Actuarial Outpost ASM 1st edition 4th Printing Q10.2
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 Long-Term Actuarial Math Old Exam MLC Forum

#1
07-02-2019, 03:50 PM
 Pro-Procrastinator SOA Join Date: Feb 2016 Posts: 18
ASM 1st edition 4th Printing Q10.2

Why is the solution essentially calculating 1q79.5 as opposed to 0.5|1q79 = .5p79 * 1q79.5, which takes into consideration the probability of survival to age 79.5?

What is the difference between this question and 10.7, which to me is the same type of question but the latter concept is used instead of the former?

Note: I understand the math being done, I just don't understand why it's 1q79.5 as opposed to 0.5|1q79 in 10.2.
#2
07-02-2019, 04:22 PM
 Seasplash Member CAS SOA Join Date: Jun 2014 Location: Arlington, VA Studying for MLC College: Cal Poly San Luis Obispo Posts: 389

Could you post the problem for us to see? Thanks.
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#3
07-02-2019, 07:12 PM
 Pro-Procrastinator SOA Join Date: Feb 2016 Posts: 18

see attachment for 10.2 and 10.7
Attached Images
 10-2 02-Jul-2019 18-39-14.pdf (1.05 MB, 23 views)
#4
07-02-2019, 07:32 PM
 Colymbosathon ecplecticos Member Join Date: Dec 2003 Posts: 6,128

Calendar year 2025 is one year long, starting 1/1/2025 and ending 12/31/2025, and you are given survival to 1/1/2025.
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#5
07-03-2019, 08:27 AM
 Pro-Procrastinator SOA Join Date: Feb 2016 Posts: 18

Quote:
 Originally Posted by Colymbosathon ecplecticos Calendar year 2025 is one year long, starting 1/1/2025 and ending 12/31/2025, and you are given survival to 1/1/2025.
Why is it 1q79.5 as opposed to 0.5|1q79 = .5p79 * 1q79.5? What's the difference between Q10.2 and Q10.7?
#6
07-03-2019, 08:35 AM
 Colymbosathon ecplecticos Member Join Date: Dec 2003 Posts: 6,128

Quote:
 Originally Posted by Pro-Procrastinator Why is it 1q79.5 as opposed to 0.5|1q79 = .5p79 * 1q79.5?
You are given that he is alive on 1/1/2025 --- he is age 79.5 on 1/1/2025.
__________________
"What do you mean I don't have the prerequisites for this class? I've failed it twice before!"

"I think that probably clarifies things pretty good by itself."

"I understand health care now especially very well."
#7
07-03-2019, 08:40 AM
 Abraham Weishaus Member SOA AAA Join Date: Oct 2001 Posts: 7,243

Quote:
 Originally Posted by Pro-Procrastinator Why is the solution essentially calculating 1q79.5 as opposed to 0.5|1q79 = .5p79 * 1q79.5, which takes into consideration the probability of survival to age 79.5? What is the difference between this question and 10.7, which to me is the same type of question but the latter concept is used instead of the former? Note: I understand the math being done, I just don't understand why it's 1q79.5 as opposed to 0.5|1q79 in 10.2.
You must distinguish between joint probability and conditional probability. Recall from probability the formula $\Pr(A\mid B)=\frac{\Pr(A\cap B)}{Pr(B)$
Q 10.2 is asking for a conditional probability: probability of dying between times 0.5 and 1.5 GIVEN survival to time 0.5. By the formula just cited, that is the same as the probability of surviving 0.5 years and dying in 1 year divided by the probability of surviving 0.5 years.

Q 10.7 is asking for a joint probability: probability of surviving 2 years and dying in the third year. That is the numerator only of Q 10.2.
#8
07-03-2019, 11:11 AM
 Pro-Procrastinator SOA Join Date: Feb 2016 Posts: 18

Quote:
 Originally Posted by Abraham Weishaus You must distinguish between joint probability and conditional probability. Recall from probability the formula $\Pr(A\mid B)=\frac{\Pr(A\cap B)}{Pr(B)$ Q 10.2 is asking for a conditional probability: probability of dying between times 0.5 and 1.5 GIVEN survival to time 0.5. By the formula just cited, that is the same as the probability of surviving 0.5 years and dying in 1 year divided by the probability of surviving 0.5 years. Q 10.7 is asking for a joint probability: probability of surviving 2 years and dying in the third year. That is the numerator only of Q 10.2.
Right. Thanks. And I just realized even if I used 0.5|1q79 = .5p79 * 1q79.5, the probability of surviving 0.5 years would be 1 in this case