This article is an extract from → Talk:Standard deviation/Archive 2#Mathematical identity proof
Variance
Mathematical identity proof
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 xi.
For the variance (i.e., the square of the standard deviation), assuming a finite population with equal probabilities at all points, we have:
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]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/4671755a557751770c9e5c2c36226b88aac16548)
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}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/69c75517cc2b313166e98738b54c44cf6ad3eef1)
Expanding the last term, we get:
So then for the standard deviation, we get:
— Loadmaster (talk) 18:12, 22 February 2013 (UTC)
This can also be written as:
