Timeline for Using all Metropolis-Hastings proposals to estimate an integral
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Oct 19, 2019 at 16:10 | comment | added | 0xbadf00d | Took me a while to determine the transition kernel, but now I've obtained $$\operatorname P\left[(X_n,Y_{n+1})\in A_n\times B_{n+1}\mid(X_{n-1},Y_n)\right]=\delta_{Y_n}\alpha(X_{n-1},Y_n)Q(Y_n,B_{n+1})+\delta_{X_{n-1}}(A_n)(1-\alpha(X_{n-1},Y_n))Q(X_{n-1},B_{n+1}).$$ Is that what you've got in mind? | |
| Oct 16, 2019 at 11:12 | vote | accept | 0xbadf00d | ||
| Oct 15, 2019 at 19:25 | comment | added | Xi'an | The usual transition on $X_{n-1}$ given $Y_{n-1}$ and $X_{n-2}$ times the proposal. | |
| Oct 15, 2019 at 18:46 | comment | added | 0xbadf00d | Can we give an expression for the transition kernel? Assuming that $(X_{n-1},Y_n)\sim\mathcal L(X_{n-1})\otimes Q$, where $Q$ is the proposal kernel? | |
| Oct 15, 2019 at 18:44 | comment | added | Xi'an | Yes indeed, the joint chain is a Markov chain, either $(𝑋_{𝑛−1},𝑌_𝑛)_{𝑛∈ℕ}$ or $(𝑋_{𝑛},𝑌_𝑛)_{𝑛∈ℕ}$. | |
| Oct 15, 2019 at 18:42 | history | edited | Xi'an | CC BY-SA 4.0 |
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| Oct 15, 2019 at 18:21 | comment | added | 0xbadf00d | Thank you for your answer. I'll check the papers and accept your answer soon. Just one quick question: Is the process $(X_{n-1},Y_n)_{n\in\mathbb N}$ again a Markov process? If I'm not missing anything, this should be the case. | |
| Oct 15, 2019 at 11:28 | history | edited | Xi'an | CC BY-SA 4.0 |
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| Oct 15, 2019 at 5:32 | history | edited | Xi'an | CC BY-SA 4.0 |
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| Oct 15, 2019 at 5:23 | history | answered | Xi'an | CC BY-SA 4.0 |