without loss of generality that $Z_B \leq Z_A$. It is easy to see that $\phi_A(i) \leq \phi_B(i)$ for $i=2,\ldots, q$ and hence $\phi_A(1) \geq \phi_B(1)$. Thus $$\|\phi_A - \phi_B\|_{TV} = \max_{R \subseteq [q]}|\phi_A(R) - \phi_B(R)| = |\phi_A(1) - \phi_B(1)| = 
\frac{\lambda^{a_1}}{Z_A} - \frac{\lambda^{b_1}}{Z_B}.$$ Given ${\bm{a}} = (a_1, \ldots, a_q) \in [\Delta]^q$, define $f({\bm{a}},\lambda,q) = \frac{\lambda^{a_1}}{Z_A} - \frac{\lambda^{b_1}}{Z_B}$, and let $g(\lambda,q)$ be the maximum of $f$ over all ${\bm{a}} \in [\Delta]^q$ subject to $a_1 + \cdots + a_q = \Delta$.

Observe that $$\begin{aligned}
{\mathbb{E}}\left(d(A',B')-1\big| (A,B)\right) 
&= (-1){\mathbb{P}}({\bm{v}}=u) + \sum_{v \in N(u)}{\mathbb{P}}({\bm{v}}=v)(1-p(v,A,B)) \\
&\leq -\frac{1}{n} + \frac{\Delta}{n} g(\lambda,q).\end{aligned}$$ We give an easy upper bound for $g(\lambda, q)$ as follows. First, for all ${\bm{a}}\in [\Delta]^q$ we have $$f({\bm{a}}, \lambda, q) \leq \frac{\lambda^{a_1}}{Z_A}.$$ The right hand side of the above is increasing in all directions of the form $e_1 - e_i$, where $e_1, \ldots, e_q$ is the standard basis for ${\mathbb{R}}^q$. Therefore the right hand side is maximised when ${\bm{a}} = (\Delta,0,\ldots, 0)$ giving $$g(\lambda,q) \leq \frac{\lambda^\Delta}{\lambda^\Delta + q-1} \leq 
    \frac{1}{\Delta+1},$$ using the lower bound on $q$ to obtain the final inequality. Therefore. $$\begin{aligned}
  {\mathbb{E}}\left(d(A',B')\big| (A,B)\right) \leq 
   1+ \frac{1}{n}\left(-1 + \frac{\Delta}{\Delta+ 1}\right)
   &= 1- \frac{1}{(\Delta+1)n}.\end{aligned}$$ Applying Lemma \[le:coup\] completes the proof.

Block dynamics {#se:BD}
