Talk:Borel–Cantelli lemma

In example: events need to be downward directed?

Can someone please check the "For example". I think the previous version was taking too much for granted. That is, the Borel–Cantelli lemma does say that the outcomes that exist in infinitely many events will themselves have probability zero. However, that doesn't meant that the probability of infinitely many events is zero. For example, consider sample space $\{0,1,2,\dots \}$ and random variable $X:\{0,1,\dots \}\mapsto \mathbb {R}$ defined with $X_{n}(0)\triangleq 0$ and $X_{n}(n)\triangleq 0$ for all n. The sequence of events $[X_{n}=0]$ certainly has 0 in infinitely many of them, and so Borel–Cantelli might imply that Pr({0})=0, but it's hard to convince me of more than that. It seems like you need $[X_{i}=0]\supseteq [X_{j}=0]$ for all i < j if you want to say that $X_{n}=0$ for only finitely many n with probability 1. —TedPavlic (talk) 16:15, 18 June 2009 (UTC)

The example in the article was fine before you changed it, and your example here makes no sense to me. You haven't made any assumptions about $P(X_{n}=0)$, so how are you invoking Borel–Cantelli? Algebraist 16:52, 18 June 2009 (UTC)

Perhaps you misread. The changes I made to the example left the assumptions about $P(X_{n}=0)$ in tact. Namely, it said, as the current wording does, that the series is finite. My argument is that the example as it stands can only claim that the

$\Pr({\text{set of outcomes that occur for infinitely many events in sequence}})=0$

You cannot conclude from here that $X_{n}=0$ for only finitely many n unless you make a stronger statement about the random variables (e.g., that $[X_{n}=0]\supseteq [X_{n+1}=0]$ for all n). (note: I'm going to restore my changes to the clarification at the end of the statement of the lemma; I can't imagine there was any argument there. I'll leave the example as it stands in its old form until this discussion has evolved a bit more) —TedPavlic (talk) 18:23, 18 June 2009 (UTC)

I agree with Algebraist that the example is correct as it is (though I think the last sentence could be better worded). I don't really understand what your objection is. Which is the first of the following sentences that you have a problem with?

1. For each n, let E_n be the event that X_n = 0.
2. We are given that P(E_n) = 1/n² for each n.
3. From the theorem we conclude that $P\left(\limsup _{n\to \infty }E_{n}\right)=0$.
4. This means that the probability that infinitely many E_n occur is 0.
5. That is, the probability that X_n = 0 for infinitely many n is 0.
6. Or, in other words, almost surely X_n =0 for only finitely many n.

--Zundark (talk) 12:49, 19 June 2009 (UTC)

I had a problem with bullet 4 (and hence 5 and 6). I think I've resolved my problem, but I think the example is missing an important step going from the lemma's very simple result to the example's conclusion. Recall that E_n is a set of outcomes. In particular,

$E_{n}\triangleq [X_{n}=0]\triangleq \{\omega \in \Omega :X_{n}(\omega )=0\}$

where Ω is the sample space (see probability space). That is, E_n is a set of all outcomes whose image under the random variable X_n is {0} (i.e., the preimage of {0} under X_n). Because all these random variables share the same probability measure, bullet 2 implies that the set E_n is evolving over time to include outcomes that are progressively less common. By bullet 3, the set of outcomes common to infinitely many E_n has probability zero, and so we can ignore those outcomes—we can remove them from each E_n. What confused me is that the lemma does not imply that the remaining E_n are empty for all but finitely many n. For example, consider the random variables defined with

$X_{n}(\omega )\triangleq {\begin{cases}0&{\text{if }}\omega =0\\0&{\text{if }}\omega =n\end{cases}}$

In this case,

$\limsup _{n\to \infty }E_{n}=\limsup _{n\to \infty }[X_{n}=0]=\limsup _{n\to \infty }\{0,n\}=\{0\}$

so, assuming that $\sum \Pr(E_{n})<\infty$, the lemma says that $\Pr({0})=0$. However, $E_{n}\setminus {0}=\{n\}\neq \emptyset$ for all n. So, it was not obvious to me that the Pr(X_n=0 for infinitely many n)=0. Now I see that the statement is equivalent to Pr(infinitely many intersections of E_n). By the lemma, this probability must be zero (because the only elements that are shared among infinitely many E_n have zero probability). —TedPavlic (talk) 14:39, 19 June 2009 (UTC)

In Alternative proof: why must the sets be ordered?

Why must the sets be reordered in decreasing order in the "Alternative proof" section? This does not seem necessary for the proof and can confuse the reader.

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