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#1




Percentiles and Probability Question
https://imgur.com/a/ES4nQlg
That link is to the question I'm having an issue with and the answer. Since we're looking for the 90th percentile and the benefit only pays between 2 and 10, wouldn't we need to find: [Integral f(t) between 2 and h]/[Integral f(t) between 2 and 10]=.1 Then solve for h to find the 90th percentile? I don't understand why we aren't adjusting for the fact that we're finding the 90th percentile in between two intervals when there's the probability of having no claim when t<2 or t>10. Thank you in advance!
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Former Disney World Cast Member, currently no idea what I'm doing "I think you should refrain from quoting yourself. It sounds pompous."  SweepingRocks 
#2




Suppose the warranty paid 5 at the first failure if before 2 or after 10. Would you ignore the probability of those when calculating the 90th percentile? Presumably you would not ignore that probability.
So why would you propose ignoring that probability when the warranty pays 0 for those failure times? You are treating the question as asking the 90th percentile of the payment, given than a nonzero payment was made. 
#3




Quote:
Initially I was confused because I thought that if we did account for the times when it pays $0, we would take the integral between 0 and h. But since we're looking for the 90th percentile of Z, we would start at 2, since the value for that 20% of time between 0 and 2 would be 0, correct?
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Former Disney World Cast Member, currently no idea what I'm doing "I think you should refrain from quoting yourself. It sounds pompous."  SweepingRocks 
#4




Yes, and the value above 10 is 0. You are looking for (either formulation is correct), the range for the cheapest 90% of all claims, or the most expensive 10% of all claims, on a present value basis. Easiest is to find the most expensive 10%, which are claims between 2 and h. And then you want the value of of a claim at h, because that will be the cheapest among the top 10%, on a present value basis.

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