I recently posted a similar question here, but I think I formulated it poorly and can now be more precise. Suppose we are given an irreducible Markov chain $(X_n)_{n\in\mathbb{N}}$ with state space $E$. Let $s,s'\in E$ be two distinct states. I am interested in the expectation of $t:=\inf\{n\in\mathbb{N}\mid (X_n,X_{n+1}) = (s,s')\}$. My intuition tells me that the smaller the transition probability from $s$ to $s'$, the larger the expected time. I would like to prove an inequality which resembles that. Here's what I have done so far, using the total law of expectation \begin{align} \mathbb{E}[t]= & \mathbb{P}(X_1=s,X_2=s')\\ &+ (1+\mathbb{E}[t\mid X_1=s])\cdot\mathbb{P}(X_1=s, s=X_2\neq s')\\ &+ (2+\mathbb{E}[t])\cdot\mathbb{P}(X_1=s,s\neq X_2\neq s')\\ &+ (1+\mathbb{E}[t])\cdot\mathbb{P}(X_1\neq s). \end{align} The terms in front of the probabilities come from the properties one can deduce about the respective conditional expectations, e.g. if $X_1\neq s$ then we have essentially done one step in the wrong direction and are still left with the expected waiting time, i.e. $1+\mathbb{E}[t]$.
I feel like I am already fairly close to my goal with this approach, but I can't seem to bound the term $(1+\mathbb{E}[t\mid X_1=s])$ from below in a clever way, so that I am able to prove something like $\mathbb{E}[t]\geq 1/\mathbb{P}(X_1=s,X_2=s')$.
Ideally, I would like to say $\mathbb{E}[t\mid X_1=s]\geq \mathbb{E}[t]$, but this is clearly false. Can someone help me out?
P.S.: I already know about the system of equations for mean hitting times and tried to use it to get a lower bound, but I essentially run into the same problem.
I managed to prove it myself using another identity that handles the term $\mathbb{E}[t|X_1=s]$. Using the total law of expectation and we get \begin{align} \mathbb{E}[t] &= \mathbb{E}[t|X_1=s]\cdot\mathbb{P}(X_1=s) + \mathbb{E}[t|X_1\neq s]\cdot\mathbb{P}(X_1\neq s)\\ &=\mathbb{E}[t|X_1=s]\cdot\mathbb{P}(X_1=s) + (1+\mathbb{E}[t])\cdot\mathbb{P}(X_1\neq s).\\ \end{align} Rearranging yields \begin{align} \mathbb{E}[t]= \mathbb{E}[t|X_1=s] + \frac{\mathbb{P}(X_1\neq s)}{\mathbb{P}(X_1=s)}. \end{align} To keep it short, combining this identity with the one from my initial post and using some basic measure theoretic calculations yields \begin{equation} \mathbb{E}[t]>\frac{1}{\mathbb{P}(X_2=s'\mid X_1=s)\cdot\mathbb{P}(X_1=s)} \end{equation} which of course also implies the desired inequality. If anyone is itnerested in the particular steps I can provide them, but these are the key-steps.