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Suppose $\{A_k\}_{k=1}^\infty$ and $\{B_k\}_{k=1}^\infty$ are measurable sets in the same probabiilty space such that $P(A_k)=P(B_k)$ for all $k$. If you can show that $P(A_k \text{ i.o.})=1$ (for example, the $A_k$'s are independent and $\sum_{k=1}^\infty P(A_k)=\infty$) then is it necessarily true that $P(B_k \text{ i.o.})=1$? (Obviously if I used Borel-Cantelli for the $A_k$'s, I am assuming that the $B_k$'s are not necessarily independent.) I don't think this is so, but I can't come up with a counterexample yet.

BCLC
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Howard F.
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1 Answers1

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The big point is this: we have no information about how the $A_k$ correlate with each other or how the $B_k$ correlate with each other. So, we can build a scenario like this:

Suppose we have a fair coin, which we flip over and over again.

For $k\in\mathbb{N}$, let $A_k$ be the event that the $k$th flip yields heads; let $B_k$ be the event that the first flip yields heads.

Then $P(A_k)=P(B_k)=\frac{1}{2}$ for all $k$, and $P(A_k\text{ i.o.})=1$, but $$ P(B_k\text{ i.o})=P(\text{first flip gives heads})=\frac{1}{2}. $$

Nick Peterson
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  • Of course. Bad question. Slightly better question: are there other standard ways of proving $P(B_k \text{ i.o.})=1$ other than Borel-Cantelli (that is, in the absence of independence of the $B_k$'s)? – Howard F. Jul 26 '13 at 02:16
  • There are VERY few bad questions, I assure you!

    For the rest: slightly more general is the Kolmogorov 0-1 law; it says that if you have a sequence of random variables $(X_n){n=1}^{\infty}$ and a _tail event for the sequence - that is, an event that is independent of any finite initial sequence of the random variables - then that event must have probability 0 or 1. To get more general is difficult - it really depends on the nature of the relationship between your variables.

     – Nick Peterson Jul 26 '13 at 02:59
  • @HowardF. I can see this question as a good True/False question in an exam. Not a bad question indeed – Jean-Sébastien Jul 26 '13 at 03:11
  • Thanks for the comments. I am familiar with the Kolmogorov 0-1 Law, but it still requires independence which is not available in the problem I'm thinking about. After doing some more homework, I found the following post, which addresses more of the generalization that I had in mind: [http://math.stackexchange.com/questions/134117/dependence-and-second-borel-cantelli-lemma](Dependence and second Borel-Cantelli lemma). – Howard F. Jul 26 '13 at 15:39