Sample variance calculation
Categories: Statistics
Why (n-1) when calculating sample variance?
Suppose sample mean $\overline{X}$ sample variance $S^2$ Overall mean $\mu$ Overall variance $\sigma^2$
\[S^2=\frac{1}{n-1}\sum_{i=1}^n{\left( X_i-\overline{X} \right)}^2\]if \(S^2=\frac{1}{n}\sum_{i=1}^n{\left( X_i-\overline{X} \right)}^2\) Then \(E\left( S^2 \right) =E\left( \frac{1}{n}\sum_{i=1}^n{\left( X_i-\overline{X} \right) ^2} \right)\) \(=E\left( \frac{1}{n}\sum_{i=1}^n{\left[ \left( X_i-\mu \right) -\left( \overline{X}-\mu \right) \right] ^2} \right)\)
\[=E\left( \frac{1}{n}\sum_{i=1}^n{\left[ \left( X_i-\mu \right) ^2-2\left( X_i-\mu \right) \left( \overline{X}-\mu \right) +\left( \overline{X}-\mu \right) ^2 \right]} \right) \\\] \[=E\left( \frac{1}{n}\sum_{i=1}^n{\left( X_i-\mu \right) ^2-\frac{2}{n}\sum_{i=1}^n{\left( X_i-\mu \right) \left( \overline{X}-\mu \right) +\frac{1}{n}\sum_{i=1}^n{\left( \overline{X}-\mu \right) ^2}}} \right) \\\]Note：
\[\left( \frac{1}{n}\sum_{i=1}^n{\left( X_i-\mu \right) ^2}=\frac{1}{n}\sum_{i=1}^n{X_i-\mu =\overline{X}-\mu =\frac{1}{n}\sum_{i=1}^n{\left( \overline{X}-\mu \right)}} \right)\] \[=E\left( \frac{1}{n}\sum_{i=1}^n{\left( X_i-\mu \right) ^2}-\frac{2}{n}\sum_{i=1}^n{\left( \overline{X}-\mu \right) ^2}+\frac{1}{n}\sum_{i=1}^n{\left( \overline{X}-\mu \right) ^2} \right) \\\] \[=E\left( \frac{1}{n}\sum_{i=1}^n{\left( X_i-\mu \right) ^2}-\frac{1}{n}\sum_{i=1}^n{\left( \overline{X}-\mu \right) ^2} \right) \\\] \[=E\left( \frac{1}{n}\sum_{i=1}^n{\left( X_i-\mu \right) ^2} \right) -E\left( \frac{1}{n}\sum_{i=1}^n{\left( \overline{X}-\mu \right) ^2} \right) \\\] \[=\sigma ^2-E\left( \frac{1}{n}\sum_{i=1}^n{\left( \overline{X}-\mu \right) ^2} \right) \\ <\sigma ^2\]posted 2023/09/28
modified 2023/09/28