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ShortTerm Actuarial Math Old Exam C Forum 

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#1




TIA Review Problem 235
How do they calculate the NelsonAalen estime of H(X), used with the survival function, for Jack to be 1/3 and Jill to be 1/3 + 1/4?
I'm interpreting this problem as 5 regular observations, plus 2 censored observations at 25 and 35. 
#3




This is the most recent post on this problem that I could find. I still don't understand how the survival for Jack is 1/3 (assuming then I could figure out Jill).
If I back out to find H from S= 1/3 I get H = 1.098612289. that is close to 1/7 + 1/6 + 1/5 + 1/4 + 1/3 which makes sense to me only if both Jack and Jill ran beyond 70 laps. I really have no clue how this problem was solved. The solution just says "the probability that jack runs 50 or more laps is e^1/3." But How? 
#4




We are given that Jack ran at least 35 laps. If we start a NelsonAalen estimate, looking only at what happens after time 35, there are 3 data points (45, 55, and 70), only one of which is below 50, giving a value of Hhat of 1/3, for Shat = e^{1/3}.
If you don't feel comfortable starting at 35, you could say that the estimated P[X>50  X>35] is Shat(50)/ Shat(35), and Hhat(50) = 1/5 + 1/4 + 1/3, while Hhat(35) = 1/5 + 1/4, so Shat(50)/Shat(35) = e^{(1/5+1/4+1/3)}/e^{(1/5+1/4)} = e^{1/3}. See near the end of C.2.4 for a comparison of these two approaches for conditional problems. There's also a lengthy discussion of this at http://www.theinfiniteactuary.com/mb...=19590&t=18866 
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