weak convergence/convergence in distribution of random variable $\sqrt{n}(\frac{\mu^2}{\bar{X}_n}-\bar{X_n})$

Question

I'm working on my problem sheet of a probability theory course and can't solve this exercise:

Let for every $k\in \mathbb{N}$ $X_k$ be positive, independently identical distributed random variables with $\mathbb{E}X_1 = \mu$ and $\mathbb{V}ar(X_1)=\sigma^2<\infty$. Let furthermore $\bar{X}_n = \frac{1}{n}\sum_{k=1}^nX_k$, that is the empirical mean.

Show that $\sqrt{n}(\frac{\mu^2}{\bar{X}_n}-\bar{X_n})$ converges in distribution to a random variable, which is N(0,$4\sigma^2$)-distributed.

So far I have thought about the following:

I think I want to use the central limit theorem and/or Etemadi's-theorem. CLT ensures that $\sqrt{n}(\bar{X}_n-\mu)$ converges to a random variable, whis is N(0,$\sigma^2$)-distributed.

I don't know how to handle the other term.

If you give solution instead of hints, I would appreciate if you wrote spoiler alert or something like it at the begining of the solution.

Thanks in advance!

Answer 1

Hint: use the Delta method.

Since $\sqrt{n}(\bar{X}_n - \mu) \overset{d}{\to} N(0, \sigma^2)$, we have $\sqrt{n}(g(\bar{X}_n) - g(\mu)) \overset{d}{\to} N(0, \sigma^2 (g'(\mu))^2)$, where $g(x) = \mu^2/x - x$.

Answer 2

$$\sqrt{n}\left(\frac{\mu^2}{\overline X_n}-\overline X_n\right)=\sqrt{n}\left(\frac{\mu^2-\left(\overline X_n\right)^2}{\overline X_n}\right)=\sqrt{n}\left(\frac{\left(\mu-\overline X_n\right)\left(\mu+\overline X_n\right)}{\overline X_n}\right)=\underbrace{\vphantom{\left(\frac{\mu}{\overline X_n}\right)}\sqrt{n}\left(\mu-\overline X_n\right)}_{\begin{array}${\Large\downarrow} d\\ N(0,\sigma^2)\end{array}}\underbrace{\left(\frac{\mu}{\overline X_n}+1\right)}_{\begin{array}${\Large\downarrow} p\\ 2\end{array}}.$$ Then use Slutsky's theorem to get answer.