Consider a Markov chain on $\{0, 1, . . . , d\}$ satisfying $$\displaystyle \sum_{y=0}^d y\cdot p(x,y)=x,\text{where $p(x,y)$ is the transition function} \quad (*)$$ and having no absorbing states other than $0$ and $d$. Show that the states $1 , . . . , d - 1$ each lead to $0$, and hence that each is a transient state.
My try: Consider that $p(0,0) = 1$ and that $p (d, d) = 1$.
It must be shown that $x$ lead to $0$ for $x \in \{1,2,...,d-1\}$
Let be $x=0$. Then $$0=\displaystyle \sum_{y=0}^d y\cdot p(0,y)=\displaystyle \sum_{y=1}^d y\cdot p(x,y)$$ that implies that $p(0,y)=0$ for $y \in \{1,2,...,d\}$
Now, let be $x=1,...,d-1$. Then observe that $$\displaystyle \sum_{y=0}^d x\cdot p(x,y)=x \quad(**) \quad$$
Then subtracting (*) and (**), we have:
$$\displaystyle \sum_{y=0 \\ y\neq x} (y-x) p(x,y)=0$$
But after I don't know what to do, someone could give me a hint or another way to do this proof?
Since states $1$, $2$, ..., $d-1$ are non-absorbing, the transition probability $P(x,x)$ is less than $1$ when $x$ is one of these states. Hence $P(x,y)$ is non-zero for at least one $y\ne x$ when $x$ is one of these states. But your condition implies that there must be at least two such states with $P(x,y)\ne0$, one with $y<x$ and one with $y>x$. Once you know that for every $x\in\{1,2,\ldots,d-1\}$ there is at least one allowed transition $x\to y$ with $y<x$, do you see how to argue that there must always be a path to state $0$?