Timeline for Zero-one law for an independence-like structure
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Mar 9, 2018 at 22:41 | answer | added | Nate Eldredge | timeline score: 3 | |
| Mar 9, 2018 at 21:20 | comment | added | Kurisuto Asutora | I was hoping that someone could find a way how this question fits into Levy's zero-one law. | |
| Mar 9, 2018 at 21:16 | comment | added | Kurisuto Asutora | The key assumption ís $E(X_n | \mathcal{F}_{n-1})=E(X_n)$. Consider the special case when $X_n$ is the indicator function of an event $A_n$ for all $n$. Then the assumption says that $E(X_n|\mathcal{F}_{n-1})=P(A_n)$, which means that $A_n$ is independent of the $\sigma$-field generated by $A_1, \dots, A_{n-1}$. So if I am not mistaken, then in the case when $X_n$ is {0,1}-valued, the question reduces to Kolmogorov's zero-one law (not fully sure about that, though). However, in my question $X_n$ are allowed to have more different values. | |
| Mar 9, 2018 at 21:16 | comment | added | Iosif Pinelis | It is a nice question, but mentioning of martingales was really misleading to me. Your condition is actually one of quasi-independence. E.g., if you strengthen it to $E(f(X_n) | \mathcal{F}_{n-1})=Ef(X_n)$ for all nonnegative Borel functions $f$, then you get the plain independence of $X_n$ from $\mathcal{F}_{n-1}$ and hence the independence of all $X_n$'s. | |
| Mar 9, 2018 at 21:14 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Mar 9, 2018 at 21:10 | comment | added | Nate Eldredge | I'm confused - people keep changing the key assumption. What exactly is $E(X_n \mid \mathcal{F}_{n-1})$ supposed to equal? Is it $X_n$? $E[X_n]$? $X_{n-1}$? or what? | |
| Mar 9, 2018 at 21:06 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
edited body; edited title
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| Mar 9, 2018 at 20:48 | comment | added | Kurisuto Asutora | PS: I realize that the structure here is not a martingale, but it is a sequence of random variables which are adapted to a filtration, and that reminded me of a martingale. As I remember it, it is possible to prove Kolmogorov's zero-one law by martingale theory. Maybe a similar thing works here? | |
| Mar 9, 2018 at 20:37 | comment | added | Kurisuto Asutora | No, I don't want a martingale, I need the answer to the question as stated and not the answer to a completely different question. This is no typo, this is just the situation that I have in my application. | |
| Mar 9, 2018 at 20:36 | history | edited | Kurisuto Asutora | CC BY-SA 3.0 |
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| Mar 9, 2018 at 19:09 | history | edited | Iosif Pinelis | CC BY-SA 3.0 |
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| Mar 9, 2018 at 18:41 | comment | added | Nate Eldredge | In fact, as written the condition is something closer to "uncorrelated". | |
| Mar 9, 2018 at 17:41 | history | asked | Kurisuto Asutora | CC BY-SA 3.0 |