Let $X_n$ be a Markov Chain on a countable, irreducible state space. Assume that the state $x$ is recurrent and that $\pi(x,y)$ >0, where $\pi(\cdot,\cdot)$ is the one-step transition probability.
Show that:
$$\mathbb{E}^{\mathbb{P}_x}[\tau_x] \geq \mathbb{P}_x(\tau_y < \tau_x) \mathbb{E}^{\mathbb{P}_y}[\tau_x]$$
Where $\tau_x = \inf(n\geq 1\, \vert X_n = x)$, and $\mathbb{P}_x$implies the chain starts at state $x$.
Showing that $0<\mathbb{P}_x(\tau_y < \tau_x)<1$ is easy enough. My question is mainly with the relationship between $\mathbb{E}^{\mathbb{P}_x}[\tau_x]$ and $\mathbb{E}^{\mathbb{P}_y}[\tau_x]$
Is it true that $\mathbb{E}^{\mathbb{P}_x}[\tau_x] = \mathbb{E}^{\mathbb{P}_y}[\tau_x]$? How would you show that?
First of all, it is not true that $\mathbb{E}_x \tau_x=\mathbb{E}_y \tau_x$. For example consider a chain with two states: $\mathcal{S}=\{x,y\}$ with the transition matrix $\pi(i,j)=1_{\{i\neq j\}}$ (i.e. changing state with probability $1$ at every step). In this case $\mathbb{E}_x \tau_x=2>1=\mathbb{E}_y \tau_x$.
One way to show what you want is by using conditional expectation: $$ \mathbb{E}_x \tau_x=\mathbb{P}(\tau_y<\tau_x)\mathbb{E}_x (\tau_x|\tau_y<\tau_x)+\mathbb{P}(\tau_y\geq \tau_x)\mathbb{E}_x (\tau_x|\tau_y\geq \tau_x) \\ \geq \mathbb{P}(\tau_y<\tau_x)\mathbb{E}_x (\tau_x|\tau_y<\tau_x)\geq \mathbb{P}(\tau_y<\tau_x)\mathbb{E}_y \tau_x. $$
The last inequality relies on the fact that regardless of the number of steps it took to reach $y$, you will need to take an expected additional number of $\mathbb{E}_y \tau_x$ steps to reach $x$ again.