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  #1  
Old 02-09-2018, 12:54 PM
ch1rontl34 ch1rontl34 is offline
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Default 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?
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  #2  
Old 02-09-2018, 01:10 PM
Colymbosathon ecplecticos's Avatar
Colymbosathon ecplecticos Colymbosathon ecplecticos is offline
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Have you considered what happens from time 0.6 to time 1.0?
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Old 02-09-2018, 01:15 PM
ch1rontl34 ch1rontl34 is offline
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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.
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Old 02-09-2018, 07:35 PM
Academic Actuary Academic Actuary is offline
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Apply your approach to 100 lives with the first decrement set to 0. You would get
2 + 98 x .03 decrements.
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  #5  
Old 02-09-2018, 07:55 PM
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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 View Post
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  
Old 02-10-2018, 08:12 PM
ch1rontl34 ch1rontl34 is offline
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Thank you Jim. This started to click a few hours after I posted this.
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