Ch 12 Competing risk

[DonnieDarkest]

Hi,

I'm having trouble understanding the derivation of the partial differential equations in Chapter 12 Competing Risks. In the simple example, a transition is possible from A to R, from R to D and from A to D.

I think I understand why we need v=0 (tranition intensity from R to D) in order for (aq)d = pAD ...otherwise pAD would also include the possibility of a transition from A to R then from R to D, with v=0 this is now not possible.

When this logic is extented, why do we also need to set v=0 in order for (aq)r = pAR ?
Is (aq)r = pAR by definition regardless of what v is?

Thanks!

 

[Mark Mitchell]

The key here is that there are two types of notation being used, and they only end up being equal in certain circumstances.

So:

t(aq)r = P(individual who is currently in the active state leaving this state due to retirement in time period t in the presence of other decrements)

tpAR = P(individual is in state R at time t given that they were in the active state at time 0)

Note that the first of these is to do (only) with the probability of leaving a state in a certain time period. The second is specifically to do with the state and individual occupies at time t.

For an individual to be in state R at time t, they must leave state A by retirement at some point during the time period t, AND they must not leave state R after they arrive.

The only way to ensure that they don't leave state R is to have v=0. This means that we are certain that once an individual gets to state R, they must stay there as there is no way out (R is then technically called an absorbing state).

This gives us the equality t(aq)r = tpAR (where v=0). We can then be certain that if the individual has left state A due to retirement between time 0 and time t (first part of the equation), they must be in state R at time t (second part of the equation), as once they get into R, they're stuck there.

Hoping this clears it up.

M.