Normal random variables: expectation of squared empirical mean divided by empirical second moment Assume $X_1,\dotsc, X_n$ are iid $\mathcal{N}(\mu, \sigma^2)$-distributed. Define $m=\frac{1}{n} \sum_{i=1}^n X_i$ the empirical mean and $z^2 = \frac{1}{n} \sum_{i=1}^n X_i^2$ the empirical second moment. I am looking for the expectation
$$\mathbf{E}\left[\frac{m^2}{z^2}\right].$$
The ultimate goal is to estimate the quantity
$$\frac{\mu^2}{\mu^2 + \sigma^2}.$$
What I know so far:


*

*Since $z^2\geq m^2$ (Jensen's inequality), the fraction is bounded and the expectation exists.

*$m\sim \mathcal{N}(\mu, \frac{\sigma^2}{n})$ and, thus, $\mathbf{E}[m^2] = \mu^2 + \frac{\sigma^2}{n}$

*$\mathbf{E}[z^2] = \mu^2 + \sigma^2$

*$n\frac{z^2}{\sigma^2} = \sum_{i=1}^n \left( \frac{X_i}{\sigma} \right)^2$ follows a non-central chi-squared distribution with $n$ degrees of freedom and non-centrality parameter $\lambda = n\frac{\mu^2}{\sigma^2}$

*$m^2$ and $z^2$ are obviously dependent, but $m$ is independent of the empirical variance $s^2 = \frac{1}{n-1} \sum_{i=1}^n (X_i - m)^2 = \frac{n}{n-1} (z^2 - m^2)$.

*It is straight-forward to derive an expression for $\mathbf{cov}(m^2, z^2)$


Does anybody have an idea how to compute $\mathbf{E}[\frac{m^2}{z^2}]$ and/or how to estimate $\frac{\mu^2}{\mu^2+\sigma^2}$?
 A: Regarding the estimation, the estimator you propose is a standard "method of moments estimator". It is relatively easy to show that it is consistent, so to show its a.s. convergence to $\mu^2/(\mu^2 + \sigma^2)$. (Just simply apply the strong law of large numbers for numerator and denominator, plus the continuous mapping theorem.)
Regarding the distribution of the fraction, my only idea would also be to relate it to the t-distribution. (Using the independence of $s^2$ and $m^2$ you mentioned, which by the way also gives the covariance you asked for easily.) But I don't know whether the exact distribution or the expectation of the ratio is known explicitly. (I doubt it.)
A: You say that "the ultimate goal is to estimate the quantity":
$$\frac{\mu^2}{\mu^2 + \sigma^2}$$
You propose an estimator and ask what its expectation value is, presumably so that you can evaluate the bias of the estimator.
The estimator that you propose is the maximum likelihood estimator for the quantity, since the maximum likelihood estimates, $\hat\mu$ and $\hat\sigma$, follow these relations (with ${1\over n}$ and not ${1\over n-1}$) and maximum likelihood estimates are invariant under parameter transformation.
The maximum likelihood estimator for $\sigma$ is biased, but the bias is small for large $n$. People are usually satisfied with the maximum likelihood estimators as they tend to have the smallest variance and small bias.
My answer to your question is that you already found a good estimator for the quantity. If you really want to evaluate the expectation value, then your question should be stated that your goal is to find the bias for the estimator that you found. Note that an unbiased estimator is not necessarily better than this one, as you also need to consider the variance of the estimator.
