# 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.

• The question is not clear. Please post the original exercise. What is the question? May 22, 2019 at 14:15
• What have you been doing? Calculation of $\mathbb E\hat\sigma^2$ maybe? Actually I don't know what is meant by Bias($\hat\sigma^2$). May 22, 2019 at 14:16
• I have edited the original question, does this now make sense? @drhab May 22, 2019 at 14:23

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.

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