# Law of Large numbers and central limit theorem

I have come across a bit of a dilemma in my exam revision, this is not a topic I am particularly strong with so help is most appreciated:

Assume that $X_1,X_2,...$ is an i.i.d sequence, where $E(X^2)=5$ and $E(X)=0$, Using central limit theorem, find the distribution of

$$\frac{\sum_{i=1}^nX_i}{\sqrt{5n}}$$

Here I have let $S_n= \sum_{i=1}^nX_i$ and we know that $E(X)=0$, then we have:

$$Yn=\frac{\sum_{i=1}^nX_i - nE(x))}{σ\sqrt{n}} = \frac{\sum_{i=1}^nX_i}{\sqrt{5n}} -> N(0,1)$$

Next I am asked to apply the strong law of large numbers to

$$\frac{\sum_{i=1}^nX_i^2}{n}$$

to show that it converges almost surely and compute its finite limit

I have had a go at this by letting $S_n= \sum_{i=1}^nX_i^2$ and $E(S_n)= \sum_{i=1}^nE(X_i^2)$ and plugging this into to the strong law of large numbers formula ( $\frac{S_n-E(S_n)}{n}-> 0$ a.s.) to find

$\frac{\sum_{i=1}^nX_i^2}{n}$ -> $E(X_i^2)$ a.s.

However I do not feel as though I am showing that $\frac{\sum_{i=1}^nX_i^2}{n}$ is converging almost surely using this method?

Any ideas would be appreciated

Also lastly I have to find, by using central limit theorem, the distribution of

$$\frac{\sum_{i=1}^nXi}{\sqrt{\sum_{i=1}^nX_i^2}}$$

I am lost on this one completely, although I am making the assumption as we are using central limit theorem with it them it will be N(0,1) but as I say that is just a guess,

Help/hints/explanations would be really appreciated

There are two basic criteria used to apply the strong law of large numbers. One of them requires that the variables are IID and that their expectation be finite. Since $X_1, ..., X_n$ are IID, so are $X_1^2, ..., X_n^2$, and $E(X_i^2) = 5$ is finite. So
$$\frac{\sum_{i=1}^nX_i^2}{n}$$
converges almost surely to the expected value of $E(X_i^2)$, that is to say 5. This illustrates the power of the strong law of large numbers. The criteria to apply it are very general and very often met in practice.
$$\frac{\sum_{i=1}^nX_i}{\sqrt{5n}}\frac{\sqrt{5}}{\sqrt{\frac{1}{n}\sum_{i=1}^nX_i^2}}.$$
The first term converges in distribution to $N(0,1)$ and the second term converges almost surely to 1. According to Slutsky's lemma (third item), the whole converges in distribution to $N(0,1)$.