Show all other states are transient in Markov chain

[xentity1x]

Let X be a Markov chain with a state s that is absorbing, i.e. pss(1) = 1. All other states
communicate with s i.e. i → s for all states i ∈ S. Show that all states in S except s are
transient.

I understand this intuitively, but I'm not really sure how to start the proof.

 

[lanedance]

write down a transition matrix (which can be reasonably arbitrary, make the absorbing state the first for simplicity) and consider the applying it again and again, and think what happens in the limit...

 

[xentity1x]

Ok so I wrote down the transition matrix with s=1.
\begin{bmatrix}
\begin{array}{cc}
1 & 0 & ... & 0 \\
p_{21} & p_{22} & ... & p_{2N} \\
\vdots & \vdots & \ddots & \vdots \\
p_{N1} & p_{N2} & ... & p_{NN} \\
\end{array}
\end{bmatrix}

I then raised it to the nth power
\begin{bmatrix}
\begin{array}{cc}
1 & 0 & ... & 0 \\
p_{21}(n) & p_{22}(n) & ... & p_{2N}(n) \\
\vdots & \vdots & \ddots & \vdots \\
p_{N1}(n) & p_{N2}(n) & ... & p_{NN}(n) \\
\end{array}
\end{bmatrix}

So I guess I have to show \lim_{n\rightarrow \infty} p_{ij}(n)=0 \hspace{3mm} \forall i,j>1

I'm not really to sure where to go from there.

 

[lanedance]

not quite p_{ij}^{(n)}will represent the probability of transitioning from state i at time 0, to state j at time n. So you would expect
p_{ij}^{(n)} = 1 / / i=1
p_{ij}^{(n)} = 0 / / i \neq 1I would start by looking at P^2, P^3 to get a feeling for the form and what happens and try and generalise from there

keep in mind each row of P, P^2 and so forth must sum to one...