This question is related somewhat to one that somebody else asked here. Aside from the fact that I don't feel as if that question was sufficiently answered, my question is not about the exact same thing.
I am being asked to determine the limiting distribution for the square of the sample mean, $\overline{X}^{2}$, where $\overline{X}$ is the sample mean from a population with finite mean and variance.
However, I, unlike the other OP, have absolutely no idea how to approach this. I am also not sure how they were supposed to get it converging to something normal-like without knowing about how $X$ is distributed in the first place. In my notes, this problem is given alongside a theorem characterizing weak convergence of measures, but I don't see how to apply that here.
Best Answer
You're looking for the Delta method here. From your lecture notes, \begin{align*} \frac{\sqrt{n}(\bar{X}_n-\mu)}{\sigma}&\overset{d}{\to}N(0,1),\text{ or}\\ \sqrt{n}(\bar{X}_n-\mu)&\overset{d}{\to}N(0,\sigma^2). \end{align*} The idea here is to expand $g(\bar{X}_n)=\bar{X}_n^2$ about $\mu$: \begin{align*} \bar{X}_n^2=\mu^2+2\mu(\bar{X}_n-\mu)+(\bar{X}_n-\mu)^2. \end{align*} Then \begin{align*} \sqrt{n}(\bar{X}_n^2-\mu^2)=2\mu\sqrt{n}(\bar{X}_n-\mu)+\sqrt{n}(\bar{X}_n-\mu)^2. \end{align*} If you can show $\sqrt{n}(\bar{X}_n-\mu)^2\overset{P}{\to}0$, then Slutsky's theorem implies that both sides tend to $N(0, 4\mu^2\sigma^2)$.
Edit: Just in case it was unclear, you must show that the second term of the right hand side in the final equation converges to 0. Then you use the central limit theorem to get that the first term of the right hand side, and therefore the left hand side, goes to $N(0, 4\mu^2\sigma^2)$.