# If $X_j$ is iid Unif$(-1,1)$ and $Y_n=\frac{\sum X_j}{\sum X_j^2+\sum X_j^3}$, show that $\sqrt{n}Y_n\rightarrow N(0,3)$ in distribution.

If $X_j$ is iid Unif$(-1,1)$ and $\displaystyle Y_n=\frac{\sum X_j}{\sum X_j^2+\sum X_j^3}$, show that $\sqrt{n}Y_n\rightarrow N(0,3)$ in distribution.

I know from central limit theorem that $\sqrt{n} \frac{1}{n}\sum X_j \rightarrow N(0,\frac{1}{3})$ but I'm not sure how to show the denominator of $Y_n$ goes to some constant in probability or something else in distribution.

• You should assume some independence otherwise it is obviously false. – Siméon Nov 25 '13 at 17:17
• Yes the $X_j$'s are independent and identically distributed – lightfish Nov 25 '13 at 17:18
• Do you know the Slutsky lemma? – Siméon Nov 25 '13 at 17:19
• Yes, so I show the denominator goes to a constant using weak law of large numbers? – lightfish Nov 25 '13 at 17:20
• Yes. To be more precise, it is the denominator of $$\sqrt{n}\cdot Y_n = \frac{\frac{1}{\sqrt{n}}\sum X_i}{\frac{1}{n}\sum X_i^2 + \frac{1}{n}\sum X_i^3}$$ – Siméon Nov 25 '13 at 17:22

First remark that: $$\sqrt{n}\cdot Y_n = \frac{\frac{1}{\sqrt{n}}\sum X_i}{\frac{1}{n}\sum X_i^2 + \frac{1}{n}\sum X_i^3}.$$
• I find that I get $\sqrt{n}Y_n\rightarrow N(0,1)$ instead of $N(0,3)$ as the question wants, since $\frac{1}{n}\sum X_i^2$ goes to $1/3$ in prob and $\frac{1}{3}\sum X_i^3$ goes to 0 in prob, and the numerator is $\sqrt{n}\frac{1}{n}\sum X_i$ goes to $N(0,1/3)$ in dist. Am I doing something wrong? – lightfish Nov 25 '13 at 17:58
• @lightfish: your mistake is that the second parameter of the normal distribution is the variance, so if $Z\sim\mathcal{N}(0,1/3)$, then $3\times Z \sim \mathcal{N}(0,3^2\times 1/3) = \mathcal{N}(0,3)$. – Siméon Nov 25 '13 at 18:44