Consider a simple random walk.Say $a$ is a vertex with only one neighbour $b$, but $b$ has various neighbours.
Can someone explain why $E(Ta|X0=b)$ not equal to $E(Tb|X0=a)$ ?
Where $X0$ is the position at time $t=0$ and $Ta$ is the time of the first return to state a.
From $a$ you have to go directly to neighbor $b$. So the time from $a$ to $b$ is always $1$. And so $E(Tb \mid X_0 = a) = 1$.
However from $b$ your first step may or may not be to $a$. So in general, the expected time to get from $b$ to $a$ is greater than $1$. That is $E(Ta \mid X_0 = b)> 1$.
