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$\exists$ this exercise in Rosenthal's A First Look at Rigorous Probability Theory:

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For letters d and e, how do you show that the ff events are pairwise independent?

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My attempt:

It suffices to show that $B_n$ and $B_{n+1}$ are pairwise independent (because Borel-Cantelli 2 can use pairwise independence). To do this, it suffices to show that the index of the last event in $B_n$ is less than the index of the first event of $B_{n+1}$, that is, $\left \lfloor n\log_2 n^2 \right \rfloor + \left \lfloor \log_2(\left \lfloor n\log_2n^2 \right \rfloor) \right \rfloor \leq \left \lfloor (n+1)\log_2 (n+1)^2 \right \rfloor + 1$.

How to show this? I tried using $\left \lfloor k \right \rfloor \leq k$, but I'm not sure that that does anything...

LHS $\leq \left \lfloor (n+1)\log_2 (n+1)^2 \right \rfloor + \left \lfloor \log_2(\left \lfloor n\log_2n^2 \right \rfloor) \right \rfloor $ < $\left \lfloor (n+1)\log_2 (n+1)^2 \right \rfloor + 1$ because...ugh...

$\left \lfloor \log_2(\left \lfloor n\log_2n^2 \right \rfloor) \right \rfloor \leq \log_2(\left \lfloor n\log_2n^2 \right \rfloor) < 1$?

I don't think the last inequality is true.

Help please?

Daniel Fischer
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1 Answers1

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From here: https://math.stackexchange.com/a/907506/140308

hint:

$\left \lfloor (n+1)\log_2 (n+1)^2 \right \rfloor + 1 =\left \lfloor (n\log_2 (n+1)^2)+\log_2 (n+1)^2 \right \rfloor +1\ge \left \lfloor (n\log_2 (n+1)^2) \right \rfloor +\left \lfloor \log_2 (n+1)^2\right \rfloor+1$

now prove:

$\left\lfloor (n\log_2 (n+1)^2) \right \rfloor \ge \left\lfloor (n\log_2 n^2) \right \rfloor$

$\left \lfloor \log_2 (n+1)^2\right \rfloor+1 \ge \left \lfloor \log_2(\left \lfloor n\log_2n^2 \right \rfloor) \right \rfloor $

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