Timeline for Spectrum of a Markov kernel acting on $L^2$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 4, 2020 at 16:31 | comment | added | Benoît Kloeckner | @0xbadf00d: I only barely know MH algorithms; in general, keywords are "mixing" and "spectral gap". | |
| Jan 3, 2020 at 15:24 | comment | added | 0xbadf00d | I'm mainly interested in sufficient conditions for Metropolis-Hastings kernels. Do you know a reference for that? | |
| Jan 3, 2020 at 14:58 | comment | added | Benoît Kloeckner | Irreducibility is not always sufficient. There are certainly several sufficient conditions, but what you want is what you want ($1$ is a simple isolated eigenvalue). | |
| Dec 30, 2019 at 18:03 | comment | added | 0xbadf00d | So, the ingredient I'm missing is irreducibility, right? I'm not too deep into this topic. Is $L^2(\mu)$-geometric ergodicity enough? | |
| Dec 30, 2019 at 18:01 | vote | accept | 0xbadf00d | ||
| Dec 30, 2019 at 17:38 | comment | added | Benoît Kloeckner | @0xbadf00d: the union need not be disjoint. If the Markov chain is reducible, you will have other eigenfunctions with eigenvalue $1$ (the simplest case is a Markov chain on at least two states, which almost surely stays where it was the step before: $\mathrm{P}$ is then the identity). | |
| Dec 30, 2019 at 15:06 | comment | added | 0xbadf00d | Regarding the question: You're right, I've missed that. Thank you very much! However, what I still don't understand is why $\operatorname{Spec}\left(P\mid L^2_0(\pi)\right)\subseteq[-1,1)$. I know that $\operatorname{Spec}\left(P\mid L^2(\pi)\right)=\operatorname{Spec}\left(P\mid L^2_0(\pi)\right)\cup\operatorname{Spec}\left(P\mid {L^2(\pi)}^\perp\right)$ though. This is a general fact for reducing subspaces. But I'm not sure if it is guaranteed that this union needs to be disjoint. Can you help out? | |
| Dec 30, 2019 at 14:11 | comment | added | 0xbadf00d | You can even argue more more probabilistically: If $f\in{L^2_0(\mu)}^\perp$, then $0=\langle f,f-\mu f\rangle_{L^2(\mu)}=\operatorname{Var}_\mu[f]$ and hence $f=\mu f$ $\mu$-almost surely. | |
| Dec 30, 2019 at 11:49 | history | answered | Benoît Kloeckner | CC BY-SA 4.0 |