Point estimator for product of independent RVs Let $X$ and $Y$ be two independent random variables. Given an (iid) random sample of size $n$ of $X$ and a random sample of size $n$ of $Y$, what is a good way to estimate the mean of their product, $E[XY]$?
The most obvious estimator is the sample mean: multiply each $X_i$ and $Y_i$, and then average over $n$.
Another unbiased estimator would be to take the average of the $X_i$'s, then take the average of the $Y_i$'s, and finally multiply these two averages.
Which of these two unbiased estimators is better (is less noisy or has lower variance)? Is there an estimator that has even smaller variance than the two proposed?
 A: I'm assuming what you want to  estimate is $E[XY]$ (you don't say, but the use of the sample mean suggests it)
Intuitively, $\overline{XY}$ would work even if $X$ and $Y$ weren't independent, so it should be less efficient under the additional assumption that they are independent.  Let's see how that goes
Let's look at the case where $X$ and $Y$ are Normal, to start off.  The maximum likelihood estimators of the means $\mu_x$ and $\mu_y$ of $X$ and $Y$ are the sample averages $\bar X$ and $\bar Y$, and the invariance principle for MLEs says that the MLE of $\mu_x\mu_y$ is $\bar X\bar Y$.
The mean of $\bar X\bar Y$ is $\mu_x\mu_u$ (by independence). Its variance is $\mu^2_x\sigma^2_y/n+\mu^2_y\sigma^2_x/n+\sigma^2_x\sigma^2/n^2$
The mean of $\overline{XY}$ is $\mu_x\mu_y$.  The variance of $XY$ is $\mu^2_x\sigma^2_y+\mu^2_y\sigma^2_x+\sigma^2_x\sigma^2$ so the variance of $\overline{XY}$ is $(\mu^2_x\sigma^2_y+\mu^2_y\sigma^2_x+\sigma^2_x\sigma^2)/n$
which is larger than the variance of $\bar X\bar Y$.
The mean and variance analysis still works when $X$ and $Y$ are not Normal, so it's still true that $\bar X\bar Y$ is more efficient. However, it's now possible that there are more efficient estimators, because the sample average is no longer the MLE.  For example, if $X$ and $Y$ have a Laplace distribution, the sample medians are the MLEs of the means of $X$ and $Y$,  so the product of the sample medians will be a more efficient estimator than $\bar X\bar Y$.
In the nonparametric model where all you know about $X$ and $Y$ is that they have finite means, the sample average is efficient (because basically anything else is inconsistent) and $\bar X\bar Y$ will be optimal again.
