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!



property of iid random variable
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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 $P=1$. 


