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course4solutions_1104

# course4solutions_1104 - Fall 2004 Course 4 Solutions...

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Fall 2004 Course 4 Solutions Question #1 Key: C ( | ) 3 , ( | ) 3 (1 ) E X q q Var X q q q = = 1 1 3 0 0 (3 ) 3 2 2 2 E q q qdq q µ = = = = 1 1 3 4 0 0 [3 (1 )] 3 (1 )2 2 1.5 0.5 v E q q q q qdq q q = = = = 1 1 2 2 2 2 4 0 0 (3 ) (9 ) 9 2 2 4.5 4 4.5 4 0.5 a Var q E q q qdq q µ = = = = = = k = v / a = 0.5/0.5 = 1 1 0.5 1 1 Z = = + The estimate is 0.5(0) + 0.5(2) = 1. Question #2 Key: D 0.35(14) = 4.9 0.35 ˆ 0.1(216) 0.9(250) 246.6 π = + = Question #3 Key: D The problem asks for the confidence interval for β 3 , which is 28 ± 1.96(38.8423) 1/2 or 28 ± 12 or [16, 40]. Question #4 Key: E At y 1 = 0.9 the risk set is r 1 = 7 and s 1 = 1. At y 2 = 1.5 the risk set is r 2 = 6 and s 2 = 1. Then, 10 6 5 (1.6) 0.7143. 7 6 S = =

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Question #5 Key: B ( ) Pr( 250 | 1)Pr( 1) Pr 1| 250 Pr( 250 | 1)Pr( 1) Pr( 250 | 2)Pr( 2) 0.5(2/3) 10 . 0.5(2/3) 0.7(1/3) 17 claim class class class claim claim class class claim class class = = = = + = = = + ( | 1) 0.5(250) 0.3(2,500) 0.2(60,000) 12,875. E claim class = + + = ( | 2) 0.7(250) 0.2(2,500) 0.1(60,000) 6,675. E claim class = + + = 10 7 ( | 250) (12,875) (6,675) 10,322. 17 17 E claim = + = Question #6 Key: D 3 ( ) (0.74) (0.81) (0.95) ( 1)0.74 ( 1)0.81 ( 1)0.95 ( 1) (0.56943) ( ) ln ( ) 3ln( 1) ln(0.56943) 3 '( ) 0.563119 0 1 3 1 5.32747 0.563119 4.32747. p p p p L p f f f p p p p l p L p p p l p p p p = = + + + = + = = + + = = + + = = = Question #7 Key: E Homogeneous nonstationary processes have the desirable property that if they are differenced one or more times, eventually one of the resulting series will be stationary. Question #8 Key: E The sample mean is 1 and therefore mq = 1. For the smoothed empirical 33 rd percentile, (1/3)(5 + 1) = 2 and the second smallest sample item is 0. For the 33 rd percentile of the binomial distribution to be 0, the probability at zero must exceed 0.33. So, (1 ) 0.33 m q > and then 1 (1 ) 0.33 m m > . Trial and error gives m = 6 as the smallest value that produces this result.
Question #9 Key: C Let X be the number of claims. 2 2 2 ( | ) 0.9(0) 0.1(2) 0.2 ( | ) 0.8(0) 0.1(1) 0.1(2) 0.3 ( | ) 0.7(0) 0.2(1) 0.1(2) 0.4 ( | ) 0.9(0) 0.1(4) 0.2 0.36 ( | ) 0.8(0) 0.1(1) 0.1(4) 0.3 0.41 ( | ) 0.7(0) 0.2(1) 0.1(4) 0.4 0. E X I E X II E X III Var X I Var X II Var X III = + = = + + = = + + = = + = = + + = = + + = 44. 2 2 2 2 (1/ 2)(0.2 0.3 0.4) 0.3 (1/3)(0.36 0.41 0.44) 0.403333 (1/3)(0.2 0.3 0.4 ) 0.3 0.006667 0.403333/ 0.006667 60.5 50 0.45249. 50 60.5 v a k Z µ = + + = = + + = = + + = = = = = + For one insured the estimate is 0.45249(17/50) + 0.54751(0.3) = 0.31810. For 35 insureds the estimate is 35(0.31810) = 11.13. Question #10 Key: A For the given intervals, based on the model probabilities, the expected counts are 4.8, 3.3, 8.4, 7.8, 2.7, 1.5, and 1.5. To get the totals above 5, group the first two intervals and the last three. The table is Interval Observed Expected Chi-square 0—500 3 8.1 3.21 500—2498 8 8.4 0.02 2498—4876 9 7.8 0.18 4876—infinity 10 5.7 3.24 Total 30 30 6.65

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Question #11 Key: C 2 3 2 2 2 2 2 2 2 2 2 2 2 1 (0.4) ( 0.4) 1 (0.4) 0.16(0.84) 0.16 0.2944. YX YX X YX R r r r R R = = = + = i Question #12 Key: E Let ˆ ˆ ( ) H H t = and ˆ ˆ ˆ ( ( )) v Var H t = . The confidence interval is ˆ HU where / 2 ˆ ˆ exp( / ) U z v H α = ± . Multiplying the two bounds gives 2 ˆ 0.7(0.357) H = for ˆ 0.49990 H = . Then, ˆ exp( 0.49990) 0.60659. S = = Question #13 Key: C 0 2 2 0 0 2 0.575 Pr( 0) Pr( 0 | ) ( ) 1 1 2(1 ) 2(1 ) 2(1 ) 1 1 2(1 ) 2 2(0.575) 1 0.15 1.90. k k k k k k k k k k k k N N d e e e e d e e e e e e e e k θ θ θ θ π θ θ θ = = = = = = − = − + + = = = = =
Question #14 Key: C The sample -1 moment is 1 1 1 1 1 1 1 0.017094.

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