Show that $(\bar{X})^2$ is not an unbiased estimator for $\mu^2$

Question

If $X_1, ... , X_n$ are $n$ identical distributed independent random variables each with mean $\mu$ and variance $1$.

A little confused by this question. Is it asking for if $(\bar{X})^2$ != $\mu^2$.

Another part of the question is, Finding the unbiased estimator for $\mu^2$.

Unsure how to expand $E[\bar{X}]^2$

Would this be correct

$E[\bar{X}]^2 = E(1/n \sum_{i=1}^n x_i)^2 = E(1/n^2 \sum_{i=1}^n x_i^2)$

Answer 1

The question asks you to verify that $E[\bar{X}^2]\neq \mu^2$. The expansion $E[\bar{X}^2]$ gives: $$ E\Big[\bar{X}^2\Big]=\frac{1}{n^2}E\Big[\Big(\sum_iX_i\Big)^2\Big]=\frac{1}{n^2}E\Big[\sum_iX_i^2+2\sum_{i<j}X_iX_j\Big]. $$ Independence only gives $E[X_iX_j]=E(X_i)E(X_j)=\mu^2$ for $i<j$. Independence doesn't make expectations of the cross terms go away; this is your mistake. In any case, we now have $$ E\Big[\sum_iX_i^2\Big]=n E\Big[X_i^2\Big]=n(\mu^2+1);\quad E\Big[2\sum_{i<j}X_iX_j\Big]=2\frac{n(n-1)}{2}\mu^2=n(n-1)\mu^2. $$ So we have $$ E\Big[\bar{X}^2\Big]=\frac{1}{n^2}(n(\mu^2+1)+n(n-1)\mu^2)=\frac{1}{n^2}(n^2\mu^2+n)=\mu^2+\frac{1}{n}\neq\mu^2. $$ Thus, $\bar{X}^2$ is not an unbiased estimator of $\mu^2$.

Answer 2

$$ E\bar{X}_n^2 = var(\bar{X}_n)+E^2\bar{X}_n = 1/n+\mu^2 >\mu^2. $$ So the simplest unbiased estimator would be $$ \bar{X}^2_n-1/n $$