Actuarial Outpost expected value problem
 Register Blogs Wiki FAQ Calendar Search Today's Posts Mark Forums Read
 FlashChat Actuarial Discussion Preliminary Exams CAS/SOA Exams Cyberchat Around the World Suggestions

 Fill in a brief DW Simpson Registration Form to be contacted when new jobs meet your criteria.

 Probability Old Exam P Forum

#1
11-20-2017, 12:52 PM
 efast Non-Actuary Join Date: Nov 2017 Posts: 9
expected value problem

Here is a problem with its solution:

An insurance policy on an electrical device pays a benefit of 4000 if the device fails during the first year. The amount of the benefit decreases by 1000 each successive year until it reaches 0. If the device has not failed by the beginning of any given year, the probability of failure during that year is 0.4. What is the expected benefit under this policy? A. 2234 B. 2400 C. 2500 D. 2667 E. 2694
Solution. The benefit of this policy is 4000 with probability 0.4, 3000 with probability 0.6⋅0.4, 2000 with probability 0.62 ⋅0.4, or 1000 with probability 0.63⋅0.4, and 0 otherwise. It follows that E Y ( )=4000⋅0.4+3000⋅0.6⋅0.4+2000⋅0.62 ⋅0.4+1000⋅0.63⋅0.4=2694. Answer E.

What I don't understand is that the solution doesn't take into account the probability that the device fails at the beginning of a given year.
#2
11-20-2017, 01:57 PM
 Vorian Atreides Wiki/Note Contributor CAS Join Date: Apr 2005 Location: As far as 3 cups of sugar will take you Studying for ACAS College: Hard Knocks Favorite beer: Most German dark lagers Posts: 62,719

"Device failing during the year" implies that the device is working at the beginning of the year.

Your question is looking at "boundary" items; it is common to view end-of-year and beginning-of-year to be the same "point" (i.e., the boundary). So failure "at the beginning of a year" is viewed as having failed "during" the prior year.
__________________
I find your lack of faith disturbing

Why should I worry about dying? It’s not going to happen in my lifetime!

Freedom of speech is not a license to discourtesy

#BLACKMATTERLIVES
#3
11-20-2017, 02:03 PM
 Z3ta Member SOA Join Date: Sep 2015 Posts: 361

The problem seems to say that $P(k\leq X < k+1)=0.6^{k}\cdot0.4$ with $k\in\{0,1,2,\dots\}$ so that takes into account all possible failure times right?
#4
11-20-2017, 02:16 PM
 Vorian Atreides Wiki/Note Contributor CAS Join Date: Apr 2005 Location: As far as 3 cups of sugar will take you Studying for ACAS College: Hard Knocks Favorite beer: Most German dark lagers Posts: 62,719

More like $P\(k\lt X\le k+1) = 0.6^k\cdot0.4$

I don't see the need to pay out an insurance benefit for a known claim up front.
__________________
I find your lack of faith disturbing

Why should I worry about dying? It’s not going to happen in my lifetime!

Freedom of speech is not a license to discourtesy

#BLACKMATTERLIVES
#5
11-20-2017, 03:03 PM
 Z3ta Member SOA Join Date: Sep 2015 Posts: 361

Quote:
 Originally Posted by Vorian Atreides More like $P\(k\lt X\le k+1) = 0.6^k\cdot0.4$ I don't see the need to pay out an insurance benefit for a known claim up front.
Obviously if you include time 0 then you would start the clock at the moment the insurance is effective and you would not have insurance if you already knew it was broken.

Whatever the intervals are, it is clear that $0.4+0.6\cdot0.4+0.6^2\cdot0.4\dots=1$ so any events not included must have probability zero.
#6
11-20-2017, 04:00 PM
 efast Non-Actuary Join Date: Nov 2017 Posts: 9

thanks Vorian Atreides and Z3ta
#7
11-20-2017, 08:45 PM
 ChuckJ Non-Actuary Join Date: Jul 2017 College: Moravian College Posts: 15

The .6 factor IS the probability that it hasn't failed yet BY the beginning of that year. That is why .6 is not in year 1. Year 2 has a factor of .6 because it didn't fail in year 1. Year 3 has .6^2 because it did not fail in year 1 or 2, etc.