Positive recurrence without aperiodicity does not imply total-variation convergence

The matrix $p$ is a transition matrix: its entries are $0$ or $1$, and every row has the single entry $p(i,i+1)=1$, so every row sums to one. By [F2] an induction on $n$ gives $p^{(n)}(x,y)=1$ when $y\equiv x+n(\mathrm{m}\mathrm{o}\mathrm{d}3)$ and $p^{(n)}(x,y)=0$ otherwise: the case $n=0$ is [F1], and $p^{(n+1)}(x,y)={\sum }_w{p}^{(n)}(x,w)p(w,y)=p^{(n)}(x,y-1)$ takes the value $1$ exactly when $y-1\equiv x+n(\mathrm{m}\mathrm{o}\mathrm{d}3)$. In particular $p^{(3)}$ is the identity matrix, $p^{(n)}(x,x)=1$ exactly when $3$ divides $n$, and the $n$-step law from $x$ is the point mass at $(x+n)<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>3$.

The process $X$ is a $p$-chain started at $0$: $X_0=0$, and for every $n\ge 0$ and $A\subseteq E$ the conditional probability $\mathbb{P}(X_{n+1}\in A\mid {\mathcal{F}}_n)$ is the almost-sure class of the constant ${\mathbf{1}}_{\{n+1<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>3\in A\}}$, while $K(X_n,A)=p(n<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>3,A)={\mathbf{1}}_{\{n+1<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>3\in A\}}$ by step 1.1; the two sides agree, and the same computation applied to the shifted process $X_n^{(x)}:=(x+n)<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>3$ shows that each shift is a $p$-chain started at $x$.

The chain is irreducible: given $x,y\in E$, the integer $k:=(y-x)<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>3$ lies in $\{0,1,2\}$ and step 1.1 gives $p^{(k)}(x,y)=1>0$, so $x\to y$; interchanging $x$ and $y$ gives $y\to x$.

The law $\pi$ is invariant: each column of $p$ has exactly one entry $1$, namely $p(y-1,y)=1$, so ${\sum }_{x\in E}\pi (x)p(x,y)=\frac{1}{3}\cdot 1=\frac{1}{3}=\pi (y)$ for every $y$, which is the criterion of [F4].

Failure of ordinary convergence: by step 1.1 the $n$-step law from $0$ is the point mass ${\delta }_{n<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>3}$, and the half-${\mathrm{\ell }}^1$ formula [F9] gives $\parallel {\delta }_j-\pi {\parallel }_{\mathrm{T}\mathrm{V}}=\frac{1}{2}(\frac{2}{3}+\frac{1}{3}+\frac{1}{3})=\frac{2}{3}$ for each $j\in \{0,1,2\}$. Hence $\parallel p^{(n)}(0,\cdot )-\pi {\parallel }_{\mathrm{T}\mathrm{V}}=2\mathrm{/}3$ for every $n\ge 0$, including $n=0$, and the sequence of laws does not converge to $\pi$; it cycles through three distinct point masses.

Every state is positive recurrent with return time three: for the chain started at $x$, step 2.1 gives $X_n^{(x)}=(x+n)<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>3$, so $X_n^{(x)}=x$ exactly when $3$ divides $n$; by [F5] this says $T_x^{+}=3$ identically, hence ${\mathbb{P}}_x(T_x^{+}<\mathrm{\infty })=1$ and ${\mathbb{E}}_x{T}_x^{+}=3<+\mathrm{\infty }$, and [F6] makes $x$ positive recurrent.

The chain is not aperiodic: by step 1.1 the positive return set of [F7] is $R_x=\{3,6,9,\dots \}$, whose greatest common divisor is $3$, so $d(x)=3$ for every $x$; since the chain is irreducible by step 2.2 the periods agree, and [F8] gives $\mathrm{per}(p)=3\mathrm{\ne }1$, so $p$ is not aperiodic.

Cesàro convergence: put $C:=p^{(0)}+p^{(1)}+p^{(2)}$. Step 1.1 gives $C(x,y)=1$ for all $x,y$, since exactly one of $k=0,1,2$ satisfies $y\equiv x+k(\mathrm{m}\mathrm{o}\mathrm{d}3)$. Writing $n=3m+r$ with $0\le r\le 2$ and $R_r:=p^{(0)}+\cdots +p^{(r-1)}$ (so $R_0=0$, $R_1=I$, $R_2=I+p$), the periodicity in step 1.1 gives ${\sum }_{k=0}^{n-1}{p}^{(k)}=mC+R_r$, hence $\frac{1}{n}{\sum }_{k=0}^{n-1}{p}^{(k)}(x,y)=\frac{m}{n}C(x,y)+\frac{R_r(x,y)}{n}\to \frac{1}{3}\cdot 1+0=\frac{1}{3}=\pi (y)$ because $m\mathrm{/}n\to 1\mathrm{/}3$ and $R_r\mathrm{/}n\to 0$; for $n$ divisible by $3$ the average equals $\pi$ exactly. The chain is irreducible and positive recurrent with invariant $\pi$ by steps 2.2–2.4, so the general supplier [F10] gives the same limit, and the present computation verifies it directly.

Boundary and scope cases: the value $n=0$ is included and step 2.4 covers it, so the divergence is present from the first term and is not an artifact of a tail; the distance is $2\mathrm{/}3=1-\frac{1}{3}=1-\frac{1}{d}$ with period $d=3$, and for the deterministic two-cycle ($d=2$) the same formula gives $1\mathrm{/}2$, so no single nonzero constant is being asserted and the example is sharp at period three; the state space is finite, so all sums in steps 2.3 and 4.1 are finite and no summation is interchanged; the chain lies outside the aperiodicity hypothesis by step 3.2, which is exactly the hypothesis whose necessity is being shown; the process is built by the explicit formula $X_n=n<mrow><mi\; mathvariant="normal">m</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">d</mi></mrow>3$ on a one-point space, so no selection is made in steps 1.1–3.1, and AC [A1] is spent only on the general Cesàro supplier [F10], whose conclusion step 4.1 also establishes directly; the item refutes only the failure direction "positive recurrence without aperiodicity implies total-variation convergence", and it claims no converse and no failure of Cesàro convergence.