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:

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