∞:B∞:Î#-:Î#:B""##/† /.- .: œ/† Ð" /Ñ .:.œ/.: /.: œ " !!∞∞:Î#:ÐB Ñ"""###B""#This is, so that the density function ofisJÐBÑ\\.Answer:B0ÐBÑ œÒ" Ó œ\."#.B#B"Ð#B"Ñ#16.Sincehas a uniform distribution on the interval from 8 to 12,'s distribution function isXXJ Ð>Ñ œ TÒX Ÿ >Ó œTÒX >Ó œ) Ÿ > Ÿ "#X>)"#>"#)"#),and, for.;V œJ Ð<Ñ œ TÒV Ÿ <Ó œ TÒŸ <Ó œ TÒX Ó œœ $ Þ"!"!"!#Þ&XX<"#)<"#V"!<The density function ofis.V0 Ð<Ñ œ J Ð<Ñ œJ Ð<Ñ œÒ$ Ó œVVVw..#Þ&#Þ&.<.<<<#Alternatively,for.0 Ð>Ñ œ) Ÿ > Ÿ "#V œœ 1ÐXÑ p X œœ 2ÐVÑX""!"!%XV.Answer:Ep 0 Ð<Ñ œ 0 Ð2Ð<ÑÑ † l2 Ð<Ñl œ† ll œVXw""!#Þ&%<<##17.The standard approximation to the sum (total) of a collection of independent random variables isthe normal approximation.The total contribution is, the sum of theX œ G G â G"##!#&2025 contributions.is the amount of the -th contribution, the's are mutually independent,G3G33and each has meanand variance.IÒG Ó œ $"#&Z +<ÒG Ó œ Ð#&!Ñ33#The mean and variance ofareandXIÒXÓ œIÒG Ó œ Ð#!#&ÑÐ$"#&Ñ œ 'ß $#)ß "#&3œ"#!#&3.Z +<ÒXÓ œZ +<ÒG Ó œ Ð#!#&ÑÐ#&! Ñ œ "#'ß &'#ß &!!3œ"#!#&3#We will denote the 90th percentile ofby .We find the approximate 90th percentile ofbyX:Xapplying the normal approximation to.We wish to findso that.X:TÒX Ÿ :Ó œ Þ*We standardize the probability:.TÒX Ÿ :Ó œ TÒŸÓ œ Þ*!X'ß$#)ß"#&:'ß$#)ß"#&"#'ß&'#ß&!!"#'ß&'#ß&!!
316PROBLEM SET 9©ACTEX 2015SOA Exam P - Probability17. continuedX'ß$#)ß"#&"#'ß&'#ß&!!is approximately standard normal (mean 0, variance 1), so that:'ß$#)ß"#&"#'ß&'#ß&!!is the 90-th percentile of the standard normal distribution.From the table for thestandard normal distribution, we see that(.Therefore we haveF"Þ#)#Ñ œ Þ*!:'ß$#)ß"#&"#'ß&'#ß&!!œ "Þ#)#: œ 'ß $%#ß &%(Þ&, from which we get.Answer:C18.For policyholder , letbe the number of claims filed in the year,.3\3 œ "ß #ß ÞÞÞß "#&!3Eachis Poisson with a mean of 2, and therefore has variance of 2 also;,\IÒ\ Ó œ #33.The total number of claims in the year is.Since the's are mutuallyZ +<Ò\ Ó œ #X œ\\3333œ""#&!independent, the distribution ofis approximately normal.The mean ofisXX, and the variance ofisIÒXÓ œ IÒ\ Ó œIÒ\ Ó œ Ð"#&!ÑÐ#Ñ œ #&!!X3œ"3œ""#&!"#&!33(since the's are independent, thereZ +<ÒXÓ œ Z +<Ò\ Ó œZ +<Ò\ Ó œ Ð"#&!ÑÐ#Ñ œ #&!!\3œ"3œ""#&!"#&!333are no covariances between's).We wish to find, by using the normal\TÒ#%&! Ÿ X Ÿ #'!!Ó3approximation for.Applying the normal approximation we getXTÒ#%&! Ÿ X Ÿ #'!!Ó œ TŸŸ#%&!IÒXÓXIÒXÓ#'!!IÒXÓZ +<ÒXÓZ +<ÒXÓZ +<ÒXÓœ TŸŸÓ œ TÒ " Ÿ ^ Ÿ ##%&!#&!!X#&!!#'!!#&!!#&!!#&!!#&!!(from the normal theœÐ#Ñ Ò" Ð"ÑÓ œ Þ*((# Ð" Þ)%"$Ñ œ Þ)")&FFtable provided with the exam).Answer:B19.The sum of independent Poisson random variables is also Poisson, so that the number of claimsoccurring in a 5 day week has a Poisson distribution with 6 claims expected.ThenTÒ\ $Ó œ " TÒ\ œ !Ó TÒ\ œ "Ó TÒ\ œ #Ó. Answer:Cœ " / /† /†œ " Þ!'#! œ Þ*$)'''''"x#x#20.Letrepresent the time until a catastrophe occurs on policy , and letrepresent the time until theX3X3first catastrophe occurs.ThenAllTÒX >Ó œ TÒX >Ó œ TÒÐX >Ñ ∩ ÐX >Ñ ∩ â ∩ ÐX >ÑÓ3"#8(this last equality follows from the independenceœ TÒX >Ó † TÒX >ÓâTÒX >Ó"#8of the's).From the exponential distribution, we have, so thatXTÒX >Ó œ /33>Îα, thushas an exponential distribution with mean. Answer:BTÒX >Ó œ Ð/Ñœ /XÎ8>Î8>8Îααα
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