Kac return-time formula for a state

By [F1] the invariant probability satisfies $\pi (b)>0$ and the chain is positive recurrent with ${\mathbb{E}}_b{T}_b^{+}\le 1\mathrm{/}\pi (b)<+\mathrm{\infty }$; hence ${\mu }_b$ is finite-valued with ${\sum }_{y\in E}{\mu }_b(y)={\mathbb{E}}_b{T}_b^{+}$, and since positive recurrence makes $b$ recurrent, [F2] gives ${\mu }_{b}p={\mu }_b$ as well as ${\mu }_b(b)=1$.

Define $\nu (y):=\pi (y)\mathrm{/}\pi (b)$ for $y\in E$; this is nonnegative and finite by step 1.1, $\nu (b)=1$, and for $y\mathrm{\ne }b$ the invariance identity [F4] gives $(\nu p)(y)={\sum }_x\nu (x)p(x,y)=\frac{1}{\pi (b)}{\sum }_x\pi (x)p(x,y)=\pi (y)\mathrm{/}\pi (b)=\nu (y)$.

By the minimality clause of [F2] applied to $\nu$, one has $\pi (b){\mu }_b(y)\le \pi (b)\nu (y)=\pi (y)$ for every $y$; hence $\eta (y):=\pi (y)-\pi (b){\mu }_b(y)$ is a well-defined nonnegative extended function with $\eta (b)=\pi (b)-\pi (b)\cdot 1=0$ and finite total mass ${\sum }_y\eta (y)=1-\pi (b){\mathbb{E}}_b{T}_b^{+}$.

The defect $\eta$ is invariant: for every $y$, ${\sum }_x\eta (x)p(x,y)={\sum }_x\pi (x)p(x,y)-\pi (b){\sum }_x{\mu }_b(x)p(x,y)=\pi (y)-\pi (b){\mu }_b(y)=\eta (y)$, using the invariance identity [F4] for $\pi$, the identity ${\mu }_{b}p={\mu }_b$ from step 1.1, and the fact that both subtracted series have finite values; iterating with the Chapman–Kolmogorov identity [F5] and the interchange of nonnegative sums [F6] gives ${\sum }_x\eta (x)p^{(n)}(x,y)=\eta (y)$ for every $n\ge 0$.

Evaluate the conservation identity of step 4.1 at $y=b$ and $n$ arbitrary: $0=\eta (b)={\sum }_{x\in E}\eta (x)p^{(n)}(x,b)$, a sum of nonnegative terms, so $\eta (x)p^{(n)}(x,b)=0$ for every $x$ and every $n$; for fixed $x$, [F3] provides $n$ with $p^{(n)}(x,b)>0$, hence $\eta (x)=0$. Therefore $\eta \equiv 0$, that is, $\pi (y)=\pi (b){\mu }_b(y)$ and so ${\mu }_b(y)=\pi (y)\mathrm{/}\pi (b)$ for every $y\in E$.

Summing the identity of step 5.1 and using ${\sum }_y{\mu }_b(y)={\mathbb{E}}_b{T}_b^{+}$ from [F2] gives $1={\sum }_y\pi (y)=\pi (b){\sum }_y{\mu }_b(y)=\pi (b){\mathbb{E}}_b{T}_b^{+}$, that is, ${\mathbb{E}}_b{T}_b^{+}=1\mathrm{/}\pi (b)$, finite and positive.

The three assertions of the statement hold: $\pi (b)>0$ by step 1.1, ${\mathbb{E}}_b{T}_b^{+}=1\mathrm{/}\pi (b)$ by step 6.1, and ${\mu }_b(y)=\pi (y)\mathrm{/}\pi (b)$ by step 5.1.

Boundary and axiom cases: if $E$ is a singleton the formulas give ${\mu }_b={\delta }_b$, ${\mathbb{E}}_b{T}_b^{+}=1$ and $\pi (b)=1$, matching step 6.1; if $\pi$ is not unique the argument applies to each invariant probability separately, since only invariance of $\pi$ and irreducibility are used, and no uniqueness is asserted; a transient or null-recurrent chain has no invariant probability by [F1], so the hypothesis cannot be vacuous in those cases; $\eta$ is nonnegative by the minimality clause, so no infinite minus infinite subtraction occurs in step 4.1, and the subtracted series there are separately finite; the identities are equalities, not implications, so there is no iff case separation; and AC [A1] enters exactly through [F2] and [F1], both of which assume it.