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Let $ (\xi_i)_{i \ge 1} $ be independent identically distributed random variables, taking values in $ (1,3]$.

Can we show:

$P( \exists N \in \mathbb{N}, \text{ s.t. } \forall k \ge 0, \prod_{i=1}^{N}\xi_{i+kN} > 6N ) =1 ?$

i.e, almost surely, for each consecutive block with length $ N$, its product is greater than its length?

I think law of large number may give clue for this problem:

the main problem is that $ \xi_i $ can be closed to 1, but we can prove $ \mathbb{E}\xi_i >1$, then roughly speaking $ \xi_i \sim \mathbb{E}\xi_i$, which may imply $\prod_{i=1}^{N}\xi_{i+kN}$ has exponential growth.

this is my idea, but do not know how to give a formal proof. Thanks!

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The answer depends on whether the support (say $S$) of the distribution of $\xi_1$ contains $1$ or not. If $1\in S$, then the answer is no; otherwise, yes.

Indeed, suppose first that $1\in S$. For each natural $N$, let \begin{equation} A_N:=\Big\{\forall k \ge 0\ \prod_{i=1}^{N}\xi_{i+kN} > 6N\Big\}=\bigcap_{k=0}^\infty B_{N,k}, \end{equation} where \begin{equation} B_{N,k}=\Big\{\prod_{i=1}^{N}\xi_{i+kN} > 6N\Big\}. \end{equation} The events $B_{N,0},B_{N,1},\dots$ are independent and for each $k$ \begin{align*} \P(B_{N,k})=\P(B_{N,0})&\le 1-\P\big(\xi_i<(6N)^{1/N}\ \forall i=1,\dots,N\big) \\ &=1-\P\big(\xi_1<(6N)^{1/N}\big)^N=:q<1, \end{align*} since $1\in S$ and $(6N)^{1/N}>1$. So, \begin{equation} \P(A_N)=\prod_{k=0}^\infty \P(B_{N,k})\le \prod_{k=0}^\infty q=0, \end{equation} for each natural $N$. So, \begin{equation} P:=\P\Big( \exists N \in \mathbb{N} \text{ s.t. } \forall k \ge 0\ \prod_{i=1}^{N}\xi_{i+kN} > 6N \Big) =\P\Big(\bigcup_{N \in \mathbb{N}} A_N\Big)=0\ne1. \end{equation}

If now $1\notin S$, then $\P(\xi_i>c)=1$ for some real $c>1$ and hence almost surely $\prod_{i=1}^{N}\xi_{i+kN}>c^N > 6N$ for all $k$ and all large enough $N$. Hence, in this case $P=1$.

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  • $\begingroup$ yeah, I got it. nice proof. Thanks! $\endgroup$
    – jason
    May 28, 2018 at 4:19
  • $\begingroup$ @jason : So, is this answer acceptable for you? $\endgroup$ Jun 3, 2018 at 18:43
  • $\begingroup$ yes, Thanks! I could not rate up your answer last month since my reputation was too low. $\endgroup$
    – jason
    Jun 4, 2018 at 4:49
  • $\begingroup$ @jason : Thanks Jason. If an answer to your question is acceptable to you, you can formally accept it by clicking on the check button at the upper left corner of the answer. That would give 15 points to the answerer and 2 points to you. $\endgroup$ Jun 4, 2018 at 11:45

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