Expected number of steps to absorbtion - Markov chain

Question

I want to calculate the expected number of steps (transitions) needed for absorbtion. So the transition probability matrix $P$ has exactly one (lets say it is the first one) column with a $1$ and the rest of that column $0$ as entries.

$P = \begin{bmatrix} 1 & * & \cdots & * \\ 0 & \vdots & \ddots & \vdots \\ \vdots & & & \\ 0 & & & \end{bmatrix} \qquad s_0 = \begin{bmatrix} 0 \\ \vdots \\ 0 \\ 1 \\0 \\ \vdots \\ \\ \end{bmatrix}$

How can I now find the expected (mean) number of steps needed for the absorbtion for a given initial state $s_0$?

EDIT: An explicit example here:

$P = \begin{bmatrix} 1 & 0.1 & 0.8 \\ 0 & 0.7 & 0.2 \\ 0 & 0.2 & 0 \end{bmatrix} \qquad s_0 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \implies s_1 = \begin{bmatrix} 0.8 \\ 0.2 \\ 0 \end{bmatrix} \implies s_2 = \begin{bmatrix} 0.82 \\ 0.14 \\ 0.04 \end{bmatrix} \ldots $

Answer 1

We conventionally read row by row, but it appears that in your matrix, the columns add to 1 instead. I'll write your matrix like this and proceed:

\begin{align*} P = \begin{bmatrix} 0 & 0.2 & 0.8 \\ 0.2 & 0.7 & 0.1 \\ 0 & 0 & 1 \end{bmatrix} \end{align*}

which is of the form

$$P=\left[ \begin{array}{c|c} Q & R \\ \hline 0 & 1 \end{array}\right] $$

and

\begin{align*} s_0 = \begin{bmatrix} 1 & 0 & 0 \end{bmatrix} \end{align*}

For the expected time to absorption, we need to calculate

$\left(I-Q\right)^{-1} \cdot \begin{pmatrix} 1 \\ 1 \end{pmatrix} \approx \left(\begin{array}{r} 1.92307692307692 \\ 4.61538461538461 \end{array}\right) $

Answer 2

Let $t_i$ denote the mean time to absorption at state $0$ starting from state $i$ then $t_0=0$, one is after $t_2$, and the usual one-step Markov property yields $$t_2=1+\frac15t_1,\qquad t_1=1+\frac7{10}t_1+\frac15t_2,$$ which is solved by $$ t_2=\frac{25}{13}=1.923\ldots $$ For $n+1$ states, one gets an $n\times n$ affine system whose solution $(t_i)$ describes the mean times to absorption at state $0$ starting from each state $i\ne0$.