stylistic:tone, linguistic:translation, structural:reformatting, quality:corruption, stylistic:style_transfer
Population variance

(The right side has duplicate terms in the sum while the middle side has only unique terms to sum.) This is true because 1 2 N 2 ∑ i, j = 1 N (x i − x j) 2 = 1 2 N 2 ∑ i, j = 1 N (x i 2 − 2 x i x j + x j 2) = 1 2 N ∑ j = 1 N (1 N ∑ i = 1 N x i 2) − (1 N ∑ i = 1 N x i) (1 N ∑ j = 1 N x j) + 1 2 N ∑ i = 1 N (1 N ∑ j = 1 N x j 2) = 1 2 (σ 2 + μ 2) − μ 2 + 1 2 (σ 2 + μ 2) = σ 2. {\displaystyle {\begin{aligned}&{\frac {1}{2N^{2}}}\sum _{i,j=1}^{N}\left(x_{i}-x_{j}\right)^{2}\\={}&{\frac {1}{2N^{2}}}\sum _{i,j=1}^{N}\left(x_{i}^{2}-2x_{i}x_{j}+x_{j}^{2}\right)\\={}&{\frac {1}{2N}}\sum _{j=1}^{N}\left({\frac {1}{N}}\sum _{i=1}^{N}x_{i}^{2}\right)-\left({\frac {1}{N}}\sum _{i=1}^{N}x_{i}\right)\left({\frac {1}{N}}\sum _{j=1}^{N}x_{j}\right)+{\frac {1}{2N}}\sum _{i=1}^{N}\left({\frac {1}{N}}\sum _{j=1}^{N}x_{j}^{2}\right)\\={}&{\frac {1}{2}}\left(\sigma ^{2}+\mu ^{2}\right)-\mu ^{2}+{\frac {1}{2}}\left(\sigma ^{2}+\mu ^{2}\right)\\={}&\sigma ^{2}.\end{aligned}}} The population variance matches the variance of the generating probability distribution. In this sense, the concept of population can be extended to continuous random variables with infinite populations.