# finiteness of moments of the stationary distribution of a Markov chain

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.

• If your transition probabilities don't have finite moments, why would you expect $\pi$ to have finite moments? – Martin Hairer Oct 13 '20 at 8:48
• 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. – Laurent Lessard Oct 13 '20 at 16:37

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.

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