If $\bar{x}$ is an unbiased estimator of the population mean $\mu$,then is $\bar{x}^2$ is a biased estimator of $\mu^2$?
I tried this. $$E(\bar{x}^2)=E(\frac{1}{n}\sum_{i=1}^nx_i)^2=E(\frac{1}{n}\sum_{i=1}^nx_i^2)+E(\frac{2}{n}\sum_{i<j}x_ix_j)$$ but I do not know what to do after this.I need to use the fact that $E(\bar{x})=\mu$. Maybe I am missing something?
[Math] $\bar{x}^2$ is a biased estimator of $\mu^2$
probability
Best Answer
Are you familiar with Jensen's inequality and when it is strict? Assuming the distribution of the $X_i$ is nontrivial (i.e. $X_i$ is not a.s. constant) $$ E[\bar x^2] > E[\bar x]^2 = \mu^2. $$
If you aren't familiar with Jensen's inequality, write $\bar x^2 = (\bar x - \mu + \mu)^2 = (\bar x - \mu)^2 + 2\mu(\bar x - \mu) + \mu^2$ so that $$ E[\bar x^2] = \mathrm{Var}(\bar x) + 2\mu E[\bar x - \mu] + \mu^2 $$ $$ =\frac 1n \mathrm{Var}(X_1) + \mu^2 $$ and again we see that if $X_1$ is not a.s. constant than this will be strictly greater than $\mu^2$.