Timeline for Support of bivariate joint distribution of stationary and ergodic sequence
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Jun 14, 2018 at 20:21 | comment | added | Algernon | The condition (with $n=1$) is mentioned here: [hairer.org/notes/Markov.pdf] as Theorem 4.29. | |
| Jun 14, 2018 at 20:19 | comment | added | Algernon | The proof is pretty simple: in order for the invariant measure of $Q$ to be unique, it is enough that the invariant measure of $Q^n$ is unique. So, we can assume $n=1$. Now for every $a$ we can write $Q(a,\cdot)=\alpha\rho(\cdot)+(1-\alpha)\tilde{Q}(a,\cdot)$, where $\tilde{Q}$ is another kernel. This means in order to choose $X_n$, we can first flip a biased coin $B_n$ with parameter $\alpha$. If $B_n=1$, we choose $X_n$ according to $\rho$, otherwise according to $\tilde{Q}(X_{n-1},\cdot)$. A standard coupling argument now shows the uniqueness (and convergence). | |
| Jun 14, 2018 at 20:18 | comment | added | Algernon | Hmm... I think it is originally due to Doeblin, but I am not sure about a reference. | |
| Jun 14, 2018 at 18:57 | comment | added | user424747 | Do you have a reference for the sufficient condition in your answer? It looks a bit like Harris' ergodic theorem. I am trying to extend your result for an arbitrary distribution whose support equals the n-ball, but after digging around on this site it seems that proving ergodicity for general markov chains (though this one doesn't seem overly general) is hard. | |
| Jun 13, 2018 at 17:11 | vote | accept | user424747 | ||
| Jun 12, 2018 at 19:18 | comment | added | Algernon | Fair enough! I added an argument for the ergodicity of $X_0,X_1,\ldots$. | |
| Jun 12, 2018 at 19:18 | history | edited | Algernon | CC BY-SA 4.0 |
Argument for the ergodicity added
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| Jun 12, 2018 at 17:17 | comment | added | user424747 | Understood. I now see the reason for the comment by @NateEldredge, whose example, like the ergodic components of your construction, satisfies properties 1 and 2 as written. I will edit the question for clarity. My apologies. | |
| Jun 12, 2018 at 15:41 | comment | added | Algernon | Just let me emphasize again that in construction above, if $\tilde{X}_0,\tilde{X}_1,\ldots$ is chosen according to one of the ergodic components, then the support of $(\tilde{X}_{n-1},\tilde{X}_n)$ will only be included in the unit disk. If you wish that the support is exactly the unit ball, you would need something extra (e.g., an argument that the original process $X_0,X_1,\ldots$ is itself ergodic). | |
| Jun 12, 2018 at 15:33 | comment | added | Algernon | Yes, you could do the same by constructing an $(n-1)$-step Markov process using an $n$-tuple $(U_1,U_2,\ldots,U_n)$ that is uniformly distributed over the unit $n$-ball. In fact, you could use any $n$-tuple $(U_1,U_2,\ldots,U_n)$ as long as the obvious constraints imposed by stationarity are satisfied, that is, the individual variables $U_1,U_2,\ldots,U_n$ have the same distribution, the pairs $(U_1,U_2), (U_2,U_3), \ldots,(U_{n-1},U_n)$ have the same distribution, and so forth. | |
| Jun 12, 2018 at 5:43 | comment | added | user424747 | Thanks again; the notes are very helpful. One more question if I may: Was there anything special about property 1 restricting only bivariate joint distributions as opposed to n-variate ones? I ask because this led to a construction exhibiting the markov property, but I wonder if I had changed the question to restrict the support of$(X_1,...,X_n)$ to equal the unit n-ball at the origin your argument would essentially still go through. This is a question of representing a modification of your construction as a dynamical system: Does the extension theorem implicitly used in your answer work here? | |
| Jun 12, 2018 at 0:05 | comment | added | Algernon | See the added note for explanation. | |
| Jun 12, 2018 at 0:04 | history | edited | Algernon | CC BY-SA 4.0 |
Added note regarding the ergodic component
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| Jun 11, 2018 at 22:08 | comment | added | user424747 | Thank you for your answer. I understand that this construction is stationary, but I'm having trouble reasoning that some component of the ergodic decomposition satisfies properties 1. and 2. from your answer. This may be a standard result. If so, can you please state it? | |
| Jun 11, 2018 at 21:10 | history | answered | Algernon | CC BY-SA 4.0 |