True or false? Explain your answer:
If $X_1, ..., X_n\sim U(-1,1)$ are independent, then $$ \frac{1}{n} \sum_{j=1}^n X_j^2$$ is asmyptotic chi-squared distributed.
I try to solve the problem with the central limit theorem or the law of large numbers but i failed..
I calculated: $ f_{X_i^2}(x)=\frac{1}{2 \sqrt{x}} 1_{[0,1]} (x) $
$E(X_j^2)= \frac{1}{2} \int_0^1 \frac{x}{x^{-\frac{1}{2}}} dx= 1/3 $
$E([X_j^2]^2)= \frac{1}{2}\int_0^1 \frac{x^2}{x^{-\frac{1}{2}}} dx = 1/5$ $\rightarrow Var(X_j^2)= \frac{1}{5} - (\frac{1}{3})^2= \frac{4}{45}$
So with the CLT:$ \frac{\sum_{j=1}^n X_j^2 - \frac{1}{3} n}{ \sqrt{n} * \frac{4}{45}}= \frac{45}{4} [\frac{1}{\sqrt{n}} \sum_{j=1}^n X_j^2 - \sqrt{n} \frac{1}{3}]$ converge weak to the standard normal distribution.
And on the other hand with the Law of large numbers: $ \frac{1}{n} \sum_{j=1}^n X_j^2 \rightarrow 1/3$ almost surely.