Unbiased, positive estimator for the square of the mean Assume we have access to i.i.d. samples from a distribution with true (unknown) mean and variance $\mu, \sigma^2$, and we want to estimate $\mu^2$.
How can we construct an unbiased, always positive estimator of this quantity? 
Taking the square of the sample mean $\tilde{\mu}^2$ is biased and will overestimate the quantity, esp. if $\mu$ is close to 0 and $\sigma^2$ is large.
This is possibly a trivial question but my google skills are letting me down as estimator of mean-squared only returns mean-squarred-error estimators

If it makes matters easier, the underlying distribution can be assumed to be Gaussian.

Solution:


*

*It is possible to construct an unbiased estimate of $\mu^2$; see knrumsey's answer

*It is not possible to construct an unbiased, always positive estimate of $\mu^2$ as these requirement are in conflict when the true mean is 0; see Winks' answer
 A: It should not be possible produce an estimator that is both unbiased and always positive for $\mu^2$.
If the true mean is 0, the estimator must in expectation return 0 but is not allowed to output negative numbers, therefore it is also not allowed to output positive numbers either as it would be biased. An unbiased, always positive estimator of this quantity must therefore always return the correct answer when the mean is 0, regardless of the samples, which seems impossible.
knrumsey's answer shows how to correct the bias of the sample-mean-squarred estimator to obtain an unbiased estimate of $\mu^2$.
A: Note that the sample mean $\bar{X}$ is also normally distributed, with mean $\mu$ and variance $\sigma^2/n$. This means that
$$\operatorname E(\bar{X}^2) = \operatorname E(\bar{X})^2 + \operatorname{Var}(\bar{X}) = \mu^2 + \frac{\sigma^2}n$$
If all you care about is an unbiased estimate, you can use the fact that the sample variance is unbiased for $\sigma^2$. This implies that the estimator
$$\widehat{\mu^2} = \bar{X}^2 - \frac{S^2}n$$
is unbiased for $\mu^2$.  
