Timeline for finiteness of moments of the stationary distribution of a Markov chain
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Oct 13, 2020 at 16:38 | vote | accept | Laurent Lessard | ||
| Oct 13, 2020 at 15:23 | comment | added | Laurent Lessard | Let us continue this discussion in chat. | |
| Oct 13, 2020 at 13:03 | comment | added | Iosif Pinelis | There was no way for me to know your particular $\Psi$ (I was going to ask you about that, but then forgot). In both versions of your question, your approach was to list a number of properties of $\Psi$, which you apparently hoped will be enough. I showed that, in both versions of your question, the lists of properties of $\Psi$ you gave were not enough. This constitutes complete answers to both versions of your question. You if want to ask further questions with your particular $\Psi$ disclosed or with more properties of it listed, then you can do it in a separate post. | |
| Oct 13, 2020 at 5:10 | comment | added | Laurent Lessard | My question was about bounding the moments for a particular $\Psi$. As stated in my question, I have an analytic expression for $\Psi$. The issue is that I cannot find $\pi$ in closed form so computing moments directly is not possible, hence the need to find bounds. The fact that $\pi$ need not always exist, or that moments can sometimes be unbounded is irrelevant. For my particular $\Psi$, the moments do exist and are bounded; I just can't compute them. I'm sorry my question wasn't clear. I rewrote it twice in an effort to clarify but evidently I failed. I removed your downvote. | |
| Oct 13, 2020 at 3:22 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Oct 13, 2020 at 2:32 | comment | added | Iosif Pinelis | @LaurentLessard : Your original question, "How can I bound the moments of $\pi$?", was answered, by showing that the very premise of the question (that is, the existence of $\pi$) is false in general under your conditions. After that, you invalidated that valid answer (which you should never do) by adding conditions guaranteeing the existence of $\pi$. That modified question was answered as well, by showing that even under those expanded conditions, the best bound on the moments is $\infty$. Finally, for all my time and effort to steer you in the right direction, you downvoted the answer? | |
| Oct 13, 2020 at 1:04 | comment | added | Laurent Lessard | Because you didn't actually answer my question. Will happily upvote once question is answered or addressed. | |
| Oct 12, 2020 at 23:43 | comment | added | Iosif Pinelis | Why the downvote for this answer? | |
| Oct 12, 2020 at 23:15 | comment | added | Laurent Lessard | ok -- I realized I don't know what is needed to ensure finite moments, so I modified the question once more. | |
| Oct 12, 2020 at 18:05 | comment | added | Iosif Pinelis | @LaurentLessard : This is still not enough. | |
| Oct 12, 2020 at 18:04 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Oct 11, 2020 at 22:12 | comment | added | Laurent Lessard | Thanks -- I was missing some crucial details in my original post. I think my statements only guaranteed irreducibility. I will edit my post to make the question more precise. | |
| Oct 11, 2020 at 20:43 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Oct 11, 2020 at 18:43 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |