Hi, I'm struggling to understand (mathematically) why the integral of [1 - CDF(x)] dx is equal to the integral of [xPDF(x)] dx. I can visualise why this is true; as the probability drops, you multiply (integrate) by x for every increment of [1 - CDF(x)] and every increment of [1 - CDF(x)] is the same as the derivative i.e. PDF(x) hence leading from [1 - CDF(x)] dx to [xPDF(x)] dx but this is the 'logical' explanation. When attempting to do the proof i.e. integrate [xPDF(x)] dx = xCDF(x) - [CDF(x)] dx ... well you can see the right hand side is not [1 - CDF(x)] dx because of the xCDF(x) term (this should be just x). Can someone please explain the missing step? Best, MoleMan
Hi MoleMan, I don't think you need to worry about general statistical results to do this question. We need E[T0], so we integrate tp0 from 0 to infinity. This is perfectly fine to use in the exam, we just add up the years that they live IF they live them. I think I might know what you're asking, which is how can we justify this from the original E[ ] definition... E[T]=integral 0 to infinity of t * tpx * mu x+t Set u = t, du/dt = 1 dv/dt = tpx * mu x+t, v = - tpx (from Kolmogorov's fwd diff eqn, which you can prove elsewhere) E[T]= t * - tpx evaluated between 0 and infinity (this is 0 - 0 = 0) + integral 0 to infinity of tpx I would say this is a proof that is very unlikely to be tested looking at past papers. Much more likely is to prove the result I've used half way through, d/dt tpx = -tpx mu x+t. AND solve this to get page 32 of Tables, also a fair chance you get asked this. Good luck! John
Thanks John! Great explanation; very CT4 focused in terms of hazard rates. So we have: E(t) = [t * PDF(t)] dt E(t) = [t * tP0 * h(t)] dt [tP0 * h(t)] dt = -tP0 [t * tP0 * h(t)] dt = t * -tP0| - [-tP0] dt We can set t * -tP0| = 0 because we integrate between 0 and ∞ (although we’re integrating the first term between 0 and 5?) Leading to: E(t) = [t * tP0 * h(t)] dt = [tP0] dt Many Thanks, MoleMan
