Law of Large Numbers for squared fractions

Question

Suppose that we have $(X_{n})_{n\in\mathbb{N}}$ iid random variables with mean $\mu$ and variance $\sigma^{2}<\infty$.And we have to prove that

$$\lim_{n\rightarrow \infty}\frac{(X_{1}+...+X_{n})^{2}}{n(X_{1}^{2}+...+X_{n}^{2})}$$

has a limit and that converges to that limit with probability $1$.

I believe that we have to use LLN.

Maybe we should rewrite it as $\dfrac{(X_{1}+...+X_{n})^{2}}{n^{2}}\dfrac{n}{(X^{2}_{1}+...+X_{n}^{2})}$ and apply LLN for each of the fractions.

for $\dfrac{(X_{1}+...+X_{n})^{2}}{n^{2}}\rightarrow m^{2}$ because from LLN we know that $\dfrac{(X_{1}+...+X_{n})}{n}\rightarrow m$.

But for the second fraction $\dfrac{n}{(X^{2}_{1}+...+X_{n}^{2})}$ I'm completely lost.

Any advise or help would be great.

Answer 1

Let $Y_i:=X_i^2$. Then $Y_i$'s are i.i.d. For the strong law of large numbers to hold for $Y_i$, we only need to assume that $E|Y_1|<\infty$. (Refer to [Durrett] for example) In this case, $$ \frac 1 n \sum_{i=1}^nY_i\to EY_1 \text{ a.s.} $$ Now $E|Y_1|=EY_1=EX_1^2=\mu^2+\sigma^2$. Then $$ \frac 1 n\sum_{i=1}^n X_i^2=\frac 1 n \sum_{i=1}^nY_i\to \mu^2+\sigma^2 \text{a.s.} $$

Answer 2

Hopefully $X_i$ is not constant zero, then: $\frac{n}{X_1^2+\ldots + X_n^2} = \left( \frac{X_1^2+\ldots + X_n^2}{n}\right)^{-1} \longrightarrow_{n\rightarrow\infty} (\Bbb E [X_1^2])^{-1}$