1
$\begingroup$

A while ago I posted the following problem:

Suppose I have a two sided stationary sequence of random variables $\ldots,X_{-1},X_0,X_1,\ldots$ such that all finite dimensional joint densities $f(x_1,\ldots,x_n)$, $n\in\mathbb{N}$ exist. I want to ensure the following:

Let $A$ and $B$ be events such that $P(\ldots,X_{-1},X_0\in A)>0$ and $P(X_1,X_2,\ldots\in B)>0$, then $$ P(\ldots,X_{-1},X_0\in A \qquad\text{and}\qquad X_1,X_2,\ldots\in B)>0. $$

Finding general conditions turned out to be hard. However, what can we say when we assume that $(X_t)_{t\in\mathbb{Z}}$ is a solution to a system $$ X_{t+1} = \phi(X_t,\varepsilon_{t+1}), $$ where $(\varepsilon_t)_{t\in\mathbb{Z}}$ is iid. In that case we only have to show $$ P(X_1,X_2,\ldots\in B \mid X_0 \in A_0)>0, $$ where $A_0$ is the projection of $A$ on the coordinate for $X_0$. Is it enough to assume $\phi(x,\varepsilon_{t+1})$ is a random variable with full support for all $x\in\mathbb{R}$?

Improve this question
$\endgroup$
1
  • $\begingroup$ Under your assumptions, $(X_t, \varepsilon_{t+1})$ is a Markov chain, and conditioning on the distribution of $(X_0, \varepsilon_1)$ should give the desired result. However, you need to be careful with the definition of $A_0$: a projection may fail to be measureable, and it could be the entire space even if $A$ has probability zero (think: $A$ is the set of constant sequences). Instead, $A_0$ should be defined as the set of those $x$ for which $P(\ldots, X_{-2}, X_{-1}, X_0 \in A | X_0 = x) > 0$ (note that this set is defined only up to a null set of the distribution of $X_0$). $\endgroup$
    – Mateusz Kwaśnicki
    Commented Jul 19, 2018 at 9:15

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .