STAT3801_2011Unit4[1]

STAT3801_2011Unit4[1] - THE UNIVERSITY OF HONG KONG...

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THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT3801 ADVANCED LIFE CONTINGENCIES Unit 4 2010-11 2 nd semester Multiple decrement models Extensions of standard mortality models Simultaneous operation of several causes of decrement A life fails because of one of these decrements 4.1 Multiple decrement examples Terminate membership of a pension plan Death Withdrawal (Termination) Disability Retirement Terminate individual life insurance policy Death Surrender (Withdrawal) Lapse Maturity (Survival)
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2 Terminate disability insurance policy Death Recovery Expiry Death and disability considered distinct claims Public health planning Analysis of mortality by cause of death Accidents and violence Cancers Cardiovascular diseases (Heart diseases) Infectious diseases Others Discontinue survivor’s pension Death Remarriage Expiry
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3 4.2 General multiple state model r + 1 states, labelled 0 , 1 , 2 , ··· , r M n = 0 means Process is in state 0 at time n Individual is in state 0 at age x + n Still member of the group under consideration M n = i, i = 1 , 2 , ··· , r Process is in state i at time n M n = 0 means Still member of the pension plan Life insurance policy continuing in force Disability benefit continuing to be paid Individual is alive Survivor’s pension is continuing to be paid
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4 r - decrement model States 1 , 2 , ··· , r represent different decrements States 1 , 2 , ··· , r are absorbing Transition from 0 to b is decrement of type b Individuals in state 0 form a closed group Conditional probabilities of transition t q ( j ) x = t Q (0 ,j ) 0 = Pr ( M t = j | M 0 = 0) t q ( τ ) x = r X j =1 t q ( j ) x Force of decrement due to cause j μ ( j ) x ( t ) or μ ( j ) x + t = lim h 0 1 h · h q ( j ) x + t = lim h 0 1 h · t + h q ( j ) x - t q ( j ) x t p ( τ ) x
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5 Standard results μ ( τ ) x = r X j =1 μ ( j ) x t p ( τ ) x = exp - Z t 0 μ ( τ ) x ( s ) ds = exp " - Z t 0 r X b =1 μ ( b ) x ( s ) ds # t p ( τ ) x = 1 - t q ( τ ) x n p ( τ ) x = p ( τ ) x · p ( τ ) x +1 ··· p ( τ ) x + n - 1 Probability of transition from state 0 to state j within a time period t t q ( j ) x = Z t 0 u p ( τ ) x μ ( j ) x ( u ) du for j = 1 , 2 , ··· , r Examples 3 p ( τ ) 20 2 q (1) 22 1 | 2 q (2) 25 Example 4.1 Suppose that in a triple-decrement model, you are given constant forces of decrement, for a person now age x, as follows: μ (1) x + t = b, for t 0 , μ (2) x + t = b, for t 0 , μ (3) x + t = 2 b, for t 0 . You are also given the probability that (
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This note was uploaded on 07/30/2011 for the course STAT 3801 taught by Professor Kc during the Fall '11 term at HKU.

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STAT3801_2011Unit4[1] - THE UNIVERSITY OF HONG KONG...

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