• We are pleased to announce that the winner of our Feedback Prize Draw for the Winter 2024-25 session and winning £150 of gift vouchers is Zhao Liang Tay. Congratulations to Zhao Liang. If you fancy winning £150 worth of gift vouchers (from a major UK store) for the Summer 2025 exam sitting for just a few minutes of your time throughout the session, please see our website at https://www.acted.co.uk/further-info.html?pat=feedback#feedback-prize for more information on how you can make sure your name is included in the draw at the end of the session.
  • Please be advised that the SP1, SP5 and SP7 X1 deadline is the 14th July and not the 17th June as first stated. Please accept out apologies for any confusion caused.

Ch12 Q12.4

B

Brett Kim

Member
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?
 
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
 
Back
Top