Of course in general the answer to my question is NO, however I am interested in two sequences of random variables that have very specific structure.
Let $a_n,b_n$ be deterministic (real-valued) sequences and let $\xi_n$ be an i.i.d sequence of random variables. Then let
$$ X_n=a_n+b_n\xi_n, \quad Y_n=a_n+b_n\xi,$$
where $\xi=\xi_n$ in distribution. If we suppose $X_n \to 0$ almost surely, can we conclude that $Y_n \to 0$ almost surely?
An idea of a possible method of proof (or a counter example) would be much appreciated.
Note we also assume that the sequence $b_n$ is non-degenerate i.e not the zero sequence (otherwise the question would not be very interesting).
Best Answer
Let $X_n\to0$ a.s. Then, for every $\varepsilon >0$ we have $P(|X_n|>\varepsilon\textrm{ i.o.})=0$. Let $\varepsilon>0$. Suppose $P(|Y_n|>\varepsilon\textrm{ i.o.})>0$. Then $$\begin{aligned}P(|Y_n|>\varepsilon\textrm{ i.o.})>0&\stackrel{\textrm{B-C I}}{\implies} \sum_nP(|Y_n|>\varepsilon)=\infty\\ &\stackrel{\xi_n\sim\xi,\forall n}\implies \sum_nP(|X_n|>\varepsilon)=\infty\\ &\stackrel{\textrm{B-C II}}{\implies}P(|X_n|>\varepsilon\textrm{ i.o.})=1 \end{aligned}$$ But this is a contradiction. Then $P(|Y_n|>\varepsilon\textrm{ i.o.})=0$. We conclude, because $\varepsilon$ was arbitrary.