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Suppose let $\displaystyle (X_n)_{n\ge 1}$ be i.i.d. with N(1,3) random variables. How can I show that:
$\displaystyle \displaystyle\lim_{n\to\infty}\frac{X_1+...+X_n}{X _1^2+...+X_n^2}=\frac{1}{4} $ (a.s)
From my understanding: $\displaystyle \displaystyle\lim_{n\to\infty}\frac{S_n}{n}=\displ aystyle\lim_{n\to\infty}\frac{1}{n}\displaystyle\s um_{j=1}^{n}X_j=\mu$
Then the nominator will go to 1.. How about the denominator? Why is its expected value 4?
Hello,
It's not exactly the numerator that will go to 1, but it's $\displaystyle X_1+\dots+X_n/n$
You have $\displaystyle \displaystyle \frac{X_1+\dots+X_n}{X_1^2+\dots+X_n^2}=\frac{X_1+ \dots+X_n}{n}\cdot\frac{n}{X_1^2+\dots+X_n^2}$
For the second term, we know - also from the SLLN, that it tends to $\displaystyle \mathbb{E}[X_1^2]$. But we know that $\displaystyle 3=Var[X_1]=\mathbb{E}[X_1^2]-(\mathbb{E}[X_1])^2$, so the result should be...
