Shifted absorption time If $(X_n)_{n\geq0}$ is a Markov chain, and $T=inf\left\{n\geq0 : X_n=1 \right\}$, why does $$\mathbb{E}_i[T-1|X_1=j]=\mathbb{E}[T|X_0=j] ?$$
This question is partially answered here : Markov chains: Expected time until absorption., but I'd like a rigorous proof of that fact. 

It's enough to prove that
$$
   \Pr[T = t \mid X_0 = i, X_1 = j] = \Pr[T = t-1 \mid X_0 = j].
$$
Then $\mathbb E[T]$ will follow by the usual expectation formula.
Both sides of this probability can be decomposed into sums over paths: on the left, paths $(i,j, \dots,1)$ of length $t$, and on the right, paths $(j,\dots,1)$ of length $t-1$. 
For any fixed $P$ on the left, we have a corresponding path $P'$ on the right by dropping the initial $i$, and vice versa. Following the path $P$ given that $X_0 = i$ and $X_1 = j$ has the same probability as following $P'$ given that $X_0 = j$, by the Markov probability (and by stationarity: we'll get terms like $\Pr[X_{k+1} = x \mid X_k = y]$ on one side and $\Pr[X_k = x \mid X_{k-1} = y]$ on the other side). So each path contributes the same to both sides, and therefore both probabilities are equal.
