Using the Central Limit Theorem to work out the approximate distribution of $\frac{1}{n}\sum_{i=1}^nX_i^2$ Suppose $X_1,X_2,\ldots,X_n$ are I.I.D $N(0,1)$. Then, what is the approximate distribution of $\frac{1}{n}\sum_{i=1}^n X_i^2\,$?
I have the solution but none of it is making any sense to me.
Thank you for your help!
 A: You can use some "heuristics" to answer this questions. If $X_{i}\sim \mathcal{N}(0,1)$, then by definition $X_i^2\sim \chi^2(1)$, where $EY=1$, and $Var(Y)=2$. Denote $Y_i = X_{i}^2$, hence $\frac{1}{n}\sum_{i}^nX_i^2 = \frac{1}{n}\sum_{i}^nY_i$. Thus, for large enough $n$, ($n\to \infty$), you can use the CLT to deduce 
\begin{align}
\frac{1}{n}\sum_{i}^nY_i \xrightarrow{D}\mathcal{N}(1, 2/n).
\end{align}
A: Standard references will tell you that $\sum\limits_{i=1}^n X_i^2$ has a chi-square distribution with $n$ degrees of freedom, and that distribution has expected value $n$ and variance $2n$.  Consequently the distribution of
$$
\frac{X_1^2+\cdots+X_n^2 - n}{\sqrt {2n}}
$$
is approximately $N(0,1)$.
To prove the assertions about the expectation and variance, it is enough to deal with $X_1^2$ and then multiply the expectation and variance by $n$.
Let $\varphi$ be the standard normal density
$$
\varphi(x) = \frac 1 {\sqrt{2\pi}} e^{-x^2/2}.
$$
Then
\begin{align}
\operatorname{E}(X_1^2) & = \int_{-\infty}^\infty x^2 \varphi(x)\,dx \\[10pt]
& = 2\int_0^\infty x^2 \varphi(x)\,dx & \text{by symmetry} \\[10pt]
& = \sqrt{\frac 2 \pi} \int_0^\infty x \left( e^{-x^2/2} (x\,dx) \right) \\[10pt]
& = \sqrt{\frac 2 \pi} \int_0^\infty \sqrt {2u}\, e^{-u}\,du \\[10pt]
& = \frac 2 {\sqrt\pi} \int_0^\infty u^{\frac 3 2 -1} e^{-u}\,du \\[10pt]
& = \frac 2 {\sqrt\pi} \Gamma\left( \frac 3 2 \right) = \frac 2 {\sqrt\pi}\,\frac 1 2 \Gamma\left( \frac 1 2 \right) = 1.
\end{align}
Similarly $\operatorname{E}(X_1^4) = 3$.  Therefore
$$
\operatorname{var}(X_1^2) = \operatorname{E}(X_1^4) - (\operatorname{E}(X_1^2))^2 = 3 - 1 = 2.
$$
