Let $S$ be a finite set, $\mu$ a probability distribution on $S$ with $\mu(x)>0$ for all $x \in S$.
Let $(q(x,y))_{x,y \in S}$ be a stochastic matrix ($q(x,y) \geq 0 \forall x,y \in S$ and $\sum_{y \in S} q(x,y)=1 \forall x \in S$).
Show that the matrix defined by $P=(p(x,y))_{x,y \in S}$ also is a stochastic matrix for
$$p(x,y)=q(x,y)\cdot \min\Big(1, \frac{\mu(y)q(y,x)}{\mu(x)q(x,y)}\Big) \text{, for } y \neq x$$
My attempt:
The condition $p(x,y)\geq 0 \forall x,y \in S$ is clear.
The second condition:
If $1$ is always the minimal value this follows from $(q(x,y))$ being stochastic. If the other value is always minimal, we have $$\sum_{y\in S}p(x,y)=\sum_{y \in S}q(x,y)\cdot \min \Big(1, \frac{\mu(y)q(y,x)}{\mu(x)q(x,y)}\Big)=\sum_{y \in S} \frac{\mu(y)q(y,x)}{\mu(x)}=\frac{1}{\mu(x)} \sum_{y \neq x} \mu(y)q(y,x)=\frac{\mu(x)}{\mu(x)}=1$$
But I do not really understand what happens when the minima are mixed.
My most promising result is the condition:
$$1=\sum_{y \in S}p(x,y)=\sum_{y \in S, \frac{\mu(y)q(y,x)}{\mu(x)q(x,y)}<1}\frac{\mu(y)q(y,x)}{\mu(x)}+\sum_{y \in S, \frac{\mu(y)q(y,x)}{\mu(x)q(x,y)} \geq 1}q(x,y)$$
$$\iff$$
$$\mu(x)=\sum_{y \in S, \frac{\mu(y)q(y,x)}{\mu(x)q(x,y)}<1}\mu(y)q(y,x)+\sum_{y \in S, \frac{\mu(y)q(y,x)}{\mu(x)q(x,y)} \geq 1}\mu(x)q(x,y)$$
We can include the equality to $1$ of the fraction in either of the sums, it does not change anything. Then I have:
$$\sum_{y \in S, \frac{\mu(y)q(y,x)}{\mu(x)q(x,y)} \leq 1}\mu(y)q(y,x)+\sum_{y \in S, \frac{\mu(y)q(y,x)}{\mu(x)q(x,y)} > 1}\mu(x)q(x,y) \leq$$
$$\sum_{y \in S, \frac{\mu(y)q(y,x)}{\mu(x)q(x,y)} \leq 1}\mu(x)q(x,y)+\sum_{y \in S, \frac{\mu(y)q(y,x)}{\mu(x)q(x,y)} > 1}\mu(x)q(x,y)=\mu(x)$$
But I cannot give $\mu(x)$ as a lower bound.
Any insight is welcome!