The sum of Two Identical Normals is Normal

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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}}

The Distribution of the Sum of Two iid Normal Variables

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}}

Conclusion: the sum is normally distributed, with mean \left.2\mu\right., and with standard deviation \sqrt{2}\sigma.


More Generally

More generally, the sum of two normals is normal, with parameters mean

\left.\mu_{X+Y}=\mu_{X}+\mu_{Y}\right.

and variance

\sigma_{X+Y}^2=\sigma_{X}^2+\sigma_{Y}^2

By induction, the sum of n normals will be normal, with parameters

\mu_{\sum_{i=1}^nX_i}=\sum_{i=1}^n\mu_{X_i}

and

\sigma_{\sum_{i=1}^nX_i}^2=\sum_{i=1}^n\sigma_{X_i}^2
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