Geometric small-set renewal convergence gives an integrable exponential hitting moment

Set $w_1(x)=ϵ1_C(x)$ and $w_0(x)=1-w_1(x)$. On $\tilde{X}=X\times \{0,1\}$ define $K_{x,1}=\nu$; define $K_{x,0}=(Q(x,\cdot )-ϵ\nu )\mathrm{/}(1-ϵ)$ for $x\in C$, and $K_{x,0}=Q(x,\cdot )$ otherwise. The split kernel is $\tilde{Q}((x,i),B\times \{j\})={\int }_B{w}_j(y)K_{x,i}(dy)\mathrm{.}$ Nonnegativity follows from [F1], measurability from [F3], and finite summation over the two flag sections defines a probability on every product-measurable event. Thus [F2] realizes the split path. If the initial flag at $x$ has probabilities $w_i(x)$, summing it gives ${\sum }_i{w}_i(x)K_{x,i}=Q(x,\cdot )$. The same summation at each subsequent time shows that the $x$-coordinate path has the original $Q$-chain law. The set $\alpha =C\times \{1\}$ is an atom: all its states have the identical outgoing transition obtained by using $\nu$.

Put $\tilde{\pi }(dx,i)=w_i(x)\pi (dx)$. Summing its incoming transitions first over $i$ gives $\pi Q=\pi$ and then assigns the new flag according to $w_j$, proving $\tilde{\pi }\tilde{Q}=\tilde{\pi }$ by [F13]. Let $p=\tilde{\pi }(\alpha )=ϵ\pi (C)>0$, and let $\kappa =\tilde{\pi }(\cdot \mid \alpha )$, an ordinary normalized restriction. Write ${\mathbb{P}}_{\alpha }={\mathbb{P}}_{\kappa }$; its future after time zero is the same from every atom state. Starting in $\alpha$, the next coordinate law is $\nu$ and subsequent coordinate transitions are $Q$, so $u_0=1,u_n={\mathbb{P}}_{\alpha }(Z_n\in \alpha )=ϵ\nu Q^{n-1}(C)(n\ge 1)\mathrm{.}$ The hypothesis implies $\mathrm{\mid }{u}_n-p\mathrm{\mid }\le c{\theta }^n$ for all $n\ge 0$ after increasing a finite $c$.

Let $\tau =\inf \{n\ge 1:Z_n\in \alpha \}$ and $f_n={\mathbb{P}}_{\alpha }(\tau =n)$ for $n\ge 1$, with $f_0=0$. Decompose a visit at time $n\ge 1$ by the first return time and use [F4] and the common atom transition to obtain $u_n={\sum }_{j=1}^n{f}_j{u}_{n-j}$. The possible event $\tau =\mathrm{\infty }$ contributes zero. For $\mathrm{\mid }z\mathrm{\mid }<1$, absolute convergence and [F8] give $U(z)={\sum }_{n\ge 0}{u}_n{z}^n=\frac{1}{1-F(z)},F(z)={\sum }_{n\ge 1}{f}_n{z}^n\mathrm{.}$ Here $\mathrm{\mid }F(z)\mathrm{\mid }\le \mathrm{\mid }z\mathrm{\mid }{\sum }_n{f}_n\le \mathrm{\mid }z\mathrm{\mid }<1$, so the denominator is legitimate. Since $u_n-p$ decays geometrically, [F9] gives $U(z)=p\mathrm{/}(1-z)+J(z)$ with $J$ holomorphic on $\mathrm{\mid }z\mathrm{\mid }<{\theta }^{-1}$. For real $t\uparrow 1$, $U(t)\to \mathrm{\infty }$; hence $F(t)\to 1$ and [F7] gives ${\sum }_n{f}_n=1$.

Let $T=\inf \{n\ge 0:Z_n\in \alpha \}$ and start the split chain from $\tilde{\pi }$. By [F6], $I_j=1_{\alpha }(Z_j)$ have mean $p$ and $\mathrm{Cov}(I_0,I_n)=p(u_n-p)$. Expanding the square of $N^{-1}{\sum }_{j=0}^{N-1}{I}_j-p$ and summing the geometric covariance bound from step 2.1 gives expectation at most $c_1\mathrm{/}N$ for some finite $c_1$. On $\{T\ge N\}$ this square equals $p^2$, so ${\mathbb{P}}_{\tilde{\pi }}(T\ge N)\le c_1\mathrm{/}(N{p}^2)\to 0$. In particular the stationary split chain hits the atom almost surely; no unproved Harris or skeleton-recurrence theorem is needed.

Near $z=1$ the expression $\widehat{F}(z)=1-\frac{1-z}{p+(1-z)J(z)}$ is holomorphic and agrees with $F$ on its overlap with $\mathrm{\mid }z\mathrm{\mid }<1$. Indeed the denominator has value $p>0$ at $1$ and remains nonzero on a sufficiently small disk; the quotient is complex differentiable there by the difference-quotient formula for the derivative of a reciprocal. By [F10], for $0<t<1$ and $k\ge 0$, $F^{(k)}(t)={\sum }_{n\ge k}{f}_n(n{)}_k{t}^{n-k}$, where $(n{)}_k=n!\mathrm{/}(n-k)!$. Letting $t\uparrow 1$ and using [F7] and derivative continuity from [F11] gives ${\sum }_{n\ge k}{f}_n(n{)}_k={\widehat{F}}^{(k)}(1)$. Choose a radius $\delta >0$ whose closed disk about $1$ lies strictly within this holomorphic neighborhood and a finite bound $B$ for $\mathrm{\mid }\widehat{F}\mathrm{\mid }$ on its circle. [F12] then bounds these nonnegative derivative values by $k!B{\delta }^{-k}$.

For bounded nonnegative $g$, decompose $g(Z_N)$ by the last atom visit at times $0,\dots ,N$. A last visit at time $N-k$ contributes $p{\mathbb{E}}_{\alpha }[g(Z_k);\tau >k]$, by stationarity and [F5]. The remaining paths have no atom visit through time $N$, and their contribution $R_N$ lies between zero and $\mathrm{\parallel }g{\mathrm{\parallel }}_{\mathrm{\infty }}{\mathbb{P}}_{\tilde{\pi }}(T>N)$. Thus $\tilde{\pi }(g)=p{\sum }_{k=0}^N{\mathbb{E}}_{\alpha }[g(Z_k);\tau >k]+R_N\mathrm{.}$ Let $N\to \mathrm{\infty }$ using step 3.2, then extend from bounded $g$ to all nonnegative measurable $g$ using [F7] and [F8]. This proves the stationary occupation identity $\tilde{\pi }(g)=p{\mathbb{E}}_{\alpha }{\sum }_{k=0}^{\tau -1}g(Z_k)$. At $k=0$ the atom state has law $\kappa$; subsequent transitions are independent of which atom state was chosen, so this argument does not require $g$ to be constant on $\alpha$.

For $0<h<\delta$, the finite binomial expansion for each integer $n$, followed by [F8], yields ${\mathbb{E}}_{\alpha }[(1+h{)}^{\tau }]={\sum }_{k\ge 0}\frac{h^k}{k!}{\sum }_{n\ge k}{f}_n(n{)}_k\le \frac{B}{1-h\mathrm{/}\delta }<\mathrm{\infty }\mathrm{.}$ Thus some $b>1$ has ${\mathbb{E}}_{\alpha }{b}^{\tau }<\mathrm{\infty }$. The return is almost surely finite by step 3.1. This argument establishes an exponential moment from positivity and analytic continuation; it does not assume in advance that the original series $F$ converges beyond the unit disk.

Fix $1<s<b$ and let $g(z)={\mathbb{E}}_z{s}^T$, allowing infinity. The bounded path functions $s^{T\wedge l}$ increase to $s^T$, so [F5] and [F7] give measurability and the Markov identity for this nonnegative function by truncation. On $\alpha$, $g=1$. For $k\ge 1$, on $\{\tau >k\}$, the residual time to $\alpha$ is $\tau -k$; the Markov identity and step 4.2 give $\tilde{\pi }(g)=p(1+{\mathbb{E}}_{\alpha }{\sum }_{k=1}^{\tau -1}{s}^{\tau -k})\le p{\mathbb{E}}_{\alpha }[\tau s^{\tau }]<\mathrm{\infty }\mathrm{.}$ The last moment is finite because ${\sup }_{j\ge 1}j(s\mathrm{/}b{)}^j<\mathrm{\infty }$ and step 5.1 bounds ${\mathbb{E}}_{\alpha }{b}^{\tau }$. For the coordinate chain initialized with the prescribed flag, its first $C$ visit ${\sigma }_C$ is no later than its first atom visit $T$. Consequently $H(x)={\mathbb{E}}_x{s}^{{\sigma }_C}$ is measurable by the same bounded-path approximation and $\pi (H)\le \tilde{\pi }(g)<\mathrm{\infty }$. This proves the conclusion, including the null-set qualification.