Ch12 Q12.4

[Brett Kim]

This question is in regard to Q12.4(i) in the ch12 end of chapter questions. In it, we're provided with a Markov jump process model of Healthy, Disease (Y&Z), Death states and corresponding transition rates, and the question asks us to calculate the probability that a healthy life aged 70 is dead by the end of the year i.e. P(hd).

Because it's a Markov jump model I tried to use the CK differential equations i.e. the probability of staying in H until time t then instantaneously jumping to D + the probability of going from H to Y in time t, then instantaneous jumping from Y to D + probability of going from H to Z in time t then instantaneously jumping from Z to D.

Mathematically:
d/dt p(hd) = mu(hd)p(hh) + mu(yd)p(hd) + mu(zd)p(zd)
p(hd) = mu(hd)int(p(hh)) + mu(yd)int(p(hd)) + mu(zd)int(p(hd))
and then using the fact that: p(ij) = exp(-mu(ij)*t)
to basically end up with
p(hd) = 0.014*int(0,1)(exp(-0.026t) dt) + 0.4*int(0,1)(exp(-0.005t) dt) + 0.7*int(0,1)(exp(-0.007t))

However, calculating this gives an answer > 1 so it's obviously wrong. I've looked at the answer and I get where it's coming from, but I don't get what is wrong with the reasoning above. Am I applying the markov chain stuff where it really shouldn't be applied?

 

[Dave Johnson]

Hi Brett

The error you are making is writing p_ij(t) = exp{-mu_ij * t}

(Sorry - I've altered your notation to make it clear this is a function of t.)

You've used the formula for a holding probability (ie remaining in a state for a period) and it doesn't generalise when there are transitions. It would be OK to write p_HH(t) = exp{-mu_HH * t} for example, because return to state H is impossible, so this probability implies no jumps, but we can't write p_HD(t) = exp{-mu_HD * t}.

The problem you have with this method more generally is that this system is sufficiently complex that we can't write p_HH(t), p_HY(t) and p_HZ(t) in terms of p_HD(t), which means we can't use our usual tools to integrate the expression.

So I would focus on the model solution - it's the only way I can see that you will solve this.

Hope that helps.

Dave