Elt 15

Discussion in 'CT4' started by snell1, Apr 4, 2012.

  1. snell1

    snell1 Member

    Hi just wondering is there any quick explanation why µx is less than qx for certain values of x in the ELT 15 Tables. Also why is qx not always equal to dx/lx? All of the x are supposed to be subscript.
     
    snell1, Apr 4, 2012
    #1
  2. John Potter

    John Potter ActEd Tutor Staff Member

    There's not really any reason to compare the absolute values of mux and qx. One is a force, the other is a probability.

    qx = dx/lx

    If you've found a counter example then it's a mistake. Please tell us about it and we will get the Tables corrected.

    John
     
    John Potter, Apr 12, 2012
    #2
  3. snell1

    snell1 Member

    It dosen't look like a mistake. qx starts to drift away from dx/lx at around age 85 and is very different at age 108 where lx = 2, dx = 1 and qx = 0.56225.
     
    snell1, Apr 12, 2012
    #3
  4. Calum

    Calum Member

    One would hope that an error like that would be noticed quicker than 10years!

    Snell, if the force of mortality were perfectly constant over several years, it would be equal to q_x. Normally, it is increasing, so mu_x is normally slightly larger than q_x.

    If mu_x is decreasing, then, you would expect lower values. This is typical of the back end of the accident hump.

    I can't find anything for the methodology behind deriving ELT15, but the usual approach these days is to estimate forces of mortality and derive everything else from that. I suspect the l_x, d_x numbers are derived and rounded in this way.
     
    Calum, Apr 13, 2012
    #4
  5. John Potter

    John Potter ActEd Tutor Staff Member

    The dx/lx drifting away at the end is just rounding. Perhaps the 100,000 lives we start with is more like 1,000,000 but we only see it to the nearest thousand. Or, I like Calum's idea that everything is worked out from mu(x) and we work backwards, again with rounding problems.

    Calum, careful with this qx = mu x idea.

    I don't know what "perfectly constant" means as opposed to just constant but, if mu is constant between x and x+1 then we would get:

    qx = integral beteen 1 and 0 of tpx mu(x+t) dt
    = mu * integral beteen 0 and 1 of tpx dt
    = mu * amount of time you expect someone to live over the next year
    = mu * 0.99ish probably
    <> mu

    Or, qx = 1 - exp(-mu) from page 32 Tables
    = mu - mu^2 / 2! + mu^3 /3! - .... (using page 2 Tables)

    Also, in ELT15 we have mu(x) < qx at age 85 and mu(x) > qx at age 86 in ELT15 males. So, what you're saying here isn't really true either :-(

    I agree that, if mu(x) is decreasing then qx - mu(x) will be bigger than it otherwise would have been. But this doesn't necessarily mean that qx - mu(x) is positive. Again, we can't really compare the absolute values of mux and qx because one is a FORCE, the other is a PROBABILITY.

    Good luck!
    John
     
    Last edited: Apr 13, 2012
    John Potter, Apr 13, 2012
    #5
  6. Calum

    Calum Member

    Hi John,

    My previous answers were without the benefit of ELT15 in front of me, so I'll try again!

    Yes, my mistake - for constant mu, q_x is slightly smaller. I was going off the q_x~=d.mu_x approximation and not thinking it through!

    q_x in the tables is the mortality of an average x-year-old, so in fact the value given is the exact mortality of a hypothetical x+0.5 year old [so q_x = 1-exp(0.5(mu_x+mu_x+1))]. At least this seems to be the case from a few sample calculations. I think this explains Snell's question - q_x<mu_x exact at most points in time but that is not the values the tables give.

    Around 85 the increase in mortality over a year starts to dominate and so the numbers reverse.
     
    Calum, Apr 14, 2012
    #6

Share This Page