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  #111  
Old 03-13-2019, 05:08 PM
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Quote:
Originally Posted by Chappy View Post
Is it because the first problem specifically states that the survival distribution follows a uniform RV? Without that assumption, I would need to assume the distribution is Beta?
Yes, the uniform in the first is important. But without knowing it was uniform, you couldn't do the problem at all. It would not necessarily be Beta. It could be almost anything. You know the distribution of the male mu and female mu at 60, but not at any other age. You wouldn't know that the ratio at every age was 60%, so you wouldn't know Beta.
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  #112  
Old 03-13-2019, 05:09 PM
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For example, they could have told you that the distribution for females was exponential, which would give yet another answer to the question.
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  #113  
Old 03-13-2019, 05:53 PM
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Originally Posted by Chappy View Post
Can somebody explain the difference between these two problems.

3-F01:33
S_0 for males is uniform(0,75)
S_0 for females follows uniform(0,w)

At age 60, the female force of mortality is 60% of the male force of mortality.

In the solution, they use .6/15 = .04 and thus w = 25 for a female age 60.

55.14 of ASM 15th edition.
Nonsmokers havea force of mortality equal to one-half that of smokers of equal age.
For nonsmokers, S_0 is uniform(0,75)
(55) is a nonsmoker
(65) is a smoker
Calculate e_65:55

In this problem, the solution uses the fact that the female is a beta distribution with A = 2 and T =10.

What is the difference in these two questions. When I first solved the first problem up above, I original thought it was a beta with A = .6 but couldn't get the right answer. When I solved the second one, I went to smokers being uniform with T = 5 (2/10 = 1/5).
In the first case, one force of mortality is 60% of the other AT ONE SPECIFIC AGE. In the second case, one force of mortality is 50% of the other AT ALL AGES.

By the way, the Uniform model has almost completely disappeared from LTAM exams since the change to the Dickson/Hardy/Waters textbook, Beta even moreso. What you see are problems using the Standard Ultimate Life Table or constant forces.
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  #114  
Old 03-14-2019, 02:47 AM
SweepingRocks SweepingRocks is offline
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Happy Pi day all!

I just finished up the material. Going to use the next 45ish days to go over practice problems and get a real grasp of it. I feel like I have a better understanding of the material at this point than I did with STAM when I was finished with the first go through.

How's everybody else doing?!
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  #115  
Old 03-16-2019, 08:18 AM
NchooseK NchooseK is offline
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ELM 10, but I need to go faster!

This thing is a race.

The more time left over for WA, the better.

I may request the tiny tables on a billboard, since they slow me down most.
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  #116  
Old 03-17-2019, 09:13 AM
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For ASM manual 1st edition, 3rd printing, on page 633, Example 34C, Abar50 = 0.2 at force of interest 0.04, Abarx = 0.16 at force of interest 0.05, I tried to calculate mortality using formula Abarx=mu/(mu+force of interest) but the two Abarx yield two different mu’s. Is there a reason why? Am I doing anything wrong? Thanks
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  #117  
Old 03-17-2019, 09:34 AM
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I don't have the manual so I can't see the problem. Does it say that mu is a constant? The formula Abarx=mu/(mu+force of interest) only applies for constant mu.
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  #118  
Old 03-17-2019, 01:05 PM
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What is the actual question?
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  #119  
Old 03-17-2019, 02:20 PM
Abraham Weishaus Abraham Weishaus is offline
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It does not state mu is constant, and you've proved that it isn't.


The question asks you to calculate P for a whole life on (50) at delta=.04 if (50)'s force of mortality is 0.01 higher than the force used for the given information at all times.
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  #120  
Old 03-18-2019, 09:22 PM
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Oooh I see!!! Thanks!!!
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