How to calculate the bias of the estimator for variance?

Question

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.

Answer 1

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.

Answer 2

Observe that $x_{i}-\overline{x}=y_{i}-\overline{y}$ where $y_{i}=x_{i}-\mathbb{E}x_{i}$.

Substituting this and working out $\hat{\sigma}^{2}$ we find: $$\hat{\sigma}^{2}=\frac{1}{n}\sum_{i=1}^{n}y_{i}^{2}-\overline{y}^{2}$$

Here $\mathbb{E}y_{i}^{2}=\sigma^{2}$ so that: $$\mathbb{E}\hat{\sigma}^{2}=\sigma^{2}-\mathbb{E}\overline{y}^{2}=\sigma^{2}-\frac{1}{n^{2}}\mathbb{E}\sum_{i=1}^{n}\sum_{j=1}^{n}y_{i}y_{j}=\sigma^{2}-\frac{1}{n}\sigma^{2}$$