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 $\int\_{\mathbb{R}^{\mathbb{N}}}f(x)d\mu$.



Support of bivariate joint distribution of stationary and ergodic sequence
Sure!


Let $(U,V)$ be a pair of random variables with values from $[-1,1]$ such that


1. The joint distribution of $(U,V)$ is supported on the unit disk,
2. $U$ 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 $\alpha:=\varepsilon\sqrt{1-\varepsilon^2}$.


