Actuarial Outpost Multiple Decrement Probability Question from ASM Manual
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#1
02-09-2018, 01:54 PM
 ch1rontl34 Member Join Date: May 2008 Posts: 145
Multiple Decrement Probability Question from ASM Manual

This is from ASM MLC Manual, 15th Edition, 4th printing, page 1024, question 50.7.

The question:

For a double-decrement table, you are given:

1.) Decrements from cause 1 are uniformly distributed over each year of age in the associated single decrement table

2.) In the associated single decrement table, 40% of decrements from cause 2 occur at time 0.25 and 60% occur at time 0.6 within each year of age

3.) q'(1) (age 45) = .1

4.) q'(2) (age 45) = .05

Calculate q(2) (age 45)

My work assumes starting with one life. Then there should be 1 - (.25)(.1) =.975 lives at time 0.25 due to cause 1.

Then cause 2 acts on the remaining .975 lives as follows: .975(.4)(.05) = .0195 decrements. So at time 0.25, after both decrements, there are .975 - .0195 = .9555 lives remaining.

Then at time 0.6, there are .9555(1-.35(.1)) = .9220575 lives due to cause 1. Cause 2 acts on those lives at follows: .9220575(0.60)(0.05)=.02766.

Therefore q(2) (age 45) = .0195 + .02766 = .04716.

The solution does this math:

"The exposure after cause 1 is 1-.25*(0.1) = .975 at time 0.25 and 1-(0.6)(0.1) = .94 at time 0.6. Therefore:

q(2)(age 45) = .05*.4*.975 + .05*.6*.94 = .0477.

Why is my method flawed?
#2
02-09-2018, 02:10 PM
 Colymbosathon ecplecticos Member Join Date: Dec 2003 Posts: 6,002

Have you considered what happens from time 0.6 to time 1.0?
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#3
02-09-2018, 02:15 PM
 ch1rontl34 Member Join Date: May 2008 Posts: 145

Doesn't cause 2 stop impacting the survivors immediately following time 0.6, at which time cause 1 is the only remaining decrement? I must be missing something, but I don't see how time 0.6 to time 1 applies for cause 2.
#4
02-09-2018, 08:35 PM
 Academic Actuary Member Join Date: Sep 2009 Posts: 8,339

Apply your approach to 100 lives with the first decrement set to 0. You would get
2 + 98 x .03 decrements.
#5
02-09-2018, 08:55 PM
 Jim Daniel Member SOA Join Date: Jan 2002 Location: Davis, CA College: Wabash College B.A. 1962, Stanford Ph.D. 1965 Posts: 2,688

The error I see in your approach is in computing the fraction of the 0.9555 alive at time 0.25 that are killed by #1 in the next 0.35. You take that fraction to be 0.35 x 0.1, but that is actually the fraction of a whole person at time 0 that would be killed in that interval. The fraction of a whole person alive at time 0.25 that would be killed in the next 0.35 by #1 would be 0.35 x .1 / [1 - 0.25 x .1].

Quote:
 Originally Posted by ch1rontl34 This is from ASM MLC Manual, 15th Edition, 4th printing, page 1024, question 50.7. The question: For a double-decrement table, you are given: 1.) Decrements from cause 1 are uniformly distributed over each year of age in the associated single decrement table 2.) In the associated single decrement table, 40% of decrements from cause 2 occur at time 0.25 and 60% occur at time 0.6 within each year of age 3.) q'(1) (age 45) = .1 4.) q'(2) (age 45) = .05 Calculate q(2) (age 45) My work assumes starting with one life. Then there should be 1 - (.25)(.1) =.975 lives at time 0.25 due to cause 1. Then cause 2 acts on the remaining .975 lives as follows: .975(.4)(.05) = .0195 decrements. So at time 0.25, after both decrements, there are .975 - .0195 = .9555 lives remaining. Then at time 0.6, there are .9555(1-.35(.1)) = .9220575 lives due to cause 1. Cause 2 acts on those lives at follows: .9220575(0.60)(0.05)=.02766. Therefore q(2) (age 45) = .0195 + .02766 = .04716. The solution does this math: "The exposure after cause 1 is 1-.25*(0.1) = .975 at time 0.25 and 1-(0.6)(0.1) = .94 at time 0.6. Therefore: q(2)(age 45) = .05*.4*.975 + .05*.6*.94 = .0477. Why is my method flawed?
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#6
02-10-2018, 09:12 PM
 ch1rontl34 Member Join Date: May 2008 Posts: 145

Thank you Jim. This started to click a few hours after I posted this.