5

I am interested in finding the distribution of time to absorption for a given absorbing Markov chain transition matrix. I've looked at first passage time (where $f_{ij}^{(n)}$ is the probability that the first passage from $i$ to $j$ occurs in exactly $n$ steps).

In my particular case I have a single absorbing state, which I'll call $j$. I think I can find the probability mass function since $P(x=0)=0\cdot f_{ij}^{(0)}, P(x=1)=1 \cdot f_{ij}^{(1)}, \ldots $, but I'm not sure where to go from here.

hoffee
  • 175
  • 5

1 Answers1

5

The distribution of a random variable representing time-to-absorption of a Markov chain with a single absorbing state is called a discrete phase-type distribution.

Summarising the wiki article: assume the last row of the $n\times n$ transition matrix corresponds to the absorbing state and let $T$ be the upper left $(n-1)\times(n-1)$ block. The pmf for the absorption time given initial distribution $\tau$ is then $$p(k) = \tau T^{k-1} T_0 $$ where $T_0$ is the vector formed by the first $n-1$ entries of the last column of the transition matrix, showing that these distributions are generalisations of the geometric distribution.

Matthew Towers
  • 10,854