I have looked around the site to see if a similar question was asked. I couldn't find one, but please refer to one, if it happens I'm mistaken.

My question is with regard to some intuition about the Borel-Cantelli lemma and almost surely convergence (a.s.).

For example take the stochastic process $X_n=1_{[0,\frac{1}{n^2}]}$ for $n\in \mathbb{N}$. The book, that I'm reading, uses this definition for almost surely convergence: $P(X_n \rightarrow X)=1$, where $(X_n \rightarrow X)=(\forall \epsilon>0 \exists N\in \mathbb{N} \forall n\geq N: |X_n-X|<\epsilon)$. In the example we have that for every N we pick, we have that $P(|X_N-0|< \epsilon)=1-\frac{1}{N^2}\neq 1$ (for $\epsilon<1$). Hence according to the definition we don't have that $X_n$ convergences a.s. to 0 (I know that this isn't correct).

By the Borel-Cantelli lemma we have that for $\epsilon<1$ that $\sum_{n=1}^{\infty} P(|X_n-0|\geq \epsilon)=\sum_{n=1}^{\infty} \frac{1}{n^2}<\infty \Rightarrow P(|X_n-0|\geq \epsilon \, i.o.)=0$. Hence $X_n$ convergences a.s. to 0, since $|X_n-0|\geq \epsilon$ only happens a finite amount of times. Therefore, there exists an N such that for $n\geq N$ we have $P(|X_n-0|\leq \epsilon)=1$.

I can't quite understand the intuition in this example. Why can we say that $X_n$ converges a.s. to 0, when we can't pick a specific N? Doesn't that go against the definition of a.s. convergence? If I'm not wrong, then the Borel-Cantelli lemma says that there exists an N, but in this example then no matter what N we pick, it won't satiesfy the a.s. convergence definition. I just think it's a bit mind boggling. Most of all I'm just curios if this can be explained in a nice fashion. If the example is trivial or wrong, I'm sorry.

  • 1
    $\begingroup$ In your second-to-last paragraph here is a fix: Since $X_n\rightarrow 0$ with prob 1, we know that, with prob 1, for all $\epsilon>0$ there exists an $N$ such that $|X_n-0|\leq \epsilon$ for all $n \geq N$. Here $N$ is random. You are incorrectly stating it as $P[|X_n-0|\leq \epsilon] = 1$ (which is never true). $\endgroup$
    – Michael
    Aug 24 '21 at 18:23

So to write your definition more precisely, I think that the book is saying the following:

Let $X_n$ be a sequence of random variables, and $X$ be another random variable on the same probability space $(\Omega, \mathcal F, \mathbb P)$. Let

$A = \{\omega \in \Omega : \text{ for all }\epsilon > 0, \text{ there exists } N(\omega, \epsilon) \in \mathbb N \text{ such that } n \geq N \Rightarrow \left|X(\omega) - X_n(\omega)\right| < \epsilon\}$.

Then we say that $X_n$ converges almost surely to $X$ if $\mathbb P[A] = 1$.

This is equivalent to the following statement: $X_n$ is said to converge to $X$ almost surely if there exists a set $N \in \mathcal F$ with $\mathbb P[N] = 0$ such that for all $\omega \in \Omega \backslash N$, and for all $\epsilon > 0$, there exists some $N(\omega, \epsilon)\in \mathbb N$ such that $n \geq N \Rightarrow \left|X(\omega) - X_n(\omega)\right| < \epsilon$.

The problem with your statement in your example is that you've gotten the order mixed up. In the definition, $N$ depends both on $\omega$ and $\epsilon$. It is true that $\mathbb P(\{|X_N - 0| : \omega \in \Omega\}) < 1$, but this $N$ does not depend on $\omega$, which it needs to.

A quick bit about intuition: $X_n$ converges almost surely to $X$ if it converges pointwise everywhere except on a set of measure 0. This convergence need not be uniform. Just like a sequence of real-valued functions $f_n:\mathbb R \rightarrow \mathbb R$ might converge for every $x \in \mathbb R$, it may require a different $N$ for each $x$. Similarly, if $X_n$ converges almost surely to $X$, you can still have that the convergence is not uniform -- i.e. for different $\omega$'s, you need a different $N$.

Hope this helps.

  • 1
    $\begingroup$ I thought about it, and I have a small follow-up question. If we have Bernoulli independent r.v. $X_1$, $X_2$,..., with $P(X_n=1)=\frac{1}{n^2}$, then we would have that $X_n \rightarrow 0$ a.s. So for all sequences $(X_1(\omega), X_2(\omega),...)$ for $\omega \in \Omega$, there would only be a finite amount of 1's. Where if we have the same setup, but with $P(X_n=1)=\frac{1}{n}$, then the 2. Borel-Cantelli lemma states that it doesn't convergence a.s. So that would mean that at least one sequence $(X_1(\omega), X_2(\omega),...)$ would have infinitely many 1's. Is that the right intuition? $\endgroup$
    – Chrisoe
    Aug 30 '21 at 17:55
  • 1
    $\begingroup$ I shouldn't have written '$\omega \in \Omega$', but have written '$\omega \in A$, where $P(A)=1$'. And not ' at least one sequence $(X_1(\omega),...)$ would have infinitely many 1's', but instead: 'there is a set, B, such that for $\omega \in B$ the sequence $(X_1(\omega),...)$ would have infinitely many 1's, where P(B)>0.' $\endgroup$
    – Chrisoe
    Aug 30 '21 at 20:17
  • $\begingroup$ @Chrisoe I think that this is not exactly correct. You are right about the sequence where $\mathbb P(X_n = 1) = \frac 1 {n^2}$. However, I don't recall there being a converse to the Borel-Cantelli lemma. I believe there are "converses" that add extra assumptions. I'm not sure if they apply to this situation. Regardless, if we are just talking about the general Borel-Cantelli lemma, even if $\sum_{n=1}^\infty \mathbb{P}[X_n = 1] = \infty$, it is still possible that $X_n$ converges almost surely to 1, though it may also not. $\endgroup$ Aug 31 '21 at 15:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.