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The Noisy voter model combines copying with spontaneous opinion changes. On a finite connected undirected graph with $n\ge 2$ vertices, give each vertex a spin in $\{0,1\}$. At each discrete step choose a vertex uniformly. With Probability $\epsilon \in (0,1]$, replace its spin by an independent fair bit; otherwise copy a uniformly chosen neighbour. This is a Markov chain on spin configurations.^[ref]

Positive noise makes the finite chain irreducible: a sequence of refreshes can produce any configuration with positive Probability. There is also a positive holding probability, so it is aperiodic and has a unique Stationary distribution. At zero noise, the two consensus configurations are absorbing; they cease to be absorbing under positive noise.

A weighted coupling explains convergence even when the graph is not regular. Let $R(v,u)$ be the simple-random-walk transition matrix and $w_v=\mathrm{deg}(v)/(2|E|)$ its Stationary distribution. For two spin configurations use the weighted Hamming distanceCouple the two chains with the same updated vertex, refresh decision, fresh bit and copied neighbour. If ${\delta }_v$ is the current disagreement indicator, the expected one-step change in $D$ issince $wR=w$. Thus $\mathbb{E}{D}_t=(1-\epsilon /n{)}^t{D}_0$. With $w_{\min }={\min }_v{w}_v>0$, a remaining disagreement costs at least $w_{\min }$, so coupling with a stationary chain yields worst-case total variation distance at most $w_{\min }^{-1}(1-\epsilon /n{)}^t$. The time here counts single-vertex updates; switching to a rate-one clock at each vertex changes time by a factor of $n$.

For two configurations $\eta ,\xi \in \{0,1{\}}^V$, the weighted Hamming distance for the noisy voter model is $D(\eta ,\xi )={\sum }_v\pi (v)1_{\{\eta (v)\ne \xi (v)\}}$, where $\pi$ is the stationary distribution of the copying walk. On a connected undirected graph, $\pi (v)=\mathrm{deg}(v)/(2|E|)$. Positive weights make this a metric, with values between zero and one.

Couple two chains using the same uniformly selected vertex, refresh decision, fresh bit and copied neighbor. If the refresh probability is $\epsilon$, stationarity $\pi P=\pi$ gives $\mathbb{E}[D_{t+1}\mid {\eta }_t,{\xi }_t]=(1-\epsilon /|V|)D_t$. Copying redistributes disagreements without increasing their stationary-weighted total in expectation; refreshes remove them. The factor $|V|$ belongs to the single-vertex update convention, not a simultaneous update of every vertex.^[ref]