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#111
03-13-2019, 05:08 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 31,605

Quote:
 Originally Posted by Chappy 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.
#112
03-13-2019, 05:09 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 31,605

For example, they could have told you that the distribution for females was exponential, which would give yet another answer to the question.
#113
03-13-2019, 05:53 PM
 Jim Daniel Member SOA Join Date: Jan 2002 Location: Davis, CA College: Wabash College B.A. 1962, Stanford Ph.D. 1965 Posts: 2,753

Quote:
 Originally Posted by Chappy 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
03-14-2019, 02:47 AM
 SweepingRocks Member SOA Join Date: Jun 2017 College: Bentley University (Class of 2019ish) Posts: 293

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
03-16-2019, 08:18 AM
 NchooseK Member SOA Non-Actuary Join Date: Nov 2012 Location: Philly area College: Swarthmore College (BA Mathematics), Villanova University (MS Applied Stat) Posts: 353

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
03-17-2019, 09:13 AM
 SunnyDale Member SOA Join Date: Jul 2018 Posts: 31

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
#117
03-17-2019, 09:34 AM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 31,605

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.
#118
03-17-2019, 01:05 PM
 Academic Actuary Member Join Date: Sep 2009 Posts: 9,660

What is the actual question?
#119
03-17-2019, 02:20 PM
 Abraham Weishaus Member SOA AAA Join Date: Oct 2001 Posts: 7,260

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.
#120
03-18-2019, 09:22 PM
 SunnyDale Member SOA Join Date: Jul 2018 Posts: 31

Oooh I see!!! Thanks!!!

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