Tutorial-10-Problems

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Nov 24, 2024

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THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT3801/3909 Advanced Life Contingencies Tutorial 10 Problems 1. The mortality of ( x ) and ( y ) follows a common shock model with components T * ( x ) , T * ( y ) and Z. T * ( x ) , T * ( y ) and Z are independent and have exponential distributions with respec- tive forces μ 1 , μ 2 and λ. The probability that ( x ) survives 1 year is 0.96. The probability that ( y ) survives 1 year is 0.97. λ = 0 . 01 . Calculate the probability that both ( x ) and ( y ) survives 5 years. 2. You want to calculate the expected present value of a last survivor whole life insurance of 1 on ( x ) and ( y ) using multi-state methodology. You defined states as State 0 = both alive State 1 = only ( x ) alive State 2 = only ( y ) alive State 3 = neither alive You assume: Death benefits are payable at the moment of death. The future lifetime of ( x ) and ( y ) are independent. μ 01 x + t : y + t = μ 02 x + t : y + t = μ 13 x + t : y + t = μ 23 x + t : y + t = 0 . 06 , t 0 μ 03 x + t : y + t = 0 , t 0 δ = 0 . 05 You later find out that the particular lives do not have independent future lifetimes. While your model correctly projects the survival function of ( x ) and ( y ) , a common shock model should be used for their joint future lifetimes. Based on this, you realize you should be using μ 0 03 x + t : y + t = 0 . 02 , t 0 . (a) You now want to calculate the expected present value of a whole life insurance of 1 payable at the first death of ( x ) and ( y ) . Calculate the expected present value for the first-to-die benefit. (b) Hence, calculate the expected present value of the last-to-die insurance, payable at the moment of the second death. (c) Calculate the probability that both have died by the end of year 3. 1
( x ) ( y ) T * ( x ) T * ( y ) Z T * ( x ) T * ( y ) Z δ t P ( t ) t = 1 , 2 , 3 , 4 , 5 . t = 0 , 1 , 2 , 3 , 4 . π N π Q 70 Q 70 = 0 . 022 0 . 025 0 . 028 Q 70 π
D ( N ) lim N →∞ p [ D ( N )] N 6 = 0 . L ( N ) lim N →∞ p [ L ( N )] N 6 = 0 .
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