How to show that $d_G(t)$ is decreasing in $t$ for a geometry mixing time?

Question

Let $P$ be the transition matrix of a Markov chain with state-space $\mathcal{X}$, $\pi$ is the stationary distribution with $\pi=\pi P$, and $Z_t$ be a geometric random variable of parameter $1/t$ taking values in $\{1,2,\dots, \}$ and independent of $x$. Define $$d_G(t):=\max_{x\in\mathcal{X}}\|P_x(X_{Z_t}=\cdot)-\pi\|_{TV}$$ where "TV" means total variation distance between two probability distributions $\mu$ and $\nu$ on $\mathcal{X}$ is defined by $$\|\mu-\nu\|_{TV}:=\max_{A\subset \mathcal{X}}|\mu(A)-\nu(A)|$$

How to show that $d_G(t)$ is decreasing in $t$?

Answer 1

This is true assuming that $Z_t$ is independent of the Markov chain. Indeed, then $$d(t):=d_G(t)=\max_x E\|P_x(Z_t)-\pi\|_{TV},$$ where $$P_x(n):=\delta_x P^n$$ and $\delta_x$ is the row matrix $([a_y]_{y\in\mathcal X})^T$ with $a_y:=1_{y=x}$. It is easy to see that for any probability measures $\mu$ and $\nu$ $$\|\mu-\nu\|_{TV}=\sup_{0\le f\le1} \int f\,d(\mu-\nu), \tag{0}$$ where $\sup_{0\le f\le1}$ is taken over all measurable functions $f$ such that $0\le f\le1$.

Take now any real $s$ and $t$ such that $1\le s\le t$. We have to show that then $d(t)\le d(s)$. So, it suffices to show that for each $x$ $$E\|P_x(Z_t)-\pi\|_{TV}\overset{\text{(?)}}\le E\|P_x(Z_s)-\pi\|_{TV}. \tag{1}$$ The random variable $Z_t$ is stochastically greater than $Z_s$. So, without loss of generality, $Z_t\ge Z_s$. Take now any column matrix $f=[f_x]_{x\in\mathcal X}$ with $f_x\in[0,1]$ for all $x$. Then the entries of the (random) column matrix $P^{Z_t-Z_s}f$ are in the interval $[0,1]$ as well and hence $$(P_x(Z_t)-\pi)f=(P_x(Z_s)-\pi)P^{Z_t-Z_s}f\le\|P_x(Z_s)-\pi\|_{TV}$$ by (0). So, again by (0), $$\|P_x(Z_t)-\pi\|_{TV}\le\|P_x(Z_s)-\pi\|_{TV}.$$ Taking now the expectations, we get (1), as desired.