Expected value of product of sample moments (from a normal sample)

Question

Let $X_1, \ldots, X_n$ be iid r.v. normally distributed, with mean $3$ and variance $1$. Define $P_n$ as the product of the first two sample moments, i.e. $$P_n = \left(\frac1n\sum_{i = 1}^n X_i\right)\left(\frac1n\sum_{i = 1}^n X_i^2\right)$$ Define $m$ and $s$ as the mean and variance of $P_n$. What values should $m$ and $s$ approximate?

Sum of normals is normal, while sum of squared normal is a chi-squared distribution (non-central in this case) $$P_n = \frac1{n^2}XY$$ with $$\begin{align*} X &\sim \mathcal N(3n, 1)\\ Y &\sim \chi^2_{n,3n} \end{align*}$$ and $X$ and $Y$ are not independent because they are calculated on the same sample.

Therefore $$\mathbb E[P_n] = \frac1{n^2}(\mathbb E[X]\mathbb E[Y] + \operatorname{Cov}[X, Y])$$ and while I can calculate the first term, I am stuck on the second one as if I use $\operatorname{Cov}[X, Y] = \mathbb E[(X - \mathbb E[X])(Y - \mathbb E[Y])]$ I am right back to $\mathbb E[XY]$, which I'm trying to calculate.

I have no idea about the variance of $P_n$. Any hints?

EDIT: I realized that I can expand the product of sums as a Cauchy product. I'm trying that.

Answer 1

A product of sums can be rewritten as a double sum. In this case, there will be $n^2$ products of the form $X_i X^2_j$.

Since the sample is iid, $\mathbb E[X_i X^2_j] = \mathbb E[X_i] \mathbb E[X^2_j] = 3 \cdot 10 = 30$. Therefore, since expectation is linear, $$\mathbb E[P_n] = \frac1{n^2} 30 n^2 \to 30$$

As for the variance, the same reasoning applies, and $$\operatorname{Var}[P_n] = \frac1{n^4} \operatorname{Var}[X_i X^2_j] n^2 = \frac{480}{n^2} \to 0$$

For the general case $X_i \sim \mathcal N(\mu, \sigma^2)$ the variance will tend to zero just the same, while the expected value can be calculated using the formulas for the first two moments (there's a table here): $$\mathbb E[P_n] = \mu(\mu^2 + \sigma^2)$$