Let $(X_n)_n\sim \mathcal{N}(0,1)$ be independent variables and $\mathbb{E}(X_1^4)=3$. Does $\frac{1}{\sqrt{n}}\sum \limits_{i=1}^{n}(X_i^2-1)$ converge to $\mathcal{N}(0,1)$ in distribution?
I think: yes. Here's my attempt:
This does hint at the strong law of large numbers, we have to square $\frac{1}{\sqrt{n}}\sum \limits_{i=1}^{n}(X_i^2-1)$.$$\left(\frac{1}{\sqrt{n}}\sum \limits_{i=1}^{n}(X_i^2-1)\right)^2=\frac{1}{n}\left(\sum \limits_{i=1}^{n}(X_i^2-1)\right)^2 $$ Is this, as $n\to \infty$, equal to $E(X_i^2-1)^2$? Or is "$\mathbb{E}(X_1^4)=3$" a trick to throw one off?
I'm quite confident, notwithstanding confusion in the comments, that the expression is intended to be read as $$\frac{1}{\sqrt{n}}\sum_{i=1}^n(X_i^2-1).$$ The $X_i^2-1$ are i.i.d. random variables with mean $0$ and variance $2$, so immediately the central limit theorem implies that the series converges in distribution to $N(0,2).$