How to show this estimator of variance is biased?

Question

It is known that the sample variance is an unbiased estimator:

$$s^2 = \frac 1{n-1} \sum_{i=1}^n (X_i - \bar X)^2$$

I would like show that $\sigma '^2 = (X_1 - X_2)^2 $ is a biased estimator.

My work:

$$E((X_1 - X_2)^2)= E(X_1^2) - 2E(X_1 X_2) + E(X_2^2)$$

I wasn't taught of how to specifically simplify these kinds of expression, but I suspect that $E(X_1^2)=E(X_2^2)$ since it's symmetrical.

I don't have any further ideas about how to show that the expected value is not the population variance. Please give me some hints to work on it. Thanks.

Answer 1

It is unbiased if $\mathsf{E}(\hat{\sigma}^2)=\sigma^2$.

$$ \begin{align} \mathsf{E}(\hat{\sigma}^2)=\mathsf{E}((X_1-X_2)^2)&=\mathsf{E}(X_1^2)+\mathsf{E}(X_2^2)-2\mathsf{E}(X_1X_2)\\ &=2(\sigma^2+\mu^2)-2\mu^2\\ &=2\sigma^2\neq\sigma^2 \end{align}$$

Answer 2

Hint: Your approach works fine. Use the fact that $X_1, X_2, \ldots , X_n$ are independently and identically distributed.

Alternatively, let $\mu=E(X_1)=E(X_2)$. The population mean, assuming they exist, will be same due to identical distribution. Now we write $E(X_1-X_2)^2$ as $E\big\{(X_1-\mu)-(X_2-\mu)\big\}^2$.

Note that $E\big\{(X_1-\mu)-(X_2-\mu)\big\}^2=E(X_1-\mu)^2+E(X_2-\mu)^2$ (The cross-product term vanishes!), which equals twice the value of the population variance.