

FlashChat  Actuarial Discussion  Preliminary Exams  CAS/SOA Exams  Cyberchat  Around the World  Suggestions 
Probability Old Exam P Forum 

Thread Tools  Search this Thread  Display Modes 
#1




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




"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 endofyear and beginningofyear 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 
#4




More like
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




Quote:
Whatever the intervals are, it is clear that so any events not included must have probability zero. 
#6




thanks Vorian Atreides and Z3ta

#7




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.

Thread Tools  Search this Thread 
Display Modes  

