Bias of Sample Variance

Theorem
Let $X_1, X_2, \ldots, X_n$ form a random sample from a population with mean $\mu$ and variance $\sigma^2$.

Let:


 * $\ds \bar X = \frac 1 n \sum_{i \mathop = 1}^n X_i$

Then:


 * $\ds \hat {\sigma^2} = \frac 1 n \sum_{i \mathop = 1}^n \paren {X_i - \bar X}^2$

is a biased estimator of $\sigma^2$, with:


 * $\operatorname{bias} \paren {\hat {\sigma ^2} } = -\dfrac {\sigma^2} n$

Proof
If $\hat {\sigma^2}$ is a biased estimator of $\sigma^2$, then:


 * $\expect {\hat {\sigma^2} } \ne \sigma^2$

We have:

We have that:

So:

So $\hat {\sigma^2}$ is a biased estimator of $\sigma^2$.

Further, we have:


 * $\operatorname{bias} \paren {\hat {\sigma ^2}} = \sigma^2 - \dfrac {\sigma^2} n - \sigma^2 = -\dfrac {\sigma^2} n$