# Help in showing that the estimator $\hat{\sigma}^2= \frac{1}{n}\sum_{i=1}^n (X_i - \overline{X})^2$ is a biased estimator of $\sigma$.

Let $X_1, \dotsc, X_n$ be a random sample of a random variable $X$ having a probability distribution with mean $\mu$ and variance $\sigma^2$.

Define the estimator $\hat{\sigma}^2 = \frac{1}{n}\sum_{i=1}^n (X_i - \overline{X})^2$.

It can be shown that

$$\mathrm{E}\big[\hat{\sigma}^2\big] = \frac{n-1}{n}\sigma^2$$

I want to determine whether or not $\hat{\sigma}$ is a biased or unbiased estimator of $\sigma$.

I believe that the estimator is biased for $\sigma$, because

$$\sqrt{\mathrm{E}\big[\hat{\sigma}^2\big]} = \sqrt{\frac{n-1}{n}} \sigma < \sigma$$

which I believe shows that the estimator underestimates $\sigma$, but I'm not sure if this is the correct way to prove it.

Any help would be appreciated. Thanks!

• It is not the correct way of proving it. To be shown is that $\mathsf E\hat{\sigma}\neq\sigma$ which is not the same thing. In general we do not have $\sqrt{\mathsf{E}X}=\mathsf{E}\sqrt{X}$. – drhab Jan 22 '18 at 13:47

As drhab mentioned, $\sqrt{\operatorname EX}$ is not necessarily equal to $\operatorname E\sqrt{X}$. However, using Jensen's inequality, $$\operatorname E\sqrt X\le\sqrt{\operatorname EX}$$ for a non-negative random variable $X$. Hence, $$\operatorname E\hat\sigma\le\sqrt{\operatorname E\hat\sigma^2}=\sigma\sqrt{\frac{n-1}n}<\sigma$$ and $\operatorname E\hat\sigma<\sigma$, which shows that the estimator is biased.