exam mlc may solutions

# exam mlc may solutions - SPRING 2005 EXAM M SOLUTIONS...

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SPRING 2005 EXAM M SOLUTIONS Question #1 Key: B Let K be the curtate future lifetime of ( x + k ) 1 :3 1 1000 1000 K k x K Lv P a + + =− × ±± When (as given in the problem), ( x ) dies in the second year from issue, the curtate future lifetime of () 1 x + is 0, so 1 :3 1 1000 1000 x LvP a 1000 279.21 1.1 629.88 630 =≈ The premium came from x x x A P a = 1 x x Ad a 1 1 279.21 x x xx da Pd aa == =

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Question #2 Key: E Note that above 40, decrement 1 is DeMoivre with omega = 100; decrement 2 is DeMoivre with omega = 80. That means () ( ) 12 40 40 20 1/ 40 0.025; 20 1/ 20 0.05 µµ == 40 20 0.025 0.05 0.075 τ µ =+ = Or from basic definition of , 2 40 60 40 2400 100 60 40 2400 t tt t t p −− + =×= 40 / 100 2 / 2400 t dp d t t =− + at 20 t = gives 60/ 2400 0.025 −= 20 40 40 40 20 40 2/3 * 1/2 1/3 20 / / 0.025/ 1/3 0.075 t p t p ττ ⎡⎤ = = ⎢⎥ ⎣⎦
Question #3 Key: B 5 35 45 5 35 5 45 5 35 45 5 35 40 5 45 50 5 35 45 40 50 5 35 40 5 45 50 5 35 5 45 40 50 5 35 40 5 45 50 5 35 5 45 40 50 1 1 09 03 08 005 09 08 1 097 095 0 01048 qq q q pq p q p p p p p pp : : :: : .. . . . . . =+− × × =+ = b g bg bgbg b g bg b g Alternatively, () ( ) ( ) 63 5 53 5 4 0 6 45 5 45 50 0.90 1 0.03 0.873 0.80 1 0.05 0.76 p p = = = = 56 5 35:45 35:45 35:45 5 35 5 45 5 35:45 6 35 6 45 6 35:45 qpp ppp =− −+ ( ) 53 5 54 5 5 64 5 63 5 p pppp + ×−+ × ( ) 0.90 0.80 0.90 0.80 0.873 0.76 0.873 0.76 ×− + −× 0.98 0.96952 0.01048 =

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Question #4 Key: D Let G be the expense-loaded premium. Actuarial present value (APV) of benefits = 35 100,000 A APV of premiums = 35 Ga ±± APV of expenses = () ( ) 35 0.1 25 2.50 100 ++ ⎡⎤ ⎣⎦ Equivalence principle: 35 35 35 35 35 100,000 0.1 25 250 100,000 0.1 275 Ga A G a A GG a =+ + + + ( ) 35 0.9 100,000 275 100 8.36 275 0.9 1234 GP G + = = Question #5 Key: D Poisoned wine glasses are drunk at a Poisson rate of 2 × 0.01 = 0.02 per day. Number of glasses in 30 days is Poisson with 0.02 30 0.60 λ = ×= 0.60 00 . 5 5 fe ==
Question #6 Key: E View the compound Poisson process as two compound Poisson processes, one for smokers and one for non-smokers. These processes are independent, so the total variance is the sum of their variances. For smokers, () ( ) 0.2 1000 200 λ == Var(losses) () () 2 Var XE X ⎡⎤ =+ ⎢⎥ ⎣⎦ 2 200 5000 100 3,000,000 = For non-smokers, ( ) 0.8 1000 800 Var(losses) 2 Var X 2 800 8000 100 14,400,000 = Total variance = 3,000,000 + 14,400,000 = 17,400,000

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Question #7 Key: E [ ] x EZ bA = since constant force / x A µ µδ =+ E ( Z ) () 0.02 /3 0.06 b b b == = + [ ] ( ) T2 2 2 2 2 2 22 Var Var Var 2 21 4 10 9 45 x x Z bv b v b A A b bb µµ δ Τ ⎡⎤ = ⎣⎦ ⎛⎞ ⎜⎟ =− ++ ⎝⎠ = ⎢⎥ () () Var Z EZ = 2 4 45 3 41 3.75 45 3 b b = =⇒=
Question #8 Key: A () 12 3 30 30 30 30:3 23 1000 500 250 1 1.53 1 1.61 1 1.70 1000 500 0.99847 250 0.99847 0.99839 1.06 1000 1.06 1000 1.06 1000 Av q v q v q =+ + ⎛⎞ + ⎜⎟ ⎝⎠ 1.4434 0.71535 0.35572 2.51447 + = ( ) 1 2 2 11 1 22 2 3 0 30:1 1 1 1 0.00153 1 0.97129 1 1.06 2 2 2 0.97129 0.999235 0.985273 aq = ±± Annualized premium 2.51447 0.985273 = 2.552 = Each semiannual premium 2.552 2 = 1.28 =

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Question #9 Key: B Ex dx d ⎡− > ⎤ ⎣⎦ is the expected payment per payment with an ordinary deductible of d It can be evaluated (for Pareto) as () ( ) () 1 1 11 1 d Ex Ex d Fd d α θθ θ αα ⎡⎤ ⎛⎞ −− ⎢⎥ ⎜⎟ + ⎝⎠ −∧ = −⎡ + 1 1 d d αθ −+ = + 1 d + = d =+ in this problem, since 2 = 5 3 5 3 100 100 50 50 100 50 Ex x x > ⎤= += + 300 3 250 5 25 + == 150 150 150 150 25 175 x > + =
Question #10 Key: D Let S = score () 75 ES EES E Var S E Var S Var E S θθ θ == = ⎡⎤ =+ ⎣⎦ 2 8 EV a r 2 64 6 100 = S

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exam mlc may solutions - SPRING 2005 EXAM M SOLUTIONS...

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