Support of bivariate joint distribution of stationary and ergodic sequence Let $\{X_t\}_{t\in \mathbb{N}}$ be a strictly stationary and ergodic sequence of real valued random variables and let the support of $X_1$ equal $[-1,1]$. Can the support of $(X_1,X_2)$ equal the unit disc centered at the origin?
(Under the stronger condition that the sequence is i.i.d., the support of the joint distribution of $(X_1,X_2)$ must equal the square $[-1,1]^2$.)
$\bf{Edit:}$ I will use the following definition of ergodic:
Let $\mu$ be the (shift-invariant) measure induced on $\left(\mathbb{R}^{\mathbb{N}},\mathcal{B}(\mathbb{R}^{\mathbb{N}})\right)$ by the stationary stochastic process $X:=\{X_{t}\}_{t\in\mathbb{N}}$. Let $T:\,\mathbb{R}^{\mathbb{N}}\rightarrow\mathbb{R}^{\mathbb{N}}$ be the left shift operator mapping sequences $\{x_{t}\}_{t\in\mathbb{N}}$ onto $\{x_{t+1}\}_{t\in\mathbb{N}}$. 
I will say $\{X_{t}\}_{t\in\mathbb{N}}$ is ergodic if for any measurable $f\in L^{1}(\mathbb{R}^{\mathbb{N}},\mathcal{B}(\mathbb{R}^{\mathbb{N}}),\mu)$, the averages $\frac{1}{T}\sum_{t=1}^{T}f(T^{t-1}X)$ converge pointwise almost everywhere to $\int_{\mathbb{R}^{\mathbb{N}}}f(x)d\mu$.
 A: Sure!
Let $(U,V)$ be a pair of random variables with values from $[-1,1]$ such that


*

*The joint distribution of $(U,V)$ is supported on the unit disk,

*$U$ and $V$ have the same individual distributions.
For instance, you can choose $(U,V)$ uniformly at random from the unit disk.
Now construct a Markov process by first choosing $X_0$ at random according to the distribution of $U$ and then choosing $X_1,X_2,\ldots$ recursively as follows: given $X_0,X_1,\ldots,X_{n-1}$, choose $X_n$ according to the distribution $\mathbb{P}(V\in\cdot\,|\,U=X_{n-1})$.
A sequence $X_0,X_1,\ldots$ constructed like this is stationary and has the property that $(X_{n-1},X_n)$ is supported on the unit disk.  Let us show that when $(U,V)$ is uniformly distributed over the unit disk, the sequence is also ergodic.
In order for the sequence $X_0,X_1,\ldots$ to be ergodic, it is enough that the Markov process with transition kernel $Q(a,\cdot):=\mathbb{P}(V\in\cdot\,|\,U=a)$ has a unique invariant measure.
A sufficient condition for the uniqueness of the invariant measure is that


*

*there is a number $n>0$, a probability measure $\rho$ on $[-1,1]$ and a constant $\alpha>0$ such that $Q^n(a,\cdot)\geq\alpha\rho(\cdot)$.


In the case where $(U,V)$ is uniformly distributed over the unit disk, it is easy to verify that the above condition is satisfied.  Namely, note that for every $a\in[-1,1]$ and measurable $B\subseteq[-1,1]$,
$$Q(a,B):= \frac{1}{2\sqrt{1-a^2}}\lambda\big(B\cap [-\sqrt{1-a^2},\sqrt{1-a^2}]\big) \;,$$
where $\lambda$ is the Lebesgue measure on $[-1,1]$.
Choose $\varepsilon>0$ small.  Then, you can verify that
\begin{align*}
   Q^2(a,B) &= \mathbb{P}(X_2\in B\,|\,X_0=a) \\
   &\geq \mathbb{P}(X_1\in[-\sqrt{1-\varepsilon^2},\sqrt{1-\varepsilon^2}]\,|\,X_0=a)\\
   & \qquad\cdot\mathbb{P}(X_2\in B\,|\,X_1\in[-\sqrt{1-\varepsilon^2},\sqrt{1-\varepsilon^2}]) \\
   &\geq \sqrt{1-\varepsilon^2}\frac{\lambda(B\cap[-\varepsilon,\varepsilon])}{2} \;.
\end{align*}
So the above uniqueness condition is satisfied with $\rho:=(2\varepsilon)^{-1}\lambda(\cdot\cap[-\varepsilon,\varepsilon])$ and $\alpha:=\varepsilon\sqrt{1-\varepsilon^2}$.
