# Stationary sequence and nonzero probabilities

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

Is it enough to assume that the finite dimensional joint densities all have full support?

No, this is not enough. Take $Y_n$ i.i.d. Gaussians, $r$ an independent Bernoulli taking values in $\{-1,1\}$, and set $X_n = r + Y_n$. Then the events $A = \{X\,:\,\limsup_{n \to \infty} {1\over n}\sum_{k=1}^n X_k = 1\}$ and $B = \{X\,:\,\limsup_{n \to \infty} {1\over n}\sum_{k=1}^n X_{-k} = -1\}$ both have probability ${1\over 2}$, but $A \cap B$ has probability $0$.
Edit In the case when $X$ is a stationary Gaussian sequence, these kind of problems are very well studied. A necessary condition which guarantees that finite-dimensional densities have full support and that the sequence is strongly mixing (and a fortiori ergodic) is that its spectral measure has a density $f$ with respect to Lebesgue measure such that $f$ is almost everywhere strictly positive and $\log f$ is integrable. The property you want to ensure however is equivalent (in the Gaussian case) to the property of the sequence being off-white (in the sense of Tsirelson) which holds if and only if $\log f$ belongs to the Sobolev space $H^{1/2}$. This shows that imposing ergodicity or mixing isn't sufficient either. See for example the old books by Dym & McKean and by Ibragimov & Rozanov for statements along these lines.
No, choose them to be iid N(-1, 1) with probability .5 and otherwise iid N(1,1), Let A be the set where $$\frac {X_1 + ... + X_n} n \rightarrow 1$$ and B be the set where $$\frac {X_{-1} + ... + X_{-n}} n \rightarrow -1$$ They are disjoint. The joint density of any finite number is a 50-50 mix of i.i.d. normal with different means and so has a density with full support.