Markov Chains: illustration that the equilibrium state does not determine the transition matrix

Andy Long
Spring, 2018

This is an example to accompany Olinick’s book An Introduction to Mathematical Models in the Social and Life Sciences.

Suppose you start with a certain SIR equilibrium; can we determine the SIR model that gave rise to it? 
No -- it’s not unique.

If the equilibrium vector (probability vector, which sums to 1) is ((Subscript[w, S]),(Subscript[w, I]),(Subscript[w, R])), then so long as
(Subscript[w, S])(Subscript[p, SI])=(Subscript[w, I])(Subscript[p, IR])=(Subscript[w, R])(Subscript[p, RS])
this Markov chain will stay in equilibrium. 

There’s only one free parameter in the transition probability matrix: let’s choose to make it (Subscript[p, SI]): without that, we have no infectivity! 

Then 
(Subscript[p, IR])=((Subscript[w, S])/(Subscript[w, I]))(Subscript[p, SI])
and
(Subscript[p, RS])=((Subscript[w, S])/(Subscript[w, R]))(Subscript[p, SI])
Here’s an example:

{ws,wi,wr}={.3,.1,.6};
(* Let this be our stable distribution: *)
ws/wi
ws/wr
equilibrium={ws,wi,wr};
Manipulate[
Module[
{prs,pir},
prs=ws/wr psi;
pir=ws/wi*psi;
probabilities={psi,pir,prs};
Print[probabilities];
TransitionMatrix={
{1-psi,psi,0     },
{0     ,1-pir, pir},
{prs,     0, 1-prs}};
Print[MatrixForm[TransitionMatrix]];
Print[MatrixForm[MatrixPower[TransitionMatrix,30]]];
equilibrium.TransitionMatrix
]
,{{psi,.1},0,Min[1,wi/ws,wr/ws],.01}
,Text["The equilibrium states don't change:"]
]