Consider infinite independent coin tossing where $H_n = \{$nth coin is heads$\}$ for $n = 1, 2, ...$.

Let $$A_n = \bigcap_{i=1}^{\left \lfloor \log_2 n \right \rfloor} H_{n+i}$$

How do you show that $(B_n = A_{f(n)})$, where $f(n) = \left \lfloor n\log_2 n^2 \right \rfloor$, is an independent subsequence?

I was able to show that $(B_n)$ is pairwise independent by showing 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, $\left \lfloor n\log_2 n^2 \right \rfloor + \left \lfloor \log_2(\left \lfloor n\log_2n^2 \right \rfloor) \right \rfloor \leq \left \lfloor (n+1)\log_2 (n+1)^2 \right \rfloor + 1$

but don't know as to how this extends to mutual independence


The family of events $(B_n)_{n\geq 1}$ is independent if for every finite $J \subset \{n \geq 1 \}$: $$P[\bigcap_{m \in J} B_m] = \prod_{m \in J} P(B_m),$$ i.e. every finite subset of $(B_n)_{n\geq 1}$ has to be independent (we view $(B_n)_{n \geq 1}$ as a collection of events).

In your case: $$\bigcap_{m \in J} B_m = \bigcap_{m\in J}\bigcap_{i=1}^{\lfloor \log_2f(m) \rfloor} H_{i+f(m)} = \bigcap_{k \in I}H_k,$$ for $I = \bigcup_{m\in J}\{k \in \mathbb{N}: f(m)+1\leq k \leq f(m) + \lfloor \log_2f(m) \rfloor \}.$

Since $(H_n)_{n\geq 1}$ are independent and the fact that $$\{k \in \mathbb{N}: f(m)+1\leq k \leq f(m) + \lfloor \log_2f(m) \rfloor \}\cap \{k \in \mathbb{N}: f(m')+1\leq k \leq f(m') + \lfloor \log_2f(m') \rfloor \} = \emptyset$$ for $m \neq m'$(this is what you showed for m'=m+1), we can conclude that $$P\left[ \bigcap_{m \in J} B_m \right] = P\left[ \bigcap_{k\in I}H_k \right] = \prod_{k\in I} P(H_k) $$ $$= \prod_{m\in J}P(\bigcap_{f(m)+1\leq k \leq f(m) + \lfloor \log_2f(m) \rfloor }H_k) = \prod_{m\in J}P(B_m).$$

Thus $(B_n)_{n\geq 1}$ is independent.

  • $\begingroup$ Thanks danwin. In short, we can extend $m \ne m+1$ to $m \ne m'$? $\endgroup$ – BCLC Nov 23 '15 at 22:27
  • 1
    $\begingroup$ Yeah, you can do that by using induction to show that the index sets are disjunct. $\endgroup$ – danwin Nov 24 '15 at 13:17

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