Consider a stationary process $(X(i), i\in\mathbb{Z})\in \{0,1 \}^\mathbb{Z}$ with the following structure; runs of 0s alternate with runs of 1s, with the length of all runs independent, and with each run of 0s having geometric distribution with parameter $\alpha$, and each run of 1s having geometric distribution with parameter $\beta$.

Put more straightforwardly, this is just a Markov chain on $\{0,1\}$ with transition matrix $p_{00}=1-p_{01}=\alpha$, $p_{11}=1-p_{10}=\beta$. This chain can be dominated by a process $(Y(i), i\in\mathbb{Z})$ with independent entries, in which each $Y(i)$ is 1 with probability $p$ and 0 with probability $1-p$; we may take $p$ to be $\max(1-\alpha, \beta)$. Here "domination" means we could couple $X$ and $Y$ with $X(i)\leq Y(i)$ for all $i$.

The domination is easy to see -- for example, notice that conditional on any configuration $(X(i), i < k)$, the probability that $X(k)=1$ is no larger than $p$, so the coupling could be built up one site at a time from left to right.

Now alter the process $X$ by changing the distribution of the runs of 1s. Now it is not geometric with parameter $\beta$, but it is stochastically bounded above by the geometric distribution with parameter $\beta$. The distribution of runs of 0s is unchanged.

Is it still true that the process $X$ is dominated by a process $Y$ with independent entries? (excluding the trivial process where every entry is 1).

Loosely, we have reduced the number of 1s in the process, so one might expect the same domination to hold. But notice that the uniform bound on the conditional probability above needn't hold any longer. For example, it could be that the run of 1s never has length exactly 2. Then $\mathbb{P}(X(3)=1|X(0)=0, X(1)=1, X(2)=1)=1$.

  • $\begingroup$ Looks like an interesting thing to think of but can somebody, please, remind me what "stochastically bounded" means? $\endgroup$ – fedja May 25 '11 at 14:03
  • $\begingroup$ I could have said "dominated" again: (the distribution of) a random variable U is stochastically bounded above by (the distribution of) another random variable V if you can find a coupling of the two such that U≤V with probability 1. Or in this case, more simply, $P(U>x)<P(V>x)$ for all $x$. $\endgroup$ – James Martin May 25 '11 at 19:09

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