Let $X_1, \ldots , X_n$ be independent random variables having a common density with mean $\mu$ and variance $\sigma^2$. The sample variance is defined as the random variance of the sample:
$$S^2=\frac{1}{n-1}\sum_{i=1}^{n}(X_i-\bar{X})^2$$
where $\bar{X}=\frac1n\sum^n_{j=1}X_j$. Calculate $E[S^2]$.
Attempt
Is $X_i-\bar{X}=(X_i-\mu)-(\bar{X}-\mu)$? Assuming it is, then
$$\frac{1}{n-1}\sum_{i=1}^{n}(X_i-\bar{X})=\frac{1}{n-1}\sum_{i=1}^{n}(X_i-\mu)-n(\bar{X}-\mu)$$
Now,
$$E\left[\frac{1}{n-1}\sum_{i=1}^{n}(X_i-\bar{X})\right]=E\left[ \frac{1}{n-1}\sum_{i=1}^{n}(X_i-\mu)-n(\bar{X}-\mu) \right]$$
Best Answer
Using the definition of variance:
$$ \begin{align} E\left(X^{2}\right)&=\sigma^{2}+\mu^{2}\\ E\left(\bar{X}^{2}\right)&=\frac{\sigma^{2}}{n}+\mu^{2} \end{align} $$
Then substitute these into the equation:
$$ \begin{align} E\left(S^{2}\right)&=E\left[\frac{1}{n-1}\sum_{i=1}^{n}{\left(X_{i}-\bar{X}\right)^{2}}\right]\\ \\ &= \frac{1}{n-1}E\left[\sum_{i=1}^{n}{X_{i}^{2}-n\bar{X}^{2}}\right]\\ \\ &=\frac{1}{n-1}\left[\sum_{i=1}^{n}{E\left(X_{i}^{2}\right)}-nE\left(\bar{X}^{2}\right)\right]\\ \\ &=\sigma^{2} \end{align} $$