A Markov process has a Gaussian stationary distribution. What is implied about the tails of the conditional distribution? Suppose that for all $t\in\mathbb{Z}$, the distribution of $x_t|x_{t-1},x_{t-2},\dots$ has probability density function $f(x_t|x_{t-1})$, where $x_t,x_{t-1}\in\mathbb{R}^n$.
Suppose further that the unconditional (stationary) distribution of $x_t$ is that of a standard $n$-dimensional Gaussian distribution (i.e. mean zero, covariance identity).
Can anything be inferred about the tails of $f(\cdot|x_{t-1})$? Intuitively, it seems like a stationary distribution ought to have at least as fat tails as the conditional distribution. Is this a theorem?
 A: I will elaborate on a more general -yet vague- following statement: a
mixture of continuous distributions has a tail which is heavier than
that of its components. For that aim, consider a mixture of
absolutely continuous distributions $f(y \vert \theta)$ with "weight"
distribution $\pi(\theta)$
\begin{equation}
 \tag{1}
f(y) = \int_{\Theta} f(y \vert \theta)\, \pi(\theta) \, \text{d} \theta,  
\end{equation}
and assume here that the support $\Theta$ of $\pi$ is a real interval.
With slight changes in the notations, this context applies to the
stationary distribution of a Markov chain, the densities $f$ and
$\pi$ then being identical.
I will focus on the case where the distribution -say of a r.v. $Y$-
  has $\infty$ as its upper end-point and is Regularly Varying (RV)
  with index $\alpha \geq 0$ which means that
  $$
  S(y) \sim y^{-\alpha} \mathcal{L}(y), \quad y \to \infty
  $$
  where $S(y) := \text{Pr}\{Y > y\}$ is the survival function, and
  $\mathcal{L}(y)$ is slowly varying function, that is: $\lim_{y \to
  \infty} \mathcal{L}(ty) / \mathcal{L}(y) = 1$ holds for every finite
  $t >0$. I use here the definition of the report by T. Mikosch cited
  below, although the tail index is usually defined as $\xi :=
  1/\alpha$.  So, the smaller $\alpha$, the thicker the tail.  Without
  loss of generality, we can assume that $Y \geq 0$.  An interesting
  characterisation of $\alpha$ is in terms of moments:
  $$
  \begin{cases} 
    \mathbb{E}[Y^{\beta}] < \infty & \text{if } \beta < \alpha, \\
     \mathbb{E}[Y^{\beta}] = \infty & \text{if } \beta > \alpha,
  \end{cases}
  $$
  see Prop 1.3.2 in Mikosch. Using Tonelli's theorem 
  $$
  \text{E}[Y^\beta]=  \int_0^\infty y^{\beta}  \left[ \int_{\Theta} 
 f(y \vert \theta) \,\pi(\theta) \, \text{d} \theta \right]  \text{d}y
 =  \int_{\Theta}
   \left\{\int_0^\infty y^{\beta}
 f(y \vert \theta)\, \text{d} y \right\} \pi(\theta) \, \text{d} \theta.
$$
Now assume that $f(y \vert \theta)$ is RV with index $\alpha(\theta)$.


*

*Assume that $\alpha(\theta)$ is constant w.r.t. $\theta$. If $\beta
> \alpha(\theta)$, then the integral between the curly brackets {} is
infinite for all $\theta$, and the left hand side is also infinite. So
$\beta > \alpha(\theta)$ implies that $\beta \geq \alpha$, which tells
that $\alpha \leq \alpha(\theta)$.

*Assume that $\alpha(\theta)$ varies smoothly with $\theta$. If
$\beta > \min_\theta \alpha(\theta)$ then there exist a real
interval $I \subset \Theta$ with positive width such that $\beta >
  \alpha(\theta)$ for every $\theta \in I$, implying that the integral
between the curly brackets is infinite, hence that the left hand
side is infinite. As before we conclude that $\alpha \leq
  \min_{\theta} \alpha(\theta)$.
So we see that the mixture has a tail which is at least as heavy as
that of the heaviest-tailed component $f(y\vert \theta)$. This could
be generalised to $\alpha = \infty$ with a few changes in the
definition for RV then. Moreover, by replacing the moments
$\text{E}[Y^\beta]$ by exponential moments $\text{E}[e^{\beta Y }]$,
thin-tailed distributions can be compared similarly.
Beside the stationary distribution of a Markov chain in the question,
there are many other examples ot this "tail-broadening" phenomenon.


*

*The heavy-tailed Lomax distribution (with $\alpha > 0$) can be
obtained by taking $\pi(\theta)$ to be gamma with shape $\alpha$ and
$f(y \vert \theta)$ to be exponential with rate $\theta$, for which
$\alpha = \infty$. So, the Lomax distribution is a mixture of
thin-tailed distributions.

*In the Bayes context, we can take $\pi(\theta)$ to be $\pi(\theta
  \vert \mathbf{y}_{\text{obs}})$ for some observed vector
$\mathbf{y}_{\text{obs}}$. We then get that the tail of the
predictive posterior distribution $f(y \vert \mathbf{y}_{\text{obs}})$
is heavier than that of the
likelihood $f(y \vert \theta)$, due to the uncertainty on $\theta$ conditional
on $\mathbf{y}_{\text{obs}}$. Most predictive distributions are
heavy-tailed as are Student and Fisher-Snedecor distributions.
Mikosch, T (1999) Regular Variation, Subexponentiality and Their
    Applications in Probability Theory, Eindhoven University of
    Technology.
A: 
Can anything be inferred about the tails of $f(\cdot|x_{t-1})$?

By stationarity
$$
\int f(x_t \mid x_{t-1}) \pi(x_{t-1}) dx_{t-1} =\pi(x_{t}).
$$
By your assumptions of normality:
$$
\int f(x_t \mid x_{t-1}) \phi^n(x_{t-1}) dx_{t-1} =\phi^n(x_{t}), \tag{2}
$$
where $\phi^n(x_t) = \prod_{i=1}^n \phi(x_{ti})$. This (probably) means that 
$$
f(x_t \mid x_{t-1}) = N(x_t ; Ax_{t-1}, \Sigma),
$$
for any $A$, $\Sigma$ such that $\Sigma + AA' = I$. So both densities have a standard kurtosis. Actually, I am not 100% certain about how multivariate kurtosis is defined.
Edit:
This won't completely satisfy what you're looking for but if you look at the LHS of (2) you get
$$
(2 \pi)^{-1} \det(\Sigma)^{-1/2} \int \exp\left[-\frac{1}{2}\left\{(x_t - A x_{t-1})'\Sigma^{-1}(x_t - Ax_{t-1})- x_{t-1}'x_{t-1} \right\} \right] dx_{t-1} \tag{3}
$$
inside the curly braces turns out to be
$$
x_t'\Sigma^{-1}x_t - 2x_{t-1}'A'\Sigma^{-1}x_t + x_{t-1}'\left[ A'\Sigma^{-1}A-I\right] x_{t-1}
$$
and you can complete the square and re-arrange this as
$$
(x_{t-1} - Bx_t)'C^{-1}(x_{t-1} - Bx_t) - x_tB'Bx_t + x_t'\Sigma^{-1}x_t 
$$
where $C^{-1} = A'\Sigma^{-1}A-I$ and $B = \left(A'\Sigma^{-1}A-I\right)^{-1}A'\Sigma^{-1} $. Completing the square like this allows us to integrate (3) by recognizing a Gaussian density. You integrate out $x_{t-1}$ and what should be left is a Gaussian density for $x_t$ with  mean $0$ and variance $I$.
This is technically only a sufficient condition for the marginals to be Gaussian, and not the other way around, which is what you're looking for, but you can start playing around with what happens if you don't have this. The integral would be pretty tough to get.
In the univariate case, the tails you're after, $E[X_t^4 \mid x_{t-1}]$, on average have to be $3$ by the law of total expectation. But with this example, we just assume these are all $3$, regardless of $x_{t-1}$. If you find an example that's different from this one, where $x_{t-1}$ influences more than just the mean of $x_t \mid x_{t-1}$, then I would be interested in hearing about it.
