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 $$\|\mu-\nu\|\_{TV}:=\max\_{A\subset \mathcal{X}}|\mu(A)-\nu(A)|$$



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> How to show that $d\_G(t)$ is decreasing in $t$?
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How to show that $d_G(t)$ is decreasing in $t$ for a geometry mixing time?
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.


