9
$\begingroup$

Given a sequence of events $A_n$ for $n\in \mathbb N$, the first Borel Cantelli lemma states that, if the sum over all probabilities $\sum_{i=1}^n P(A_n)$ is finite, then the probability of the limit supremum of the $A_n$ is zero. Or more intuitively, if the sum $\sum_{i=1}^n P(A_n)$ is finite, then the probability for $\{ A_n$ happens infinitely often } is zero.

Now my question is: is there a continuous version of this?

i.e. is there any statement like: given $A_t$ with $t>0$, if $ \int_0^{\infty} P(A_t) dt<\infty$ then $P\{ A_t$ happens infinitely often }$=0$?

$\endgroup$
  • $\begingroup$ Yeah, you modified the post and now it makes sense. But consider that if $A_t=\Omega$ for every $t\lt1$ and $A_t=\varnothing$ for every $t\gt1$ then the integral converges but $(A_t$ happens infinitely often$)=\Omega$. This indicates that the setting must be made more precise for a positive answer to be possible. $\endgroup$ – Did Oct 14 '14 at 7:45
  • $\begingroup$ Yes you are right and that's exactly why I wrote "something like" because I just don't know if there is a similar formulation. Maybe you could replace the "P($ A_t$ happens infinitely often)$=0$" by "there is a T in $\mathbb{R}$ s.t. $P(A_t$ does not happen for any $t>T)>0$" $\endgroup$ – Barkas Oct 14 '14 at 13:00
  • $\begingroup$ @Did Would you agree that my answer below resolves the question? $\endgroup$ – user93511 Dec 31 '16 at 16:51
  • $\begingroup$ @Did For reasons beyond me it got downvoted. $\endgroup$ – user93511 Dec 31 '16 at 16:54
  • 1
    $\begingroup$ @AidanRocke Indexed by the reals instead of, by the integers. $\endgroup$ – Did Dec 31 '16 at 17:20
3
$\begingroup$

Let $T$ be the collection of all the $t$ that $A_t$ happens. That is, for any $\omega$, an element in the probability space which your are working with, define $$ T(\omega) := \{t\geq 0: \mathbf 1_{A_t} (\omega)=1\}. $$ Denote $l(T)$ as the Lebesgue measure of this random set $T$. Then, by Fubini's theorem, $$ \int_0^\infty \mathbf P[A_t]dt = \mathbf E[l(T)]. $$ Therefore, if the left side is finite, we surely have that $$ l(T)<\infty $$ happens as surely.

$\endgroup$
  • $\begingroup$ An essential and unfortunate limitation of this 'extension' is that $T$ having finite measure here does imply that it is bounded, as it does with normal Borel-Cantelli. $${}$$ In other words, one thing that makes the usual Borel-Cantelli lemma so useful is that when $t$ is countable, '$T$'s measure finite' immediately implies '$A_t$ eventually stops happening.' But you don't get that when $t$ is in a space with a larger topology. $\endgroup$ – enthdegree Aug 18 '17 at 5:31
  • $\begingroup$ Zhenyao Sun, Barkas, et al, actually is this specifically a Borel-Cantelli issue? 1. Does continuity of measures hold for uncountable unions? 2. https://math.stackexchange.com/questions/106102/use-of-sum-for-uncountable-indexing-set $\endgroup$ – BCLC Apr 15 '18 at 20:37
  • $\begingroup$ Is $$\int_0^\infty \mathbf P[A_t]dt$$ well-defined? Or do you define $$\int_0^\infty \mathbf P[A_t]dt \color{red}{:=} \mathbf E[l(T)]$$? $\endgroup$ – BCLC Apr 15 '18 at 21:00
2
$\begingroup$

This is just a clarification of Zhenyao Sun's answer (measurability and why this generalization is natural)

Let $A \subset [0,\infty)\times \Omega$ be measurable and $A_t = \{\omega\in \Omega: (t, \omega) \in A\} $. If $$ \int_0^\infty \mathrm P(A_t) dt<\infty, $$ then $$\lambda(\{t\ge 0: A_t \text{ happens} \}) <\infty$$ almost surely.

Proof By Fubini's theorem, $$ \int_0^\infty \mathrm P(A_t) dt = \mathrm E[\lambda(\{t\ge 0: A_t \text{ happens} \})] <\infty, $$ whence the statement follows.


Why is it natural? If $[0,\infty)$ is replaced by $\mathbb N$ and $\lambda$, by the counting measure, then it transforms to the usual Borel-Cantelli lemma.

$\endgroup$
  • 2
    $\begingroup$ Sorry but this duplicates Zhenyao Sun's answer, doesn't it? $\endgroup$ – Did Aug 17 '17 at 20:26
  • $\begingroup$ @Did, yes. I wanted ro clarify some points, but forgot to acknowledge that. $\endgroup$ – zhoraster Aug 18 '17 at 5:12
  • $\begingroup$ You might want to "clarify" the points from ZS's previous answer that your post "clarifies"... $\endgroup$ – Did Aug 18 '17 at 8:05
2
$\begingroup$

This discussion is independent of my first answer, which I still think is the right form of the continuous version of the Borel-Cantelli lemma.

However, it seems that people are interested in the following question:

"What is the right condition making a family of random events $(A_t)_{t\geq 0}$ as surely stop happening while $t$ is large enough?"

Unfortunately, as far as I know, there is no universally applicable theory that addresses this.

Even if each event $A_t$ has zero probability, it is still not enough to allow us to make the conclusion that $(A_t)_{t\geq 0}$ will stop happening eventually.

For example, let us consider the one demensional Brownian motion $B_t$, and define that $A_t:=\{B_t=0\}$.

It's easy to see that, for each $t>0​$, event $A_t​$ has probability 0. Which seems to imply that $A_t​$ should never happen.

However, as a matter of fact, a classical property of one demensional Brownian motion is that $$ \limsup_{t\to\infty} B_t =\infty, \quad \liminf_{t\to\infty} B_t=-\infty. $$ This says that $A_t$ will not stop happenning.

So my opinion is that those type of problems should be discussed case by case.

And often, those discussions will be related to the regularity property of the path of the indicated process: $$ X_t:= \mathbf 1_{A_t}, \quad t\geq 0. $$

In the Brownian motion example, the indicated process $X_t$ is very irregular. And as a consequence, the probability of events $A_t$ gives no information in answering the question.

In the case, that $X_t$ is a continuous process. We see that $A_t$ happens for all $t$ if and only if $A_0$ happens. So the desired property is solely determined by the probability of $A_0$.

In the case, that $X_t$ is a cadlag process. If we know that the probability of $A_t$ decay to 0 fast enough, then we should have that $(A_t)_{t\geq 0}$ will stop happening eventually. (Warning: I am not very sure about this last assertion.)

$\endgroup$
2
$\begingroup$

$\newcommand{\R}{\mathbb{R}}$ $\newcommand{\N}{\mathbb{N}}$ $\newcommand{\set}[2]{ \left\{ #1 \,\middle|\, #2 \right\} }$ $\newcommand{\P}{\operatorname{\mathbb{P}}}$ $\newcommand{\E}{\operatorname{\mathbb{E}}}$ $\newcommand{\abs}[1]{\left|#1\right|}$ $\newcommand{\wrap}[1]{ \left\{ #1 \right\} }$

This question can make sense if we frame it in the form of random variables. A stochastic process $\wrap{ X_t }_{t\in\R_+}$ is said to converge $X_t\overset{as}{\longrightarrow}X$ if $\P\Omega_0=1$ for: $$\Omega_0\triangleq\set{\omega\in\Omega}{\lim_{t\rightarrow\infty}X_t(\omega)=X(\omega)}$$

We ask about Borel-Cantelli in the continuous case. If we have for all $\epsilon>0$:

$$ \int_{\R_+}dt\, \P\wrap{\abs{X_t-X}>\epsilon}<\infty $$

Then do we have $X_t\rightarrow X$?

First, let's note the implications of this formulation for events. If $X_t=1_{A_t}$ for some events $A_t$ and $X=0$ then the above formulation is equivalent for any $\epsilon$ to OP's condition $\int_{\R_+}dt\,\P A_t<\infty$. And the conclusion $1_{A_t}\rightarrow 0$ indeed says that with probability 1 for sufficiently large $T$, $\bigcup_{t>T}A_t$ doesn't happen.

Unfortunately, this certainly does not hold. First, consider $X=0$ and: $$ X_t=\begin{cases} 1&t\in\mathbb{Q}\\ 0&\text{o/w} \end{cases} $$

Certainly for any $\epsilon$ the above integral is 0, though $X_t\not\rightarrow X$. One could require continuity in $t$. Still not enough. Let $g$ be some smooth unit window function with compact support (e.g., the Hann window). Then set $f_a(x)=g(ax)$, which has area $1/a$, and define: $$ f(t)=\sum_{n=1}^\infty f_{n^2}(t-n) $$

Then setting $X_t=f(t)$ and $X=0$ we have the same issue as above.

The issue is that the finite integral condition obscures information about Lebesgue-negligible sets in $\R$, but these sets can misbehave with the topological condition $\lim_t X_t=X$.

$\endgroup$
  • $\begingroup$ The spurious $x$ should have been a $t$; thanks for the catch. $\endgroup$ – VF1 Jun 10 '18 at 3:10

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.