I got this question asked: A celltype can exist in four states: A growth state, a resting state, a reproduction
state, or it can be dead. It starts in the growth state. From the growth state it moves to the
resting state with a rate of 0.3 and to the reproduction state with a rate 0.1, or it dies with a
rate of 0.1. From the resting state it moves to the growth state with a rate of 0.2, or it dies
with a rate of 0.05. From the reproduction state it moves to the resting state at a rate 0.5,
or it dies at a rate of 0.3.
What is the expected life time of the cell? Compute the answer numerically, or de-
scribe in detail how such computations are done.
So, the way I imagine you do this is by calculating the fundamental matrix which would look something like this:
With our transition matrix looking like:
$P = \begin{bmatrix}
0 & 0.3 & 0.1 & 0.1 \\
0.2 & 0 & 0 & 0.05 \\
0 & 0.5 & 0 & 0.3 \\
0 & 0 & 0 & 1
\end{bmatrix} $
$F = (I-Q)^{-1} = (I – \begin{bmatrix}
0 & 0.3 & 0.1 \\
0.2 & 0 & 0 \\
0 & 0.5 & 0 \\
\end{bmatrix})^{-1} $
And then you could sum the rows to see the respective life time starting at different states. However, the given answer to the question is very different I do not understand why. The answer to the question is given as:
Can anyone explain this given answer please, would be kindly appreciated.
Best Answer
From the answer we may infer that this is not discrete, but a continuous-time process. So the matrix $Q$ is not a transition matrix, but a $Q$-matrix, sometimes called an intensity matrix. In that sense, the value $Q_{i, j}$ is not a probability that we will transfer from the state $i$ to the state $j$, but the intensity of such transition. Generally, the higher is the intensity, the faster we transit from $i$ to $j$ - note that time is continuous, we do not transit in "turns".
Now, by convention, the intensity matrix has transition intensities on $Q_{i, j}$ when $i \neq j$ and $$Q_{i, i} = \sum_{j \neq i} Q_{i, j}$$ You can observe it in the matrix $Q$ in the solution - the diagonal entries are exactly negative sums of the non-diagonal entries. Specifically, we can obtain a transition matrix $T$ from the intensity matrix as: $$T = \exp(tQ) = I + (tQ)^1+\frac{(tQ)^2}{2!}+\frac{(tQ)^3}{3!}+\cdots,$$ and note that the transition matrix may be dependent on time.
Now, to the proper part of the solution. You are calculating the fundamental matrix as $(I-M)^{-1}$, where $M$ is the transient matrix derived from $T$ using the rows and columns corresponding to the non-absorbing states. In continuous-time Markov process we can as well calculate the fundamental matrix of the process based on the $Q$-matrix. However, without going into much detail, note that the $Q$-matrix already has entries on the diagonal that cause the rows of the matrix to have sums equal to $0$. Therefore, instead of calculating $(I-M)^{-1}$, we simply calculate $-V^{-1}$, where $V$ is the transient matrix derived from $Q$. I am aware that this is highly non-formal explanation, but continuous-time Markov chains are significantly different from their discrete-time variants and this post would need to get much longer to properly explain all the differences. Nonetheless I hope you'll understand the topic a bit better.