Timeline for Integrable version of the Borel-Cantelli theorem?
Current License: CC BY-SA 4.0
28 events
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| Apr 23, 2019 at 20:15 | comment | added | Nate Eldredge | I don't know a book that studies this issue explicitly and in depth, but you see examples of the phenomenon in many places. For instance, I like the construction of Brownian motion in Durrett's Probability: Theory and Examples, which really only constructs it at a countable dense set of times, and shows that it is uniformly continuous on this set. Other books try to really construct an uncountable family of random variables and end up with awkward steps involving "modifications" of the process. | |
| Apr 23, 2019 at 15:49 | comment | added | Aidan Rocke | @NateEldredge Thank you for the additional comment. Measure theory was one of my favourite courses as an undergrad but I only took one course. These types of pathological cases weren't considered. Might there be particular books which develop this "right" perspective? Eager to learn more. | |
| Apr 23, 2019 at 15:42 | comment | added | Nate Eldredge | Not offhand. Frankly, I think this question illustrates a general principle in probability theory that no good ever comes of considering an uncountable family of independent events. For that matter, if you think about it "right", no good ever comes of really considering uncountable families of events; in contexts like continuous-time stochastic processes when you think you want to do this, you want assumptions like cadlag that mean you can learn everything by studying countable families. | |
| Apr 23, 2019 at 15:39 | comment | added | Aidan Rocke | @NateEldredge That's a very good point that I didn't consider. I would also be happy to look into this in order to improve the question. I found this: emis.de/journals/GMJ/vol6/v6n3-1.pdf Might you have other references in mind? | |
| Apr 23, 2019 at 15:33 | comment | added | Nate Eldredge | Which actually raises another issue with this question: given an uncountable family of events $E_t$, the set $\{E_t \text{ i.o.}\}$ need not be measurable, so without more assumptions it does not even make sense to talk about its probability. | |
| Apr 23, 2019 at 15:23 | history | edited | Aidan Rocke | CC BY-SA 4.0 |
small clarification
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| Apr 23, 2019 at 15:22 | comment | added | Aidan Rocke | @LSpice Good question. i.o. means infinitely often. I'll edit the question to make this clear. | |
| Apr 23, 2019 at 15:07 | comment | added | LSpice | What does "i.o." mean? | |
| Apr 23, 2019 at 14:03 | history | edited | Aidan Rocke | CC BY-SA 4.0 |
improved grammar
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
replaced http://math.stackexchange.com/ with https://math.stackexchange.com/
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| Dec 31, 2016 at 19:26 | vote | accept | Aidan Rocke | ||
| Dec 31, 2016 at 18:26 | answer | added | John Dawkins | timeline score: 4 | |
| Dec 31, 2016 at 18:25 | history | edited | Aidan Rocke | CC BY-SA 3.0 |
fixed question
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| Dec 31, 2016 at 18:23 | comment | added | James Martin | (but let's not pursue more iterations of the question -- at least, this is not the right place to do it.) | |
| Dec 31, 2016 at 18:22 | comment | added | James Martin | Both of these now look wrong. For the second, aren't we back where we started? Without some condition of independence, or correlation decay, this one must be doomed. For the first, let $E_t$ be independent with $P(E_t)=1$ if $t$ is an integer, and $P(E_t)=0$ otherwise. Then the integral is zero, but from the standard B-C lemma, w.p.1 there will be infinitely many integers $t$ such that $E_t$ occurs. Or, alternatively, let $E_t$ be independent with $P(E_t)=e^{-t}$. Then the integral is finite, but for any $0<a<b<1$ there are uncountably many events $E_t$ with probability in $(a,b)$. | |
| Dec 31, 2016 at 17:56 | history | edited | Aidan Rocke | CC BY-SA 3.0 |
clarified question
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| Dec 31, 2016 at 5:07 | history | edited | Aidan Rocke | CC BY-SA 3.0 |
changed formatting
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| Dec 31, 2016 at 4:34 | history | edited | Aidan Rocke | CC BY-SA 3.0 |
added motivation
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| Dec 31, 2016 at 2:20 | history | edited | Aidan Rocke | CC BY-SA 3.0 |
added second case
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| Dec 30, 2016 at 17:30 | comment | added | Aidan Rocke | @JamesMartin I think I now have both cases of the Borel-Cantelli theorem, and this result is more general than what I wanted earlier. Would you agree? | |
| Dec 30, 2016 at 15:54 | comment | added | James Martin | "Clarified" as in "changed" :) Now it is implied by the normal Borel-Cantelli lemma, since you can find a countable sequence $t_k$ such that $\sum P(E_{t_k})=\infty$. | |
| Dec 30, 2016 at 14:23 | history | edited | Aidan Rocke | CC BY-SA 3.0 |
simplified title
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| Dec 30, 2016 at 13:43 | history | edited | Aidan Rocke | CC BY-SA 3.0 |
clarified definitions
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| Dec 30, 2016 at 13:34 | history | edited | Aidan Rocke | CC BY-SA 3.0 |
assume integrability rather than continuity
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| Dec 30, 2016 at 12:06 | comment | added | Aidan Rocke | @JamesMartin Ok. I clarified the question. Is the statement now true? | |
| Dec 30, 2016 at 12:06 | history | edited | Aidan Rocke | CC BY-SA 3.0 |
clarified the question
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| Dec 30, 2016 at 12:02 | comment | added | James Martin | As written, this must be false: for example, let all $E_t$ be the same event, with probability $1/2$. Then $P(E_t \, i.o.)=P(E_1)=1/2$. You would need to add some condition of independence, or decay of correlations, or the like... | |
| Dec 30, 2016 at 10:11 | history | asked | Aidan Rocke | CC BY-SA 3.0 |