Unbiasedness Identity

$$\sum _i(X_i-\mu {)}^2=\sum _i(X_i-\overline{X}{)}^2+n(\overline{X}-\mu {)}^2$$

Proof

• ${\sum }_i(x_i-\mu {)}^2={\sum }_i(x_i-\overline{x}+\overline{x}-\mu {)}^2$
• $={\sum }_i(\overline{x}-\mu {)}^2+2{\sum }_i(x_i-\overline{x})(x-\mu )$
• Note that with respect to $i$, $x_i$ is changing, $\overline{x},\mu$ are constants
• $={\sum }_i(x_i-\overline{x}{)}^2+n(\overline{x}-\mu {)}^2+2(\overline{x}-\mu ){\sum }_i(x_i-\overline{x})$
• Note that ${\sum }_i(x_i-\overline{x})=0$
• $={\sum }_i(x_i-\overline{x}{)}^2+n(\overline{x}-\mu {)}^2$
• Re-arrange to get ${\sum }_i(x_i-\overline{x}{)}^2={\sum }_i(x_i-\mu {)}^2-n(\overline{x}-\mu {)}^2$
• $⟹E[{\sum }_i(x_i-\overline{x}{)}^2]=E[{\sum }_i(x_i-\mu {)}^2]-nE[(\overline{x}-\mu {)}^2]$
• $⟹E[{\sum }_i(x_i-\overline{x}{)}^2]={\sum }_{i}E[(x_i-\mu {)}^2]-nE[(\overline{x}-\mu {)}^2]$
• $⟹E[{\sum }_i(x_i-\overline{x}{)}^2]=V[x_i]-nV[\overline{x}]$
• $⟹E[{\sum }_i(x_i-\overline{x}{)}^2]=(n-1)({\sigma }^2)$

Example

• Unbiasedness Identity for Sample Variance Statistic