2.1. Quasi-stationary distributions

A QSD $\mu=(\mu(i)\,:\, i\in I)$ associated with $P_I$ is a probability measure $\mu$ on I such that $\text{for all } i\in I, \ \mathbb{P}_\mu({\mathcal{X}}_n=i \mid \tau_{\mathcal{E}}>n)=\mu(i)$. By writing this equality for $n=1$, we check that the row vector $\mu^\top$ is a strictly positive left eigenvector of $P_I$ properly normalized (the sum of its components is 1), with eigenvalue $\gamma=\mathbb{P}_\mu(\tau_{\mathcal{E}}>1)\in (0,1)$, that is,

(1) \begin{equation}\mu^\top P_I={\gamma\mu}^{\!\top}, \hbox{ with }\gamma=\sum_{i,j\in I}\mu(i)P(i,j)=\mathbb{P}_\mu(\tau_{\mathcal{E}}>1).\end{equation}

It follows that $\mathbb{P}_\mu(\tau_{\mathcal{E}}>k)=\gamma^k$ for all $k\ge 0$. So, if $\mu$ is a QSD then the survival time is Geometric($1-\gamma$) distributed; see Lemma 2.2 in [Reference Ferrari, Martínez and Picco8]. In the finite case there is a unique QSD (see [Reference Darroch and Seneta4]); it corresponds to the normalized left Perron–Frobenius eigenvector, and $\gamma$ is the associated eigenvalue. The properties of QSD depend on the killed trajectory ${\mathcal{X}}^{(K)}=({\mathcal{X}}_n\,:\, 0\le\break n<\tau_{\mathcal{E}})$. In the infinite case QSDs can exist or not (because the positive left eigenvectors can be of infinite mass), and when they exist there could be more than one (even a continuum of them). From now on we fix some QSD $\mu$ which, as just discussed, exists and is unique in the finite case and we assume its existence in the infinite case.

Let us give some independence properties between the time of killing and the absorption state. In Theorem 2.6 in [Reference Collet, Martínez and San Martín3] the independence relation $\mathbb{P}_\mu({\mathcal{X}}_n=i, \tau_{\mathcal{E}}>n)=\mu(i) \gamma^n$ for all $i\in I$ and $n\ge 0$ was stated. Let us prove that when starting from $\mu$, the pair $({\mathcal{X}}_{{\tau_{\mathcal{E}}}-1},{\mathcal{X}}_{\tau_{\mathcal{E}}})$, consisting of the last visited state before absorption and the absorption state, is independent of the random time $\tau_{\mathcal{E}}$. For $n\ge 1$, $i\in I$, and $\varepsilon\in {\mathcal{E}}$, we have

\begin{align*}\mathbb{P}_\mu({\mathcal{X}}_{{\tau_{\mathcal{E}}}-1}=i, {\mathcal{X}}_{\tau_{\mathcal{E}}}=\varepsilon, \tau_{\mathcal{E}}=n) & =\mathbb{P}_\mu({\mathcal{X}}_{n-1}=i, {\mathcal{X}}_n=\varepsilon, \tau_{\mathcal{E}}=n) \\& = \mathbb{P}_\mu({\mathcal{X}}_n=\varepsilon \mid {\mathcal{X}}_{n-1}=i)\mathbb{P}_\mu({\mathcal{X}}_{n-1}=i,\tau_{\mathcal{E}}>n-1) \\& = P(i,\varepsilon)\, \mu(i) \, \mathbb{P}_\mu(\tau_{\mathcal{E}}>n-1)= P(i,\varepsilon) \, \mu(i) \, \gamma^{n-1}.\end{align*}

Then, the independence relation follows. We can be more precise: we have

\begin{equation*}\mathbb{P}_\mu({\mathcal{X}}_{{\tau_{\mathcal{E}}}-1}=i, {\mathcal{X}}_{\tau_{\mathcal{E}}}=\varepsilon)=P(i,\varepsilon)\, \mu(i) \Bigg(\sum_{l\ge 1}\gamma^{l-1}\Bigg)=P(i,\varepsilon) \, \mu(i) \, (1-\gamma)^{-1}.\end{equation*}

Since $\mathbb{P}_\mu(\tau_{\mathcal{E}}=n)=(1-\gamma)\gamma^{n-1}$, the desired relation holds:

\begin{equation*}\mathbb{P}_\mu({\mathcal{X}}_{{\tau_{\mathcal{E}}}-1}=i, {\mathcal{X}}_{\tau_{\mathcal{E}}}=\varepsilon, \tau_{\mathcal{E}}=n)=\mathbb{P}_\mu({\mathcal{X}}_{{\tau_{\mathcal{E}}}-1}=i, {\mathcal{X}}_{\tau_{\mathcal{E}}}=\varepsilon)\mathbb{P}_\mu(\tau_{\mathcal{E}}=n).\end{equation*}

The above computations also show that the exit law of I when starting from $\mu$ satisfies

(2) \begin{equation} \begin{aligned} \mathbb{P}_\mu({\mathcal{X}}_{\tau_{\mathcal{E}}} & = \varepsilon)=(1-\gamma)^{-1}\Bigg(\sum_{i\in I}\mu(i)P(i,\varepsilon)\Bigg) , \\ \mathbb{P}_\mu({\mathcal{X}}_{\tau_{\mathcal{E}}} & = \varepsilon,\tau_{\mathcal{E}}=n)=\Bigg(\sum_{i\in I}\mu(i)P(i,\varepsilon)\Bigg) \gamma^{n-1}. \end{aligned}\end{equation}

These properties depend on the absorbed trajectory ${\mathcal{X}}^{(A)}=({\mathcal{X}}_n\,:\, 0\le n\le \tau_{\mathcal{E}})$.

4.1. Decoupling the stationary chain

Let $\tau_I=\inf\{n\ge 1\,:\, X_n\in I\}$ be the first return time of ${\mathbb{X}}$ to I. Now, consider the stochastic matrix $Q=(Q(i,j)\,:\, i,j\in I)$ given by $Q(i,j)=\mathbb{P}_i(X_{\tau_{\,I}}=j)$. By using that $\mathbb{P}_\varepsilon(X_{\tau_{\,I}}=j)=\pi(j \mid I)=\mu(j)$ for all $\varepsilon\in {\mathcal{E}}, j\in I$, we get

(6) \begin{align}\nonumber Q(i,j) & = \mathbb{E}_i \Big({\bf 1}_{\{X_{\tau_{\,I}}=j, {\tau_I}=1\}}\Big)+\mathbb{E}_i\Big({\bf 1}_{\{X_{\tau_{\,I}}=j,{\tau_I}>1\}}\Big)\nonumber\\ & = P(i,j)+ \mathbb{P}_i({\tau_I}>1)\mathbb{P}_i\left(X_{\tau_{\,I}}=j \mid X_{\tau_{\,I}-1}, {\tau_I}>1 \right)\nonumber\\ & = P(i,j)+P(i,{\mathcal{E}}) \mu(j).\end{align}

Let ${\mathbb{Y}}=(Y_n\,:\, n\in \mathbb{Z})$ be a Markov chain with transition matrix Q. It is straightforwardly checked that $\mu$ is a stationary measure for ${\mathbb{Y}}$.

The chain ${\mathbb{Y}}$ can be constructed as follows. Let $\Xi=\{\xi_l\,:\, l\in \mathbb{Z}\}$ be the ordered sequence given by

\begin{equation*}\{\xi_l\,:\, l\in \mathbb{Z}\}=\{n\in \mathbb{Z}\,:\, X_n\in I\} \hbox{ with }\xi_{l-1}<\xi_{l}\, , \; \xi_{-1}<0\le \xi_0.\end{equation*}

Then, $(\xi_{l}-\xi_{l-1}\,:\, l\in \mathbb{Z})$ is a renewal stationary sequence with interarrival times distributed as $\mathbb{P}(\xi_{l}-\xi_{l-1}=\bullet)=\mathbb{P}_\mu(\tau_I=\bullet)$, $l\neq 0$. By definition, $(X_{\xi_l}\,:\, l\in \mathbb{Z}\}$ is a stationary sequence distributed as ${\mathbb{Y}}=(Y_n\,:\, n\in \mathbb{Z})$, so $(X_{\xi_l}\,:\, l\in \mathbb{Z})$ is a copy of ${\mathbb{Y}}$.

The random sequence ${\bf{b}}=(b_l\,:\, l\in \mathbb{Z}, \, l\neq 0)$ defined by $b_l=1$ if $\xi_{l}-\xi_{l-1}=1$ and $b_l=0$ if $\xi_{l}-\xi_{l-1}>1$ is a collection of independent and identically distributed (i.i.d.) Bernoulli random variables, with

\begin{equation*}\mathbb{P}(b_l=1)=\pi(I) \hbox{ and }\mathbb{P}(b_l=0)=\sum_{i\in I} \mu(i) P(i,{\mathcal{E}})=\pi({\mathcal{E}}).\end{equation*}

(Recall that $\pi(I)+\pi({\mathcal{E}})=1$). When $X_0\in I$, we find $\tau_{\mathcal{E}} =\inf\{l\ge 1\,:\, b_l=0\}$.

Remark 2. Every irreducible matrix stochastic matrix Q with stationary distribution $\mu$ can be written as in (6). In fact, let $\chi=(\chi(i)\,:\, i\in I)$ be a non-null vector, $\chi\neq {\vec 0}$, that satisfies

\begin{equation*}\text{for all } i\in I: \quad 0\le \chi(i)<1 \hbox{ and } \chi(i)\le \min\{Q(i,j)\mu(j)^{-1}\,:\, j\in I\}.\end{equation*}

This can be achieved because $\mu$ is strictly positive. Define $P_I=Q-\chi \mu^\top$, so

(7) \begin{equation}\text{for all } i,j\in I: \quad P(i,j)=Q(i,j)-\chi(i) \mu(j).\end{equation}

To avoid the trivial situation we can assume that the vector $\chi$ also satisfies that for every $i\in I$ and for some (or for all) $j\in I$ for which $Q(i,j)>0$, we have $P(i,j)>0$. This allows us to take $\chi$ ensuring that $P_I$ is irreducible. From the construction, $P(i,j)\in [0,1)$ and, since $\chi\neq 0$, we get

\begin{equation*}\sum_{j\in I} P(i,j)=1-\chi(i)\in (0,1], \qquad \sum_{i, j\in I} (1-P(i,j))>0.\end{equation*}

Hence, P is strictly substochastic, it is not trivial, and when adding the cemetery $\partial$ we have $\chi(i)=P(i,\partial)$. So,

\begin{equation*}\mu^\top P=\mu^\top\big(Q-\chi\mu^\top\big)=\Bigg(1-\sum_{i\in I} \mu(i)P(i,\partial)\Bigg)\mu^\top ; \end{equation*}

that is, $\mu^\top$ is the Perron–Frobenius left eigenvector of P with eigenvalue $\pi(I)=\sum_{i, j\in I} \mu(i) P(i,j)$ (see (1) and (4)). From (7) it follows that $Q(i,j)=P(i,j)+P(i,\partial)\mu(j)$, so (6) is satisfied.

The restriction of $P=P^\pi$ to the absorption states ${\mathcal{E}}$ satisfies

\begin{equation*}\text{for all } \varepsilon,\delta\in {\mathcal{E}}\,:\, \quad P(\varepsilon,\delta)=\pi(\delta),\hbox{ so }\mathbb{P}(X_1= \delta \mid X_0=\varepsilon, X_1\in {\mathcal{E}})=\pi(\delta \mid {\mathcal{E}}).\end{equation*}

The transition law to an absorbing point after being in $X_{t-1}=i\in I$ is given by

\begin{equation*}\text{for all } \delta\in {\mathcal{E}}\,:\, \quad\mathbb{P}(X_{t}=\delta \mid X_{t}\in {\mathcal{E}}, X_{t-1}=i)=P(i,\delta)/P(i,{\mathcal{E}}).\end{equation*}

So, if $X_{-1}\in I$ and $X_0\in {\mathcal{E}}$, the total sojourn time at ${\mathcal{E}}$ is $\tau_I$, and it is distributed as a $\hbox{Geometric}(\pi(I))$. Then, immediately after the entrance to ${\mathcal{E}}$ the chain ${\mathbb{X}}$ makes a walk on ${\mathcal{E}}$ of length $\tau_I-1$ (a quantity that could vanish). To describe it, take ${\mathbb{G}}=(G_n\,:\, n\in \mathbb{Z})$ a Bernoulli chain with probability vector $\pi(\bullet \mid {\mathcal{E}})$. Let us consider the finite random sequence $V=(G_l\,:\, 1\le l<\tau_I)$ (with V empty if $\tau_I=1$), which is distributed as

(8) \begin{align}\mathbb{P}(V=\emptyset) & = \mathbb{P}(\tau_I=1)=\pi(I) ,\\ \mathbb{P}(V=(\delta_1, \ldots, \delta_{k-1})) & = \mathbb{P}(G_1 = \delta_1, \ldots, G_{k-1} = \delta_{k-1},\tau_I = k)\nonumber\\ & = \Bigg(\prod_{l=1}^{k-1}\pi(\delta_l \mid {\mathcal{E}})\Bigg)\pi({\mathcal{E}})^{k-1}\pi(I)=\Bigg(\prod_{l=1}^{k-1}\pi(\delta_l)\Bigg)\pi(I)\nonumber\\ &\quad \hbox{ for } k\ge 2, \ (\delta_1, \ldots, \delta_{k-1})\in {\mathcal{E}}^{k-1}.\nonumber\end{align}

Notice that the last equality also holds when $\tau_I=k=1$ because an empty product satisfies $\prod_{l=1}^{k-1}=1$. We have $(X_t, 1\le t < \tau_I \mid X_{-1}\in I, X_0\in {\mathcal{E}})\sim V$, and V is called a walk on ${\mathcal{E}}$. Note that $\tau_I-1 \mid \tau_I>1$ is equally distributed as $\tau_I$. The exit law from ${\mathcal{E}}$ is $\mathbb{P}(X_{\tau_{\,I}}\in \bullet)\sim \mu$. In fact, for all $\delta\in {\mathcal{E}}$,

\begin{align*}\mathbb{P}_\delta(X_{\tau_{\,I}}=i) & = \sum_{\varepsilon\in {\mathcal{E}}}\mathbb{P}_\delta(X_{\tau_{\,I}}=i, X_{\tau_{\,I}-1}=\varepsilon)\\ & = \sum_{\varepsilon\in {\mathcal{E}}} \mathbb{P}_\delta(X_{\tau_{\,I}-1}=\varepsilon)\mathbb{P}_\varepsilon(X_1=i \mid X_1\in I)\\ & = \pi(i \mid I)=\mu(i).\end{align*}

Notice that

(9) \begin{equation}\text{for all } \delta\in {\mathcal{E}}, \ \mathbb{E}_\delta(\tau_I)=\pi(I)^{-1} . \end{equation}

We consider a sequence of i.i.d. random variables ${\bf{{\mathcal{T}}}}=({\mathcal{T}}_n\,:\, n\in \mathbb{Z})$ which are $\hbox{Geometric}(\pi(I))-1$ distributed, that is, $\mathbb{P}({\mathcal{T}}_n=l)=\pi(I) (1-\pi(I))^{l}$ for $l\ge 0$. The construction of i.i.d. walks on ${\mathcal{E}}$ is made as follows. One takes an increasing sequence of times $(t_n\,:\, n\in \mathbb{Z})$ with $t_{n+1}-t_n={\mathcal{T}}_n$ and such that $t_n\to \infty$ if $n\to \infty$ and $t_n\to -\infty$ if $n\to -\infty$. We define $V^n=(G_{t_n}, \ldots, G_{t_{n+1}-1})=(V^n_1, \ldots, V^n_{{\mathcal{T}}_n})$. So, ${\mathbb{V}}=\left(V^n\,:\, n\ge \mathbb{Z}\right)$ is an i.i.d. sequence of walks on ${\mathcal{E}}$. The walk $V^n$ is empty when ${\mathcal{T}}_n=0$.

When ${\mathcal{E}}=\{\partial\}$ is a singleton, we have $\pi(\partial \mid {\mathcal{E}})=1$, ${\mathbb{G}}=(G_n\,:\, n\in \mathbb{Z})$ is the orbit with the unique symbol $G_n=\partial$ for all n, and the random sequence $V=(G_l\,:\, 1\le l<\tau_I)$ has the symbol $\partial$ repeated $|\tau_I|-1$ times.

4.2. Retrieving the stationary chain

Let ${\mathbb{Y}}=(Y_n\,:\, n\in \mathbb{Z})$ be a stationary Markov chain with transition matrix Q. Our purpose is to construct a copy of ${\mathbb{X}}$ from ${\mathbb{Y}}$ by adding a series of random operations.

Let $\mathbb{P}$ be a probability measure governing the law of ${\mathbb{Y}}$ when it starts from the stationary distribution $\mu$, the sequences ${\mathbb{G}}$, ${\mathcal{T}}$ and so ${\mathbb{V}}$, and also the random element ${\mathbb{B}}^{I,I}$ and ${\mathbb{D}}^I$ defined below.

Let ${\mathbb{B}}^{I,I}=\big((B^{i,j}_l\,:\, l\in \mathbb{Z});\ i,j\in I \big)$ be an independent array of Bernoulli random variables such that $B^{i,j}_l\sim B^{i,j}$ for $l\in \mathbb{Z}$, where

(10) \begin{equation}\begin{aligned} \mathbb{P}(B^{i,j}=1) & = \theta_{i,j}, \qquad \mathbb{P}(B^{i,j}=0)={\overline{\theta}}_{i,j}=1-\theta_{i,j} , \\ \theta_{i,j} & = \frac{P(i,j)}{P(i,j)+P(i,{\mathcal{E}})\mu(j)}=\frac{P(i,j)}{Q(i,j)}.\end{aligned}\end{equation}

Let $\tau_\partial=\inf\{l\ge 1\,:\, B^{Y_{l-1},Y_l}_l=0\}$. For $k\ge 1$, $i_0, \ldots, i_{k-1}\in I$ we have

\begin{multline*}\mathbb{P}(Y_0=i_0, Y_1=i_1, \ldots, Y_{k-1}=i_{k-1}, \tau_\partial=k) \\ =\mu(i_0)\Bigg(\prod_{l=1}^{k-1} P(i_{l-1},i_l)\Bigg)\Bigg(\sum_{j\in I}P(i_{k-1},{\mathcal{E}})\mu_j\Bigg)=\mu(i_0)\Bigg(\prod_{l=1}^{k-1} P(i_{l-1},i_l)\Bigg)P(i_{k-1},{\mathcal{E}}).\end{multline*}

Hence, the sequence $Y^{(K)}=(Y_l\,:\, 0\le l<\tau_\partial)$ is distributed as a killed chain ${\mathcal{X}}^{(K)}$ starting from $\mu$.

Now take an independent array ${\mathbb{D}}^I=\left((D^{i}_l\,:\, l\in \mathbb{Z});\, i\in I \right)$ of random variables taking values in ${\mathcal{E}}$ and with law

(11) \begin{equation}\text{for all } \delta \in {\mathcal{E}}\,:\,\quad \mathbb{P}(D^{i}_l=\delta)=P(i,\delta)/P(i,{\mathcal{E}}).\end{equation}

For $k\ge 1$, $i_0, \ldots, i_{k-1}\in I$, $\delta\in {\mathcal{E}}$ we set

\begin{align*}\mathbb{P}(Y_0=i_0, & Y_1=i_1, \ldots, Y_{k-1}=i_{k-1}, D^{i_{k-1}}_k=\delta,\tau_\partial=k)\\[4pt] & =\mu(i_0)\Bigg(\prod_{l=1}^{k-1} P(i_{l-1},i_l)\Bigg)P(i_{k-1},{\mathcal{E}})\big(P(i_{k-1},\delta)/P(i_{k-1},{\mathcal{E}})\big) \\[4pt] & = \mu(i_0)\Bigg(\prod_{l=1}^{k-1} P(i_{l-1},i_l)\Bigg)P(i_{k-1},\delta).\end{align*}

Then, the sequence $Y^{(A)}=(Y_0, \ldots, Y_{\tau_\partial-1},D^{Y_{\tau_{\partial}-1}}_{\tau_\partial})$ is distributed as an absorbed chain ${\mathcal{X}}^{(A)}$ starting from $\mu$.

Let us construct a chain ${\mathbb{S}}^{\text{s}}=({\mathbb{S}}^{\text{s}}_t\,:\, t\in \mathbb{Z})$ from ${\mathbb{Y}}$, ${\mathbb{B}}^{I,I}$, ${\mathbb{D}}^Y$, ${\mathbb{G}}$, and ${\mathcal{T}}$ (and so also ${\mathbb{V}}$), having the same distribution as ${\mathbb{X}}$. First, define a random sequence ${\mathbb{S}}=({\mathbb{S}}_t\,:\, t\in \mathbb{Z})$ as follows. We set $T_0=0$, ${\mathbb{S}}_0=Y_0$ (so ${\mathbb{S}}_0=Y_0\in I$ is distributed as $\mu$), and:

1. I In a sequential way on $n\ge 0$ we make the following construction. Assume at step n that $T_n$ has been defined; then, put ${\mathbb{S}}_{T_n}=Y_n$ and go to step $n+1$.

2. Ia If $B^{Y_n,Y_{n+1}}_{n+1}=1$ put $T_{n+1}=T_n+1$, ${\mathbb{S}}_{T_{n+1}}=Y_{n+1}$, and go to step $n+2$.

3. Ib If $B^{Y_n,Y_{n+1}}_{n+1}=0$ put $T_{n+1}=T_n+{\mathcal{T}}_n+2$, define ${\mathbb{S}}_{T_{n}+1}=D^{Y_n}_{n+1}$, ${\mathbb{S}}_{T_{n}+1+l}=V^n_l$ for $1\le\break l<{\mathcal{T}}_n$ (it is empty when ${\mathcal{T}}_n=0$), and ${\mathbb{S}}_{T_{n+1}}=Y_{n+1}$. Then continue with step $n+2$.

4. II Similarly, in a sequential way on $n<0$ we make the following construction for step n:

5. IIa If $B^{Y_{n},Y_{n+1}}_{n+1}=1$ put $T_{n}=T_{n+1}-1$, ${\mathbb{S}}_{T_{n}}=Y_n$, and continue with step $n-1$.

6. IIb If $B^{Y_n,Y_{n+1}}_{n+1}=0$ put $T_{n}=T_{n+1}-({\mathcal{T}}_n+2)$, ${\mathbb{S}}_{T_n+1}=D^{Y_n}_{n+1}$, ${\mathbb{S}}_{T_{n}+1+l}=V^n_l$ for $1\le l<{{\mathcal{T}}_n}$, and ${\mathbb{S}}_{T_n}=Y_{n}$. Then, continue with step $n-1$.

Let ${\mathbb{S}}=({\mathbb{S}}_t\,:\, t\in \mathbb{Z})$ be the random sequence resulting from this construction, and let $\mathbb{T}=(T_n\,:\, n \in \mathbb{Z})$, recalling that $T_0=0$. By an abuse of notation we also denote by $\mathbb{T}=\break \{T_n\,:\, n\in \mathbb{Z}\}$ the set of these values. By definition, $\mathbb{T}=\{t\in \mathbb{Z}\,:\, {\mathbb{S}}_t\in I\}$ is the set of random points where ${\mathbb{S}}$ is in I. In Proposition 2 we will prove that $({\mathbb{S}}, \mathbb{T})$ is a regenerative process (see [Reference Asmussen2], pp. 169–170), that is, for all $l\ge 0$ the process $({\mathbb{S}}_{\bullet + T_{l}}\,:\, \bullet \ge 0;\ T_{n+1}-T_n,\ n\ge l)$ has the same distribution as $({\mathbb{S}}_{\bullet}\,:\, \bullet \ge 0;\ T_{n+1}-T_n,\ n\ge 0)$ and it is independent of $(T_n\,:\, n \le l)$.

The cycles of this regenerative process, $({\mathbb{S}}_{T_n}, \ldots, {\mathbb{S}}_{T_{n+1}-1})$, $n\in \mathbb{Z}$, are i.i.d., and so all of them have the same distribution as $({\mathbb{S}}_0, \ldots, {\mathbb{S}}_{T_1-1})$. By shifting the process ${\mathbb{S}}^{\text{s}}=({\mathbb{S}}^{\text{s}}_t\,:\, t\in \mathbb{Z})$ by a random time chosen uniformly in $\{0, \ldots, T_1-1\}$ and conditionally independent of the rest of the process, we get a stationary process ${\mathbb{S}}^{\text{s}}=({\mathbb{S}}^{\text{s}}_t\,:\, t\in \mathbb{Z})$ (see Theorem 6.4 in [Reference Asmussen2]). So, ${\mathbb{S}}^{\text{s}}_0$ takes values in $I\cup {\mathcal{E}}$ and, from the next proposition, it is distributed as $\pi$ (different than ${\mathbb{S}}_0$, which takes values in I and is distributed as $\mu$).

The proof can be found in the Appendix.

5.1. The killed chain

The stationary representation of the killed chain will be a stationary Markov chain on the set of states $I^2\times \{0,1\}=\{(i,j,a)\,:\, (i,j)\in I^2,\ a\in \{0,1\}\}$. Prior to defining the transition matrix we introduce the function

\begin{equation*}\varphi(i,j,a)=(\theta_{i,j}{\bf 1}(a=1)+{\overline{\theta}}_{i,j}{\bf 1}(a=0)),\qquad (i,j,a)\in I^2\times \{0,1\}.\end{equation*}

The transition matrix ${\mathcal{K}}=\left({\mathcal{K}}((i,j,a),(l,k,b))\,:\, (i,j,a), (l,k,b)\in I^2\times \{0,1\}\right)$ is defined by

\begin{equation*}{\mathcal{K}}((i,j,a), (l,k,b)) = \begin{cases}\!\!\!\!\! &0 \hbox{ if } l \neq j , \\[5pt]\!\!\!\!\! & Q(l,k)\varphi(l,k,b) = P(l,k){\bf 1}(b = 1)+ P(l,{\mathcal{E}})\mu(k){\bf 1}(b = 0)\hbox{ if } l = j.\end{cases}\end{equation*}

It can be straightforwardly checked that this is a stochastic matrix: we claim its stationary distribution $\zeta=(\zeta(i,j,a)\,:\, (i,j,a)\in I^2\times \{0,1\})$ is given by

(13) \begin{align}\zeta(i,j,a)& = \mu(i)Q(i,j)\varphi(i,j,a)\nonumber\\ & = \mu(i)P(i,j){\bf 1}(a=1) + \mu(i)P(i,{\mathcal{E}})\mu(j){\bf 1}(a=0).\end{align}

By using $\sum_{i\in I} \mu(i) Q(i,l)=\mu(l)$ we get the desired property,

\begin{align*}\sum_{(i,j,a)\in I^2\times \{0,1\}}\!\!\!\!\zeta(i,j,a) {\mathcal{K}}((i,j,a), (l,k,b)) & =\Bigg(\sum_{i\in I} \mu(i) Q(i,l)\Bigg) Q(l,k)\varphi(l,k,b) \\& = \mu(l)Q(l,k)\varphi(l,k,b)=\zeta(l,k,b),\end{align*}

so the claim follows.

The killed Markov chain presented in its stationary form is denoted

\begin{equation*}{\mathbb{Y}}^{({\mathcal{K}})}=((Y_{n-1},Y_n,B_n)\,:\, n\in \mathbb{Z});\end{equation*}

it takes values in $I^2\times \{0,1\}$ and has transition matrix ${\mathcal{K}}$. The component $B_n$ is called the label at n. By hypothesis, $P_I$ is irreducible so also ${\mathcal{K}}$ is an irreducible matrix. Then, the Markov shift ${\mathbb{Y}}^{({\mathcal{K}})}$ is ergodic (see Proposition 8.12 in [Reference Denker, Grillenberger and Sigmund5]).

It is straightforward to check that the mapping

(14) \begin{equation}\Upsilon^{({\mathcal{K}})}\,:\, {\mathbb{Y}}^{({\mathcal{K}})}\to {\mathbb{Y}},\ ((Y_{n-1},Y_n,B_n)\,:\, n\in \mathbb{Z})\to (Y_n\,:\, n\in \mathbb{Z}) \end{equation}

is a factor, which means that it is measure preserving and commutes with the shift on $\mathbb{Z}$.

Remark 3. We show that ${\mathbb{Y}}^{({\mathcal{K}})}$ models the killed Markov chain. Let ${\mathcal{N}}=\{n\in \mathbb{Z}\,:\, B_n=0\}$ and write it as ${\mathcal{N}}=\{n_l\,:\, l\in \mathbb{Z}\}$, where $n_l$ is increasing with l, and $n_{-1}<0\le n_0$. Note that $\mathbb{P}(0\in {\mathcal{N}})=\pi({\mathcal{E}})$.

The orbit $((Y_{n-1},Y_n,B_n)\,:\, n\in \mathbb{Z})$ in ${\mathbb{Y}}^{({\mathcal{K}})}$ is denoted in the simpler form $((Y_{n-1},B_n)\,:\break n\in \mathbb{Z})$ and we can divide it into the disjoint connected pieces $(Y,B)^{(K)}_{l}=((Y_{n_l},1), \ldots,$ $(Y_{n_{l+1}-2},1), (Y_{n_{l+1}-1}, 0))$, $l\in \mathbb{Z}$. The component $Y_{n_l}$ is distributed with law $\mu$ for all l, and one can identify $(Y,B)^{(K)}_{l}$ with $Y^{(K)}_{l}=(Y_{n_l}, \ldots, Y_{n_{l+1}-1})$, a piece of the orbit $Y=(Y_n\,:\, n\in \mathbb{Z})$ starting from $\mu$ at $n_l$ and killed at $n_{l+1}-1$. We get that $Y^{(K)}_{l}\sim {\mathcal{X}}^{(K)}$ for all $l \neq -1$ when ${\mathcal{X}}^{(K)}$ starts from $\mu$. In fact, for $s\ge 0$, $i_0, \ldots, i_s\in I$, we have

\begin{multline*}\mathbb{P}\big({\mathcal{X}}^{(K)} = (i_0, \ldots, i_s)\big) = \mu(i_0)\prod_{r=0}^{s-1} P(i_r,i_{r+1})P(i_s,\partial) \\ = \mu(i_0)\Bigg(\prod_{r=0}^{s-1} Q(i_r,i_{r+1})\theta_{i_r,i_{r+1}}\Bigg)\Bigg(\sum_{m\in I}Q(i_s,m){\overline{\theta}}_{i_s,m}\Bigg)=\mathbb{P}\Big(Y_{l}^{(K)}=(i_0, \ldots, i_s)\Big).\end{multline*}

Let $s\ge 0$, $(i_0, \ldots, i_s)\in I^{s+1}$. For almost all the orbits $Y\in {\mathbb{Y}}^{({\mathcal{K}})}$ we have

\begin{equation*}\mathbb{P}(n_0=0, Y_0^{(K)}=(i_0, \ldots, i_s))=\pi({\mathcal{E}})\mu(i_0)\prod_{r=0}^{s-1} P(i_r,i_{r+1}) P(i_s,\partial)>0.\end{equation*}

Since the killed trajectories are finite, the class of killed trajectories is countable. From the ergodic theorem, and since ${\mathbb{Y}}^{({\mathcal{K}})}$ is ergodic, it follows that $\mathbb{P}$-a.e. (almost everywhere) the orbits of ${\mathbb{Y}}^{({\mathcal{K}})}$ contain all the killed trajectories of the chain.

The entropy of the killed chain satisfies

\begin{align*} h\big({\mathbb{Y}}^{({\mathcal{K}})}\big) & = -\!\!\!\sum_{(i,j)\in I^2} \!\!\! \mu(i) Q(i,j) \sum_{k\in I}Q(j,k)\left(\theta_{j,k} \log (Q(j,k)\theta_{j,k})+ {\overline{\theta}}_{j,k}\log (Q(j,k){\overline{\theta}}_{j,k})\right) \\& = - \!\!\sum_{(j,k)\in I^2} \!\!\! \mu(j) Q(j,k) \log Q(j,k)+\sum_{(j,k)\in I^2}\mu(j)Q(j,k) H(B^{j,k}) ,\end{align*}

where $H(B^{j,k})=-\left(\theta_{j,k} \log \theta_{j,k}+{\overline{\theta}}_{j,k} \log {\overline{\theta}}_{j,k}\right)$ is the entropy of the Bernoulli random variable $B^{j,k}$. Hence,

(15) \begin{equation}h\big({\mathbb{Y}}^{({\mathcal{K}})}\big)=h({\mathbb{Y}})+\Delta(B) , \quad \hbox{with }\Delta(B)=\sum_{(j,k)\in I^2}\mu(j)Q(j,k) H(B^{j,k}).\end{equation}

The quantity $\Delta(B)=h({\mathbb{Y}}^{({\mathcal{K}})})-h({\mathbb{Y}})$ is the conditional entropy of ${\mathbb{Y}}^{({\mathcal{K}})}$ given the factor ${\mathbb{Y}}$ (see Lemma 2 and Definition 3 in [Reference Downarowicz and Serafin6]). To be more precise, given an orbit $Y=(Y_n\,:\, n\in \mathbb{Z})$ of ${\mathbb{Y}}$, the fiber given by (14) satisfies $(\Upsilon^{({\mathcal{K}})})^{-1}\{Y\}=\{(B^{Y_{n-1},Y_{n}}_n\,:\, n\in \mathbb{Z})\in\{0,1\}^\mathbb{Z}\}$, and it is distributed as a sequence of independent Bernoulli variables given by (10); we denote it by $(\bf P)_{Y}$. We have

(16) \begin{equation}H_{(\bf P)_{Y}}\big(B_1^{Y_0,Y_{1}}\big)=-\big(\theta_{Y_0,Y_1} \log \theta_{Y_0,Y_1}+ {\overline{\theta}}_{Y_0,Y_1}\log{\overline{\theta}}_{Y_0,Y_1}\big).\end{equation}

Let us summarize the results on the entropy of ${\mathbb{Y}}^{({\mathcal{K}})}$.

Proposition 3. The entropy of the stationary representation ${\mathbb{Y}}^{({\mathcal{K}})}$ of the killed chain satisfies

(17) \begin{align}h\big({\mathbb{Y}}^{({\mathcal{K}})}\big) & = h({\mathbb{Y}})+\Delta(B),\\ \nonumber \Delta(B) & = \int H_{(\bf P)_{Y}}\big(B_1^{Y_0,Y_{1}}\big) \, {\text{d}} \mathbb{P}(Y)=\sum_{(i,j)\in I^2}\mu(i)Q(i,j) H(B^{i,j}) , \end{align}

and

(18) \begin{align} h\big({\mathbb{Y}}^{({\mathcal{K}})}\big) & = -\sum_{i\in I} \mu(i) P(i,{\mathcal{E}})\log P(i,{\mathcal{E}})-(1-\pi(I))\sum_{j\in I} \mu(j)\log \mu(j)\\ \nonumber& \quad -\sum_{i,j\in I} \mu(i) P(i,j)\log P(i,j). \end{align}

Proof. From (15) and (16), and by using the Markov property, we retrieve the Abramov–Rokhlin formula (see [Reference Abramov and Rokhlin1,Reference Downarowicz and Serafin6]),

\begin{equation*}\Delta(B)=h\big({\mathbb{Y}}^{({\mathcal{K}})}\big)-h({\mathbb{Y}})=\int H_{(\bf P)_{Y}}\big(B_1^{Y_0,Y_1}\big) \, {\text{d}}\mathbb{P}(Y)=\sum_{i,j\in I} \mu(i) Q(i,j)H(B^{i,j}).\end{equation*}

This gives (17). The only thing left to prove is (18). By using

\begin{equation*}\sum_{i\in {\mathcal{E}}}\mu(i)P(i,{\mathcal{E}}) = 1\!-\!\pi(I), \quad \sum_{j\in I}P(i,j) = 1\!-\!P(i,{\mathcal{E}}), \quad \sum_{i\in I}\mu(i) P(i,j) = \pi(I) \mu(j) \end{equation*}

and (12), we get

\begin{align*}\Delta(B) & =-\sum_{i,j\in I} \mu(i) \left(P(i,{\mathcal{E}})\mu(j)\log(P(i,{\mathcal{E}})\mu(j))+P(i,j)\log P(i,j)\right) \\& \quad + \sum_{i,j\in I} \mu(i) Q(i,j)\log Q(i,j) \\& = -\sum_{i\in I} \mu(i) P(i,{\mathcal{E}}) \log P(i,{\mathcal{E}})-(1-\pi(I)) \sum_{j\in I} \mu(j)\log \mu(j) \\& \quad +\sum_{i,j\in I} \mu(i) P(i,j)\log P(i,j)-h({\mathbb{Y}}).\end{align*}

This shows (18).

Remark 4. From (13) we get that there are, in mean,

\begin{equation*}\sum_{i,j\in I}\zeta(i,j,1) = \sum_{i,j\in I}\mu(i)P(i,j)=\pi(I)\end{equation*}

sites in $\mathbb{Z}$ where ${\mathbb{Y}}^{({\mathcal{K}})}$ has made a transition with label 1, and a mean

\begin{equation*}\sum_{i,j\in I}\zeta(i,j,0)=\sum_{i\in I}\mu(i)P(i,{\mathcal{E}})=\pi({\mathcal{E}})\end{equation*}

of sites in $\mathbb{Z}$ where ${\mathbb{Y}}^{({\mathcal{K}})}$ has made a transition with label 0, and so resurrects with distribution $\mu$.

5.2. The absorbed chain

Let us construct a stationary representation of the absorbed chain in a similar way as we did for the killed chain. Define ${\mathcal{E}}^*={\mathcal{E}}\cup \{o\}$ with $o\not\in {\mathcal{E}}\cup I$. The absorbed chain will take values on the set of states $I^2\times {\mathcal{E}}^*=\{(i,j,\delta)\,:\, i\in I,\ j\in I,\ \delta\in {\mathcal{E}}^*\}$. The matrix ${\mathcal{A}}=({\mathcal{A}}((i,j,\delta), (l,k,\varepsilon))\,:\,(i,j,\delta), (l,k,\varepsilon)\in I^2\times {\mathcal{E}})$ defined by

\begin{equation*}{\mathcal{A}}((i,j,\delta), (l,k,\varepsilon))=\begin{cases} & \!\!\!\!\! 0 \hbox{ if } l\neq j , \\[3pt] &\!\!\!\!\!Q(l,k)\theta_{l,k}=P(l,k) \hbox{ if } l=j,\ \varepsilon=o , \\[3pt] &\!\!\!\!\!Q(l,k) {\overline{\theta}}_{l,k} P(l,\varepsilon)/P(l,{\mathcal{E}}) = P(l,\varepsilon) \mu(k)\hbox{ if } l = j,\ \varepsilon\in {\mathcal{E}} \end{cases}\end{equation*}

is a stochastic matrix whose stationary distribution $\eta=(\eta(i,j,\delta)\,:\, (i,j,r)\in I^2\times {\mathcal{E}}^*)$ is given by

\begin{align*}\eta(i,j,\delta) & = \mu(i)Q(i,j)\left(\theta_{i,j} {\bf 1}(\delta=o)+{\overline{\theta}}_{i,j} P(i,\delta)/P(i,{\mathcal{E}}) {\bf 1}(\delta\in {\mathcal{E}})\right)\\ & = \mu(i) P(i,j){\bf 1}(\delta=o)+\mu(i)P(i,\delta)\mu(j){\bf 1}(\delta\in {\mathcal{E}}).\end{align*}

In fact, since $\sum_{\delta\in {\mathcal{E}}^*}(\theta_{i,l} {\bf 1}(\delta=o)+{\overline{\theta}}_{i,l} P(i,\delta)/P(i,{\mathcal{E}}) {\bf 1}(\delta\in {\mathcal{E}}))=1$, we get the stationarity property

\begin{eqnarray*}&{}&\sum_{(i,j,\delta)\in I^2\times {\mathcal{E}}^*} \eta(i,j,\delta){\mathcal{A}}((i,j,\delta), (l,k,\varepsilon))\\&{}& \, =\Bigg(\sum_{i\in I} \mu(i) Q(i,l)\Bigg)Q(l,k)\big(\theta_{l,k}{\bf 1}(\varepsilon=o)+{\overline{\theta}}_{l,k} P(l,\varepsilon)/P(l,{\mathcal{E}}){\bf 1}(\varepsilon\in {\mathcal{E}})\big)\\&{}& \, =\mu(l) Q(l,k) \big(\theta_{l,k}{\bf 1}(\varepsilon=o)+{\overline{\theta}}_{l,k}P(l,\varepsilon)/P(l,{\mathcal{E}}) {\bf 1}(\varepsilon\in {\mathcal{E}})\big)=\eta(l,k,\varepsilon).\end{eqnarray*}

We denote by ${\mathbb{Y}}^{({\mathcal{A}})}=((Y_{n-1},Y_n, D^*_n)\,:\, n\in \mathbb{Z})$ the absorbed Markov chain presented in its stationary form, and taking values in $I^2\times {\mathcal{E}}^*$ with transition matrix ${\mathcal{A}}$. Since ${\mathcal{A}}$ is irreducible, the Markov shift ${\mathbb{Y}}^{({\mathcal{A}})}$ is ergodic.

It is straightforward to check that the mapping

(19) \begin{eqnarray}\nonumber&{}&\Upsilon^{({\mathcal{A}})}\,:\, {\mathbb{Y}}^{({\mathcal{A}})}\to {\mathbb{Y}}^{({\mathcal{K}})},((Y_{n-1},Y_n,D^*_n)\,:\, n\in \mathbb{Z})\to (Y_{n-1},Y_n,B_n)\,:\, n\in \mathbb{Z})\\&{}& \hbox{ with } B_n= {\bf 1}(D^*_n=o) \end{eqnarray}

is a factor between ${\mathbb{Y}}^{({\mathcal{A}})}$ and ${\mathbb{Y}}^{({\mathcal{K}})}$.

Remark 5. Let us see that the stationary chain ${\mathbb{Y}}^{({\mathcal{A}})}$ models the absorbed Markov chain. First, denote ${\mathcal{N}}^*=\{n\in \mathbb{Z}\,:\, D^*_n\in {\mathcal{E}}\}$ and write it by ${\mathcal{N}}^*=\{n_l\,:\, l\in \mathbb{Z}\}$ with $n_l$ increasing in l and $n_{-1}<0\le n_0$. We have $\mathbb{P}(0\in {\mathcal{N}}^*)=\pi({\mathcal{E}})$. Similarly to Remark 3, an orbit $((Y_{n-1},Y_n,D^*_n)\,:\, n\in \mathbb{Z})$ of ${\mathbb{Y}}^{({\mathcal{A}})}$ is denoted in the form $((Y_{n-1},D^*_n)\,:\, n\in \mathbb{Z})$ and is partitioned into the disjoint connected pieces

\begin{equation*}(Y,D^*)^{(A)}_{l}=((Y_{n_l},o), \ldots, (Y_{n_{l+1}-2},o),(Y_{n_{l+1}-1}, D^*_{n_{l+1}})) \hbox{ with } l\in \mathbb{Z} .\end{equation*}

The component $Y_{n_{l}}$ is distributed with law $\mu$ for all l, and we can identify $(Y,D^*)^{(A)}_l$ with $Y^{(A)}_l=(Y_{n_l}, \ldots, Y_{n_{l+1}-1}, D^*_{n_{l+1}})$ starting from $\mu$. Since the events $\{n\in \mathbb{Z}\,:\, D^*_n\in {\mathcal{E}}\}$ have the same distribution as $\{n\in \mathbb{Z}\,:\, B_n=0\}$ in ${\mathbb{Y}}^{({\mathcal{K}})}$, it can be checked that, for all $l\neq -1$, $Y^{(A)}_{l}\sim {\mathcal{X}}^{(A)}$, where ${\mathcal{X}}^{(A)}$ starts form $\mu$. In fact, for $s\ge 0$, $i_0, \ldots, i_s\in I$, $\varepsilon\in {\mathcal{E}}$, we have

\begin{equation*}\mathbb{P}\big({\mathcal{X}}^{(A)}=(i_0, \ldots, i_s, \varepsilon)\big) =\mu(i_0)\left(\prod_{r=0}^{s-1}P(i_r,i_{r+1})\right)P(i_s,\varepsilon) = \mathbb{P}\big(Y_{l}^{(A)}=(i_0, \ldots, i_s,\varepsilon)\big).\end{equation*}

Let $s\ge 0$, $(i_0, \ldots, i_s)\in I^{s+1}$, $\varepsilon\in {\mathcal{E}}$. For almost all the orbits $Y\in {\mathbb{Y}}^{({\mathcal{A}})}$ we have

\begin{equation*}\mathbb{P}\big(n_0=0, Y_{0}^{(A)}=(i_0, \ldots, i_s,\varepsilon)\big)=\pi({\mathcal{E}})\mu(i_0)\prod_{r=0}^{s-1} P(i_r,i_{r+1}) P(i_s,\varepsilon)>0.\end{equation*}

Since the absorbed trajectories are finite, the class of absorbed trajectories is countable. Then, since ${\mathbb{Y}}^{({\mathcal{A}})}$ is ergodic, we get from the ergodic theorem that, $\mathbb{P}$-a.e., the orbits of ${\mathbb{Y}}^{({\mathcal{A}})}$ contain all the absorbed trajectories of the chain.

The entropy of the absorbed chain satisfies

\begin{align*} h\big({\mathbb{Y}}^{({\mathcal{A}})}\big) & = - \!\!\!\! \sum_{(i,j)\in I^2}\!\!\!\! \mu(i) Q(i,j) \sum_{k\in I} \!Q(j,k) \theta_{j,k} \log (Q(j,k)\theta_{j,k}) \\ & \quad - \!\!\!\! \sum_{(i,j)\in I^2}\!\!\!\! \mu(i) Q(i,j) \!\!\!\sum_{k\in I,\varepsilon\in {\mathcal{E}}}\!\!\!\!Q(j,k) {\overline{\theta}}_{j,k} P(j,\varepsilon)/P(j,{\mathcal{E}}) \log(Q(j,k){\overline{\theta}}_{j,k}P(j,\varepsilon)/P(j,{\mathcal{E}}))\\& = -\sum_{(j,k)\in I^2}\!\! \mu(j) Q(j,k) \log Q(j,k)-\sum_{j\in I}\mu(j)Q(j,k)(\theta_{j,k} \log \theta_{j,k}+{\overline{\theta}}_{j,k} \log {\overline{\theta}}_{j,k})\\& \quad -\sum_{(j,k)\in I^2}\!\! \mu(j) Q(j,k){\overline{\theta}}_{j,k}\sum_{\varepsilon\in {\mathcal{E}}} P(j,\varepsilon)/P(j,{\mathcal{E}}) \log (P(j,\varepsilon)/P(j,{\mathcal{E}})).\end{align*}

Then,

(20) \begin{align}\nonumber h\big({\mathbb{Y}}^{({\mathcal{A}})}\big) & = h\big({\mathbb{Y}}^{({\mathcal{K}})}\big)+\sum_{i\in I}\mu(i) P(i,{\mathcal{E}}) H(D^i),\quad \hbox{where} \\ H(D^i) & = -\sum_{\delta\in {\mathcal{E}}}P(i,\delta)/P(i,{\mathcal{E}}) \log (P(i,\delta)/P(i,{\mathcal{E}}))\end{align}

is the entropy of a random variable in ${\mathcal{E}}$ distributed as the transition probability from $i\in I$ to a state conditioned to be in ${\mathcal{E}}$. Note that the above expression can also be written as

\begin{equation*}h\big({\mathbb{Y}}^{({\mathcal{A}})}\big)=h\big({\mathbb{Y}}^{({\mathcal{K}})}\big)+\sum_{i\in I}\mu(i) (P(i,{\mathcal{E}}) H(D^i)+P(i,I)H(o)),\end{equation*}

$H(o)=0$ being the entropy of a constant.

Define

\begin{equation*}\Delta(D)=h\big({\mathbb{Y}}^{({\mathcal{A}})}\big)-h\big({\mathbb{Y}}^{({\mathcal{K}})}\big)=\sum_{i\in I}\mu(i) P(i,{\mathcal{E}}) H(D^i).\end{equation*}

This is the conditional entropy of ${\mathbb{Y}}^{({\mathcal{A}})}$ given the factor ${\mathbb{Y}}^{({\mathcal{K}})}$. To see this, take an orbit $Y^{({\mathcal{K}})}=((Y_{n-1},Y_{n},B_n)\,:\, n\in \mathbb{Z})$ of ${\mathbb{Y}}^{({\mathcal{K}})}$. The fiber given by (19) satisfies $(\Upsilon^{({\mathcal{A}})})^{-1}\{Y^{({\mathcal{K}})}\}=\{(D^{Y_n, B_n}_n\,:\, n\in \mathbb{Z})\in ({\mathcal{E}}^*)^\mathbb{Z}\}$ with $D^{Y_n, B_n}_n\in {\mathcal{E}}$ when $B_n=0$ and $D^{Y_n, B_n}_n=o$ when $B_n=1$. These variables are independently distributed as a Bernoulli $D^{Y_n}_n$ given in (11) if $B_n=0$ and the constant variable o if $B_n=1$. This probability measure is denoted by $(\bf P)_{Y^{({\mathcal{K}})}}$. Thus,

(21) \begin{equation}H_{(\bf P)_{Y^{({\mathcal{K}})}}}\big(D^{Y_0,B_0}_0\big)=\begin{cases}& \!\!\!\!\! -\sum_{\delta\in {\mathcal{E}}} P(Y_0,\delta)/P(Y_0,{\mathcal{E}}) \log(P(Y_0,\delta)/P(Y_0,{\mathcal{E}})) \hbox{ if } B_0=0,\\[4pt] & \!\!\!\!\! 0 \hbox{ if } B_0=1.\end{cases}\end{equation}

Proposition 4. The entropy of the stationary representation ${\mathbb{Y}}^{({\mathcal{A}})}$ of the absorbed chain satisfies

\begin{align*} h\big({\mathbb{Y}}^{({\mathcal{A}})}\big) & = h\big({\mathbb{Y}}^{({\mathcal{K}})}\big)+\Delta(D) \quad \hbox{with} \\ \Delta(D) & = \int H_{(\bf P)_{Y}}\big(D^{Y_0,B_0}_0\big) \, {\text{d}}\mathbb{P}(Y^{({\mathcal{K}})})=-\sum_{i\in I}\mu(i) P(i,{\mathcal{E}}) H(D^i) \end{align*}

and

(22) \begin{equation} \begin{aligned} h\big({\mathbb{Y}}^{({\mathcal{A}})}\big) & = -(1-\pi(I))\sum_{j\in I}\mu(j)\log \mu(j)-\sum_{i,j\in I} \mu(i) P(i,j)\log P(i,j) \\& \quad -\sum_{i\in I, \delta\in {\mathcal{E}}}\mu(i) P(i,\delta) \log P(i,\delta).\end{aligned}\end{equation}

Proof. From (20) and (21), and by using the Markov property,

\begin{eqnarray*}\Delta(D)&=&\int H_{(\bf P)_{Y^{({\mathcal{A}})}}}\big(D^{Y_0,B_0}_0\big) \, {\text{d}}\mathbb{P}(Y^{({\mathcal{K}})})\\&=&-\sum_{i\in I}\mu(i)P(i,{\mathcal{E}})\sum_{\delta\in {\mathcal{E}}} P(i,\delta)/P(i,{\mathcal{E}}) \log (P(i,\delta)/P(i,{\mathcal{E}}))\\&=& -\sum_{i\in I,\delta\in {\mathcal{E}}} \mu(i) P(i,\delta) \log P(i,\delta)+\sum_{i\in I} \mu(i) P(i,{\mathcal{E}}) \log P(i,{\mathcal{E}}).\end{eqnarray*}

We then use (18) to get the expression in (22).

5.3. Entropy balance

The associated stationary chain ${\mathbb{X}}$ with transition kernel $P=P^\pi$ is retrieved from the stationary chain ${\mathbb{Y}}$ with transition kernel Q, a collection of Bernoulli variables ${\mathbb{B}}^{I,I}$ that assign 0 or 1 between the connections of ${\mathbb{Y}}$, a set of Bernoulli variables ${\mathbb{D}}^I$ giving the transition from I to ${\mathcal{E}}$, and a family of walks ${\mathbb{V}}$ whose components are Bernoulli variables $(G_n)$ distributed as $\pi(\bullet \mid {\mathcal{E}})$. The length of these walks is Geometric$(\pi(I))-1$ distributed, and so they could be empty.

It is straightforward to prove the following equality, relating $h({\mathbb{X}})$ given by (5) to the entropies of the elements forming the chain ${\mathbb{X}}$.

Proposition 5.

\begin{equation*}h({\mathbb{X}})=\pi(I) h\big({\mathbb{Y}}^{({\mathcal{A}})}\big)+\pi({\mathcal{E}})^2 h({\mathbb{G}})+\pi(I)\pi({\mathcal{E}}) \log \pi(I) +\pi({\mathcal{E}})^2 \log \pi({\mathcal{E}}).\end{equation*}

Below we discuss the way this equality appears. We have reduced the elements forming the chain ${\mathbb{X}}$ to only two, the absorbed chains ${\mathbb{Y}}^{({\mathcal{A}})}$ and the walks ${\mathbb{V}}$ with Bernoulli variables $G_n$. From (22), we have

\begin{align*}h\big({\mathbb{Y}}^{({\mathcal{A}})}\big) = & -\pi({\mathcal{E}})\sum_{j\in I}\mu(j)\log \mu(j)- \sum_{i,j\in I} \mu(i) P(i,j)\log P(i,j)\\ & -\sum_{i\in I, \delta\in {\mathcal{E}}}\!\!\!\! \mu(i) P(i,\delta) \log P(i,\delta),\end{align*}

and the Bernoulli sequence ${\mathbb{G}}=(G_n)$ has entropy

\begin{equation*}h({\mathbb{G}})=-\sum_{\delta\in {\mathcal{E}}} \pi(\delta \mid {\mathcal{E}}) \log\pi(\delta \mid {\mathcal{E}})=-\pi({\mathcal{E}})^{-1}\sum_{\delta\in {\mathcal{E}}} \pi(\delta) \log \pi(\delta)+\log \pi({\mathcal{E}}).\end{equation*}

Taking some N, we divide the sequence $(X_1, \ldots, X_N)$ into the set of absorbed chains ${\mathcal{X}}^{(A)}$ and the set of nonempty walks V in ${\mathcal{E}}$. The proportion of elements in I approaches $\pi(I)$ as $N\to \infty$. Therefore, from (9) we obtain that for every time $t\in \mathbb{T}$ there are in mean $\sum_{i\in I} \mu(i) P(i,{\mathcal{E}}) (\pi(I)^{-1}-1)$ points belonging to a walk in ${\mathcal{E}}$. Since the set of points in $\mathbb{T}$ has a weight $\pi(I)$, we obtain that the proportion of points in $\mathbb{Z}$ belonging to a walk in ${\mathcal{E}}$ is $\pi(I)\cdot \sum_{i\in I} \mu(i) P(i,{\mathcal{E}}) (\pi(I)^{-1}-1)=\pi({\mathcal{E}})^2$. Hence, the proportion of sites in $(X_1, \ldots, X_N)$ with symbols in ${\mathbb{G}}$ arising from a walk V in ${\mathcal{E}}$ approaches $\pi({\mathcal{E}})^2$ as $N\to \infty$. We have

(23) \begin{equation}\pi({\mathcal{E}})^2 h({\mathbb{G}})=-\pi({\mathcal{E}}) \sum_{\delta\in {\mathcal{E}}} \pi(\delta) \log \pi(\delta)+\pi({\mathcal{E}})^2\log \pi({\mathcal{E}}).\end{equation}

Let us compute $\pi(I) h\big({\mathbb{Y}}^{({\mathcal{A}})}\big)$. Since $\mu(i)=\pi(i)/\pi(I)$ for $i\in I$, we have

\begin{equation*}-\pi(I)\sum_{j\in I} \mu(j)\log \mu(j)=-\sum_{j\in I} \pi(j)\log \pi(j)+\pi(I)\log \pi(I),\end{equation*}

and so, using (22), we get

\begin{eqnarray*}\pi(I) h\big({\mathbb{Y}}^{({\mathcal{A}})}\big)&=&-\pi({\mathcal{E}})\sum_{j\in I}\! \pi(j)\log \pi(j)+\pi({\mathcal{E}}) \pi(I)\log \pi(I)\\&{}&\;\; -\sum_{i,j\in I}\! \pi(i) P(i,j)\log P(i,j)-\sum_{i\in I, \delta\in {\mathcal{E}}}\!\! \pi(i) P(i,\delta) \log P(i,\delta).\end{eqnarray*}

Then, the equality given in Proposition 5 has been proved:

\begin{equation*}h({\mathbb{X}})-(\pi({\mathcal{E}})^2 h({\mathbb{G}})+\pi({\mathcal{E}})^2 \log \pi({\mathcal{E}}))=\pi(I)h\big({\mathbb{Y}}^{({\mathcal{A}})}\big)+\pi({\mathcal{E}}) \pi(I)\log \pi(I).\end{equation*}

The term $\pi(I)\pi({\mathcal{E}}) \log \pi(I)$ has an origin similar to the last term in (23). In fact, from Remark 4, the resurrection weights $\mu(j)$, $j\in I$, appear with frequency $\pi({\mathcal{E}})$ in the sequence ${\mathbb{Y}}$ because this occurs at the sites where there is a jump to ${\mathcal{E}}$. Since the sequence ${\mathbb{Y}}$ appears with frequency $\pi(I)$, then the term $-\sum_{i\in I}\mu(i)\log \mu(i)$ appears with frequency $\pi(I)\pi({\mathcal{E}})$. Hence, as in (23), we have

\begin{equation*}-\pi(I)\pi({\mathcal{E}})\sum_{i\in I}\mu(i)\log \mu(i)=-\pi({\mathcal{E}})\sum_{i\in I}\pi(i)\log \pi(i)+\pi(I)\pi({\mathcal{E}}) \log \pi(I),\end{equation*}

and since $-\pi({\mathcal{E}})\sum_{i\in I}\pi(i)\log \pi(i)$ is the term present in (5), the extra term given by $\pi(I)\pi({\mathcal{E}}) \log \pi(I)$ appears.