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Consider a sequence $(X_n)_{n\geq 1}$ of real random variables satisfying: $\rho_n X_n\to 0$ in probability as $n\to\infty$, for any sequence $(\rho_n)_{n\geq 1}$ of positive numbers diverging to $\infty$.

In general, such a sequence needs not converge almost surely to $0$ (which is quite a surprising fact). For instance, let $X_n$ have the Poisson distribution with parameter $1/n$ for all $n\geq 1$, and assume that $X_1,X_2,\ldots$ are independent. In that case, it is easy to check that infinitely many $X_n$'s are non-zero, with probability $1$, hence, $X_n$ being integer valued, it cannot converge almost surely to $0$.

However, I am wondering the following: If, on top of the above assumption ($\rho_n X_n$ converges in probability to $0$ for all sequences $(\rho_n)_{n\geq 1}$ diverging to $\infty$), we know that $X_n$ does converge almost surely to $0$, must it hold that with probability $1$, $X_n=0$ for all large enough $n$?

TrivialPursuit
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  • @TheOtherTerry The question isn't whether $X_n$ has a nonzero probability of being nonzero for every $n$. It is whether with probability 1, only a finite number of values are nonzero. – btilly Oct 02 '25 at 00:25

1 Answers1

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The answer is no. High level thought: the hypothesis $\rho_n X_n \to 0$ for all sequences $\rho_n$ is equivalent to $P(|X_n| > 0) \to 0$ (show this). This is stronger than convergence in probability, but not that much, as your Poisson example showed. The a.s. condition kills this example because Poissons jump a whole lot when they exceed zero, but this is easy to fix: we just scale them down to jump only a little bit. In other words, we can modulate your example a little to get a counterexample for the extended statement.

To this end, let $a_n$ be a positive deterministic sequence going to $0$. Also let $B_n$ be a sequence of independent Bernoullis (or Poissons work just as well) with the chance that $B_n = 1$ being $1/n$. Set $X_n = a_n B_n$.

Then first note that $0 \le X_n \le a_n B_n \le a_n,$ so $X_n \to 0$ almost surely (in fact, surely). Secondly, for any $\rho_n \ge 0,$ $P(\rho_n X_n > 0) = P(X_n > 0) = P(B_n = 1) \to 0,$ so $\rho_n X_n \overset{p}\to 0.$ But $\sum P(X_n > 0) = \infty,$ and the $X_n$s are independent, so by the second Borel-Cantelli lemma, $P(X_n > 0 \textrm{ infinitely often}) = 1$.

stochasticboy321
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  • I guess maybe a more conceptual way to put this is that the statement "$X_n \to 0$ a.s." is nowhere near close to a statement like "$X_n > 0$ only finitely often" (indeed, the deterministic sequence $1/n$ goes to $0$, but is always positive). A couple more in-probability like conditions are not going to change this... – stochasticboy321 Oct 02 '25 at 00:56
  • Thank you, great counter-example! I do not fully agree with your last comment though, because it is really about the fact that convergence (here, in probability) happens at any rate. For instance, if almost sure convergence happens at any rate (that is, $\rho_n X_n$ goes to 0 almost surely for any diverging sequences $(\rho_n)$), does it hold that $X_n$ must be $0$ for all large enough $n$ with probability $1$? – TrivialPursuit Oct 02 '25 at 10:00
  • Sure, but notice that this is a sample path statement (and also a strenghtening of $X_n \to 0$ a.s.). Anyway, a clarification question: do you mean "for all sequences $\rho_n, P( \rho_n X_n \to 0) = 1,$" or "$P( \forall \textrm{ seq.} \rho_n, \rho_n X_n \to 0) = 1$". These are not the same because the number of (rational) sequences is uncountable, so some care is needed in setting the question up. – stochasticboy321 Oct 02 '25 at 14:09
  • Under the second interpretation, this is easy in that if for some $\omega, X_n(\omega) > 0$ i.o., then $\limsup \rho_n X_n(\omega) > 1$ for $\rho_n = 2/\sup_{j \ge n} X_n(\omega)$. So ${ \omega: X_n(\omega) > 0 \textrm{ i.o. }} \subset { \omega : \exists \rho_n \textrm{ s.t. } \rho_n X_n(\omega) \not\to 0}$. For the first interpretation, I'll have to think a bit. – stochasticboy321 Oct 02 '25 at 14:12
  • Sorry for the misunderstanding, I meant the first interpretation of course: $(\rho_n)_{n\geq 1}$ should not be random. – TrivialPursuit Oct 02 '25 at 15:49
  • There's a deleted answer from last night which gets at this, due to @The Other Terry. Here's the approach: let $A_N = { \omega : \textrm{ at infintely many } n \ge N, X_n(\omega) > 0}.$ Note that $P(A_N)$ are all identical. Assume $P(A_1) > 0$ for the sake of contradiction. Then we can find a $N_1$ such that $P( \exists n \le N_1 : X_n > 0 ) \ge P(A_1)/2$ (as you increase $N_1$, this event dominates $A_1$), and so, some $\epsilon_1$ such that $P( \exists n \le N_1 : X_n > \epsilon_1) \ge P(A_1)/4$ (again, consider the sequence of events as you shrink $\epsilon_1$). (contd.) – stochasticboy321 Oct 02 '25 at 23:41
  • Now, cut off the first $N_1$ time steps, and repeat. You find a $N_2 > N_1,$ and a $\epsilon_2$ such that $P(\exists n \in [N_1 + 1: N_2] : X_n > \epsilon_2) \ge P(A_{N_1})/4 = P(A_1)/4$. Iterating this, we end up with a deterministic sequence of $(N_i, \epsilon_i)$ such that the chance that $X_n > \epsilon_i$ over $[N_i+1:N_{i+1}]$ is at least $P(A_1)/4$. But then the sequence $\rho_n = 1/\epsilon_i$ for all $n \in [N_i + 1 : N_{i+1}]$ has $\limsup \rho_n X_n > 1$ a.s. (Borel-Cantelli), breaking a.s. convergence for it. – stochasticboy321 Oct 02 '25 at 23:43