0
$\begingroup$

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?

$\endgroup$

2 Answers 2

8
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ Thank you for the reply. I note that both your and @micheal's answer rely on mixing random variables such that the process has different time average than average over the probability space. Does ergodicity ensure what I want, or does it only complicate finding a counter example? $\endgroup$
    – Marc
    Apr 23, 2018 at 13:42
  • $\begingroup$ @Marc Ergodicity (or even mixing) is still not enough, see the edit to my answer. $\endgroup$ Apr 24, 2018 at 7:14
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.