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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, considersample space${\displaystyle \{0,1,2,\dots \}}$ and random variable${\displaystyle X:\{0,1,\dots \}\mapsto \mathbb{R}}$ defined with${\displaystyle X_n(0)\triangleq 0}$ and${\displaystyle X_n(n)\triangleq 0}$ for alln. The sequence of events${\displaystyle [X_n=0]}$ 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${\displaystyle [X_i=0]\supseteq [X_j=0]}$ for alli < j if you want to say that${\displaystyle X_n=0}$ for only finitely manyn with probability 1. —TedPavlic (talk)16:15, 18 June 2009 (UTC)[reply]

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${\displaystyle P(X_n=0)}$, so how are you invoking Borel–Cantelli?Algebraist16:52, 18 June 2009 (UTC)[reply]

Perhaps you misread. The changes I made to the example left the assumptions about${\displaystyle P(X_n=0)}$ 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

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

You cannot conclude from here that${\displaystyle X_n=0}$ for only finitely manyn unless you make a stronger statement about the random variables (e.g., that${\displaystyle [X_n=0]\supseteq [X_{n+1}=0]}$ for alln). (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)[reply]

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 eachn, letE_n be the event thatX_n = 0.
2. We are given thatP(E_n) = 1/n² for eachn.
3. From the theorem we conclude that${\displaystyle P\left(\underset{n\to \mathrm{\infty }}{lim sup}{E}_n\right)=0}$.
4. This means that the probability that infinitely manyE_n occur is 0.
5. That is, the probability thatX_n = 0 for infinitely manyn is 0.
6. Or, in other words, almost surelyX_n =0 for only finitely manyn.

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

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 thatE_n is a set of outcomes. In particular,

${\displaystyle E_n\triangleq [X_n=0]\triangleq \{\omega \in \mathrm{\Omega }:X_n(\omega )=0\}}$

where Ω is thesample space (seeprobability space). That is,E_n is a set of all outcomes whoseimage under therandom variableX_n is {0} (i.e., thepreimage of {0} underX_n). Because all these random variables share the same probability measure, bullet 2 implies that the setE_n is evolving over time to include outcomes that are progressively less common. By bullet 3, the set of outcomescommon to infinitely manyE_n has probability zero, and so we can ignore those outcomes—we can remove them from eachE_n.What confused me is that the lemma doesnot imply that the remainingE_n are empty for all but finitely manyn. For example, consider the random variables defined with

In this case,

${\displaystyle \underset{n\to \mathrm{\infty }}{lim sup}{E}_n=\underset{n\to \mathrm{\infty }}{lim sup}[X_n=0]=\underset{n\to \mathrm{\infty }}{lim sup}\{0,n\}=\{0\}}$

so, assuming that, the lemma says that${\displaystyle Pr(0)=0}$. However,${\displaystyle E_n\setminus 0=\{n\}\ne \mathrm{\varnothing }}$ for alln. So, it was not obvious to me that the Pr(X_n=0 for infinitely manyn)=0. Now I see that the statement is equivalent to Pr(infinitely many intersections ofE_n). By the lemma, this probability must be zero (because the only elements that are shared among infinitely manyE_n have zero probability). —TedPavlic (talk)14:39, 19 June 2009 (UTC)[reply]

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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