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### Part6-Notes-431-2011-F

Course: ACTSC 431, Fall 2011
School: Waterloo
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Word Count: 1137

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Notes Review for Loss Models 1 - ACTSC 431/831, FALL 2011 Part 6 The Classical Continuous Time Ruin Model In this part, times are measured in years, unless stated otherwise. 1. Poisson Process: Let Nt be the number of claims up to time t or the number of claims occurring in the time interval (0, t], t &gt; 0. The process {Nt , t 0} is said to be a Poisson process with rate &gt; 0 if the following three...

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Notes Review for Loss Models 1 - ACTSC 431/831, FALL 2011 Part 6 The Classical Continuous Time Ruin Model In this part, times are measured in years, unless stated otherwise. 1. Poisson Process: Let Nt be the number of claims up to time t or the number of claims occurring in the time interval (0, t], t > 0. The process {Nt , t 0} is said to be a Poisson process with rate > 0 if the following three conditions hold: (a) N0 = 0. (b) The number of claims in any time interval of length t has a Poisson distribution with mean t, namely, for all s 0 and t > 0, Pr{there are n claims in (s, s + t]} = Pr{Nt+s Ns = n} = (t)n et . n! (c) The process {Nt , t 0} has stationary and independent increments. Stationary increments mean that for all n = 1, 2, ..., 0 t0 < t1 < < tn and h 0, the distribution of the random vector (Nt1 +h Nt0 +h , Nt2 +h Nt1 +h , , Ntn +h Ntn1 +h ) does not depend on h. In particular, Nt+h Nh has the same distribution as Nt . Independent increments mean that for all n = 1, 2, ..., 0 t0 < t1 < < tn , random variables Nt1 Nt0 , Nt2 Nt1 , , Ntn Ntn1 are independent. For the Poisson process {Nt , t 0}, (a) For any t > 0, Nt has a Poisson distribution with mean t. (b) For any 0 < s < t, Ns and Nt Ns are independent. Note that Ns and Nt are not independent. 2. Inter-claim times and waiting times: Let Tn be the time of the nth claim in a Poisson process {Nt , t 0} with rate > 0. Then 0 < T1 < T2 < T3 < . Let Tn Tn1 = Wn be the time between the (n 1)th claim and the nth claim for n = 1, 2, ... with T0 = 0. Then W1 , W2 , W3 ,... are called inter-claim times or waiting times. (a) An important result about the inter-claim times: The inter-claim times W1 , W2 , W3 ,... are independent and identically distributed exponential random variables with the same mean 1/. 1 3. Compound Poisson Process: Assume that the number of claims up to time t is a Poisson process Nt with rate > 0. The size of the j th claim is Xj , j = 1, 2, .... Then the aggregate claims up to time t are Nt St = Xj j =1 with St = 0 if Nt = 0. The process {St , t 0} is called a compound Poisson process. Unless stated otherwise, for a compound Poisson process {St , t 0}, we assume that the Poisson process {Nt , t 0} is independent of the claim sizes {X1 , X2 , } and the claim sizes {X1 , X2 , } are independent and identically distributed nonnegative random variables and have the same distributions as X . In particular, for any t > 0, E (St ) = tE (X ), V ar(St ) = tE (X 2 ). 4. Surplus Process: Assume that the initial surplus at time 0 is u 0, the premium rate per year is c > 0 collected continuously, the aggregative claims up to time t is a compound Poisson process Nt Xj . Then surplus the at time t is given by j =1 Nt Ut = u + ct Xj j =1 with U0 = u, where {X1 , X2 , ...} have the same distribution function F (x) and the same mean > 0 and the Poisson process {Nt , t 0} has rate > 0. (a) Ruin probabilities of the surplus process: Ruin is said to occur when the surplus Ut is negative. Note that ruin can occur only at claim times in the surplus process. Let (u) be the innite-time ruin probability or (u) = Pr{ruin occurs at some time t > 0 | U0 = u} and let n (u) be the probability that ruin occurs on or before the nth claim, n = 1, 2, ..., namely, n (u) = Pr{ruin occurs on or before the nth claim | U0 = u} 2 (b) We assume that c > in the surplus process. This condition guarantees that the ruin probabilities are less than one. (c) The relative security loading factor > 0 of a surplus process Ut = u + ct Nt j =1 Xj is dened by = c E (Xj ) . E (Xj ) (d) The adjustment coecient R > 0 of a surplus process Ut = u + ct is the smallest positive solution to the following Lundberg equation: Nt j =1 1 + (1 + )t = E (etX1 ) = MX1 (t). Note that the adjustment coecient R > 0 satises E (eR(X1 c T1 ) ) = E (eR X1 ) E (eR c T1 ) = 1, where X1 and T1 are the amount and the time of the rst claim, respectively. (e) Some important results about ruin probabilities: i. For any u 0, 1 (u) 2 (u) 3 (u) and lim n (u) = (u). n ii. Recursion formula for n (u): for any u 0 and n = 1, 2, ..., n+1 (u) = 0 0 n (u + ct x)f (x)dx + 0 = u+ct et f (x)dx dt u+ct u+ct et n (u + ct x)f (x)dx + S (u + ct) dt 0 with 1 (u) = et S (u + ct)dt, 0 where f (x) and S (x) are the pdf and the survival function of X1 . 3 Xj iii. Inequalities for the survival function S (x): for any x 0, S (x) eR x E (eRX1 ), where R > 0 is the adjustment coecient. iv. The Lundberg upper bounds for the ruin probabilities: Let R > 0 be the adjustment coecient. Then for any u 0, 1 (u) eR u , n (u) eR u , for all n = 2, 3, ..., and hence (u) eR u . v. If the initial surplus u = 0, then 1 . 1+ (0) = vi. The ruin probability (u) satises the following integro-dierential equation u (u x)f (x)dx S (x) , u 0. (u) c 0 vii. The ruin probability (u) satises the following defective renewal equation (u) = (u) = Fe (x) 1 + 1+ 1+ u (u t)dFe (t), u 0, 0 where Fe (x) = 1 x S (t)dt, x 0 0 is the equilibrium distribution of F (x) and 1 Fe (x) = 1 Fe (x) = S (t)dt, x 0, x is the survival function of Fe (x). (f) The general solution to the rst order linear dierential equation of d y (t) = (t)y (t) + (t) dt can be expresses as t y (t) = e 0 t (s)ds e s 0 (u)du (s) ds + C , 0 where C is an arbitrary constant and can be determined from the boundary conditions on y (t). 4
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Waterloo - ACTSC - 431
ACTSC 431/831 LOSS MODELS 1, FALL 2011Instructor: Jun Cai, M3 4012, 519-888-4567 ext. 36990, jcai@uwaterloo.caInstructors Oce Hours: Monday, 14:0016:00.Lectures: 11:3012:20 on Monday, Wednesday, and Friday in MC 2066.Course Webpage: UWACE. Course info
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Parameter estimationActSCi 432/8321 / 31Distributions SummaryX is Binomial with parameters n (positive integer) andp (0, 1)p .f . f (x ) =nxp (1 p )nx for x = 0, 1, 2, 3, .nxE (X ) = np ,Var (X ) = np (1 p )X is Bernoulli with parameters p (0,
Waterloo - ACTSC - 432
Credibility IActSCi 432/8321 / 16Distributions SummaryX is Binomial with parameters n (positive integer) andp (0, 1)p .f . f (x ) =nxp (1 p )nx for x = 0, 1, 2, 3, .nxE (X ) = np ,Var (X ) = np (1 p )X is Bernoulli with parameters p (0, 1) X i
Waterloo - ACTSC - 432
Credibility IIActSCi 432/8321 / 22Greatest Accuracy Credibility Theory (GACT)Bayesian approach developed by H. BhlmannuSuppose we have a risk r.v. X that depends on a xed butunknown parameter . (Here can be a vector in general)For example, represe
Waterloo - ACTSC - 432
Credibility IIIActSCi 432/8321 / 20Linear Exponential Family - A quick reviewDenition (LEF)X LEF if the density is given byf (x ; ) =p (x )e r ()xq ()r () is called the canonical parameter.It turns out that the LEF class contains well-known dist
Waterloo - ACTSC - 432
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Waterloo - ACTSC - 432
1!!&quot;#&quot;\$%!#&amp;'()*&amp;+%!)*&quot;%!,*&amp;(*\$!!-&amp;%.!/0)1!'&quot;*&quot;%-!!This Rate Manual has been approved by theSuperintendent of Insurance of the State of New Yorkand is effective September 1, 1993.FOURTH REPRINT: May 1, 2007Fourth Revision: March 3, 2010This r
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ActSc 432/832: Loss Models II (Fall 2011)1Course outlineInstructor: Joseph H.T. Kim (M3 3126, ext. 35539, jhtkim@uwaterloo.ca) Best to email me since I do notcheck my voice mail often.Lecture: 2:30-3:20 MWF (MC 4059)Teaching Assistants and oce hours
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Waterloo - ACTSC - 432
Name:ID:1. The random variable has pdf . It is known that =1000. You are given the following fiveobservations: 43 145 233 396 775 (sample mean = 318.4).Determine the method-of-moments estimate of .A.B.C.D.E.&lt;44&lt;4.24.2&lt;4.44.4&lt;4.6&gt;4.62. Suppo
Waterloo - ACTSC - 432
Name:ID:1. A random variable X is assumed to satisfy X|( = ) ~ Pareto (, = 1200), and thus= , x &gt; 0 and the prior distribution for is Exponential with mean 5, i.e. , &gt;0. Determine theposterior distribution of given X = 155 (i.e. say which type of dist
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Waterloo - ACTSC - 463
1. Use the following data to answer the questions below. Assume this company does not reserve for salvage or subrogation recoveries.Claim # Acc. Year Rep. Year1200520052200520063200620064200720075200720076200720087200820088200620
Waterloo - ACTSC - 463
1. Use the transactional data to create the cumulative data:Paid Triangle20052006200720081250070072050024700900900362300105048230012120020015755002417005001000365002004850012170090022951000242400140019003628001
Waterloo - ACTSC - 463
1. Use the transactional data (as of December 31, 2008) to develop cumulative accident year paidand incurred loss triangles.Claim #1112233455566777891010Acc Date12-Jan-0512-Jan-0512-Jan-0510-Mar-0510-Mar-0501-Nov-0501-Nov-05
Waterloo - ACTSC - 463
1. SolutionNote that there is a lot of extaneous information.Since we are interested in the value of 2006 at 24 months we need to look at data from 2008 which iscurrently at 24 months of development.Calculate the average case outstanding for 2008 at 2
Waterloo - ACTSC - 463
1. You are given the following information:AY20052006200720082009Incurred Paid Claims Paid ClaimsClaims @@@ Open Claim Open Claim12/31/2009 12/31/2009 12/31/2007 Counts @ Counts @(000's)(000's)(000's) 12/31/2009 12/31/2007117,160100,75864
Waterloo - ACTSC - 463
1. XYZ Insurance Company has been writing business since 1/1/2005 and has provided you withfollowing incurred loss data evaluated at 12/31/2009:Report YearAccident Year20052006200720082009DevelopmentInterval1-2 Years2-3 Years3-4 Years4-5 Yea
Waterloo - ACTSC - 463
1. SolutionCalculate ultimate AY and RY loss development factors:Cumulative LDFsDevelopmentInterval1-2 Years2-3 Years3-4 Years4-5 Years5-UltimateAccident CumulativeYearAY1.5002.3821.2501.5881.1001.2711.1001.1551.0501.050ReportYear
Waterloo - ACTSC - 463
1. XYZ Insurance Company has been writing business since 1/1/2005 and has provided you withfollowing incurred loss data evaluated at 12/31/2009:Report YearAccident Year2005200620072008200910,0004,0002,0001,000200512,0005,0002,5001,500200
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ACTSC 463/863Property and CasualtyLoss Reserving1Case OutstandingDevelopment Technique2Case Outstanding Development&gt; Attempts to develop ultimate incrementalclaim payments based on historic levels ofcase reserve adequacy&gt; Assumptions Similar t
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ACTSC 463/863Property and CasualtyLoss Reserving1Recoveries2Recoveries&gt; Salvage/Subrogation&gt; Reinsurance Quota Share Treaty with pro-rata sharing of premiums and losses Per-Risk Excess of Loss Per-risk treaty where losses above a certain reten
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ACTSC 463/863Property and CasualtyLoss Reserving1ALAE &amp; ULAE2ALAE&gt; Allocated Loss Adjustment Expense Expense that can be directly associated with aparticular claim Legal Expert witness Police reports External adjuster fees Availability of da
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ACTSC 463/863Property and CasualtyLoss Reserving1Discounting Loss Reserves2Discounting&gt; Process of recognizing the time value ofmoney Required in Canadian reporting Outstanding reserves can be invested \$1 of future payments requires &lt;\$1 of rese
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ACTSC 463/863Property and CasualtyLoss Reserving1Examples From PreviousCalculate the ULAE reserve using both the traditional approach and the Kittel Refinement and thefollowing information:CY 2010 Paid LossesTotal Reserves at 12/31/2009Total Rese
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Property and Casualty Insurance Fall 2011ACTSC 463/8631Property and Casualty InsuranceACTSC 463/863Fall 2011InstructorE-MailOffice HoursChris Van Kooten, FCAS,FCIAvankootenc@gmail.comBy appointment onlyCourse DescriptionThis course is focuse
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