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This is supposed to be related to the 2nd Borel-Cantelli Lemma (my justification for the independence tag). In Williams' Probability with Martingales, 2BCL is proven and then the following is given as an exercise:

Prove $S \doteq \sum_{n=1}^\infty p_n < \infty \implies \prod_{n=1}^\infty (1-p_n) > 0$ assuming $0 \leq p_n < 1$.

Hint: Show that $S < 1 \implies \prod_{n=1}^\infty (1-p_n) \geq 1 - S$.

Proving the hint:

I tried to find $\{a_n\}$ s.t.

$$\prod_{n=1}^\infty (1-p_n) \geq \prod_{n=1}^\infty e^{a_n} = e^{\sum_{n=1}^\infty a_n} \geq 1 - \sum_{n=1}^\infty a_n \geq 1 - S > 0$$

That is, find $\{a_n\}$ s.t.:

1a. $1 - p_n \geq e^{a_n}$

1b. $a_n \leq p_n$

I thought of $a_n = \ln(1-p_n)$ (a reason why $p_n < 1$, I guess)

Is that right?

Edit: Actually, assuming my proof is right, is $S < 1$ used in the proof? If so, where?

From Williams book:

enter image description here

Community
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BCLC
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1 Answers1

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Hint for the case $S<1$: Prove this by induction for a finite number of probabilities using that $(1-a)(1-b)\ge 1-a -b$ and then proceed to the limit.

Hint for the rest of proof: If $\sum_{n=1}^\infty p_n<\infty$, then there exists $N$ such that $\sum_{n=N}^\infty p_n<1$.

zhoraster
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  • You mean $\forall m \in \mathbb{N}, \sum_{n=1}^{m} p_n < 1 \to \prod_{n=1}^{m} (1-p_n) \geq (1- \sum_{n=1}^{m} p_n) \to \sum_{n=1}^{\infty} p_n < 1 \to \prod_{n=1}^{\infty} (1-p_n) \geq (1- \sum_{n=1}^{\infty} p_n)$ ? Are you sure? $H_n \infty \forall n \in \mathbb{N}$ but $H_n \to \infty$ – BCLC Sep 13 '15 at 16:06
  • @BCLC You're already given (in the hint) that the sum converges to a value less than 1, and the product is trivially bounded from above by 1. – Steven Stadnicki Sep 13 '15 at 17:03
  • @StevenStadnicki Creepy coincidence. Just edited wondering where $S < 1$ is used in the proof. Wait, so is my proof wrong? Actually... – BCLC Sep 13 '15 at 17:05
  • @StevenStadnicki Actually I don't get why your proof is right. Sure we can take limit of both the inequality in the hypothesis and the inequality in the conclusion but how do you know that the new hypothesis implies the new conclusion? – BCLC Sep 13 '15 at 17:09
  • It's not my proof, but: do you see why you can prove by induction using the algebraic formula in this hint that $\prod_{i=1}^n(1-p_i)\geq 1-\sum_{i=1}^np_i$ (i.e., the 'finite version')? – Steven Stadnicki Sep 13 '15 at 17:17
  • Once you have that, then you have $\prod_{i=1}^n+\sum_{i=1}^n\geq 1$ for all $n$ by trivial algebraic manipulation. But now you can pass to the limit in $n$. – Steven Stadnicki Sep 13 '15 at 17:19
  • @StevenStadnicki Oh lol it's zhoraster's. Sorry. I get why the finite conclusion is true. 1 Where is the 'sum converges to a value less than 1' included in the proof? 2 Is my proof wrong? – BCLC Sep 13 '15 at 17:38
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    @BCLC 'The sum converges to a value less than 1' is included (a) in allowing you to pass to the limit (since you need convergence of the sum for that to be possible) and (b) in making $\prod_{i=1}^\infty(1-p_i)\geq 1-S$ being a non-trivial statement - if $S$ is greater than one then this says nothing. – Steven Stadnicki Sep 13 '15 at 17:42
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    Your proof looks correct at heart but it needs more additional machinery than this one - you need inequalities on exp and ln, for instance. – Steven Stadnicki Sep 13 '15 at 17:43
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    @BCLC, I've updated the answer. I also suggest to open a chat for discussion. – zhoraster Sep 13 '15 at 17:59
  • Thanks @StevenStadnicki ! zhoraster, did that here. Landon Carter gave a more creative proof. – BCLC Sep 13 '15 at 18:32