Suppose we have the following random variables $X_1$, $X_2$,….$X_n$,.., that are $iid$ but we dont know what distribution they follow.
I know that the sample mean $\bar{X}$ is an unbiased estimator of the population mean. But, how can i prove that the square of the sample mean is an biased (or maybe unbiased) estimator of the variance?
My particular doubt is how to continue this:
$E[\bar{X}^2] = E[(\frac{\sum_{i=1}^nX_i}{n})^2] = E[\frac{\sum_{i=1}^nX_i}{n} \times\frac{\sum_{i=1}^nX_i}{n}] = \frac{1}{n^2} E[\sum_{i=1}^nX_i \times \sum_{i=1}^nX_i] = …..$
I think the estimator is biased, but i want to confirm it…
Best Answer
You have $X_1, X_2, \dots, X_n$ are iid from an unknown distribution with mean (say) $\mu$ and variance (say) $\sigma^2$.
$\bar{X}$ is an unbiased estimator of the mean, and thus $E(\bar{X}) = \mu$. Also, $Var(\bar{X}) = \sigma^2/n$. Thus since, \begin{align*} E[\bar{X}^2] & = Var(\bar{X}) + E[\bar{X}]^2\\ & = \dfrac{\sigma^2}{n} + \mu^2. \end{align*}
You can now figure out what the bias is. Clearly, $\bar{X}^2$ is a horrible estimator for $\sigma^2$. As wolfies pointed you, you will do better with $n\bar{X}^2$.