Couple two <noisy voter model> chains by choosing the same update vertex, the same refresh coin and refreshed spin, and, for a voter update, the same chosen neighbor. Let $D_t$ be their <Hamming distance>. If $G$ is $r$-regular, conditioning on the current disagreement set gives
$$
\begin{aligned}
\mathbb E[D_{t+1}\mid D_t]
&=D_t-\frac{D_t}{n}
+\frac{1-p}{n}\sum_{v\in V}
\frac{|N(v)\cap D_t|}{r}\\
&=\left(1-\frac pn\right)D_t,
\end{aligned}
$$
because every disagreeing vertex is counted in exactly $r$ neighbor sets. Therefore
$$
\mathbb E D_t\leq n\left(1-\frac pn\right)^t.
$$
The <coupling inequality for total variation> and $\mathbf1_{\{\sigma_t\ne\sigma_t'\}}\leq D_t$ give
$$
\boxed{\max_{\sigma,\sigma'}
\|P^t(\sigma,\cdot)-P^t(\sigma',\cdot)\|_{\mathrm{TV}}
\leq n\left(1-\frac pn\right)^t.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 215 / 3 / i / Solution

The noisy voter model updates a chosen vertex either by replacing its spin with fresh random noise or by copying a neighboring spin. A positive refresh probability destroys conserved opinions and permits contractive coupling bounds even on nonregular graphs.


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