[proofplan]
The proof is a telescoping argument corrected by the lagged coupled chain. For finite terminal time $n$, we write $\mathbb{E}[h(X_n)]$ as $\mathbb{E}[h(X_m)]$ plus successive increments, then use the equality in law of $Y_{k-1}$ and $X_{k-1}$ to replace each backward increment by a coupled difference $h(X_k)-h(Y_{k-1})$. Lag-faithfulness makes these differences vanish after the meeting time, so the infinite correction series is exactly the finite random sum defining $H_m$. The stated integrability condition justifies taking expectations of the random sum and passing to the limit $n\to\infty$.
[/proofplan]
[step:Define the coupled correction terms and prove their integrability]
For each integer $k\geq m+1$, define the real-valued [random variable](/page/Random%20Variable) $D_k:\Omega\to\mathbb{R}$ by
\begin{align*}D_k(\omega)=h(X_k(\omega))-h(Y_{k-1}(\omega)).\end{align*}
By lag-faithfulness and the definition of $\tau$, if $k\geq \tau(\omega)$, then $X_k(\omega)=Y_{k-1}(\omega)$, hence $D_k(\omega)=0$. Therefore, for every $N\geq m+1$,
\begin{align*}\sum_{k=m+1}^{N}|D_k|\leq \sum_{k=m+1}^{\tau-1}|D_k|.\end{align*}
The assumed integrability condition implies that $h(X_m)$ is integrable and that each $D_k$ is integrable. It also implies that
\begin{align*}H_m=h(X_m)+\sum_{k=m+1}^{\infty}D_k\end{align*}
as an absolutely integrable random variable, because all terms with $k\geq\tau$ vanish $\mathbb{P}$-a.s.
[/step]
[step:Convert the finite telescoping identity into coupled increments]
Fix an integer $n>m$. Since $h(X_k)$ is integrable for every $k\geq 0$, the ordinary finite telescoping identity for [real numbers](/page/Real%20Numbers) may be applied inside expectation:
\begin{align*}\mathbb{E}[h(X_n)] = \mathbb{E}[h(X_m)] + \sum_{k=m+1}^{n}\left(\mathbb{E}[h(X_k)]-\mathbb{E}[h(X_{k-1})]\right).\end{align*}
For each $k\geq m+1$, the random variables $Y_{k-1}$ and $X_{k-1}$ have the same law. Since $h:E\to\mathbb{R}$ is measurable and $h(X_{k-1})$ is integrable, this equality in law gives
\begin{align*}\mathbb{E}[h(Y_{k-1})]=\mathbb{E}[h(X_{k-1})].\end{align*}
Substituting this into the preceding display yields
\begin{align*}\mathbb{E}[h(X_n)] = \mathbb{E}[h(X_m)] + \sum_{k=m+1}^{n}\mathbb{E}[D_k].\end{align*}
[guided]
The finite identity is the place where the lagged construction enters. We start with a deterministic telescoping sum applied to the sequence of numbers $\mathbb{E}[h(X_k)]$:
\begin{align*}\mathbb{E}[h(X_n)] = \mathbb{E}[h(X_m)] + \sum_{k=m+1}^{n}\left(\mathbb{E}[h(X_k)]-\mathbb{E}[h(X_{k-1})]\right).\end{align*}
This is legitimate because the theorem assumes $h(X_k)$ is integrable for every $k$, so each expectation is finite.
The goal is to rewrite the increment from time $k-1$ to time $k$ using the coupled pair at lag one. For a fixed $k\geq m+1$, the hypothesis says that $Y_{k-1}$ and $X_{k-1}$ have the same law on $(E,\mathcal{E})$. Since $h:E\to\mathbb{R}$ is measurable, the real-valued random variables $h(Y_{k-1})$ and $h(X_{k-1})$ also have the same law. Because $h(X_{k-1})$ is integrable, equality in law implies integrability of $h(Y_{k-1})$ and equality of expectations:
\begin{align*}\mathbb{E}[h(Y_{k-1})]=\mathbb{E}[h(X_{k-1})].\end{align*}
Therefore each increment satisfies
\begin{align*}\mathbb{E}[h(X_k)]-\mathbb{E}[h(X_{k-1})] = \mathbb{E}[h(X_k)]-\mathbb{E}[h(Y_{k-1})] = \mathbb{E}[h(X_k)-h(Y_{k-1})].\end{align*}
By the definition
\begin{align*}D_k=h(X_k)-h(Y_{k-1}),\end{align*}
the finite telescoping identity becomes
\begin{align*}\mathbb{E}[h(X_n)] = \mathbb{E}[h(X_m)] + \sum_{k=m+1}^{n}\mathbb{E}[D_k].\end{align*}
This is the algebraic core of the estimator: the bias between time $m$ and stationarity is represented as a sum of lagged coupled differences.
[/guided]
[/step]
[step:Pass from finite coupled sums to the random meeting-time sum]
For every $n>m$, linearity of expectation gives
\begin{align*}\mathbb{E}[h(X_m)] + \sum_{k=m+1}^{n}\mathbb{E}[D_k] = \mathbb{E}\left[h(X_m)+\sum_{k=m+1}^{n}D_k\right].\end{align*}
Since $D_k=0$ for $k\geq\tau$ $\mathbb{P}$-a.s. and
\begin{align*}\left|h(X_m)+\sum_{k=m+1}^{n}D_k\right| \leq |h(X_m)|+\sum_{k=m+1}^{\tau-1}|D_k|,\end{align*}
the dominating random variable on the right is integrable by hypothesis. Moreover,
\begin{align*}h(X_m)+\sum_{k=m+1}^{n}D_k \to h(X_m)+\sum_{k=m+1}^{\infty}D_k = H_m\end{align*}
$\mathbb{P}$-a.s. as $n\to\infty$. Hence dominated convergence gives
\begin{align*}\lim_{n\to\infty}\mathbb{E}\left[h(X_m)+\sum_{k=m+1}^{n}D_k\right] = \mathbb{E}[H_m].\end{align*}
[/step]
[step:Take the terminal-time limit and identify the target expectation]
Combining the finite identity with the preceding limit gives
\begin{align*}\mathbb{E}[H_m] = \lim_{n\to\infty}\mathbb{E}[h(X_n)].\end{align*}
By the assumed convergence of the marginal expectation of $X_n$,
\begin{align*}\lim_{n\to\infty}\mathbb{E}[h(X_n)]=\pi(h).\end{align*}
Therefore
\begin{align*}\mathbb{E}[H_m]=\pi(h).\end{align*}
This proves that the coupled estimator $H_m$ is unbiased for $\pi(h)$.
[/step]
