The variance article presents a proof of the standard deviation equivalent formula in terms of E(X). Here is another proof, given in terms of the sample points x_i.

For the variance (i.e., the square of the standard deviation), assuming a finite population with equal probabilities at all points, we have:

${\displaystyle \sigma ^{2}={\frac {1}{N}}\sum _{i=1}^{N}(x_{i}-{\overline {x}})^{2}}$

Expanding this, we get:

${\displaystyle \sigma ^{2}={\frac {1}{N}}\left[(x_{1}-{\overline {x}})^{2}+(x_{2}-{\overline {x}})^{2}+\cdots +(x_{n}-{\overline {x}})^{2}\right]}$

Simplifying, we get:

${\displaystyle {\begin{aligned}\sigma ^{2}&={\frac {1}{N}}\left[(x_{1}^{2}-2x_{1}{\overline {x}}+{\overline {x}}^{2})+(x_{2}^{2}-2x_{2}{\overline {x}}+{\overline {x}}^{2})+\cdots +(x_{n}^{2}-2x_{n}{\overline {x}}+{\overline {x}}^{2})\right]\\&={\frac {1}{N}}\left[x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}\,-\,2x_{1}{\overline {x}}-2x_{2}{\overline {x}}-\cdots -2x_{n}{\overline {x}}\,+\,{\overline {x}}^{2}+\cdots +{\overline {x}}^{2}\right]\\&={\frac {1}{N}}\left[x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}\,-\,2{\overline {x}}(x_{1}+x_{2}+\cdots +x_{n})\,+\,N{\overline {x}}^{2}\right]\\&={\frac {1}{N}}\left[x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}\right]-2{\overline {x}}{\frac {1}{N}}(x_{1}+x_{2}+\cdots +x_{n})+{\overline {x}}^{2}\\&={\frac {1}{N}}\left(\sum _{i=1}^{N}x_{i}^{2}\right)-2{\overline {x}}\left({\frac {1}{N}}\sum _{i=1}^{N}x_{i}\right)+{\overline {x}}^{2}\\&={\frac {1}{N}}\left(\sum _{i=1}^{N}x_{i}^{2}\right)-2{\overline {x}}^{2}+{\overline {x}}^{2}\\&={\frac {1}{N}}\left(\sum _{i=1}^{N}x_{i}^{2}\right)-{\overline {x}}^{2}\end{aligned}}}$

Expanding the last term, we get:

${\displaystyle \sigma ^{2}={\frac {1}{N}}\sum _{i=1}^{N}x_{i}^{2}\,-\,\left({\frac {1}{N}}\sum _{i=1}^{N}x_{i}\right)^{2}}$

So then for the standard deviation, we get:

${\displaystyle \sigma ={\sqrt {{\frac {1}{N}}\sum _{i=1}^{N}(x_{i}-{\overline {x}})^{2}}}={\sqrt {{\frac {1}{N}}\left(\sum _{i=1}^{N}x_{i}^{2}\right)-{\overline {x}}^{2}}}={\sqrt {{\frac {1}{N}}\sum _{i=1}^{N}x_{i}^{2}-\left({\frac {1}{N}}\sum _{i=1}^{N}x_{i}\right)^{2}}}\;.}$

— Loadmaster (talk) 18:12, 22 February 2013 (UTC)

This can also be written as:

${\displaystyle \sigma ={\sqrt {\frac {N\sum _{i=1}^{N}x_{i}^{2}-\left(\sum _{i=1}^{N}x_{i}\right)^{2}}{N^{2}}}}\;.}$