Find steps to reach absorbing markov chain state

How can I find the steps it takes or days or whatever the time variable is till the matrix reaches the absorbing state.

e.g. take the matrix (The probability of each row adds to 1) $$\left[ \begin{array}{ccc} 0.3 & 0.5 & 0.2 \\ 0.6 & 0.4 & 0 \\ 0 & 0 & 1 \\ \end{array} \right]$$ please show the process

• you claim the column add to 1 but it's the rows in your matrix that sum to 1 – gota Oct 25 '17 at 11:56

You can't find "the steps it takes" because that's a random quantity. But you can find the expected number of steps it takes, starting in a given state. In your example, let $u_1$ and $u_2$ be the expected number of steps until reaching the absorbing state, starting in states $1$ and $2$ respectively.
Suppose, for example, you start in state $1$.
The first column of the matrix says that after the first step, with probability $0.3$ you stay in state $1$ (and then the expected number of additional steps until reaching the absorbing state is again $u_1$); with probability $0.6$ you are in state $2$ (and then the expected number of additional steps is $u_2$), and with probability $0.1$ you are in the absorbing state $3$ so no additional steps are needed. This says $u_1 = 1 + 0.3 u_1 + 0.6 u_2$. Do a similar analysis for starting in state $2$, getting another equation, and solve the system of two equations for $u_1$ and $u_2$.
For a chain with an absorbing state, the time until absorption is asymptotically distributed according to $1 - \alpha^n$ where $\alpha$ is the Perron-Frobenius eigenvalue corresponding the the sub-stochastic matrix obtained by deleting the row and column corresponding to the absorbing state: in your case, $3$ is absorbing, so let $\tau_3$ be the time to hit state $3$. Then $$\lim_{n \to \infty} P[\tau_3 > n]^{1/n} = \alpha$$ where $\alpha$ is the largest eigenvalue of the matrix $$\left[ \begin{array}{cc} 0.3 & 0.5 \\ 0.6 & 0.4 \\ \end{array} \right].$$
• Well, the matrix you have is like that. The upper left $2\times 2$ matrix $A$ is the same as the one I gave, to the right of that you have a $2 \times 1$ zero matrix, below that a $1\times 1$ identity matrix, and take $B$ to be what's left over. $A$ governs how the chain moves about the non-absorbing states of the space, $B$ gives the probability of being absorbed from a given state, and the identity matrix is there because there are no transitions $i \to j$ for $i \neq j$ once its absorbed. Hope that helps – Rookatu Mar 25 '14 at 18:21