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 $n=2^{30}$).


  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @BCLC Yes, Rosenthal. $3.4$ $\endgroup$ – majmun Aug 13 '16 at 3:12

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.