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

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  • $\begingroup$ Can you state the definition of "ergodic" that you want to use here? $\endgroup$ – Nate Eldredge Jun 11 '18 at 19:02
  • $\begingroup$ @NateEldredge Done. $\endgroup$ – user424747 Jun 11 '18 at 23:36
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    $\begingroup$ Also, regarding the answer below, I assume that "supported on" means "the support equals". If it means "the support is contained in" then this is trivial; let $X_i$ be iid $U(-1/2, 1/2)$. $\endgroup$ – Nate Eldredge Jun 12 '18 at 3:45
  • $\begingroup$ Yes, it means "the support equals". I was looking for an example like that provided by Algernon. $\endgroup$ – user424747 Jun 12 '18 at 5:51
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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 $\alpha:=\varepsilon\sqrt{1-\varepsilon^2}$.

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  • $\begingroup$ Thank you for your answer. I understand that this construction is stationary, but I'm having trouble reasoning that some component of the ergodic decomposition satisfies properties 1. and 2. from your answer. This may be a standard result. If so, can you please state it? $\endgroup$ – user424747 Jun 11 '18 at 22:08
  • $\begingroup$ See the added note for explanation. $\endgroup$ – Algernon Jun 12 '18 at 0:05
  • $\begingroup$ Thanks again; the notes are very helpful. One more question if I may: Was there anything special about property 1 restricting only bivariate joint distributions as opposed to n-variate ones? I ask because this led to a construction exhibiting the markov property, but I wonder if I had changed the question to restrict the support of$(X_1,...,X_n)$ to equal the unit n-ball at the origin your argument would essentially still go through. This is a question of representing a modification of your construction as a dynamical system: Does the extension theorem implicitly used in your answer work here? $\endgroup$ – user424747 Jun 12 '18 at 5:43
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    $\begingroup$ The proof is pretty simple: in order for the invariant measure of $Q$ to be unique, it is enough that the invariant measure of $Q^n$ is unique. So, we can assume $n=1$. Now for every $a$ we can write $Q(a,\cdot)=\alpha\rho(\cdot)+(1-\alpha)\tilde{Q}(a,\cdot)$, where $\tilde{Q}$ is another kernel. This means in order to choose $X_n$, we can first flip a biased coin $B_n$ with parameter $\alpha$. If $B_n=1$, we choose $X_n$ according to $\rho$, otherwise according to $\tilde{Q}(X_{n-1},\cdot)$. A standard coupling argument now shows the uniqueness (and convergence). $\endgroup$ – Algernon Jun 14 '18 at 20:19
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    $\begingroup$ The condition (with $n=1$) is mentioned here: [hairer.org/notes/Markov.pdf] as Theorem 4.29. $\endgroup$ – Algernon Jun 14 '18 at 20:21

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