I dont quite understand the second part of the proof of this theorem:
The vector of hitting probabilities $h_A = (h_A(x) : x ∈ S)$ is the smallest nonnegative solution to the system of equations
$f(x)= \sum_{y \in S}P(x,y)f(y), x\notin A$
$f(x)=1 , x \in A$
So the proof goes like this:
Let us first verify that the hitting probabilities satisfy the equations. Again we denote conditional probabilities given $X_0 = x$ by $P_x$. Then $h_A(x) = \Bbb P_x(T_A < ∞)$, where $T_A$ is the passage time of the chain into set $A$. If the initial state $x ∈ A$, then the chain surely visits $A$, so that $h_A(x) = 1$. Assume next that $x \notin A$. Then by applying the Markov property we may conclude that
$\Bbb P_x(T_A < ∞ | X_1 = y) = \Bbb P(T_A < ∞ | X_1 = y, X_0 = x) = \Bbb P(T_A < ∞ | X_1 = y) = h_A(y)$,
so that ,
$h_A(x) = \Bbb P_x(T_A < ∞) = \sum_{y \in S} \Bbb P_x(X_1=y) \Bbb P_x(T_Q \lt \infty | X_1=y) = \sum_{y \in S} P(x,y) h_A(y)$ Hence $(h_A(x) : x ∈ S)$ is a nonnegative solutions to the equation.
Now here goes the part which I don't understand:
Assume next that $f = (f(x) : x ∈ S)$ is some nonnegative solutions to the equations and let us show that then $f(x) ≥ h_A(x)$ for all $x$. Now obviously $f(x) = hA(x) = 1$ for all$ x ∈ A$. If $x \notin A$, then
$f(x)= \sum_{y \in S}P(x,y)f(y)= \sum_{y \in A}P(x,y) + \sum_{y \notin A} P(x,y)f(y)= \Bbb P_x(X_1 \in A) + \sum_{y \notin A}P(x,y)f(y)$
**Question :**How is $\sum_{y \in S}P(x,y)f(y)$ equal to $\sum_{y \in A}P(x,y) + \sum_{y \notin A} P(x,y)f(y)$? I don't understand it. The "$ y \in A$" and "$y \notin A$" parts??
Also, how is $\sum_{y \in A}P(x,y) $ equal to $\Bbb P_x(X_1 \in A) $ ? Why exactly $X_1$? Why not $X_0$ I'm confused by this. The proof continues which part I don't obviously understand, because I don't understand this part. If someone could exaplin me that would be great!
You didn't introduce any of the notation you used, but from what I gather, $S$ is the set of all states, $A$ is the set of states to be hit, and $P(x,y)$ is the transition probability from state $x$ to state $y$.
$\sum_{y\in S}P(x,y)f(y)=\sum_{y\in A}P(x,y)f(y)+\sum_{y\not\in A}P(x,y)f(y)$ simply because every $y\in S$ is either in $A$ or not in $A$. In the first sum, you can drop $f(y)$ because $f$ is assumed to be a solution to the equations, which include $f(x)=1$ for $x\in A$.
$\sum_{y\in A}P(x,y)=\mathbb P_x(X_1\in A)$ because by definition $P_x(X_1\in A)$ is conditional on $X_0=x$; then $P(x,y)$ is the probability that the first step leads from $X_0=x$ to $X_1=y$; and then summing over $y\in A$ yields the probability that $X_1\in A$.