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
Best Answer
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 $\mu^2$.