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!
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.