Some questions about CMP CS2 CH5 to CH7

Discussion in 'CS2' started by ykai, Aug 5, 2023.

  1. ykai

    ykai Ton up Member

    1.CMP CS2-CH5-question5.10-(iv)
    Does the basis of first or last jump in the formula refer to mu?

    2.What is the first principle meaning in CMP CS2-CH5-questions5.8-(ii),5.9-(i),5-10-(i)?
    I don't know how to deal with this kind of problem.
    I don't know what it ask for.
    What is first principle of each questions?

    3.CMP CS2-CH7-3.2-p11-c
    Did the death happen before or after the cendord?

    4.CS2 Assignments X2-2.7-(a)
    What is logic in solution?
    I can't understand logic of "The quickest way to do this is to multiply 0.11765 by 18, then by 17 etc, until you get very close to an integer. Here we get 0.11765× 17 = 2.00005 so we conclude that there must have been 2 jockeys falling off their horses out of 17 ".
    Why we use the closest value to find it?
    Is there any provement to prove it?

    5.CS2 Assignments X2-2.8-(i)-(b)
    Why the value of first row and forth column in generator matrix is 0?
    Why can't two lives die at the same time?
     
  2. Andrew Martin

    Andrew Martin ActEd Tutor Staff Member

    Hello

    Please see some comments below:

    1.

    I'm not entirely sure what you mean here. The equations are derived by considering different ways of breaking up all the possible paths from i to j. For example, we can condition on where the first jump out of state i went and what time this happened. By considering all possible times the first jump could take place and all the possible places it could go, we get the backwards equations.

    Another way is to condition on the last jump, where it came from and what time it occurred. Considering all these possibilities is the basis of the forward equations.

    Finally, we should also consider the process staying in the same state for the entire duration, but only if i = j.

    2.

    It means deriving the differential equations (as opposed to, for example, stating them). Have a look at sections 2 and 4 in this chapter.

    3. We assume deaths happen before the censorings (so that those lives censored are assumed to have been at risk of that death event).

    4. Here we've been presented with the cumulative hazard function. Someone else calculated it from some data on these 18 jockeys and we need to 'back out' or 'reverse engineer' their workings.

    As the cumulative hazard function changes at time 2, there must have been a death event at time 2. The estimate of the discrete hazard at that time must be the 0.11765 (as the numbers presented are the cumulative hazards and, at this time, there is only one).

    So, we must have that:

    d1 / n1 = 0.11765

    where d1 is the number of jockeys who fell off at time 2 and n1 is the number at risk of falling off at time 2. (Note that the 1 index refers to the fact that this is the first death event, which happens at time 2).

    We must have that n1 <= 18, as there are 18 total in the study. The decimal is just bigger than 1/10, so we could try 2/18 as a first guess. This is a bit low, so we try 2/17 and this seems to work.

    We can continue in this fashion, noting that, for example, the second death event occurs at time 5 and the estimate of the hazard at this point is 0.34842 - 0.11765 and so we again have to find a fraction equal to this difference.

    5. Two continuous random variables (here the survival times) have probability 0 of taking the same value. The model allows for as small a time period between the deaths as you like (eg there is a non-zero probability of a transition from 1 to 2 and then 2 to 4 in very quick succession) but just doesn't allow for them to occur at the exact same time.

    Hope this helps!

    Andy
     
  3. ykai

    ykai Ton up Member

    Thank you for your response, but I still have some questions.

    1.What I told is that mu in backward method seems to mean stay in original state before jump.
    backward method:
    d/dt pij(s,t)=sum(mu_ik(s)*p_kj(s,t))
    p_ij(s,t)=sum(inegral(p_ii(s,s+w)*mu_ik(s+w)*p_kj(s+w,t))), for all i not j

    mu in forward method seems to mean stay in end state after jump.
    forward method:
    d/dt p_ij(s,t)=sum(p_ik(s,t)*mu_kj(s))
    p_ij(s,t)=sum(inegral(p_if(s,t-w)*mu_kj(t-w)*p_jj(t-w,t))), for all i not j

    5.Why model don't allow it?

    Can this condition be inferred from the question?

    It's possible for two people to die at the same time, isn't it?
     
  4. Andrew Martin

    Andrew Martin ActEd Tutor Staff Member

    Hello

    1. Again, sorry but not entirely sure what you're asking here, could you try rephrasing it? In both cases, mu_ij refers to the transition rate from state i to state j. The mu's that appear in each equation are different and this does relate to nature of how the forward and backward equations are constructed.

    Just to note that your backward and forward equations are not quite right. For example the backwards differential equations should be differentiated with respect to the start time, s, in the time inhomogeneous case (you appear to be using the notation for this case). It should also have a negative sign. It may be worth looking up the formulae for these again in the notes.

    5. I would always assume that, in any of these types of questions about a jump process, it is never possible for two events to happen at the exact same time.

    Two people dying at the exact same time has probability 0 (think about measuring time in milliseconds or nanoseconds or even smaller units of time, two people dying at exactly the same time has 0 probability).

    Hope this helps!

    Andy
     
  5. ykai

    ykai Ton up Member

    Thank you for your response.
    I have totally understood!
     

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