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I stumbled upon the sequential characterization of probability convergence.

$ X_n \xrightarrow{\mathbb{P}} X$ if and only if for each subsequence $X_{n_k}$, there exists a subsequence $X_{n_{k_j}}$ such that $X_{n_{k_j}} \xrightarrow{a.s.} X$.

Now, normally, for some idea of convergence $\to$ in a space $\Omega$, one defines (sequential) compactness as follows:

For all sequences $X_n$ in $\Omega$ there exist a convergent subsequence $X_{n_k} \to X$ for $X\in\Omega$.

This does very much sound like the characterization above, if not for the fact that X is fixed a priori. Thus, here comes the question: is there some way to interpret the notion of $\mathbb{P}$ convergence as the compactness of the space $X_n$ with respect to a.s. convergence?

Thank you.

Simone Licciardi
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    Unfortunately, almost-sure convergence is not topologizable in general. (See Remark of Theorem 2.3.3 in Durrett, for instance.) So it doesn't make sense to discuss "the space of random variables w.r.t. a.s.-convergence". – Sangchul Lee May 16 '24 at 09:42
  • To me, it more sounds like the following fact: in any topological space $\Omega$, we have $X_n$ converging to $X$ in $\Omega$ if and only if every subsequence of $X_n$ has a further subsequence converging to $X$ in $\Omega$. In your setting, the fact that the convergence of this sub-sub-sequence can be strengthened into an a.s. convergence just comes from the Borel-Cantelli lemma. Besides, a.s. convergence cannot even be topologized. – nejimban May 16 '24 at 09:42
  • Thanks everyone. – Simone Licciardi May 16 '24 at 15:38

2 Answers2

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First, the mentioned result actually implies that there is no "topology of almost sure convergence." Indeed, let $(X_n)$ be a sequence of random variables that converges in probability to a random variable $X$, but does not converge almost surely. If there would be a topology of almost sure convergence, there would be a neighborhood of $X$ that excludes infinitely many terms of $(X_n)$. Take the subsequence with these terms. As the subsequence of a sequence converging in probability, it still converges in probability to $X$. But by the mentioned result, a further subsequence converges almost surely to $X$, which would require these terms to eventually lie in the neighborhood of $X$, which is impossible.

Second, you might want to have a notion of a set of random variables to be "sequentially compact" under almost sure convergence in the sense that every sequence of random variables in the set has an almost surely converging subsequence. The result tells you then that a set of random variables is sequentially compact in this sense under almost sure convergence if and only if it is (sequentially, here the same) compact under convergence in probability.

Michael Greinecker
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Actually something weaker is true. $ X_n \xrightarrow{\mathbb{P}} X$ if and only if for each subsequence $X_{n_k}$, there exists a subsequence $X_{n_{k_j}}$ such that $X_{n_{k_j}} \xrightarrow{\mathbb{P}} X$. This is merely a consequence of convergence properties for sequences in the real line (or more generally metric spaces).

If one reads through the proof then it becomes amply clear.

For an arbitrary fixed $t>0$, you need to focus on the real sequence sequence $a_{n,t}:=P(|X_{n}-X|>t)$

Then, what the condition gives you is that given any subseuqence $a_{n_{k},t}$ of $a_{n,t}$, there exists a further subsequence $a_{n_{k_{j}},t}$ which converges to $0$. This means that the whole sequence $a_{n,t}\to 0$. Which means that $P(|X_{n}-X|>t)\to 0$. Since this holds for each fixed $t>0$, you have that $X_{n}\xrightarrow{\mathbb{P}}X$.

So you see that this is a simple consequence of this fact rather than it having a deeper topological significance.

On the otherhand, for convergence in distribution, you do have that $X_{n}\xrightarrow{d}X$ if and only if for each subsequence there exists a further subsequence which converges in distribution to $X$. This can be thought of having a topological significance in the sense that convergence in distribution can be metrized using the Levy-Prokhorov metric.

But do note that this subsequential characterization is not enough for almost sure convergence. Take $X_{n}$ to be independent $Bernoulli(\frac{1}{n})$ variates and then you can check that each subsequence has a further subsequence which converges almost surely to $0$ but the whole sequence cannot converge almost surely to $0$ due to Borel-Cantelli Lemma.

Mr. Gandalf Sauron
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