Consider the discrete time simple random walk $X_t$ on a finite connected graph with transition matrix $P$. At each step the walk moves to a neighbor uniformly at random. Assume that $X_t$ has a unique stationary distribution $\pi$ with mixing time $t_{\text {mix}}$.
let's fix $\alpha \in(0,1)$ and a vertex $v \in V$ (set of vertices) and define a new Markov Chain $\boldsymbol{Y}=\left(Y_t\right)_{t \geqslant 0}$, where at each step with probability $1-\alpha$ the walk moves to a neighbor uniformly at random and with probability $\alpha$ it moves to vertex $v$.
- I want to show $\boldsymbol{Y}$ has a unique stationary distribution $\pi_v=\left(\pi_v(x)\right)_{x \in V}$ and its form.
Let's consider a sequence of graphs ($G_n$) with size of the vertex set. Also, assume that $\alpha=\alpha_n$ depends on $n$. If $t_{\text {mix }}(n) \cdot \alpha_n \rightarrow 0$ as $n \rightarrow \infty$,
- Does the stationary distributions also converge to that of its limit? $$ \max _{v \in V_n} d_{\mathrm{TV}}\left(\pi_v, \pi\right) \rightarrow 0 \text { as } n \rightarrow \infty . $$
Showing that there is a stationary distribution in 1 is easy: the graph is undirected and connected, so it's irreducible, and there's a self-loop from $v$ to itself, so it's aperiodic, so there is a unique stationary distribution that is approached in the limit.
In terms of showing what its form is, I don't really see how that's possible at this level of generality. It is not just a trivial modification of the original stationary distribution. For example, notice that if you have any regular graph, the stationary distribution of the random walk on the graph is uniform, but now as you move $v$ around a regular graph you will see stationary distributions that are biased in favor of $v$ and nodes not too far from $v$.
2 seems plausible intuitively but I don't see how you'd set up a proof for it analytically.