0

Apparently for $p_i\in [0,1)$, it's true that $\prod_i (1-p_i)=0 \Leftrightarrow \sum_i p_i=\infty$.

How can I prove this using Borel-Cantelli? I've said that $p_i = P(X_i=1)$ for some sequence $X_i$ where $X_i$'s are all $0-1$ valued. If $\sum_i p_i=\infty$, then $X_i=1$ infinitely often, so it's impossible for $X_i=0$ for all $i$. Thus $\prod_i (1-p_i)=0.$

But how do I prove the reverse step?

Martin Sleziak
  • 53,687
Kashif
  • 1,497
  • 1
    Your forward direction argument is seriously flawed. First you are attempting to use the converse of the Borel=Cantelli lemma, not the lemma itself. These are not equivalent. Secondly, you seem to be mistaking probability $= 1$ to mean "it WILL happen". Where infinite different outcomes are possible (as here), this is not true. Third you appear to think that if some $X_i = 1$, then $p_i = 1$, which is also not how probability works. And finally, that concept violates your initial statement that $p_i < 1$ for all $i$. – Paul Sinclair Dec 03 '16 at 19:15
  • 1
    I meant to say I proved the reverse step first. – Kashif Dec 03 '16 at 21:35
  • http://math.stackexchange.com/questions/1433567/prove-s-doteq-sum-n-1-infty-p-n-infty-to-prod-n-1-infty-1-p-n – BCLC Dec 06 '16 at 18:16

0 Answers0