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Consider infinite, independent, fair coin tossing, and let $(H_i)$ be the event that the $n$th coin is heads (equal to 1 while 0 means tails). Determine $\begin{aligned} \Bbb{P}\big(\cap_{i=1}^{\log_2(n)}H_{n+i} \ \ \ i.o.\big) \end{aligned}$

My attempt

Define $A_n = \cap_{i=1}^{\log_2(n)}H_{n+i}$. We want to determine $\Bbb{P}\big( \limsup_n A_n \big)$. It is clear that

$\Bbb{P}\big(A_n\big)=\big(\frac12\big)^{\log_2(n)}=\frac1n$ for every $n\geq1$.

I cannot apply Borel-Cantelli lemma on

$\sum_{n=1}^{\infty}\Bbb{P}\big(A_n\big)=\sum_{n=1}^{\infty}\frac1n=+\infty$

as the $A_n$ are not independent. Hence I consider $(n^2)_{n=1}^\infty$ as a subsequence of $\Bbb{N}$, and define $B_n = \cap_{i=1}^{\log_2(n^2)}H_{n^2+i}=\cap_{i=1}^{2\log_2(n)}H_{n^2+i}$.

My argument is that the $B_n$ are independent, as

$n^2 > (n-1)^2+2\log_2(n) \; \forall n\geq 1$.

Obviously $\Bbb{P}(B_n)=\big(\frac12\big)^{\log_2(n^2)}=\big(\frac1n\big)^2$ and

$\sum_{n=1}^{\infty}\Bbb{P}\big(B_n\big)=\sum_{n=1}^{\infty}\big(\frac1n\big)^2<\infty$.

Then, by Borel-Cantelli lemma, $\Bbb{P}\big(B_n \; i.o.\big)=\Bbb{P}\big(\limsup_nB_n\big)=0$.

Then, I would conclude that

$\Bbb{P}\big(B_n \; i.o.\big)=0 \Rightarrow \Bbb{P}\big(A_n \; i.o.\big)=0$,

although I cannot prove (or disprove) such implication.

But let us consider a different subsequence, $\big(n_k\big)_{k=1}^\infty$, where $n_k=k\log_2(k^2)$, and define the events $C_n=A_{n\log_2(n^2)}=\cap_{i=1}^{\log_2(n\log_2(n^2))}H_{n\log_2(n^2)+i}$.

Just like the $A_n$, the $C_n$ are independent too, as some tedious algebra shows. But we have

$\Bbb{P}\big(C_n\big)=\big(\frac12\big)^{\log_{2}\big(n\log_{2}(n^2)\big)}=\frac{1}{2n\log_2(n)}$, and $\sum_{i=1}^\infty\Bbb{P}\big(C_n\big)=\infty$

because $\frac{1}{2n\log_2(n)}=O\big(\big(\frac1n\big)^{1+\varepsilon}\big) \; \forall \varepsilon >0$.

Hence, I can apply Borel-Cantelli lemma on $\big(C_n\big)$ and get that $\Bbb{P}\big(C_n \; i.o.\big)=1$ which should imply (again I cannot prove or disprove it) that $\Bbb{P}\big(A_n \; i.o.\big)=1$.

So, this is apparently contradictory, as there are two subsequences of $\Bbb{N}$ that lead to opposite results, and I cannot see why I should prefer one over another, as in my view both should allow to extend the result to the whole sequence $\big(A_n\big)$. The second subsequence (the one with the $C_n$ is supposed to provide the right answer, and I would like to understand what is wrong with the attempt using the $B_n$. Any help would be much appreciated!

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1 Answer 1

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There is no contradiction here. $\newcommand{\io}{\text{ i.o. }}$

$P(B_n\io)=0$ does not imply $P(A_n\io)=0$. If $P(B_n\io)=0$, then that means there will only be finitely many perfect square indices for which $A_{n^2}$ occurs. However, it is possible to have finitely many occurrences in the squares, and infinitely many occurrences elsewhere.

$P(A_{n\log_2(n^2)}\io)=1$ does imply that $P(A_n\io)=1$. If there are infinitely many occurrences in the the indices of the form $n\log_2(n^2)$ alone, there there will certainly be infinitely many occurrences in general.

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  • $\begingroup$ thank you, this is want I sensed and could not prove. So I can believe that is the case, but I cannot understand it. Can you maybe make these implications more explicit, please? $\endgroup$
    – Fran712
    Apr 23, 2023 at 11:46
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    $\begingroup$ @Fran712 (1) Show that if $C_n$ occurs i.o., then $A_n$ occurs i.o. (2) For any events $E,F$ if $E$ occurs implies $F$ occurs, then $E\subseteq F$. (3) Combining (1) and (2), you know the event $\{C_n\text{ i.o.}\}\subseteq \{A_n\subseteq\text{ i.o.}\}$. This clearly implies $P(C_n\text{ i.o.})\le P(A_n\text{ i.o})$. Since $P(C_n\text{ i.o})=1$, this implies $P(A_n\text{ i.o.})=1$. $\endgroup$ Apr 24, 2023 at 15:50

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