Let $X=(X\_n)\_{n\in\mathbb N}$ be a stochastic process in $\{0,1\}^{\mathbb N}$ with distribution $\mu$. I do not at first make any assumptions about $X$ being stationary or having any kind of correlations or associations. Let $Y=(Y\_n)\_{n\in\mathbb N}$ be an iid Bernoulli process with distribution product measure $\pi\_p$, i.e. $\mathbb P(Y\_n=1)=p$ for all $n$ and independently of all other $Y\_j$.


I want to study the conditions under which $\pi\_p\preceq \mu$, that $X$ stochastically dominates $Y$. The order $X(\omega)\geq Y(\omega)$ uses the natural ordering of their sample paths for each $\omega\in\Omega$ a common probability space upon which they are constructed, for which I just say $X\geq Y$.


From Lemma 1.1 in Liggett et al. (1997), I can conclude that a sufficient condition for $\pi\_p\preceq \mu$ is that
$$
\mathbb P(X\_n=1\mid X\_{s\_j}=\epsilon\_{s\_j}, \text{ for all } j=1,2,\ldots,m)\geq p 
\tag{1}\label{eq1}
$$
for any $n$ and any $\{s\_1,s\_2,\ldots,s\_m\}\subset \{1,2,\ldots,n-1\}$ and any $\epsilon=(\epsilon\_n)\_{n\in\mathbb N}$. Of course we must be conditioning on an event of positive probability, so I always assume that.


My question is on why this condition isn't also necessary.


I know that $\pi\_p\preceq \mu$ if and only if there exists a coupling of $X$ and $Y$ such that $\mathbb P(X\geq Y)=1$. (I hope it is clear that I am now talking about coupled "copies of" $X$ and $Y$ with the appropriate marginals.) Such a coupling must also satisfy (\ref{eq1}) yes?


I use $\mathbb P^\epsilon$ to refer to the conditional joint distribution of $(X,Y)$, conditioned on $X\_{s\_j}=\epsilon\_{s\_j}$ for $j=1,2,\ldots,m$:
$$\mathbb P^\epsilon(X\_n=1)=\mathbb P^\epsilon(X\_n=1,Y\_n=0)+\mathbb P^\epsilon(X\_n=1,Y\_n=1)$$
Since the coupling satisfies $\mathbb P(X\geq Y)=1$, we also get $\mathbb P^\epsilon(X\geq Y)=1$ (assuming we condition on an event of positive probability). Hence $\mathbb P^\epsilon(X\_n=1,Y\_n=1)=\mathbb P^\epsilon(Y\_n=1)=p$ (since $\{X\_n=0,Y\_n=1\}$ has probability zero under $\mathbb P$ and $\mathbb P^\epsilon$), therefore we clearly have
$\mathbb P^\epsilon(X\_n=1)\geq p$ as desired.


So I have (\ref{eq1}, for all $n,\epsilon,m$) if and only if $\pi\_p\preceq\mu$.


Another way I look at it is from a simulation point of view. Assuming I have complete access to the distribution of $X$ (simulating $Y$ is easy), I can simulate coupled $(X\_1,Y\_1)$ with $\mathbb P(X\_1=1)\geq p$. Then simulate $X\_2$ conditioned on the realized value of $X\_1$ (which obviously has positive probability for nontrivial $p$), etc. At every step we must have the probability that $X\_n=1$ is at least $p$ (conditioned on an event of positive probability). Of course, maybe the structure of $X$ could possibly prevent this type of simulation? Though as long as this is possible, then (\ref{eq1}), for all $n,m,\epsilon$, should be necessary and sufficient for $\pi\_p\preceq \mu$.


Am I misunderstanding something?


Liggett & Steif (2006) explores the situation when $X$ has certain associations, and there are many other papers exploring special cases for certain types of processes as well.


Primary References:  

Liggett et al (1997) <https://www.jstor.org/stable/2959530>  

Liggett & Steif (2006) <https://dx.doi.org/10.1016/j.anihpb.2005.04.002>  

(and follow any references from those and papers that cite them)



Stochastic process on $\{0,1\}^{\mathbb N}$ domination of product measures, necessary and sufficient conditions
Ok, here is a counterexample with the index set $\{1,2\}$ (you can easily extend it to whole $\mathbb N$ if you wish).


Let $(Y\_1,Y\_2)$ be independent Bernoulli($1/2$) and set
$$
(X\_1,X\_2) = \begin{cases}
(Y\_1,Y\_2), & Y\_1 + Y\_2>0,\\
(0,1), & Y\_1 = Y\_2 = 0.
\end{cases}
$$


In this case $$
\mathrm P(X\_1 = 1\mid X\_2 = 1) = \frac{P(X\_1 = 1, X\_2 = 1)}{P(X\_2 = 1)} = \frac{1/4}{3/4} = \frac13<\frac12.
$$


