0
$\begingroup$

Let $(S(t), I(t), R(t))$ be a continuous time Markov chain SIR model with discrete space, where

  • $S(t)$ stands for the number of susceptible people at time $t$;
  • $I(t)$ stands for the number of infected people at time $t$;
  • $R(t)$ stands for the number of removed people at time $t$.

Suppose $(S(0), I(0), R(0))=(2,1,0)$ and we are interested to investigate the process $(S,I,R)$ only at the transition times (which is a version of the imbedded discrete time Markov chain). Also assume that the contact between people occur with a Poisson process with parameter $\lambda$ and each contact is between two people randomly selected from the three and independent of the Poisson process and any prior or later contacts. Further, assume that in each meeting, if one of the two people is infected and the next one has not yet infected (just susceptible) then the disease will be passed to the new person with probability $p$ and is independent of all other times of contact and transformation of disease. Each infected person will remain infected for an exponentially distributed length of time with rate $\mu$ independently of all other infected people.

Now, my confusion are to fix:

  1. How many states are gonna be there?
  2. What is the transition rate matrix of this continuous time Markov chain?
  3. If the first transition occurs then what is the probability that the process goes to state $(2,0,1)$?

So far, i am thinking that the state space of this process gonna be $$ S=\{(2,1,0),(0,1,2),(0,2,1),(1,0,2),(2,0,1),(0,0,3),(1,2,0),(0,3,0),(1,1,1)\} $$ But I am completely lost in fixing the transition rate matrix of this process. Also, if the state space is as above then, does the transition from the state $(2,1,0)$ to the state $(0,3,0)$ occur? Why or why not?

$\endgroup$
0
$\begingroup$

This is the discrete SIR model. Transitions occur when an individual of type S becomes of type I or when an individual of type I becomes of type R. Transitions S$\to$I occur at rate $\beta S(t)I(t)$ and transitions I$\to$R occur at rate $\gamma I(t)$, thus, if $I(t)=0$ then no further transition occurs. Encoding each state $(S(t),I(t),R(t))$ by the three corresponding integers, this yields the diagram $$\begin{array}{c} 210&\rightarrow&201\\ \downarrow\\ 120&\rightarrow&111&\rightarrow&102\\ \downarrow&&\downarrow\\030&\rightarrow&021&\rightarrow&012&\rightarrow&003 \end{array}$$ Each "horizontal" move occurs at rate $\gamma$ (in three cases) or $2\gamma$ (in two cases) or $3\gamma$ (in one case). Each "vertical" move occurs at rate $\beta$ (in one case) or $2\beta$ (in two cases). The states $201$, $102$ and $003$ are absorbing since they all correspond to $I(t)=0$.

Edit: The mechanism described in the question leads to $$\beta=\tfrac13p\lambda,\qquad\gamma=\mu.$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ This, dear OP, is almost written down explicitely in my answer (but not completely, on purpose), and is absolutely direct from said answer if you understand it, even in the slightiest way. One minute of thought should allow to write it down... unless you are so lacking in the basics of the field that I am at a loss to imagine what any useful answer could look like. Where are you stopped when you try to write down the transition rates matrix? $\endgroup$ – Did Dec 13 '14 at 8:03
  • $\begingroup$ "I don't understand what is β and γ." See Edit. "what is the rate the process will leave the state (2,1,0) as well as (1,1,1)?" This can be deduced from my answer. Please read. Note that you nowhere explain "Where are you stopped when you try to write down the transition rates matrix?" but are just asking again and again for the pieces of the solution not explicitely written down. $\endgroup$ – Did Dec 13 '14 at 20:02
  • $\begingroup$ Yes, I have a suggestion: do your (minimal) part of the job and stop asking for full solutions. Thanks in advance. $\endgroup$ – Did Dec 15 '14 at 10:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy