A stopping time $T$ is a if
$$
\mathbb P_x(X_T=y,T=t)=\pi(y)\mathbb P_x(T=t)
$$
for every $x,y,t$. Thus $X_T\sim\pi$ and is independent of $T$. The <separation distance> is
$$
s(t)=\max_{x,y}\left\{1-\frac{P^t(x,y)}{\pi(y)}\right\}.
$$
For any strong stationary time,
$$
\begin{aligned}
P^t(x,y)
&\geq\mathbb P_x(X_t=y,T\leq t)\\
&=\sum_{j\leq t}\sum_z
\mathbb P_x(T=j,X_j=z)P^{t-j}(z,y)\\
&=\pi(y)\mathbb P_x(T\leq t).
\end{aligned}
$$
Hence $1-P^t(x,y)/\pi(y)\leq\mathbb P_x(T>t)$, and maximizing proves the claim.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 215 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-215/1/a/solution