# Find the limit distribution of $Y_n =\frac{\Sigma_{i=1}^{n}X_i}{\sqrt{\Sigma_{i=1}^{n}|X_i|^2} }$.

Let $X_1,X_2,X_3,...$ be i.i.d. with uniform distribution on $[-1,1]$. Find the limit distribution of $Y_n =\frac{\Sigma_{i=1}^{n}X_i}{\sqrt{\Sigma_{i=1}^{n}|X_i|^2} }$.

I think this is just a direct application of central limit theorem and I need to do some transformation. I got stuck when I had following results: $Y_n =\frac{\Sigma_{i=1}^{n}X_i}{\sqrt{\Sigma_{i=1}^{n}|X_i|^2} } =\frac{\Sigma_{i=1}^{n}X_i-n0}{\sqrt{n}\sigma }\times \sigma \times \sqrt{\frac{n}{\Sigma_{i=1}^{n}|X_i|^2}}$

I'm not sure how to solve the third term when $n$ goes to infinity.

Thanks guys!I think weak law of large numbers is the one I need.

• Do you know Slutsky theorem? – Zhanxiong Aug 17 '15 at 3:48

You decomposition is absolutely right. Now you need to figure out to which $\frac{1}{n}\sum_{i = 1}^n X_i^2$ converges, by weak law of large numbers, $$\frac{1}{n}\sum_{i = 1}^nX_i^2 \to E(X_1^2) = \text{var}(X_1) + (E(X_1))^2 = \frac{1}{12}(1 - (-1))^2 + 0^2 = \frac{1}{3} \text{ in probability}.$$ Based on your decomposition, by Slutsky's theorem and continuous mapping theorem, $$Y_n \Rightarrow N(0, \sigma^2) \times \sqrt{3} = N(0, 1)$$ as $n \to \infty$ since $\sigma^2 = \text{var}(X_1) = \frac{1}{3}$.