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I think title is pretty self explanatory. Consider infinite fair coin tossing. Let $H_n$ be the the event that the $n^\text{th}$ coin comes up heads. Define $$B_n = H_{n+1} \cap H_{n+2} \cap \dots \cap H_{n+ [\log_2\log_2 n]}$$ Why are the $B_n$ not independent of each other?

Also, related to this (and perhaps why I am misunderstanding), does it make sense to consider values of $n$ where $\log_2\log_2 n$ is not an integer? If so, how is this interpreted?

I mean, I think I get why: The $B_n$ overlap each other (to varying extents). But that the overlap is not apparent to me right now.

More specifically, $\log_2(\log_2 n)$ tends to be pretty small, even for large $n$. If we consider only values of $n$ where $\log_2(\log_2 n)$ is a whole number, the dependence of $\{ B_n\}$ is not obvious to me, because such values of $n$ seem to be very far apart (far enough to prevent overlap, I think?).

If we allow $\log_2(\log_2 n)$ to not be an integer, then I can see how the $B_n$ are not independent (because then we have $n+1, n+2,\dotsc, n+4$ for one value of $n$ and then $(n+1) +1 = n+2,\dotsc, (n+1) +4.xxxx$ for the next), but then I don't know how to interpret, for example, $H_{n+4.90689}$ (when $n=2^{30}$).

Thanks.

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    $\begingroup$ 1) The [] notation indicates some sort of rounding to an integer, possibly floor but it depends on whether you've transcribed it accurately, so $H_{n+[4.90689]}$ is really either $H_{n+4}$ or $H_{n+5}$. 2) It is immediately obvious from this interpretation that $B_n$ and $B_{n+1}$ overlap so long as $\log \log n$ is large enough to be $>1$. $\endgroup$ – Erick Wong Aug 13 '16 at 2:13
  • $\begingroup$ Is this from Rosenthal? If not where? $\endgroup$ – BCLC Aug 13 '16 at 2:45
  • $\begingroup$ @ErickWong I believe I transcribed it correctly, but perhaps not. Now that I think about it though, given that I'm looking at a pdf scan perhaps the character recognition changed the floor or ceiling notation to regular square brackets, $[$ and $]$. (In other words, perhaps I transcribed it correctly from an incorrect transcription...) $\endgroup$ – majmun Aug 13 '16 at 3:12
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    $\begingroup$ @BCLC Yes, Rosenthal. $3.4$ $\endgroup$ – majmun Aug 13 '16 at 3:12
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I'll assume the $H_n$'s are independent with $0 < P(H_n) = p < 1$

You can show the $B_n$'s are not independent by showing that they are not pairwise independent.

$$B_n = H_{n+1} \cap H_{n+2} \cap \cdots \cap H_{n+ [\log_2\log_2 n]}$$

$$B_{n+1} = H_{n+2} \cap H_{n+3} \cap \cdots \cap H_{n+1+ [\log_2\log_2 (n+1)]}$$

$$P(B_n) = p^{n+ [\log_2\log_2 n] - (n+1) + 1} = p^{[\log_2\log_2 n]}$$

$$P(B_{n+1}) = p^{n+1+ [\log_2\log_2 (n+1)] - (n+2) + 1} = p^{[\log_2\log_2 (n+1)]}$$

Now just show that

$$P(B_n, B_{n+1}) \ne p^{[\log_2\log_2 (n)]} p^{[\log_2\log_2 (n+1)]} $$

I believe that $\{B_n, B_{n+1}\} = H_{n+2} \cap \cdots \cap H_{n+ [\log_2\log_2 n]}$

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