stochastic processes – How to bound the mean displacements of Markov chains

This is a question about some fundamental estimates for continuous time Markov chains.

Let $(V,E)$ be a locally finite connected graph. $V$ is the vertex set, and $E$ the edge set.
For $x,y in V,$ we write $d(x,y)$ for the length of the shortest path from $x$ to $y$. For $x,y in V$, we shall write $x sim y$ if ${x,y} in E$. We assume that the graph is endowed with a weight $mu_{xy}$ ($x,y in V$). The weight is symmetric and nonnegative function on $V times V$. We assume that $mu_{xy}>0$ iff $x sim y$. For $x in V$, we write $mu_{x}=sum_{z sim x}mu_{xz}$

We denote by ${X_n}_{n ge 0}$ the discrete time Markov chain on $V$ whose heat kernel $p_n(x,y)$ is given by
begin{align*}
p_n(x,y)=P(X_n=y mid X_0=x)/mu_y,quad x,y in V,, n ge 0.
end{align*}

We can obtain a canonical (continuous time) Markov chain ${X_t}_{t ge0 }$ on $V$ from ${X_n}_{n ge0}$. That is ${X_t}_{t ge 0}$ is the Markov chain whose heat kernel $p_t(x,y)$ is given by begin{align*}
p_t(x,y)=sum_{n=0}^infty e^{-t} frac{t^n}{n!}, p_n(x,y),quad t ge 0,,x,y in V.
end{align*}

My question.

I would like to know how to estimate the mean displacement of ${X_t}_{t ge 0}$. For the mean displacement, I mean $E(d(X_t,x)mid X_0=x)$ ($t>0,xin V$). It might be sufficient to obtain some estimates for the heat kernel $p_t(x,y)$. However, I want to know other ways.

Isn’t it quite possible that some estimates on the mean displacement can be obtained without relying on such heat kernel estimates? I think this is especially true when the graph is given fairly concretely. For example, the BDG inequality is valid for SDEs with bounded coefficients, and we do not necessarily have to use the heat kernel estimates (well, we are now thinking about discrete objects, so this example may be inappropriate, though). Although the answer to this question may depend on what Markov chains I am thinking of, I am mainly interested in methods itself for bound the mean displacement of the continuous time Markov chains.