Showing that $p_{k}=\frac{k}{k-1}p_{k-1}$ forall $k$ implies $p_{k}=\frac{k}{n}$

Question

Suppose I have a Markov process over $n+1$ states $s_0,\ldots,s_n$. For $k\notin\{0,n\}$ the transition probabilities from $s_k$ are $\frac{1}{2}$ to both $s_{k-1}$ and $s_{k+1}$ and for $k\in\{0,n\}$ there is probability 1 of staying in place (absorbing states).

Denote by $p_k$ for $0<k<n$ (the transient states) the probability of ending up in $s_n$ when starting at $s_k$. Iv'e managed to show that for all such $k$

$$p_{k}=\frac{k}{k-1}p_{k-1}$$

And now need to show that this implies that $p_{k}=\frac{k}{n}$. This seemed easy enough (the question was phrased as "conclude that $p_{k}=\frac{k}{n}$") but for some reason I'm not able to. So far I only got

$$p_{k}=\frac{k}{k-1}p_{k-1}=\frac{k}{k-1}\frac{k-1}{k-2}p_{k-1}=\ldots=\frac{k}{k-1}\frac{k-1}{k-2}\frac{k-2}{k-3}\ldots\frac{2}{1}p_{1}=kp_{1}$$

I also showed, earlier in the same question, that the stationary distribution vector of this process has to be of the form $\pi = \alpha e_0 + (1-\alpha)e_n$ for some $\alpha\in [0, 1]$, a fact which may or may not be relevant.

Answer 1

This is the symmetric gambler's ruin problem and there are many ways to solve this. (If you know anything about Renewal Theory, applying the Wald Equation, which is also a very simple martingale, is my vote.)

A nice finish that makes use of your existing work, i.e. with knowledge that

$p_{k}=\frac{k}{k-1}p_{k-1}=\frac{k}{k-1}\frac{k-1}{k-2}p_{k-1}=\ldots=\frac{k}{k-1}\frac{k-1}{k-2}\frac{k-2}{k-3}\ldots\frac{2}{1}p_{1}=kp_{1}$

is to assign $p_1 = c$ so $p_k =c\cdot k$ and determine $c$ via a suitable boundary condition.

Easy finish:
supposing your gambler starts in transient state $k\in\{1,2,..., n-1\}$, you can make use of 'first step analysis'. That is the probability of being absorbed in state $n$ is $p_k$ but if we condition of the first 'step' we have (this is total probability, or total expectation if you use indicators)

$p_k = \frac{1}{2}p_{k-1} + \frac{1}{2}p_{k+1}$
or
$p_{k+1} = 2\cdot p_k- p_{k-1}$
(note: solving this recurrence directly is yet another way to get the solution to Gambler's Ruin.)

Now select $k:= n-1$ and this reads
$1 = p_n= 2 p_{n-1}- p_{n-2} = 2 \cdot c \cdot (n-1) - c \cdot (n-2) = c\cdot\big(2n-2 -(n-2)\big) = c\cdot \big(n\big)\longrightarrow c = \frac{1}{n}$
which is the boundary information you needed.

Answer 2

Let $\{X_n:n=0,1,\ldots\}$ be a Markov chain on $\{0,1,\ldots,n\}$ with your given transition probabilities. Then in fact $X_n$ is a martingale, for clearly $$ \mathbb P(X_n\in\{0,1,\ldots,n\})=1 \implies \mathbb E[|X_n|]<\infty $$ and $$ \mathbb E[X_{n+1}\mid X_n] = \begin{cases}s_n,& X_n=s_n\\ s_0,& X_n=s_0\\ \frac12(X_n+1) + \frac12(X_n-1) = X_n,& X_n\notin\{s_0,s_k\} \end{cases}. $$ See my answer here for showing that $p_k = \frac kn$, $k=1,\ldots,n-1$ when $X_n$ is a martingale: Markov Chain Martingales Dembo 6.1.18