Show that the distribution of $\frac{1}{\sqrt{n}}\sum_{i=1}^n(X_i^2-3)$ is normal Let $X_1,\ldots,X_n$ be i.i.d. variables with $\mathbb{E}[X_i]=0$ and $\mathbb{V}[X_i]=3$ and assume that $\mathbb{E}[X^4_i]<\infty$, show that
$$
\frac{1}{\sqrt{n}}\sum_{i=1}^n(X_i^2-3)
$$
follows a normal distribution.
Textbook says this is TRUE, but I don't see how. I can show that the quantity is bounded above by a normal (via delta method+fourth moment) or I can show that part of this quantity is normal (again by dm) but then the quantity $-\sqrt{N}3$ should always diverge.
 A: This is not true in general, but as $n$ grows, the distribution of this random variable can be made as close as desired to normal, by the central limit theorem. Probably the author had in mind that for large values of $n,$ this can for practical purposes be taken to be normal.
Since $\operatorname E(X)=0$ you have $\operatorname{var}(X) = \operatorname E(X^2),$ so $\operatorname E(X^2)=3.$
And
$$
\frac{ \sum_{i=1}^n (X_i^2-\operatorname E(X_i^2))}{\operatorname{sd}\left(  \sum_{i=1}^n (X_i^2-\operatorname E(X_i^2))\right) } = \frac 1 {\operatorname{sd}(X_1) \sqrt n} \sum_{i=1}^n (X_i^2 - 3), 
$$
so the central limit theorem can then be applied.
The fact that $\operatorname E(X^4)<+\infty$ is used by seeing that it entails that $\operatorname{var}(X_1^2) < +\infty.$
To see that it's not exactly normal, consider the case were $X_i= \pm \sqrt 3,$ each with probability $1/2.$ You get a discrete distribution; hence not exactly a normal distribution.
A: according to Lindeberg-Levy CLT 
https://en.wikipedia.org/wiki/Central_limit_theorem
let $ S_n = \frac{\sum_{k=1}^n X_k}{n}$

Suppose {X1, X2, …} is a sequence of i.i.d. random variables with E[Xi] = µ and Var[Xi] = σ2 < ∞. Then as n approaches infinity, the random variables √n(Sn − µ) converge in distribution to a normal N(0,σ2)

${\displaystyle {\sqrt {n}}\left(S_{n}-\mu \right)\ {\xrightarrow {d}}\ N\left(0,\sigma ^{2}\right).} $

so let's mark $X_i^2 = Y_i$
$$\frac{1}{\sqrt{n}}\sum_{i=1}^n(X_i^2-3) = \frac{1}{\sqrt{n}}\sum_{i=1}^n(Y_i-3)= \sqrt{n}(\sum_{i=1}^nY_i-3)$$
now let's show that $EY_i = 3$ and $VarY_i = 1$
$VarX_i = EX_i^2 - E^2X_i => EY_i = EX_i^2 = VarX_i + E^2X_i = 3$
$VarY_i = VarX_i^2 = EX_i^4 - E^2X_i^2 = EX_i^4 - 9 < \infty$
thus 
$$\sqrt{n}(\sum_{i=1}^nY_i-3)\xrightarrow {d} N(0, VarY_i)$$
