Applying CLT and Law of Large Numbers to a non-standard sequence involving normal distribution

Question

Suppose that $\{X_n\}_{n=1}^{+\infty}$ is an i.i.d sequence distributed as $N(0, \sigma^2)$.

Let $$U_n := \dfrac{1}{n}\sum\limits_{j=1}^{n} |X_j|$$ $$V_n := \dfrac{1}{n}\sum\limits_{j=1}^{n} X_j^2$$

The question is where does $P_n := \sqrt{n}\left(\sqrt{\frac{2}{\pi}}\frac{V_n}{U_n} - \sigma\right)$ converge by distribution?

What I did:

1. By the law of large numbers, $U_n \to E|X_1| = \sqrt{\frac{2}{\pi}}\sigma$ and $V_n \to EX_1^2 = \sigma^2$ (convergence in probability and, hence, weak convergence as well).
2. I managed to represent $P_n$ as $\alpha A_n + \beta B_n$ where $\alpha$ and $\beta$ are constants, and $A_n$, $B_n$ both converge to $A, B \sim N(0, 1)$ by CLT. However, that doesn't mean that the sequence converges to $\alpha A + \beta B$, as far as I understand (in general, we can't add it like that).
3. Maybe the delta method could help? But not sure, how to apply it here.

I would greatly appreciate any help.

Answer 1

For simplicity's sake, let's denote the symbol $\sim$ as "be asymptotically equal to ... in distribution" for $n \to +\infty$. Example: $X_n \sim Y_n$ means $X_n$ follows asymptotically the distribution of $Y_n$ when $n \to +\infty$.

Let's recall the distribution of $U_n$ and $V_n$ (by using the CLT):

$$U_n \sim E(|X|)+\frac{1}{\sqrt{n}} \mathcal{N}(0,V(|X|))$$ $$V_n \sim E(X^2)+\frac{1}{\sqrt{n}} \mathcal{N}(0,V(X^2))$$

Applying the Taylor developpement on $\frac{V_n}{U_n}$ \begin{align} \frac{V_n}{U_n} &\sim \left(E(|X|)+\frac{1}{\sqrt{n}} \mathcal{N}(0,V(|X|))\right) \left(E(X^2)+\frac{1}{\sqrt{n}} \mathcal{N}(0,V(X^2))\right)^{-1} \\ &\sim \frac{1}{E(X^2)}\left(E(|X|)+\frac{1}{\sqrt{n}} \mathcal{N}(0,V(|X|))\right) \left(1+\frac{1}{\sqrt{n}} \frac{\mathcal{N}(0,V(X^2))}{E(X^2)}\right)^{-1} \\ &\sim \frac{1}{E(X^2)}\left(E(|X|)+\frac{1}{\sqrt{n}} \mathcal{N}(0,V(|X|))\right) \left(1-\frac{1}{\sqrt{n}} \frac{\mathcal{N}(0,V(X^2))}{E(X^2)}\right) \\ &\sim \frac{1}{E(X^2)}\left(E(|X|)+\frac{1}{\sqrt{n}} \left( \mathcal{N}(0,V(|X|)) +\frac{\mathcal{N}(0,V(X^2))}{E(X^2)}\right) +\mathcal{O}(\frac{1}{n}) \right) \\ &\sim \frac{E(|X|)}{E(X^2)}+\frac{1}{\sqrt{n}} \left(\frac{ \mathcal{N}(0,V(|X|)) }{E(X^2)}+\frac{\mathcal{N}(0,V(X^2))}{E^2(X^2)}\right) +\mathcal{O}(\frac{1}{n}) \\ \end{align}

Hence, $$P_n = \sqrt{n}\left(\sqrt{\frac{2}{\pi}}\frac{V_n}{U_n}-\sigma \right) \sim \sqrt{\frac{2}{\pi}}\mathcal{N}\left(0,\frac{V(|X|)}{E^2(X^2)} +\frac{V(X^2)}{E^4(X^2)} \right) $$ or

$$P_n \xrightarrow{n\to +\infty} \mathcal{N}\left(0,\frac{2}{\pi}\left(\frac{V(|X|)}{E^2(X^2)} +\frac{V(X^2)}{E^4(X^2)} \right)\right) $$