How to calculate the bias of the estimator for variance? Question: For observations $x_1$, $x_2$, . . . , $x_n$ with sample average $\bar{x}$, we can use an estimator for the population variance: 
$\hat\sigma^2$ = $\frac1n\cdot $ $\sum\limits_{i=1}^n (x_i - \bar{x})^2$. What is the bias of this estimator?
Supposedly the answer is -$\frac{\sigma^2}n$. The reason this confuses me too is because this question is a one minute question on a multiple choice paper. Is there a shortcut I'm missing or I'm supposed to see?
$\text{Bias}\left(\hat\sigma^2\right)$ = $E\left(\hat\sigma^2\right)$ - $\left(\sigma^2\right)$ is the formula I tried to use.

You probably know that the expectation of the unbiased estimator is
$$E\left[\frac{1}{n-1}\sum\limits_{i=1}^n (X_i-\overline X )^2\right]=\sigma^2$$
To obtain the expectation of the biased estimator we just have to multiply both sides by $(n-1)$ and divide them by $n$
$$E\left[\frac{1}{n}\sum\limits_{i=1}^n (X_i-\overline X )^2\right]=\sigma^2\cdot \frac{n-1}{n}$$
Now we can calculate the difference: $\sigma^2\left(\frac{n-1}{n}-1\right)=\sigma^2\left(\frac{n-1}{n}-\frac{n}{n}\right)=-\frac{\sigma^2}{n}$
This is the fastest way I can think of.
