Strong law of Large numbers exercise

Question

I'm not sure if it's an algebraic manipulation trick or a change in our $X_i$ iid where $i$ is 1 to n but how can I use the SLLN to approach the form $\lim_{n\to \inf} |\frac{1}{n}\sum_{i=1}^n X_i-E[X_i]|=0$

Answer 1

This is called as Weak Law of Large Numbers (WLLN). Let $X_1,X_2, ....., X_n$ be n iid repetitions of Random Variable $X$, with mean $E[X]$ and Variance $Var[X]$.

Define $S_n = X_1 + X_2+....+X_n$

then $E[\dfrac{S_n}{n}]=E[X]$

and $Var[\dfrac{S_n}{n}] = \dfrac{Var(X)}{n}$.

Using Chebyschev's Inequality, for any Random Variable $Y$, and any $\eta>0$

$P[|Y-E(Y)|>\eta]\leq \dfrac{Var(Y)}{\eta^2}$

Substituting $Y = \frac{S_n}n{}$, you get

$P[|\dfrac{S_n}{n}-E(X)|>\eta] \leq \dfrac{Var(X)}{n\eta^2}$. Considering a very very small $\eta$,

$\lim_{n\to\infty}P[|\dfrac{S_n}{n}-E(X)|>\eta] = 0 $

Answer 2

The SLLN directly implies $$ \lim_{n\to\infty} \frac{1}{n}\Big(\sum_{i=1}^n X_i-E[X_i]\Big)=0. $$ holds almost surely. Combine this with the following result: $$ \lim_{n\to\infty} a_n=a\implies \lim_{n\to\infty}|a_n|=|a|. $$