Probability of getting infinitely often $\log_2(n)$ consecutive heads on fair coin tossing is equal to 1

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!

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.
• @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 $P(A_n\text{ i.o.})=1$. Apr 24, 2023 at 15:50