Let $(X_n)_{n \geq 1}$ be a sequence of iid random variables such that $\mathbb P(X_n = 0) = \frac{1}{2}$ and $\mathbb P(X_n = 1) = \frac{1}{2}$.

For all $n \geq 1$, let $S_n := \sum_{k=1}^n 2^{k-n-1}X_k$ and $T_n := \sum_{k=1}^n 2^{-k}X_k$.

I have a question: Why is there no random variable $S$ such that $S_n \to S$ almost sure?

I already showed that $S_{n+1} = \frac{1}{2} (X_{n+1} + S_n)$. Now my plan is to continue by contradiction that $S_n \to S$ almost sure for some r.v. $S$, but I'm struggling a bit. Could anybody help me out? (And tell me if the solution is correct, until now?)

And why does $S_n \to S$ in probability?


1 Answer 1


Suppose $S_n \to S$ a.s. Then $X_{n+1}=2S_{n+1}-S_n \to 2S-S=S$ a.s. But an i.i..d sequence can converge almost surely only when the r.v.'s are a.s constant but that is not the case here.

Here is an elementary argument: Suppose $X_n \to S$ a.s. Then $X_{n+1}-X_n \to 0$ a.s and this implies $P(|X_{n+1}-X_n| >\frac 1 2) \to 0$. But $P(|X_{n+1}-X_n| >\frac 1 2)\geq P(X_{n+1}=1,X_n=0)=\frac 1 4$ by independence.

  • $\begingroup$ Thanks a lot for your answer. So am I right that $S_n \to S$ not almost sure, but in probability? $\endgroup$
    – Stanisla
    May 21, 2021 at 9:29
  • $\begingroup$ Can you prove that $S_n$ converges in probability? @Stanisla $\endgroup$ May 21, 2021 at 9:30
  • $\begingroup$ I tried to, but I'm not sure... I would say the following: Let's suppose $S_n \to S$ in probability. Then $\mathbb P(|S_n - S| \leq \epsilon)$, we can use $\epsilon = \frac{1}{2}$ and this converges to 1. But this is exaxtly the definition of "convergence in probability". Any correction / comment is appreciated... $\endgroup$
    – Stanisla
    May 21, 2021 at 9:37
  • $\begingroup$ I did an edit since I realized that my solution isn't correct. I can't assume what I'd like to show... $\endgroup$
    – Stanisla
    May 21, 2021 at 9:51

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