1. ## Almost sure convergence

$Y_n=\left\{\begin{array}{cc}0,&\mbox{ with probability }
1-1/n^2 \\n, & \mbox{ with probability } 1/n^2\end{array}\right.$

Show that $Y_n$ converges almost surely to 0.

I dont know this problem. please can you help.

sarah

2. You can see that as n -> infinity, the probability of Y = 0 approaches 1. This is because as n gets bigger, 1/(n^2) gets smaller and smaller.

3. Originally Posted by eigenvex
You can see that as n -> infinity, the probability of Y = 0 approaches 1. This is because as n gets bigger, 1/(n^2) gets smaller and smaller.
Thanks but this is obvious. how can you prove this formally?? this is not just probability convergence but almost sure convergence.

4. Hmm....

Almost sure convergence is defined as

So it can be seen that P(lim n->inf (Yn) = 0) = 1 because lim(1/(n^2)) = 0. I'm not sure how much more you have to prove there, I think all you should have to do is plug in what Yn is to that equation and show what happens as n->inf.

5. you have something stronger here
you have complete convergence
Complete Convergence and the Law of Large Numbers
which via Borel-Cantelli implies a.s. convergence
http://www.ncbi.nlm.nih.gov/pmc/arti...01695-0003.pdf

6. Originally Posted by eigenvex
You can see that as n -> infinity, the probability of Y = 0 approaches 1. This is because as n gets bigger, 1/(n^2) gets smaller and smaller.
This proves convergence in probability, not almost-sure.

If you prove that, almost-surely, $Y_n=0$ for all $n$ large enough, then of course this implies that almost-surely $Y_n\to 0$.

Let $A_n=\{Y_n\neq 0\}$. Prove that Borel-Cantelli lemma applies to the sequence of events $(A_n)_n$. This will imply what I said above.