edu-exam-mlc-0507-sol - SPRING 2007 EXAM MLC SOLUTIONS...

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SPRING 2007 EXAM MLC SOLUTIONS Question # 1 Key: E 37 0 70 27 1 0.95 0.9896 0.96 p p p === 75 71 0.107 47 1 0.8985 x dx pe e μ == = 57 0 0.9896 0.8985 0.889 p =×= Question # 2 Key: B ()( ) / / 0.08 0.3443 x A μμδ μμ =+ = 0.042 = 2 0.2079 2 x A μδ + () 22 2 0.2079 0.3443 0.08 xx T AA Var a δ −− 13.962 =
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Question # 3 Key: D [] () ( ) 60 60 1 / 1 0.359 / 0.06/1.06 11.3243 aA d =− = && [] [] 60 60 60 1000 1000 / 359/11.3243 31.70 PA a == = 56 5 6 5 60 60 1000 1000 1000 VA P a × 439.80 31.70 9.8969 126.06 × = Alternatively, 65 60 5 60 60 1000 1000 1 AA V A ⎛⎞ ⎜⎟ = ⎝⎠ 0.4398 0.359 1000 1 0.359 = 126.05 =
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Question # 4 Key: E Let K = curtate future lifetime random variable () 1000 L Var L σ = 1 1 150000 150000 K x K Lv P a + + =− && 1 150000 1 K x x PP v dd + ⎡⎤ ⎛⎞ =+ ⎜⎟ ⎢⎥ ⎝⎠ ⎣⎦ 1 150000 1 K xx Var L Var v + 2 1 150,000 1 K x P Var v d + because ( ) 2 Var aX b a Var x += if a and b are constants x P is a constant, a numeric value, not a random variable. There are various ways to evaluate 1 x P d + . One possibility is 1 x P d + 1 1 x d a d 1 11 x da 1 x da = 1.0699 1 1 0.0653 x A == = −− Var L 2 1 150,000 1 K x P Var v d + ( ) 2 22 150,000 1.0699 x x AA 2 25,755,400,000 0.0143 0.0653 258,478,900 = Standard deviation 16,077
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Question # 5 Key: C Number of claims in 2 months is Poisson with 2 λ = n P ( N = n ) 0 0.1353 1 0.2707 2 0.2707 () Pr 3 1 0.1353 0.2707 0.2707 N ≥= 0.3233 = Question # 6 Key: A Donations = compound Poisson 71 00 . 8 5 6 =× × = Withdrawals = compound Poisson . 2 1 4 = ×× = expected donations = 56 15 840 ×= where 15 = mean expected withdrawals = 14 40 560 where 40 = mean expected net = 840 560 280 −= Variance of donations 2 56 75 15 + 16,800 = Variance of withdrawals 2 14 533 40 =× + 29,862 = Variance of net change 16,800 29,862 =+ 46,662 = [] 600 280 Pr 600 Pr 0,1 Pr 0,1 1.48 1 0.93 0.07 46,662 SN N ⎡⎤ >= > = > = ⎢⎥ ⎣⎦
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Question # 7 Key: E All have the same benefits, so retrospectively the one with the highest accumulated value of premiums wil have the highest reserve. All have total premiums of 10, so the one with the premiums earliest will have the highest accumulated value. That is E. Alternatively, all have the same benefits, so all have the same actuarial present value of premiums. E has the highest APV of the first 5 premiums, so it has the lowest APV (at both time 0 and time 5) of premiums for years 6 and later. Thus E has the highest prospective reserve. Question #8 Key: E Let tx p = probability Kevin still there ty p = probability Kira still there Kira’s expected playing time = () 0 1 p pd t 0.7 0.6 0 0.6 1.3 00 1 10 0.89744 0.6 1.3 tt ee d t ed t t −− ∞∞ =− =−= ∫∫ Alternatively and equivalently yx y oo Alternatively, Kira has 7 13 chance of “surviving” Kevin, and (memoryless) 1 1.667 0.6 = future lifetime then if she does 7 1.667 0.89744 13 ⎛⎞ = ⎜⎟ ⎝⎠
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Question # 9 Key: E For decrement 1 in its associated single decrement table: () 1 25 10 . 1 t p t ′ =− × Thus 1 0.2 25 0.98 p = ; 1 0.6 25 0.94 p = ; ( ) 1 0.4 25.2 0.94/ 0.98 0.9592 p == For decrement 2 in its associated single decrement table: No one age 25 dies before age 25.2.
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This note was uploaded on 11/14/2011 for the course MAT 2070 taught by Professor S.g. during the Spring '11 term at Université du Québec à Montréal.

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edu-exam-mlc-0507-sol - SPRING 2007 EXAM MLC SOLUTIONS...

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