The Borel-Cantelli lemma and the rule of product (multiplication principle)

256 Views Asked by At

Consider the following random experiment:

We play a game where in each round we either win or lose.Let $\{A_n\}$,$n\in\mathbb N$, be a sequence of events where $A_n$ is the event that we win in the $n$-round.Let $P\left(A_{n}\right)=\frac{1}{n^2}$, for each $n$.Finally, the events are independent (since the game is played successively in independent rounds).

This is a typical example of a random experiment where the Borel-Cantelli lemma can be applied.

We have $$\sum_{n=1}^{\infty} P\left(A_{n}\right)=\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6} < \infty,$$so $$P\left(\limsup\limits_{n\to\infty} A_n\right) = 0$$

Therefore, the probability that infinitely many of $A_n$ occur is $0$. Only a finite number of $A_n$ occur. Let $k$ be this finite number.

My question is the following:

Since the events are independent,we can apply the rule of product.We will apply it for the first $k+1$ events.So, we have $$P\left(\bigcap_{n=1}^{k+1} A_n\right)=\prod _{n=1}^{k+1} P(A_n)=\prod _{n=1}^{k+1} \frac{1}{n^2} =\frac{1}{((k+1)!)^2}\neq 0$$ This means that there is a non-zero probability that both $A_1,A_2,...,A_{k+1}$ occur.Thus,it is possible that $k+1$ occur.But,we assumed that only $k$ of $A_n$ occur.In fact,we can always consider more and more events to the rule of product,finding smaller and smaller probabilities (but still non-zero) for each new intersection of events.

Obviously,I am doing something wrong here.How is the Borel-Cantelli lemma connected to the rule of product in this case? Thank you very much in advance!

1

There are 1 best solutions below

1
On

The Borel-Cantelli Lemma applies to tail events "at infinity", where by the Kolmogorov 0-1 law such events have possible probability 1 or 0 only. So any finite projection up to $k$ samples (however large $k$ is) does not capture what happens at infinity.