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I have a Markov chain $\{X_k\}_{k\geq 0}$ on $\mathbb{R}$. The corresponding probability density functions satisfy $$ f_{k+1}(t) = \int_{-\infty}^\infty \Psi(t,\tau)f_k(\tau)\,d\tau,\qquad k=0,1,2,\dots $$ I have an analytic expression for the transition kernel $\Psi$, and let's suppose for the moment that the Markov chain is irreducible, positive recurrent, aperiodic, and Harris. And of course, $\int_{-\infty}^\infty \Psi(t,\tau)\,dt=1$.

I am interested in characterizing the moments of the stationary distribution $\pi$. Specifically:

  • What are sufficient conditions that would ensure the moments of $\pi$ are finite?

  • Is there a way to compute bounds on the moments of $\pi$ if they are finite? I can't do this numerically because $\Psi$ is parameterized; I'm interested in how the moments of $\pi$ vary as a function of these parameters. My first instinct was to try to write $\int_{-\infty}^\infty t^mf_{k+1}(t)\,dt$, substitute the recurrence from above and try to simplify and maybe use Holder's inequality, but I ran into a roadblock: it turns out that $\int_{-\infty}^\infty t^m \Psi(t,\tau)\,dt = \infty$ for all $m\geq 1$, even though the integral is finite for $m=0$. So at this point I have no idea how to proceed.

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    $\begingroup$ If your transition probabilities don't have finite moments, why would you expect $\pi$ to have finite moments? $\endgroup$ – Martin Hairer Oct 13 '20 at 8:48
  • $\begingroup$ Now that I think of it, I guess that doesn't make much sense... My numerical simulations suggested that $\pi$ exists and has finite moments, but you bring up a good point. I will have to think about this more. $\endgroup$ – Laurent Lessard Oct 13 '20 at 16:37
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You wrote:

I can verify that $\Psi$ is continuously differentiable, $\Psi(t,\tau)>0$ for all $t,\tau\in\mathbb{R}$, and of course, $\int_{-\infty}^\infty \Psi(t,\tau)\,dt=1$.

[...] these properties should be sufficient to guarantee that a stationary distribution $\pi$ exists and is unique, and that $f_k \to \pi$ (in the T.V. sense) for any initial $f_0$.

Of course, this is not so. E.g., if $\Psi(t,s)=g(t-s)$, where $g$ is the standard normal pdf, then (considering, for instance the Fourier transform, one can easily see that) there is no stationary distribution. Also, then for any initial $f_0$ and each real $t$ we have $f_k(t)\to0$ as $k\to\infty$.


You have now added more conditions:

let's suppose for the moment that the Markov chain is irreducible, positive recurrent, aperiodic, and Harris. And of course, $\int_{-\infty}^\infty \Psi(t,\tau)\,dt=1$

saying then the following:

These properties should be sufficient to guarantee that a stationary distribution $\pi$ exists and is unique, and that $f_k \to \pi$ (in the T.V. sense) for any initial $f_0$. Moreover, all moments of $\pi$ are finite and the $m^\text{th}$ moment of $f_k$ converges to the $m^\text{th}$ moment of $\pi$ as $k\to\infty$.

However, the latter conclusion will still fail to hold in general -- because the the state space of the chain can be nonlinearly transformed in an arbitrary manner.

More specifically, suppose (say) that the support set of the stationary distribution $\pi$ of an (irreducible positive recurrent aperiodic Harris) Markov chain $(X_k)$ is not bounded from above, so that $$G(x):=\pi\big((x,\infty)\big)>0$$ for all real $x$. Let then $$Y_k:=f(X_k),$$ where $$f(x):=\int_0^x\frac{du}{G(u)}$$ for real $x$, with $\int_0^x:=-\int_x^0$ for real $x<0$. Then $(Y_k)$ is an (irreducible positive recurrent aperiodic Harris) Markov chain with stationary distribution $\pi_f:=\pi f^{-1}$, the pushforward of $\pi$ under the map $f$. Moreover, \begin{align} \int_{[0,\infty)}y\,\pi_f(dy)&=\int_{[0,\infty)}f(x)\,\pi(dx) \\ &=\int_{[0,\infty)}\pi(dx)\,\int_0^x\frac{du}{G(u)} \\ &=\int_0^\infty\frac{du}{G(u)}\,\int_{(u,\infty)} \pi(dx) \\ &=\int_0^\infty\frac{du}{G(u)}\,G(u)=\infty. \end{align} So, the first moment of $\pi_f$ cannot be finite.

Similarly one can deal with the case when the support set of the stationary distribution $\pi$ has a finite limit point.

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  • $\begingroup$ 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. $\endgroup$ – Laurent Lessard Oct 11 '20 at 22:12
  • $\begingroup$ @LaurentLessard : This is still not enough. $\endgroup$ – Iosif Pinelis Oct 12 '20 at 18:05
  • $\begingroup$ ok -- I realized I don't know what is needed to ensure finite moments, so I modified the question once more. $\endgroup$ – Laurent Lessard Oct 12 '20 at 23:15
  • $\begingroup$ Why the downvote for this answer? $\endgroup$ – Iosif Pinelis Oct 12 '20 at 23:43
  • $\begingroup$ Because you didn't actually answer my question. Will happily upvote once question is answered or addressed. $\endgroup$ – Laurent Lessard Oct 13 '20 at 1:04

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