Drift and minorization certify geometric ergodicity

The transition rows sum to one. Every adjacent move has positive probability, so every state reaches every other state, proving counting-measure irreducibility. Each state has a positive holding probability, so no positive-mass cyclic family can move deterministically to a disjoint successor. The proposed masses sum to one. At state $0$, the incoming mass is ${\pi }_{0}P(0,0)+{\pi }_{1}P(1,0)=3\mathrm{/}8+1\mathrm{/}8=1\mathrm{/}2={\pi }_0$. For $n\ge 1$, it is ${\sum }_j{\pi }_{j}P(j,n)={\pi }_{n-1}\mathrm{/}4+{\pi }_n\mathrm{/}4+{\pi }_{n+1}\mathrm{/}2=2^{-n-2}+2^{-n-3}+2^{-n-3}=2^{-n-1}={\pi }_n\mathrm{.}$ Thus $\pi P=\pi$. Also ${\pi }_{n}P(n,n+1)=2^{-n-1}\mathrm{/}4=2^{-n-3}={\pi }_{n+1}P(n+1,n)$; these detailed-balance equalities and ${\sum }_{n\ge 0}{2}^{-n-1}=1$ provide a second check.

Let $C=\{0\}$, $\nu =P(0,\cdot )$, $m=1$, and $ϵ=1$. Then $P(0,\cdot )=\nu$, so $C$ is small. Set $V(n)=(\sqrt{2}{)}^n$. For $n\ge 1$, $\frac{PV(n)}{V(n)}=\frac{1}{4}\sqrt{2}+\frac{1}{2}\frac{1}{\sqrt{2}}+\frac{1}{4}=\frac{1}{4}+\frac{\sqrt{2}}{2}<\frac{24}{25}\mathrm{.}$ Since $\sqrt{2}\mathrm{/}2<71\mathrm{/}100$ (squaring gives $1\mathrm{/}2<5041\mathrm{/}10000$), the preceding ratio is less than $96\mathrm{/}100=24\mathrm{/}25$. At $0$, $PV(0)=3\mathrm{/}4+\sqrt{2}\mathrm{/}4<5\mathrm{/}4<49\mathrm{/}25=24\mathrm{/}25+b$ with $b=1$. Hence $PV\le (24\mathrm{/}25)V+1_C$.

The hypotheses of [F5] now hold, so there are $M<\mathrm{\infty }$ and $r<1$ with $\mathrm{\parallel }{P}^n(x,\cdot )-\pi {\mathrm{\parallel }}_{\mathrm{T}\mathrm{V}}\le MV(x)r^n$ for all $x,n$. The example supplies a concrete certificate: the target is invariant and the drift/minorization conditions are checked directly.