Estimation of the population variance Let $X_1,X_2,...$ be uncorrelated random variables each with mean $\mu$ and variance $\sigma^2$. If $\overline{X}=n^{-1}(X_1+X_2+...+X_n)$, how do I show that $\mathbb{E}(\frac{1}{n-1}\sum_{i=1}^n(X_i-\overline{X})^2)=\sigma^2$?
Apparently this is used to estimate the population variance.
 A: Consider the random variable $Y = \displaystyle \sum_{k=1}^{n} (X_{k}-\overline{X})^{2}$.
By expanding the square, you obtain :
$$ \begin{align*}
\mathbb{E}[ Y ] &= {} \mathbb{E}\left[ \sum_{k=1}^{n} \big( X_{k}^{2} - 2 \overline{X}X_{k} + \overline{X}^{2}) \right] \\[2mm]
 &= \mathbb{E}\left[ \Big( \sum_{k=1}^{n} X_{k}^{2} \Big) - 2 \overline{X} \underbrace{\Big( \sum_{k=1}^{n} X_{k} \Big)}_{= \, n\overline{X}} + n \overline{X}^{2} \right] \\[2mm]
 &= \mathbb{E}\left[  \Big( \sum_{k=1}^{n} X_{k}^2 \Big) - n\overline{X}^{2} \right] \\[2mm]
 &= \Big( \sum_{k=1}^{n} \mathbb{E}\left[ X_{k}^2 \right] \Big) - n \mathbb{E}\left[ \overline{X}^{2} \right] \\[2mm]
\end{align*}
$$
where, for all $k$, $\mathbb{E}\left[ X_{k}^{2} \right] = \mathrm{Var}\left[ X_{k} \right] + \mathbb{E}\left[X_{k}\right]^{2} = \sigma^{2} + \mu^{2}$ and $\mathbb{E}\left[ \overline{X}^{2} \right] = \mathrm{Var}\left[ \overline{X} \right] + \mathbb{E}\left[\overline{X}\right]^{2}=\displaystyle \frac{\sigma^{2}}{n} + \mu^{2}$. Therefore :
$$
\begin{align*}
\mathbb{E}\left[ Y \right] &= {} n\sigma^{2} + n \mu^{2} - n \Big( \frac{\sigma^{2}}{n} + \mu^2 \Big) \\[2mm]
 &= (n-1)\sigma^{2}. \\[2mm]
\end{align*}
$$
