Random sequence converging to zero with random indexes I need a opinion for this. The random sequence $X_{2^n + y_n}(\omega) \rightarrow 0$ a.s. for any non-stochastic $y_n \in \{0,1,...,2^n\}$. Can be concluded that  $X_{2^n + Y_n(\omega)}(\omega) \rightarrow 0$ a.s. for some random variable $Y_n(\omega) \in \{0,1,...,2^n\}$?
R. Simmons
 A: For arbitrary $Y_n$s, no.  Let $Z$ be uniform in $[0,1)$, and let
$$
X_{2^n+y}:=\left\{\begin{array}{ll}
1, &  \frac{y}{2^n} \le Z < \frac{y+1}{2^n},\\
0, & \text{otherwise,}
\end{array}\right.
$$
where
$$
n=0, 1, 2, \dots, \qquad y=0, \dots, 2^n-1.
$$
Then, for any deterministic sequence of nonnegative $y_n$s, the probability
that $X_{2^n+y_n}$ is nonzero for any $n\ge N$ is at most 
$$
2^{-N} + 2^{-(N+1)} + \cdots = 2^{-(N-1)},
$$
so $X_{2^n+y_n}\to 0$ a.s.  However, if
$$
Y_n:=\lfloor 2^n Z \rfloor, \qquad n=0, 1, 2, \dots
$$
then each $X_{2^n+Y_n}$ is identically $1$, so $X_{2^n+Y_n}$ does not converge to $0$ a.s.
Re the comment below, let $\cal Y$ be the set of all sequences ${\bf y}=(y_n)$, and let $\Omega$ be the state space.  Then for any given sequence ${\bf y}=(y_n)$, we can find a set $\Omega_{\bf y}\subseteq \Omega$ with ${\Bbb P}(\Omega_{\bf y})=1$ such that $X_{2^n+y_n}\to 0$ everywhere on $\Omega_{\bf y}$.  However we cannot set
$$
{\bar \Omega}:=\bigcap_{{\bf y}\in\cal Y}  \Omega_{\bf y}
$$
to get a single set $\bar \Omega$ with ${\Bbb P}(\bar \Omega)=1$ on which everything converges, because there are uncountably many ${\bf y}$s.  In fact, if we identify the state space with $[0,1)$, then for any $r\in [0,1)$, $$r\notin \Omega_{\bf y}, \qquad \text{  where  } {\bf y}=(y_n),\qquad y_n=\lfloor r 2^n \rfloor \text { for all } n\ge 0.$$
Therefore, $${\bar \Omega}=\bigcap_{{\bf y}\in\cal Y}  \Omega_{\bf y}=\emptyset.$$
