Question: If $(X_i, i \geq 1)$ is a sequence of independent random variables satisfying $P(X_i = 1)=P(X_i = -1)$ then $\frac{{({\sum_{i=1}^{n}{X_i}}})^2}{n}$ converges in distribution to a random variable having the chi-sqaured distribution with 1 degree of freedom.
Is this true?
So far I have: $P(X_i = 1)=P(X_i = -1)= \frac{1}{2}$
So, $M_{X_i}(t)=\frac{e^t +e^{-t}}{2}$.
Then letting $S_n= ({\sum_{i=1}^{n}{X_i}})^2$, we have that $M_{S_n}(t)= \prod_i^n E(e^{t{X_i}^2}) = e^t \dots e^t = e^{nt}$
Now let ${S_n}^*= \frac{({\sum_{i=1}^{n}{X_i}})^2}{n}$ and so $M_{{S_n}^*}(t)= e^{{n}{\frac{t}{n}}}= e^t$
Is what I have so far correct? If not, what is the correct way of working this out and how do I conclude if it does converge in distribution to a chi-squared random variable with 1 degree of freedom?
Best Answer
The answer is positive.
It will be more convenient to use the next approach.
It follows from the Central limit theorem that $$\xi_n = \frac{\sum_{i=1}^n X_i - n EX_1}{\sqrt{n DX_1}} = \frac{\sum_{i=1}^n X_i}{\sqrt{n}} \to \xi \sim N(0,1)$$ in distribution. Thus $\frac{(\sum_{i=1}^n X_i)^2}{\sqrt{n}} = \xi_n^2 \to \xi^2 \sim \chi_1^2$, where we imply convergence in distribution.