# Time to absorption and fraction of time spent in a state in a CTMC

I consider a Markov chain with a single absorbing state with $N$ transient states and I would like to find the expected time to absorption, given an initial state. I write the following equation to characterise the time to absorption from different states.

$$\left(\begin{array}{c|c}0& \mathbf 0_{1\times N} \\ \hline \\ \mathbf {\hat{t}}_{N\times 1} & \mathbf P_{N\times N}\end{array}\right)\begin{pmatrix}1_{1\times1} \\ \mathbf t_{N\times 1}\end{pmatrix}=\begin{pmatrix}1_{1\times1} \\ \mathbf t_{N\times 1}\end{pmatrix}$$

• The first row and column stand for transition to and from the absorbing state in the CTMC.
• The vector $\mathbf {\hat{t}}$ represents the time to go from different states to state 0 in the next transition multiplied by the time spent in that state. For example, if you have a state 1 with a transition rate to 0 of $\mu_1$ and a transition rate to 2 as $\lambda_{12}$, then the entry in $\mathbf {\hat{t}}$ will be $\frac{\mu_1}{\mu_1+\lambda_{12}}\times \frac1{\mu_1}$.
• $\mathbf P$ stands for transition probability matrix among non-absorptive states.
• $\mathbf t$ is the vector of expected times to reach the absorbing state from different states. Continuing with the example in the second bullet, if it is possible to only move to 2 and 0 from 1, the equation will be $t_1=\frac{1}{\mu_1+\lambda_{12}}+\frac{\lambda_{12}}{\mu_1+\lambda_{12}}t_2$.

I have two questions:

1. Is the above fixed point equation correct? Can I say anything about its solution?
2. How do I find the fraction of time spent in different states prior to absorption, given the initial state? If I consider a reduced state space of $N$ states and find it to be irreducible and positive recurrent, is it possible to relate this to the stationary probabilities and the expected time to absorption?
• Are you interested in the expected absorbing time or in the expected time spent in each state before absorption? Your post seems to invoke both so you might want to explain. Even more importantly, you should add HOW you reached the matrix identities you are asking to check.
– Did
Commented Mar 28, 2013 at 10:41
• Thanks @Did. I have added some explanation. The 1st point is about expected time to absorption and the second is on expected fraction of time spent in each state prior to absorption (given an initial state). Commented Mar 28, 2013 at 13:29
• @Did: Could you pl. let me know if this question needs to be edited any better? Commented Apr 7, 2013 at 18:22
• It seems the first line of the product of matrices in the LHS is 0 instead of 1 in the RHS?
– Did
Commented Apr 7, 2013 at 20:05
• Got something from my answer below?
– Did
Commented Jan 13, 2014 at 22:08

Call $o$ the absorbing state. For every state $x\ne o$, let $\mu_x$ denote the rate of transition from $x$ to $o$, $\lambda_{xy}$ the rate of transition from $x$ to $y\ne o$ and $t_x$ the mean time before the absorption at $o$ starting from $x$.
Starting at $x$, the Markov chain leaves $x$ after an exponential time of parameter $\lambda_x=\mu_x+\sum\limits_{y\ne o}\lambda_{xy}$, and goes to a new state which is $o$ with probability $\mu_x/\lambda_x$ and $y\ne o$ with probability $\lambda_{xy}/\lambda_x$. Thus, $$t_x=1/\lambda_x+\sum_{y\ne o}t_y\lambda_{xy}/\lambda_x.$$ This can be summarized as the fact that $(\Lambda\mathbf t)_x=1$ for every $x\ne o$, where $\mathbf t=(t_x)_{x\ne o}$, $\Lambda_{xx}=\lambda_x$ and $\Lambda_{xy}=-\lambda_{xy}$ for every $x\ne y$, from which $\mathbf t$ can be deduced.
Regarding your question (2), a useful trick to compute the mean time $t_{xy}$ spent at $y$ starting from $x$ before absorption at $o$ is to add a transition from $o$ to $x$ at rate $1/\alpha$. Then the Markov chain becomes positive recurrent and it performs some successive cycles $x\to\mathrm{states}\ne o\to o\to x$. The mean length of each cycle is $\alpha+t_x$. Considering its unique stationary distribution $(\pi^\alpha_z)_z$ (which one can compute by the usual procedure), one gets $\pi^\alpha_y=t_{xy}/(\alpha+t_x)$ for every $x\ne o$, that is, $$t_{xy}=(\alpha+t_x)\pi^\alpha_y.$$