Proof of spatial Markov property I have to prove the spatial Markov property. Here is what I mean by this.
Let $G$ be a graph, and let $I$ be an independent set in $G$, chosen from the hard-core model (that is, the probability of choosing any independent set $I$ is proportional to $\lambda^{|I|}$ for some activity parameter $\lambda$). Let $U\subseteq V(G)$ be a set of vertices in $G$ and let $S_U\subseteq U$ be a subset of $U$. Then I want to prove that
$$ 
   \mathbb{P}(I\cap U=S_U \mid I\cap U^c) = \mathbb{P}(I\cap U=S_U \mid I\cap N(U)).
$$
This is shorthand for saying that for any independent set $I_0$ disjoint from $U$,
$$ 
   \mathbb{P}(I\cap U=S_U \mid I\cap U^c = I_0) = \mathbb{P}(I\cap U=S_U \mid I\cap N(U) = I_0 \cap N(U)).
$$

The case in which $U$ is a vertex, i.e., $U=\{v\}$, for $v\in V(G)$ is proved as follows. We want to show that $\mathbb{P}(v\in I\,|\,I\setminus\{v\})=\mathbb{P}(v\in I\,|\,I\cap N(v))$.
$$\mathbb{P}(v\in I\,|\,I\setminus\{v\}=J)=\frac{\lambda^{|J|+1}}{\lambda^{|J|+1}+\lambda^{|J|}}=\frac{\lambda}{\lambda+1},$$
since conditioned on $I\setminus\{v\}=J$, there are two possibilities for the independent set $I$, either $I=J$ or $I=J\cup\{v\}$
On the other hand, if we know the set $I\cap N(v)$, the only case in which the right-hand side probability is not zero is when $I\cap N(v)=\varnothing$, therefore
$$\mathbb{P}(v\in I\,|\,I\cap N(v)=\varnothing)=\frac{\lambda}{1+\lambda},$$
which gives the desired equality.
The problem is that I don't know how to show the general case, i.e., for a general vertex set $U$. Any help would be very much appreciated. Thanks!
 A: We can partition $V(G)$ into three sets: $U$ (which is given), $U' = N(U)$, and $U'' = V(G) - U - N(U)$. The goal is to show that $I \cap U$ and $I \cap U''$ are conditionally independent given $I \cap U'$. You have stated this as
$$
   \mathbb P(I \cap U = S \mid I \cap U' = S', I \cap U'' = S'') = \mathbb P(I \cap U = S \mid I \cap U' = S')
$$
but it is actually easier to show in the following equivalent form:
$$
   \mathbb P(I \cap U = S, I \cap U'' = S'' \mid I \cap U' = S') = \\ \mathbb P (I \cap U = S\mid I \cap U' = S') \cdot \mathbb P(I\cap U'' = S'' \mid I \cap U' = S')
$$
In fact, we'll see that it's enough to show that $\mathbb P(I \cap U = S, I \cap U'' = S'' \mid I \cap U' = S')$ factors as $f(S) g(S'')$ for some functions $f,g$ that only depend on $S'$ and of course the graph $G$.
Conditional on $I \cap U' = S'$, we have:

*

*A fixed set of possibilities for $I \cap U$: the independent subsets of $U - N(S')$. Call this set of possibilities $\mathcal S$.

*A fixed set of possibilities for $I \cap U''$: the independent subsets of $U'' - N(S')$. Call this set of possibilities $\mathcal S''$.

*For any $S \in \mathcal S$ and $S'' \in \mathcal S''$, $S \cup S' \cup S''$ is an independent set. That's because there are no edges between $U$ and $U''$ to rule anything out.

For every $S \in \mathcal S$ and $S'' \in \mathcal S''$,
$$
    \mathbb P(I \cap U = S, I \cap U'' = S'' \mid I \cap U' = S') = \frac{\lambda^{|S| + |S'| + |S''|}}{\sum_{S \in \mathcal S} \sum_{S'' \in \mathcal S''} \lambda^{|S| + |S'| + |S''|}}.
$$
The $\lambda^{|S'|}$ cancels and the probability factors as
$$
    \frac{\lambda^{|S|}}{\sum_{S \in \mathcal S} \lambda^{|S|}} \cdot \frac{\lambda^{|S''|}}{\sum_{S'' \in \mathcal S''} \lambda^{|S''|}}
$$
which is the factorization we wanted to show. From here, we can use the law of total probability summing over $\mathcal S''$ to show that $$\mathbb P (I \cap U = S\mid I \cap U' = S') = \frac{\lambda^{|S|}}{\sum_{S \in \mathcal S} \lambda^{|S|}}$$ and similarly sum over $\mathcal S$ to show that  $$\mathbb P (I \cap U'' = S''\mid I \cap U' = S') = \frac{\lambda^{|S''|}}{\sum_{S'' \in \mathcal S''} \lambda^{|S''|}}$$ which implies the independence statement we wanted.
