The laziness of $P=(Q+I)/2$ makes all its eigenvalues nonnegative. Write
$$
1=\lambda_1>\lambda_2\geq\cdots\geq\lambda_{|S|}\geq0.
$$
The <relaxation time> is
$$
t_{\mathrm{rel}}=\frac1{1-\lambda_2}.
$$
If $(f_i)$ is an orthonormal eigenbasis of $L^2(\pi)$ with $f_1=1$, the spectral decomposition is
$$
\frac{P^t(x,y)}{\pi(y)}
=\sum_{i=1}^{|S|}\lambda_i^tf_i(x)f_i(y).
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 215 / 1 / a / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/iii/paper-215/1/a/solution