Bias of $\sigma^2$ estimator I need to find the bias of $\frac{\sum(x_{i}-\bar{x})^2}{n+1}$ for $\sigma^2$. To do so, one must take its expectation but add and minus $\mu$ from the summation part so we can bring $\sigma^2$ into the expression.
Giving,
$$E\left[\frac{\sum(x_{i}-\mu+\mu-\bar{x})^2}{n+1}\right]$$
$$E\left[\frac{1}{n+1}\sum(x_{i}-\mu)^2-2\frac{(\bar{x}-\mu)}{n+1}\sum(x_{i}-\mu)+\frac{\sum(\bar{x}-\mu)^2}{n+1}\right]$$
Does the latter part of this expression simplify to just $\frac{n(\bar{x}-\mu)^2}{n+1}]$ 
Thence we'd have $E\left[\frac{1}{n+1}\sum(x_{i}-\mu)^2-\frac{n(\bar{x}-\mu)^2}{n+1}\right]$
$$= E\left[\frac{n}{n+1}\sigma^2-\frac{n(\bar{x}-\mu)^2}{n+1}\right]$$
$$= \frac{n}{n+1}\sigma^2-E\left[\frac{n(\bar{x}-\mu)^2}{n+1}\right]$$
Thus,
$$\text{Bias} = \frac{n}{n+1}\left(\sigma^2-E[(\bar{x}-\mu)^2]\right)$$
So basically the variance is underestimated. Does this look correct?
 A: If you know the frequently mentioned result that
$$
\operatorname{E}\left( \frac 1 {n-1} \sum_{i=1}^n (X_i - \bar X)^2 \right) = \sigma^2
$$
then by multiplying both sides by $(n-1)/(n+1)$ you get
$$
\operatorname{E}\left( \frac 1 {n+1} \sum_{i=1}^n (X_i - \bar X)^2 \right) = \frac{n-1}{n+1} \sigma^2.
$$
Thus the bias is
$$
\frac{n-1}{n+1} \sigma^2 - \sigma^2 = \frac{-2}{n+1}\sigma^2.
$$
Your middle term involving $\dfrac{(\bar X - \mu)^2}{n+1}\sum\limits_i (X_i - \mu)$ does not have expectation zero, although either of the two factors being multiplied has expectation zero.  Note that $\operatorname{cov}(\bar X, X_i)= \sigma^2/n$, and you can use that.  In your bottom line, notice that
$$
\operatorname{E}((\bar X-\mu)^2) = \frac{\sigma^2} n,
$$
so your bias would come out positive, although it should be negative.
Perhaps it is an interesting exercise to show that among all constants $c$, the one for which
$$
c\sum_{i=1}^n (X_i-\bar X)^2
$$
has the smallest mean squared error as as estimator of $\sigma^2$ is $1/(n+1)$, if the population is normally distributed.  To show that, recall that the mean squared error is the variance plus the square of the bias.  Also recall that
$$
\frac 1 {\sigma^2} \sum_{i=1}^n (X_i-\bar X)^2 \sim\chi^2_{n-1}
$$
and remember the variance of the chi-square distribution.
