# How to show this estimator of variance is biased?

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.

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}

• Thanks for the answer. I would like to ask why $E(X_1^2)= \sigma^2 + \mu^2$. Since $X_1$ is not simply taken from the population, it is choosen from a sample of sample size at least $2$. I may be completely misunderstanding though. – lEm Dec 19 '16 at 6:28
• $E(X_1^2)= E((X_1-\mu)+\mu)^2=E(X_1-\mu)^2+\mu^2+2E(X_1-\mu)\mu=E(X_1-\mu)^2+\mu^2$, which is $\sigma^2 + \mu^2$. Note that the cross-product term term vanishes and Expectation of a constant is the same constant. – user399978 Dec 19 '16 at 6:31
• @Bubububu: When we say $X_1, X_2, \dots, X_n$ is a random sample from a population with mean $\mu$ and variance $\sigma^2$, we mean that each $X_i$ has $E(X_i) = \mu$ and $Var(X_i) = \sigma^2.$ Also, that the $X_i$ are all indep. – BruceET Dec 19 '16 at 8:24

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.

• Alternatively: $E(X_1^2) - 2E(X_1)E(X_2) + E(X_2^2) \\= E(X_1^2) - 2\mu^2 + E(X_2^2) = [E(X_1^2) - \mu^2] + [E(X_2^2 - \mu^2] = 2\sigma^2.\\$ $E(X_1 X_2) = E(X_1)E(X_2) = \mu^2$ by independence. – BruceET Dec 19 '16 at 6:19
• I was giving the same hint @BruceET – msm Dec 19 '16 at 6:23
• Happens frequently. I see your contemporaneous Answ. Great minds... (+1) – BruceET Dec 19 '16 at 6:28