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

Question

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

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: $$Y_n =\frac{\sum_{i=1}^{n}X_i}{\sqrt{\sum_{i=1}^{n}|X_i|^2} } =\frac{\sum_{i=1}^{n}X_i-n0}{\sqrt{n}\sigma }\times \sigma \times \sqrt{\frac{n}{\sum_{i=1}^{n}|X_i|^2}}$$

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.

Answer 1

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 $\sigma^2 = \text{var}(X_1) = \frac{1}{3}$.

Answer 2

Hint : In probability theory, Slutsky’s theorem extends some properties of algebraic operations on convergent sequences of real numbers to sequences of random variables.

The theorem was named after Eugen Slutsky. Slutsky’s theorem is also attributed to Harald Cramér.