# Thread: Convergence question

1. ## Convergence question

Suppose for a sequence of random variables $X_1,X_2,...$ is considered as completely convergence to $X$ if

$\displaystyle\sum_{n=1}^{\infty}P(|X_n-X|>\epsilon)<\infty$ for each $\epsilon>0$

How can I show that if the sequence $X_n$ is independent then the complete convergence becomes convergence almost surely?

Though some searching, the definition of converge almost surely is

$P(\displaystyle\lim_{n\to\infty}X_n=X)=1$.

Then

$P(\displaystyle\lim_{n\to\infty}|X_n-X|<\epsilon)=1$

Do I start from here?

2. Hello,

By Borel-Cantelli's lemma (part I), we have $\displaystyle P(\limsup_n \{|X_n-X|<\epsilon\})=0$. By taking the complement, we get that $\displaystyle \forall \epsilon>0,P(\liminf_n \{|X_n-X|<\epsilon\})=1$

This means that $\displaystyle\forall \epsilon>0,P(\exists N\in\mathbb N,\forall n\geq N,|X_n-X|<\epsilon)=1$.

Then you have to get the epsilon into the probability, and I think you have to assume first that epsilon is rational (so that there is a countable number of epsilon's).
I don't really remember how to do and I have to go out right now, I'll try to think about it, unless you do before me

3. Originally Posted by Moo
Hello,

By Borel-Cantelli's lemma (part I), we have $\displaystyle P(\limsup_n \{|X_n-X|<\epsilon\})=0$. By taking the complement, we get that $\displaystyle \forall \epsilon>0,P(\liminf_n \{|X_n-X|<\epsilon\})=1$

This means that $\displaystyle\forall \epsilon>0,P(\exists N\in\mathbb N,\forall n\geq N,|X_n-X|<\epsilon)=1$.

Then you have to get the epsilon into the probability, and I think you have to assume first that epsilon is rational (so that there is a countable number of epsilon's).
I don't really remember how to do and I have to go out right now, I'll try to think about it, unless you do before me
I thought u got it correct?

From Borel-Cantelli's lemma,
$\displaystyle\sum_{n=1}^{\infty}P(|X_n-X|>\epsilon)<\infty$ becomes

$\displaystyle \forall \epsilon>0,P(\liminf_n \{|X_n-X|<\epsilon\})=1$

which means

$\displaystyle \forall \epsilon>0,P(\lim\bigcup \{|X_n-X|<\epsilon\})=1$

Since it's independent, then

$\displaystyle \forall \epsilon>0,\prod P(\lim \{|X_n-X|<\epsilon\})=1$

which implies that

$P(\displaystyle\lim_{n\to\infty}|X_n-X|<\epsilon)=1$

4. It's not independent, because of X

5. Hsu-Robbins original paper is http://www.ncbi.nlm.nih.gov/pmc/arti...01695-0003.pdf
there are many others on complete convergence like http://www.math.uregina.ca/~volodin/adv.pdf