# Use Stein's Identity to Calculate Variance of X Bar Squared

Suppose that I have a random sample where each random variable is iid normally distributed with mean $\mu$ and variance $\sigma^2$. Suppose I want to calculate the variance of $\bar{X}^2$ "using Stein's idendity." I'm not quite sure that I understand how Stein's identity applies to this situation. I know that the normal distribution is part of the linear exponential family, but I don't understand how that can help me find properties about $\bar{X}^2$. Can somebody please show how this is supposed to work? Thank you.

Write $V(\bar{X}^2)=E(\bar{X}^4)-[E(\bar{X}^2)]^2$ and apply Stein's lemma iteratively to get the $2$nd and $4$th non-centered moments of $\bar X$.
• Okay, so I end up calculating that the variance is given by: $$\frac{4\mu^4\sigma^2}{n}+\frac{2\sigma^4}{n^2}$$ Does this look right? Commented Feb 26, 2018 at 2:39
• @jippyjoe4 - Almost ... but the exponent of $\mu$ should be $2$. BTW, $V(\bar{X}^2)$ is also readily found by noting that $\bar X\overset{distr}{=}\mu+{\sigma\over \sqrt{n}}Z$ with $Z\sim N(0,1),$ so we can just expand $(\mu+{\sigma\over \sqrt{n}}Z)^2$ and take its variance, using the known moments of the standard normal distribution (the covariance term being $0$). Commented Feb 26, 2018 at 4:22
• (cont'd) Also, note that a quick check for dimensional consistency will often catch that kind of error. E.g., each term in a sum equal to $V(\bar{X}^2)$ must have the same dimensions as $\bar{X}^4$, and since $\mu$ and $\sigma$ each have the same dimensions as $\bar X$, a term with dimensions determined by a product like $\mu^a\sigma^b$ must have $a+b=4$. Commented Feb 26, 2018 at 14:39