Suppose that nodes arrive to the two sides (left and right) of a bipartite graph $G$ dynamically according to independent Poisson processes with rate $1$ and $n>1$, respectively. Every node departs the graph with rate 1 (i.e., there's an independent exponential distribution for every node that gives how long after the node's arrival it departs the graph). A node on the left is adjacent to a node on the right independently with probability $p$. Nodes are matched to each other greedily: upon the arrival of a node $v$, if there's another node in the graph adjacent to $v$, then the two nodes are matched and removed from the graph immediately.
Let $D_p$ denote the steady state distribution of the number of nodes on the right side. Then, $D_p$ stochastically dominates $D_1$.
The above claim is very intuitive: if $p=1$, nodes will be matched more quickly, thus there'll be less nodes in the graph. I have tried coupling the Markov chains to prove it, but I face a difficulty: if $p<1$, more nodes will be accumulated on the left side, which can later lead to a higher number of matches (and thereby a lower number of nodes!). This breaks the couplings that I could think of. I'd appreciate any suggestions or ideas.