# problem on strong law of large numbers

I came across this problem. If $X_1,X_2, \ldots, X_n$ is an iid sequence with uniform distribution on [-1,1] and we define a new random variable like this:

$Y_n = \frac{\sum X_k}{\sum(X_k^2 + X_k^3)}$

How do I use the strong law of large numbers to show that $Y_n$ is normal and find its mean and variance? The uniform distribution $X_k$ will converge to a normal distribution with mean 0 and variance 3. But I am having trouble finding expectation and variance for $Y_n$.

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I think that $Y_n \to 0$ almost surely. Maybe some typo? –  Shai Covo Dec 9 '10 at 21:10
How do you get this result? –  user957 Dec 9 '10 at 21:30
I'll give an answer corresponding to $\sqrt{n}Y_n$, and then you'll see. –  Shai Covo Dec 9 '10 at 21:35
@user957: the intuitive reason is that E(X^2)>0, while E(X) and E(X^3) are both zero. So as we sum more terms, the denominator increases linearly but the numerator only increases as $\sqrt{n}$ –  Ross Millikan Dec 9 '10 at 22:18

In order to have the desired convergence to the ${\rm N}(0,3)$ distribution, we must multiply $Y_n$ by $\sqrt{n}$ (otherwise, $Y_n \to 0$ almost surely).
Suppose that $X_1,X_2,\ldots$ is a sequence of i.i.d. uniform$[-1,1]$ rv's. Define a sequence $(Y_n)$ by $$Y_n = \sqrt{n} \frac{{\sum\nolimits_{k = 1}^n {X_k } }}{{\sum\nolimits_{k = 1}^n {(X_k^2 + X_k^3 )} }}.$$ Then $Y_n$ can be written as $$Y_n = \frac{{n^{ - 1/2} \sum\nolimits_{k = 1}^n {X_k } }}{{n^{ - 1} \sum\nolimits_{k = 1}^n {X_k^2 } + n^{ - 1} \sum\nolimits_{k = 1}^n {X_k^3 } }}.$$ Now, $X_k^3$ has mean zero, and so by the strong law of large numbers (SLLN), $n^{ - 1} \sum\nolimits_{k = 1}^n {X_k^3 } \to 0$ almost surely. $X_k^2$, on the other hand, has mean equal to $\int_{ - 1}^1 {x^2 (1/2){\rm d}x} = 1/3$; hence, by SLLN, $n^{ - 1} \sum\nolimits_{k = 1}^n {X_k^2 } \to 1/3$ almost surely. Next, $X_k$ has mean zero and (hence) variance equal to ${\rm E}(X_k^2) = 1/3$. So, by the central limit theorem, $$\frac{{\sum\nolimits_{k = 1}^n {X_k } }}{{\sqrt {1/3} \sqrt n }} \to {\rm N}(0,1)$$ (in distribution). Combining it all, we see that $Y_n$ converges in distribution to $(3/\sqrt{3})Z$, where $Z \sim {\rm N}(0,1)$. Put it another way, $Y_n \to {\rm N}(0,3)$.