Consider 2 graphs G1 and G2.
G1: Any non-regular graph.
G2: Same graph but with added self-loops such that degree of each node is the same (either some $\Delta$, or maximum '$n$', where $n$ is the number of nodes in the graph).
We run a lazy random walk on G1, such that the random walk at a node $v$ either stays at $v$ with probability $1/2$ or moves out to a neighbor with probability $1/2d_{v}$, where $d_{v}$ is the degree of the node $v$. Consider, $t_{mix}$ as the mixing time of this random walk. Additionally, we maintain a queue where the starting node of the random walk is added to a queue, and thereafter each time the random walk jumps to a neighbor, the node it jumps to is added to the queue. Let the queue contain $k$ nodes when the random walk has mixed $(k<= t_{mix})$. Note that since the graph is non-regular the stationary distribution that it converges to is non-uniform.
Now, consider another random walk on G2, which chooses each edge with probability $1/d_{v}$, where $d_v$ is the degree of node $v$ including the self-loops. Assume that this random walk exactly follows the previous random walk (on G1) in the sense that each time there is a jump to the neighbor, it jumps to the same neighbor as the previous walk, i.e., it will push the exact same nodes to its maintained queue. When we consider time, of course, this would be slower as, in low-degree nodes, the random walk would have to undertake many self-loops before being able to jump to a neighbor. What I want to show that, is when the queue contains $k$ nodes or ($O(k)$ nodes), then this random walk has also mixed. Note that, as opposed to before, the stationary distribution here with respect to which we want the mixing is uniform (due to the added self-loops).
Is there any existing work that shows this? Or if not, what can be a good coupling argument to prove this? Any reference or help would be greatly appreciated.
(Observe that, the random walk on G2, can also be considered as another (biased) random walk on G1 where the random walk stays at the current node $v$ with probability $1-(deg(v)/n)$ and, with probability $1/n$, moves to a uniformly at random chosen neighbor.)
