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Let $A_n$ be independent events with $P(A_n) \neq 1$. Show that $\prod_{n=1}^{\infty} (1- P(A_n))=0$ iff $\sum P(A_n) = \infty$

It kind of looks obvious but I really have no idea how to prove it. Can someone give me help?

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    $\begingroup$ I still don't understand how I can derive this directly from Borel-Cantelli lemma. Thanks $\endgroup$ – user179454 Sep 29 '14 at 5:47
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    $\begingroup$ On second thoughts, I think only the second one is useful here. Please give me some time to type an answer. $\endgroup$ – karakusc Sep 29 '14 at 5:53
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    $\begingroup$ Posted the answer. Turns out both Borel-Cantelli lemmas are useful :) $\endgroup$ – karakusc Sep 29 '14 at 6:48
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    $\begingroup$ Hint: I was able to answer it without directly using either BCL, but I followed the BCL2 proof to prove it. $\endgroup$ – BCLC Sep 29 '14 at 7:10
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    $\begingroup$ @MichaelHardy For the reason why Borel-Cantelli lemma is NOT the best way to prove this, see my comment on karakusc's answer. $\endgroup$ – Did Sep 29 '14 at 15:54
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The proof of this result has nothing to do with probability theory in general or with Borel-Cantelli lemmas in particular, it is only plain-old-deterministic real analysis:

Let $(x_n)$ denote a real valued sequence such that $0\leqslant x_n\lt1$ for every $n$, then $\prod\limits_n(1-x_n)=0$ if and only if $\sum\limits_nx_n$ diverges.

Proof: If $\sum\limits_nx_n$ diverges, note that $\log(1-x)\leqslant-x$ for every $x\lt1$ to deduce that $$\prod\limits_n(1-x_n)=\exp\left(\sum_n\log(1-x_n)\right)\leqslant\exp\left(-\sum_nx_n\right)=0.$$ If $\sum\limits_nx_n$ converges, note that $x_n\leqslant\frac12$ for every $n$ large enough, say every $n\geqslant N$, and that $\log(1-x)\geqslant-2x$ for every $x\leqslant\frac12$ to deduce that $$\prod\limits_{n\geqslant N}(1-x_n)=\exp\left(\sum_{n\geqslant N}\log(1-x_n)\right)\geqslant\exp\left(-2\sum_{n\geqslant N}x_n\right),$$ hence $$\prod\limits_{n\geqslant N}(1-x_n)\ne0.$$ Since $x_n\lt1$ for every $n$, in particular for every $n\lt N$, this implies the desired result, namely, that $$\prod\limits_n(1-x_n)\ne0.$$

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    $\begingroup$ Did: nice proof! +1 $\endgroup$ – Olivier Oloa Sep 29 '14 at 21:29
  • $\begingroup$ @Did this related? I used ln(1-x) too math.stackexchange.com/q/524353/140308 $\endgroup$ – BCLC Oct 3 '14 at 16:12
  • $\begingroup$ If it is indeed related, then you should know I followed the BC2 proof. This proof also looks like the BC2 proof sooo....yeah...it's not used, but the proof does help. Well for YOU, you didn't need it but yeah others might :)) $\endgroup$ – BCLC Oct 3 '14 at 16:14
  • $\begingroup$ @BCLC ?? Of course this is "related" since this is the same question! Hence one can copy the approach in the present answer and it will work, yes. (Sorry but what "should I know" that I would not, exactly?) $\endgroup$ – Did Oct 3 '14 at 16:34
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    $\begingroup$ @BCLC No. What I actually wrote is that "Borel-Cantelli lemmas are not involved in the proof of this result'. This does not mean that knowing a proof of BCL cannot help to find a proof of this and, indeed, it does since the proof of BCL is basically a proof of this result. But to use the conclusion of BCL in a proof of this result is absurd. (Your insistence about this point is surprising, the matter seems settled since my comment to karakusc's answer Sep 29 at 14:06. But if your point is that the inequality $\log(1-x)\leqslant-x$ is useful, then I fully agree.) $\endgroup$ – Did Oct 3 '14 at 16:52
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By the second Borel-Cantelli lemma, if $A_n$ are independent and $\sum P(A_n) = \infty$, then $P(A_n \text{ infinitely often})=1$. Then

$$ 0=P \left( \bigcap_{n=1}^\infty A_n^c\right) = \prod_{i=1}^\infty (1- P \left( A_n\right)) $$

since this is the probability that none of $A_n$ would happen.

Conversely, if $\sum P(A_n) < \infty$, by first Borel-Cantelli lemma, $P(A_n \text{ infinitely often})=0$, meaning a.s. there exists $N<\infty$ such that $A_n$ does not happen for $n >N$. Therefore

$$ \begin{split} \prod_{i=1}^\infty (1- P \left( A_n\right)) &=P \left( \bigcap_{n=1}^\infty A_n^c\right) \\ &= P \left( \bigcap_{n=1}^N A_n^c\right) \\ & =E_N \left[ E\left[ \left.1_{\bigcap_{n=1}^N A_n^c}\right| N\right]\right] \\ & =E_N \left[ E\left[ \left. \prod_{n=1}^N 1_{A_n^c}\right| N\right]\right] \\ & =E_N \left[ \prod_{n=1}^N E\left[ \left. 1_{A_n^c}\right| N\right]\right] \\ & =E_N \left[ \prod_{n=1}^N P(A_n^c)\right] > 0\\ \end{split} $$ since $P(A_n^c)>0$ for all $n$, and $N<\infty$. ($1_{(\cdot)}$ denotes indicator function in the above).

Perhaps cleaner and a bit more rigorous solutions are possible.

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    $\begingroup$ Funny: this is using Borel-Cantelli lemma to prove a crucial step of the proof of Borel-Cantelli lemma. $\endgroup$ – Did Sep 29 '14 at 14:06
  • $\begingroup$ I don't quite catch the implication that $P(A_n \text{infinitely often}) = 0$ implies the existence of a specific finite $N$ for which $A_n$ a.s. doesn't occur beyond $N$. Is it a compactness argument? $\endgroup$ – Erick Wong Sep 29 '14 at 16:09
  • $\begingroup$ @ErickWong : If there is no value $N$ such that $A_n$ is false for all $n\ge N$, then there are infinitely many values of $n$ for which $A_n$ is true. $\endgroup$ – Michael Hardy Sep 29 '14 at 19:13
  • $\begingroup$ @MichaelHardy I completely agree for a given state, but maybe I'm misinterpreting the first two lines of the chain of equalities. Is $N$ meant to be fixed or is it a random variable? $\endgroup$ – Erick Wong Sep 29 '14 at 19:49
  • $\begingroup$ @ErickWong $N$ is a random variable in general, but we know that $P(N<\infty)=1$. $\endgroup$ – karakusc Sep 29 '14 at 20:33

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