The squared $L^2(\pi)$ distance has the diagonal identity
$$
\left\lVert\frac{P^t(x,\mathord\cdot)}{\pi(\mathord\cdot)}-1\right\rVert_{2,\pi}^2
=\frac{P^{2t}(x,x)-\pi(x)}{\pi(x)}.
$$
The diagonal excess is nonnegative and decreases with time. Therefore
$$
(2t+1)\{P^{2t}(x,x)-\pi(x)\}
\leq\sum_{k=0}^\infty\{P^k(x,x)-\pi(x)\}.
$$
Using the identity supplied in the question gives
$$
\left\lVert\frac{P^t(x,\mathord\cdot)}{\pi(\mathord\cdot)}-1\right\rVert_{2,\pi}^2
\leq\frac{\mathbb E_\pi\tau_x}{2t+1}.
$$
At $t=8\mathbb E_\pi\tau_x$ the right side is at most $1/16$, up to the immaterial integer rounding. Thus
$$
\boxed{t_{\mathrm{mix}}^{(2)}(x,1/4)\leq8\mathbb E_\pi\tau_x}.
$$


Solution

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

 Solution

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