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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.

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

First remark that: $$ \sqrt{n}\cdot Y_n = \frac{\frac{1}{\sqrt{n}}\sum X_i}{\frac{1}{n}\sum X_i^2 + \frac{1}{n}\sum X_i^3}. $$

The numerator converges in law to a normal random variable by the Central Limit Theorem.

The denominator converges almost surely (in probability is sufficient) to some constant by the Law of Large Numbers.

Determine the variance of the normal random variable, the limit constant for the denominator, and finally conclude with Slutsky's lemma.

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  • $\begingroup$ 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? $\endgroup$ – lightfish Nov 25 '13 at 17:58
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    $\begingroup$ @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 $3\times Z \sim \mathcal{N}(0,3^2\times 1/3) = \mathcal{N}(0,3)$. $\endgroup$ – Siméon Nov 25 '13 at 18:44

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