edu-exam-c-0507-sol - Question #4 Key: A The random...

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Question #4 Key: A The random variable 0.5 ln( / 0.25) S has a normal distribution with parameters 2 22 (0.015 0.35 / 2)(0.5) 0.044375 0.35 (0.5) 0.6125. μ σ =− = == The upper limit for the normal random variable is 0.044375 1.645 0.6125 0.45149 += . The upper limit for the stock price is 0.45149 0.25 0.39266 e = . Question # 5 Key: E 2 20 20 20 2 11 1 ( 50) 0.02 100 20(50) 50 0.02[51,850 100(1,000) 50,000] 37 j jj j O OO χ = ⎡⎤ + ⎢⎥ ⎣⎦ + = ∑∑ With 19 degrees of freedom, the critical value at a 0.01 significance level is 36.191 and therefore the null hypothesis is rejected at this level. Question # 6 Key: A 2 2 2 2 2 () () 0 . 5 () ()1 . 5 0.5 1.5(1 ) 1.5 0.5 ) 1.5 (0.5) (1.5) (1 ) (1.5 ) 1 1.5 1.5 1 Iv I II v II v a Z μθ θ θθ =+− = = = +
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Question # 7 Key: E The second moment is 50 100 200 222 0 50 100 350 400 22 200 350 50 100 200 350 400 33 3 3 2 350 0 50 100 200 30 36 18 100(50) 100(50) 100(100) 16 16 350 100(200) 100(200) 30 36 18 16 16 350 5,000 3 5,000 3 10,000 3 20,000 3 20,000 250 2,10 xd x x x d x x d x xx x x x ++ =+ + + + ∫∫ 0 4,200 9,300 4,900 20,750. +++= Question # 8 Key: B Using the recursive formula 3 Pr( 0) 0.0498 3 Pr( 1) (0.4)(0.0498) 0.0598 1 3 Pr( 2) [(0.4)(0.0598) 2(0.3)(0.0498)] 0.0807 2 3 Pr( 3) [(0.4)(0.0807) 2(0.3)(0.0598) 3(0.2)(0.0498)] 0.0980. 3 Se S S S == = = + = + + = Then, Pr( 3) 0.0498 0.0598 0.0807 0.0980 0.2883. S ≤= + + + = Question # 9 Key: B Because 0.9610 < 0.9810 < 0.9827 there are 4 simulated claims. To simulate a claim, solve 2.8 36 1 36 u x ⎛⎞ =− ⎜⎟ + ⎝⎠ for 1/2.8 36[(1 ) 1] xu = −− . For the first four uniform random numbers, the simulated losses are 12.704, 58.029, 4.953, and 9.193. With the deductible of 5 and maximum of 30, the payments are 7.704, 30, 0, and 4.193. The total of these four payments is 41.897.
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Question # 10 Key: E Shifting adds δ to the mean and median. The median of the unshifted exponential distribution is the solution to / 0.5 ( ) m Sm e θ == for ln(2) m = . The equations to solve are 300 240 ln(2) . =+ Subtracting the second equation from the first gives 60 [1 ln(2)], 195.53 =− = . From the first equation, 300 195.53 104.47. = Question # 11 Key: D 222 131 5 ˆ 33 12 0 2 0 2 0 5 / 3 ˆ 596 3 4 / 9 31 3 3 3 3 3 306 0.8718 5/3 351 3 34/9 v a Z ++ ⎡⎤ ⎛⎞ + + = ⎢⎥ ⎜⎟ ⎝⎠ ⎣⎦ = + Question # 12 Key: A 10 10 10 10 10 10 10 10 ˆ (9) 0.16 0.16(200) 32 0.04045 0.02625 () 0.0142 10. 32(32 ) Sr s rr s s s s =⇒ = = −= =
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Question # 13 Key: C 1 /10,000 30,000 1 ( 30,000)/10,000 0 3 2 2 1 /10,000 30,000 2 1 ( 30,000)/10,000 0 32 ( ) ( 30,000)(10,000) (10,000) (10,000) 497.87 ( ) ( 30,000) (10,000) (10,000) (2)(10,000) x y x y EY x e dx ye d x e x e d x e −− −−+ =− = == = = 2 9,957,413.67 ( ) 9,957,413.67 497.87 9,709,539.14 ( ) 3,116.01 3,116.01/ 497.87 6.259. Var Y SD Y CV = = = Question # 14 Key: E 20 20 20 1 1 1 20 1 () , (,) ln ( , ) 20ln( ) 20 ln( ) ( 1) ln( ). j j j j fx L x x Lx αα α αθ α θ θ θα + + = = + + =+ + + Under the null hypotheses, = 2 and = 3.1. The loglikelihood value is 20ln(2) 20(2)ln(3.1) 3(39.30) 58.7810. +− = Under the alternative hypotheses, = 2 and = 7.0. The loglikelihood value is 20ln(2) 20(2)ln(7.0) 3(49.01) 55.3307. = Twice the difference of the loglikelihood values is 6.901. There is one degree of freedom (no estimated parameters in the null hypothesis versus one estimated parameter in the alternative hypothesis). At the 0.01 significance level the critical value is 6.635 and so the null hypothesis is rejected.
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Question # 15 Key: D For one year, 11 1 1 81 1 6 18 1 (| ) ( 1 ) ( ( 1 ) . x xx a x a qx q q q q q q π −+ ∝− = This is a beta distribution with parameters x 1 + a and 17 – x 1 . The mean is the Bayesian credibility estimate of q and 8 times that value is the estimate of the expected number of claims in year 2. Then, with x 1 = 2, 2 2.5455 8 17 a a + = + which implies a = 5.
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edu-exam-c-0507-sol - Question #4 Key: A The random...

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