{-\infty}^{\infty}e^{\frac{-(x-\mu)^2-(s-x-\mu)^2}{2\sigma^2}}dx$
	$=\frac{1}{\left(\sqrt{2\pi}\sigma\right)^2}\int_{-\infty}^{\infty}e^{\frac{2(x^2+\mu^2)-2(x-\mu)s-s^2}{-2\sigma^2}}dx$
	$=\frac{1}{\left(\sqrt{2\pi}\sigma\right)^2}\int_{-\infty}^{\infty}e^{\frac{2(x-s/2)^2+(s-2\mu)^2/2}{-2\sigma^2}}dx$
	$=\frac{1}{\left(\sqrt{2\pi}\sigma\right)^2}\int_{-\infty}^{\infty}e^{\frac{-2(x-s/2)^2}{2\sigma^2}}e^{\frac{-(s-2\mu)^2/2}{2\sigma^2}}dx$
	$=\frac{1}{\sqrt{2\pi}\left(\frac{\sigma}{\sqrt{2}}\right)} \int_{-\infty}^{\infty}e^{\frac{-(x-s/2)^2}{2\left(\frac{\sigma}{\sqrt{2}}\right)^2}}dx \left( \frac{1}{\sqrt{2\pi}\left(\sqrt{2}\sigma\right)} e^{\frac{-(s-2\mu)^2}{2\left(\sqrt{2}\sigma\right)^2}} \right)$
	$= \frac{1}{\sqrt{2\pi}\left(\sqrt{2}\sigma\right)} e^{\frac{-(s-2\mu)^2}{2\left(\sqrt{2}\sigma\right)^2}}$