Understanding the "first step analysis" of absorbing Markov chains Consider a time-homogeneous Markov chain $\{X_n\}_{n=0}^\infty$ with the state space state space $S=\{0,1,2\}$ and the following transition probability matrix:
\begin{pmatrix} 
  1 & 0 & 0 \\ 
  \alpha & \beta & \gamma \\ 
  0 & 0 & 1
\end{pmatrix}
where $\alpha,\beta,\gamma>0$. Note that state 0 and 2 are absorbing. 
Let
$
T=\min\{n\geq 0\mid X_n=0\textrm{ or }X_n=2\}
$
be the time of absorption of the process. It is intuitively true that
$$
P(X_T=0\mid X_1=1)=P(X_T=0\mid X_0=1)\tag{*}
$$
which is the key point of the so called "first step analysis". See for instance Chapter 3 in Karlin and Pinsky's Introduction to Stochastic Modeling. But the book does not bother giving a proof of it. 
Here is my question: 

How can one prove (*) using the definition of conditional probability and the Markov property?

 A: Here is a formal proof. For any sequence ${\bf x}=(x_0,x_1,x_2,\dots)$ in $\{0,1,2\}^{\mathbb{N}}$  define
$$H({\bf x})=\sum_{n=0}^\infty \,\prod_{j=0}^{n-1} {\bf 1}[x_j=1]\,{\bf 1}[x_n=0],$$
so that $H({\bf x})=1$ if the sequence $\bf x$ hits "0" before hitting "2", $H({\bf x})=0$ otherwise. Note that an initial value of "1" can be dropped, i.e.,
$$H(1,x_1,x_2,\dots )=H(x_1,x_2,\dots).\tag1$$
Since $\mathbb{P}(X_0=1\mid X_1=1)=1$, 
under the measure $\mathbb{P}(\,\cdot \mid X_1=1)$
equation (1) gives
$$H(X_0(\omega), X_1(\omega),\dots)=H(X_1(\omega), X_2(\omega),\dots),$$
almost surely. 
Therefore,
$$E(H(X_0,X_1,\dots)\mid X_1=1)=
E(H(X_1,X_2,\dots)\mid X_1=1)=
E(H(X_0,X_1,\dots)\mid X_0=1),$$
where the last equation follows since $(X_n)$ is a time homogeneous Markov process.
That is, 
$$P(X_T=0, T<\infty\mid X_1=1)=P(X_T=0, T<\infty\mid X_0=1).$$
A: Here is a proof I learn from a note on "first step analysis", which is essentially the same as Byron's answer. 
Consider the function $f:\{0,1,2\}\to\{0,1\}$ with
$$
f(i)=\begin{cases}1, &i=0\\
0,&i\neq 0.
\end{cases}
$$
Note that $f(X_0)=\cdots=f(X_{T-1})=0$ because $X_0,\cdots,X_{T-1}$ are not absorbing states. On the other hand, by the time homogeneous property (well, yes, this step is a little too slick),
$$
E(f(X_1)+f(X_2)+\cdots+f(X_T)\mid X_1=1)=E(f(X_0)+f(X_1)+\cdots+f(X_T)\mid X_0=1).
$$
But the LHS is $$E(f(X_T)\mid X_1=1)=P(X_T=0\mid X_1=1)$$ and the RHS is $$E(f(X_T)\mid X_0=1)=P(X_T=0\mid X_0=1).$$
A: Note that the event $X_T=0$ can be decomposed as a union of disjoint events:
$$
A_0:=\{X_0=0\}, A_1:=\{X_0\neq 0,X_1=0\},A_2:=\{X_0\neq 0,X_1\neq 0,X_2= 0\},\\
A_k:=\{X_0\neq 0,\cdots,X_{k-1}\neq 0,X_k=0\},\cdots.
$$
Now, we claim that we have the following, which of course implies ($*$).
\begin{align}
\tag{1} P(A_0\mid X_1=1)&=0 \\
\tag{2} P(A_k\mid X_1=1)&=P(A_{k-1}\mid X_0=1), \quad k\geq 1
\end{align}
(1) is trivial. Now we show (2). For $k=1$, 
$$
P(X_0\neq 0,X_1=0\mid X_1=1)=0=P(X_0=0\mid X_0=1).
$$
For $k>1$, 
\begin{align}
&P(X_0\neq 0,X_1\neq 0,\cdots,X_{k-1}\neq 0,X_k=0\mid X_1=1)\\
=&\begin{cases}
P(X_2=0\mid X_1=1)&k=2\\
P(X_2\neq 0,\cdots,X_k=0\mid X_1=1)&k>2
\end{cases}\tag{3}\\
=&\begin{cases}
P(X_1=0\mid X_0=1)&k=2\\
P(X_1\neq 0,\cdots,X_{k-1}=0\mid X_0=1)&k>2
\end{cases}\tag{4}\\
=&P(A_{k-1}\mid X_0=1)
\end{align}
where in (3) we use the fact that $\{X_1=1\}\subset\{X_0\neq 0,X_1\neq 0\}$ and in (4) we apply the basic properties of Markov chains. 
