Timeline for Stochastic process on $\{0,1\}^{\mathbb N}$ domination of product measures, necessary and sufficient conditions
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Apr 20, 2022 at 17:48 | vote | accept | jdods | ||
| Apr 20, 2022 at 17:19 | answer | added | zhoraster | timeline score: 2 | |
| Apr 20, 2022 at 17:15 | comment | added | zhoraster | I guess it does not. I'll post a counterexample soon. | |
| Apr 20, 2022 at 17:11 | comment | added | jdods | Ok, let $E=\{X_j=\epsilon_j\}$ for some arbitrary set of indices $j$ all less than $n$. Assume the coupling probability measure satisfies $\mathbb P (X\geq Y)=1$. Can I actually claim that $\mathbb P(X_n=1,Y_n=1\mid X\in E)=\mathbb P(Y_n=1\mid X\in E)$? My thinking is that the event $Y_n=1$ (whether conditioned on $X\in E$ or not) implies that $X_n=1$ almost surely due to the coupling. Either way, we should have $\mathbb P(X_n=0,Y_n=1\mid X\in E)=0$. Does that make sense? | |
| Apr 20, 2022 at 17:00 | comment | added | zhoraster | That is true, about $\mathbb P$. But $\mathbb P^\epsilon$ is different, it is the conditional probability. And while indeed $Y_n$ is independent of $Y_i, i\neq n$, in general it does depend on $X_i, i\neq n$. | |
| Apr 20, 2022 at 16:56 | comment | added | jdods | You could be hinting at where I am going wrong. $(X,Y)$ are coupled, and a coupling always preserves the marginal distributions. Hence, under the coupling, is $\mathbb P(Y_n=1)=p$ since $Y=(Y_n)_{n\in\mathbb N}$ is an iid Bernoulli process with fixed parameter $p$, i.e. $\pi_p(Y_n=1)=p$ for all $n$. For a coupling $\mathbb P$, we must have $\mathbb P(Y_n=1)=\pi_p(Y_n=1)$. | |
| Apr 20, 2022 at 16:37 | comment | added | zhoraster | I still don't get it. "$\mathbb P^\epsilon$ is a coupling probability measure which respects $Y$'s marginal" sounds quite cryptic. You started with $\mathbb P$, and $\mathbb P^\epsilon$ originates from it, so why should it respect anything? | |
| Apr 20, 2022 at 16:16 | comment | added | jdods | I've also made some progress here I think in that, the condition given is necessary and sufficient for processes on $\{0,1\}^{\mathbb N}$ (essentially by my argument here, if correct) but not on $\{0,1\}^{\mathbb Z}$ or more complicated lattices. | |
| Apr 20, 2022 at 16:15 | history | edited | jdods | CC BY-SA 4.0 |
typo fix & clarification due to comment
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| Apr 20, 2022 at 16:10 | comment | added | jdods | @zhoraster oops, yes, I will fix the inequalities. $\mathbb P^\epsilon(Y_n=1)=p$ for any $\epsilon$ and $n$ because $Y$ is an iid Bernoulli process and $\mathbb P^\epsilon$ is the coupling probability measure for $(X,Y)$ which respects $Y$'s marginal. I know I've abused notation a bit which made that unclear -- or I could be misunderstanding something of course. | |
| Apr 20, 2022 at 12:30 | comment | added | zhoraster | 1) I believe the inequality $X\le Y$ should be reversed everywhere. 2) "$\mathbb P^\epsilon(Y_n=1)=p$" why? | |
| Apr 18, 2022 at 14:44 | history | edited | jdods | CC BY-SA 4.0 |
deleted 1 character in body
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| Apr 18, 2022 at 13:20 | history | asked | jdods | CC BY-SA 4.0 |