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.
 A: 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.
