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  #1  
Old 11-20-2017, 12:52 PM
efast efast is offline
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Default 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.
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Old 11-20-2017, 01:57 PM
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Vorian Atreides Vorian Atreides is offline
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"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.
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Old 11-20-2017, 02:03 PM
Z3ta Z3ta is offline
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The problem seems to say that with so that takes into account all possible failure times right?
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Old 11-20-2017, 02:16 PM
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Vorian Atreides Vorian Atreides is offline
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I don't see the need to pay out an insurance benefit for a known claim up front.
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Old 11-20-2017, 03:03 PM
Z3ta Z3ta is offline
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Quote:
Originally Posted by Vorian Atreides View Post
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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 so any events not included must have probability zero.
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Old 11-20-2017, 04:00 PM
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thanks Vorian Atreides and Z3ta
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Old 11-20-2017, 08:45 PM
ChuckJ ChuckJ is offline
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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.
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