# Strong law of large number

• Oct 30th 2010, 06:26 AM
noob mathematician
Strong law of large number
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?
• Oct 31st 2010, 03:16 AM
Moo
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...