Let $P$ be a Markov kernel on a measurable space $(E,\mathcal E)$ admitting an invariant probability measure $\pi$. $P$ acts on $L^2(\pi)$ via $$Pf:=\int\kappa(\;\cdot\;{\rm d}y)f(y).$$ The invariance means that $\int\kappa f\:{\rm d}\pi=\int f\:{\rm d}\pi$. Let $L^2_0(\pi):=\left\{f\in L^2(\pi):\int f\:{\rm d}\pi=0\right\}$. I've read the following:

enter image description here

The first equality in (22.2.3) holds since they've argued that $L^2_0(\pi)$ is a reducing subspace for $P$. But how does the second equality follow? Moreover, I've often read that $\operatorname{Spec}\left(P\mid L^2_0(\pi)\right)\subseteq[-1,1)$ (so, $1$ is excluded from the spectrum when restricting to $L^2(\pi)$. How does this follow?

Note that $$U:L^2(\mu)\to L^2(\mu)\;,\;\;\;f\mapsto\langle1,f\rangle_{L^2(\mu)}1$$ is an orthogonal projection with $\mathcal N(U)=L^2_0(\mu)$. So, $1-U$ is the orthogonal projection of $L^2(\mu)$ onto ${\mathcal R(U)}^\perp=L^2_0(\mu)$. Now, if $\lambda\in\mathbb R$, then $\lambda-\left.P\right|_{{L^2_0(\mu)}^\perp}$ is injective if and only if \begin{equation}\begin{split}\{0\}&=\mathcal N\left(\lambda-\left.P\right|_{{L^2_0(\mu)}^\perp}\right)\\&=\left\{g\in\mathcal R(U):(\lambda-P)g=0\right\}\\&=\left\{Uf:f\in L^2(\mu)\text{ and }(\lambda-P)Uf=0\right\}\\&=\left\{Uf:f\in L^2_0(\mu)\text{ and }(\lambda-P)Uf=0\right\}\\&\;\;\;\;\;\;\;\;\;\;\;\;\uplus\left\{Uf:f\in L^2(\mu)\setminus L^2_0(\mu)\text{ and }(\lambda-P)Uf=0\right\}\\&=\left\{0\right\}\uplus\left\{\langle1,f\rangle_{L^2(\mu)}1:f\in L^2(\mu)\setminus L^2_0(\mu)\text{ and }\lambda=1\right\}\\&=\left\{0\right\}\uplus\left\{c:c\in\mathbb R\setminus\{0\}\text{ and }\lambda=1\right\}\\&=\begin{cases}\mathbb R&\text{, if }\lambda=1\\\{0\}&\text{, otherwise}\end{cases},\end{split}\tag1\end{equation} where we've used that $P1=1$ (and we treat $c\in\mathbb R$ as the constant function $E\ni x\mapsto c$).

So, we can conclude that $\lambda\in\mathbb R$ is contained in the point spectrum of $\left.P\right|_{{L^2_0(\mu)}^\perp}$ if and only if $\lambda=1$. How can we conclude?

  • $\begingroup$ Could you please tell us where you read this ? $\endgroup$ – M. Dus Dec 30 '19 at 11:56
  • $\begingroup$ @M.Dus Sure. You can find it on page 531 of this book: springer.com/gp/book/9783319977034. $\endgroup$ – 0xbadf00d Dec 30 '19 at 13:59

It seems the piece you missed, which is implicit after (22.2.2) is that $L^2_0(\pi)^\perp=\langle \boldsymbol{1}\rangle$ (indeed $L^2_0(\pi)$ is a hyperplane, so that $L^2_0(\pi)^\perp$ is a line, and since $\int f\boldsymbol{1} \,\mathrm{d}\pi = \int f \,\mathrm{d}\pi = 0$ for all $f\in L^2_0(\pi)$ we have $L^2_0(\pi)^\perp=\langle \boldsymbol{1}\rangle$).

Then the restriction of $\mathrm{P}$ to this space is the identity, and its spectrum is $\{1\}$.

| cite | improve this answer | |
  • 1
    $\begingroup$ 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. $\endgroup$ – 0xbadf00d Dec 30 '19 at 14:11
  • $\begingroup$ 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? $\endgroup$ – 0xbadf00d Dec 30 '19 at 15:06
  • 2
    $\begingroup$ @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). $\endgroup$ – Benoît Kloeckner Dec 30 '19 at 17:38
  • $\begingroup$ So, the ingredient I'm missing is irreducibility, right? I'm not too deep into this topic. Is $L^2(\mu)$-geometric ergodicity enough? $\endgroup$ – 0xbadf00d Dec 30 '19 at 18:03
  • $\begingroup$ 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). $\endgroup$ – Benoît Kloeckner Jan 3 at 14:58

Two remarks too long to be posted as comments.

  1. To claim that the spectrum of a Markov operator with respect to a finite stationary measure is real one has to assume that it is self-adjoint (equivalently, that the chain is reversible). For such operators geometric ergodicity is indeed equivalent to the fact that 1 does not belong to the spectrum in $L^2_0$ (since the spectrum is closed, it means that there is a spectral gap separating the spectrum from 1).

  2. If the state space is infinite, then it is well possible that there is no spectral gap in spite of irreducibility of the operator. There are examples like this even for countable state spaces.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.