# Simulating a Continuous time markov chain

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### Accepted Answer

Dana
on 27 Aug 2020

See here:

##### 20 Comments

susman
on 2 Feb 2021

Thank you Dana. I have to work with non-homogeneous continuous time markov chains. But as a first looking a homogeneous version gives a very clear picture through your code. Please let me know if you know any good reference for non-homogeneous CTMC especially the coding part.

Secondly, regarding your code, I have following Q matrix with one absorpbing state as death and I set the parameters as T = 65 and nsims = 1000. But my simulations remain a straight line and there are no transitions. All simullations remain in state zero from the very begining. However, if I remove the absorbing state, it works well. Can you explain the reason for that?

Q = [-2.79760000000000 0.628400000000000 0 0 2.16920000000000;

0 -4.40770000000000 0.829800000000000 1.44820000000000 2.12970000000000;

0 0.672400000000000 -2.56400000000000 0 1.89160000000000;

0 0.413900000000000 0 -2.53480000000000 2.12090000000000;

0 0 0 0 0]

Basma Bargal
on 11 Sep 2021

Hi Susman,

Wondering whether you found any code or references on how to go about non-homogenous MArkov Chains in Matlab?

Thanks

### More Answers (1)

Dana
on 3 Feb 2021

Edited: Dana
on 3 Feb 2021

I don't know offhand of any non-homogeneous CTMC references, sorry.

Regarding your case, this part of the help section regarding ths inputs of simCTMC.m is relevant:

% nsim: number of simulations to run (only used if instt is not passed in)

% instt: optional vector of initial states; if passed in, nsim = size of

% instt; otherwise, nsim draws are made from the stationary

% distribution of the Markov chain (if there are multiple stationary

% distributions, an error is returned).

So if you don't pass in a vector of initial states in the variable instt, the program randomly draws nsim initial states from the stationary distribution of the MC. But when you have an absorbing state, the stationary distribution of the MC has probability 1 associated with the absorbing state, and 0 for every other state. If you draw the initial state randomly from this stationary distribution, you're always going to start with the absorbing state, and then of course you'll stay in that state forever, which obviously isn't very interesting.

Long story short, for cases like these you probably want to pick your own initial vector instt. How to pick it is up to you.

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