3
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

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.