# Random walk origin return monotinicity

Consider a Markov chain on $\mathbb{Z}^d$ with transition kernel $P$ for adjacent vertices (non-diagonal). Essentially this is a $d$ dimensional random walk with the probability of a transition dependent on it's location in the grid. This comes from a random conductance model. The theorem that concerns me is a general result for Markov chains, but I leave this motivation to assist in its proof (see below).

Let $P^{2k}(0,0)$ be the probability of going from the origin and back in $2k$ steps. Moreover, suppose $P$ is reversible. The theorem that concerns me is:

$P^{2n}(0,0)$ is decreasing in $n$.

I am interested in a probabilistic proof of this. The proof that I know is of a spectral nature:

Define $\langle f,g\rangle:= \sum_{X\in\mathbb{Z}^d} \pi(x)f(x)g(x)$,

where $\pi(x)$ is the stationary measure. This gives an inner product on $L^2(\mathbb{Z}^d)$. In the case of a random conductance model, $\pi(x)$ would be the sum of random edge weights at $x$.

Then

$P^{2k}(0,0)=\langle \delta_0,P^{2k}\delta_0\rangle$,

and since $P$ is self adjoint with $\|P\|_2\leq 1$, the desired result follows.

I have tried various approaches such as conditioning on the hitting times of the origin and as well trying to prove the result by induction. I would like to see a proof that showcases a probabilistic argument. For example, is it possible to show the result from the machinery of evolving sets of Morris and Peres?

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Let $\phi_n(x)=p_n(o,x)/\pi(x)$ be the density of the $n$-step transition probability from the reference point $o$ with respect to the stationary measure $\pi$ (I use a slightly different notation). I will normalize the measure $\pi$ in such a way that $\pi(o)=1$. Then for any $n,m\ge 0$ $$p_{n+m}(o,o) = \sum_x p_n(o,x) p_m(x,o) = \sum_x p_n(o,x) p_m(o,x) \frac{1}{\pi(x)} = \langle \phi_n,\phi_m\rangle \;,$$ where $\langle\cdot,\cdot\rangle$ denotes the scalar product with respect to the stationary measure $\pi$. In particular, for $n=k-1$ and $m=k+1$ $$p_{2k}(o,o) = \langle\phi_{k-1},\phi_{k+1}\rangle_\pi \le \|\phi_{k-1}\| \|\phi_{k+1}\| \;,$$ whence $$\frac{p_{2k}(o,o)}{p_{2k-2}(o,o)} \le \frac{p_{2k+2}(o,o)}{p_{2k}(o,o)} \;.$$ The above inequalities are strict unless $\phi_{k-1}=\phi_{k+1}$, in which case also $\phi_{k+3}=\phi_{k+5}=\dots=\phi_{k-1}$, which implies that all the return probabilities $p_{2k}(o,o), p_{2k+2}(o,o),\dots$ are the same, which is only possible for a positive recurrent chain. Thus, discarding the positive recurrence case, the sequence of ratios $p_{2k+2}(o,o)/p_{2k}(o,o)$ is strictly increasing. Obviously, its limit can not be greater than 1, whence the claim.