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This is from David Williams' book Probability using Martingales. I'm self-studying.

Question

Prove that if $$0\leq p_n < 1 \quad\text{ and }\quad S:=\sum p_n < \infty$$ then $$\prod (1-p_n) > 0$$ Hint: First show that if $S<1$, then $\prod (1-p_n)\geq 1-S$.

I was able to prove the hint using induction. Assume $\prod\limits_{n=1}^N (1-p_n) \geq 1-\sum\limits_{n=1}^N p_n$. Consider $\prod\limits_{n=1}^{N+1}(1-p_n) \geq (1-\sum\limits_{n=1}^N p_n)(1-p_{N+1})=1-\sum\limits_{n=1}^{N+1}p_n+p_{N+1}\sum\limits_{n=1}^{N}p_n \geq 1-\sum\limits_{n=1}^{N+1}p_n$.

But I'm unable to use this to prove the general result for arbitrary $S$. Any guidance would be appreciated. I'm also surprised that he asks this question after stating the 2nd Borel Cantelli lemma, I don't see the connection.

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  • $\begingroup$ in the induction proof you have written where have you used the fact that $S<1$? $\endgroup$
    – happymath
    Mar 2, 2014 at 6:18
  • $\begingroup$ I haven't. But the proof proves my original claim of $\prod(1-p_n)>0$ only if $S<1$. Its useless otherwise. $\endgroup$
    – elexhobby
    Mar 2, 2014 at 6:29
  • $\begingroup$ then you have actually shown by induction for all possible S right? $\endgroup$
    – happymath
    Mar 2, 2014 at 6:30
  • $\begingroup$ I have shown that $\prod (1-p_n)>1-S$ for all possible $S$. That doesn't prove $\prod(1-p_n)>0$, because $1-S$ is negative if $S > 1$. $\endgroup$
    – elexhobby
    Mar 2, 2014 at 6:35
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    $\begingroup$ In your induction proof, I think you should replace $1-\sum_{n=1}^{N+1}p_n+\prod_{n=1}^{N+1}p_n$ by $1-\sum_{n=1}^{N+1}p_n+p_{N+1}\sum_{n=1}^{N}p_n$. $\endgroup$ Mar 8, 2014 at 11:31

2 Answers 2

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If $\sum_{n=1}^\infty p_n<\infty$, then you can find for each $\epsilon>0$ some index $N$ such that $\sum_{n=N}^\infty p_n<\epsilon$. In particular, you can do this for some $0<\epsilon<1$. By what you have shown, $$\prod_{n=N}^M (1-p_n)>1-\epsilon>0,$$ and since $\prod_{n=N}^M (1-p_n)$ converges as $M\to\infty$ (it is a decreasing sequence bounded below), we have $$\prod_{n=N}^\infty (1-p_n)\geq 1-\epsilon>0.$$

Clearly, $\prod_{n=1}^{N-1}(1-p_n)>0$. Now $$\prod_{n=1}^\infty(1-p_n)=\prod_{n=1}^{N-1}(1-p_n) \prod_{n=N}^\infty (1-p_n)\geq (1-\epsilon)\prod_{n=1}^{N-1}(1-p_n)> 0.$$


Here is the connection to the second Borel-Cantelli lemma. The lemma says that if $(E_n)$ is a sequence of independent events satisfying $\sum_{n=1}^\infty p_n=\infty$ with $p_n=P(E_n)$, then a random point in the sample space lies with probability $1$ in infinitely many of the $E_n$. This exercise shows that this automatically fails if $\sum_{n=1}^\infty p_n<\infty$, since $\prod_{n=1}^\infty (1-p_n)$ is the probability that an element lies in none of the $E_n$.

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  • $\begingroup$ Thanks! So, this implies that if $E_n$ is a sequence of independent events, then $\sum\limits_{n=1}^{\infty}p_n=\infty$ is not only sufficient (by BC2), but also necessary for $P(E_n i.o)=1$. Am I right? $\endgroup$
    – elexhobby
    Mar 8, 2014 at 15:02
  • $\begingroup$ @elexhobby Exactly. $\endgroup$ Mar 8, 2014 at 18:22
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Without loss of generality we can take $0<p_n<1$, and by the condition $\sum p_n<\infty$ follows that $p_n\to 0$. Denote $$p=\prod_{n=1}^{\infty} (1-p_n)\in[\,0,1).$$ Since $$\lim_{n\to\infty}\frac{\log(1-p_n)}{-p_n}=1,$$ then the series $\sum\log(1-p_n)>-\infty$ converges as $\sum p_n$ converges and hence

$$\log p=\lim_{N\to\infty}\sum_{n=1}^N \log (1-p_n)=\sum_{n=1}^{\infty}\log (1-p_n)>-\infty.$$ Thus we get $$-\infty<\log p<0,$$ which implies $p>0.$

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