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Unformatted text preview: C.4 Solutions Last updated June 16, 2010 Video solutions at http://www.theinfiniteactuary.com/?page=exams&id=144 1 . D (600) = F n (600) F * (600) = 3 4 F (600) F (100) 1 F (100) = 3 4 ( 6 7 ) 2 ( 1 2 ) 2 1 ( 1 2 ) 2 = . 104 2 . Because of the truncation, we want F * (600) = P[ X ≤ 600  X > 100] = F (600) F (100) 1 F (100) = ( 6 7 ) 2 ( 1 2 ) 2 1 ( 1 2 ) 2 = . 646 3 . The xcoordinate is the smoothed empirical percentile (and in particular doesn’t need to be adjusted for the deductible), which is 3 / (4 + 1) = . 6 4 . D (100) = F n (100) F * (100) so since 0 > D (100) we have F * (100) > F n (100), which means that the fitted model places greates weight than the observed between 0 and 100. < D (200) D (100) = [ F n (200) F n (100)] [ F * (200) F * (100)], so F * (200) F * (100) < F n (200) F n (100) and the fitted model places less weight than the observed between 100 and 200. Finally, 0 > D (300) D (200) so F * (300) F * (200) > F n (300) F n (200) and the fitted model places more weight than the observed from 200 to 300. THis gives us D as our answer. 5 . By the definition of a p p plot, s is the smoothed empirical percentile so s = 4 / (7 + 1) = 0 . 5 while t = F (3 , 000). Meanwhile, D (3 , 000) = F n (3 , 000) F (3 , 000) = (4 / 7) F (3 , 000) so ( s t ) D (3 , 000) = 1 2 F (3 , 000) 4 7 F (3 , 000) = . 07 c 2010 The Infinite Actuary, LLC C.4 Solutions 6 . Our hypothesized density is that of a Pareto with θ = 1 and α = 4 and there is no truncation so F * ( x ) = F ( x ) = 1 (1+ x ) 4 . Our standard KolmogorovSmirnov table is then x F * ( x ) F n ( x ) F n ( x ) max difference . 1 . 317 . . 2 . 317 . 2 . 518 . 2 . 4 . 318 . 5 . 802 . 4 . 6 . 402 . 7 . 880 . 6 . 8 . 280 1 . 3 . 964 . 8 1 . . 164 The maximum value in the right hand column is our answer of . 402 7 . We are estimating the mean using the method of moments, so ˆ θ = ¯ X = 100. Note that this will reduce our critical values, but that isn’t asked. There is no truncation, so F * ( x ) = F ( x ) = 1 e x/ 100 and we obtain x F * ( x ) F n ( x ) F n ( x ) max difference 29 . 25 . . 2 . 25 64 . 47 . 2 . 4 . 27 90 . 59 . 4 . 6 . 19 135 . 74 . 6 . 8 . 14 182 . 83 . 8 1 . . 17 The max in the right hand column is . 27 8 . With no truncation, F * ( x ) = F ( x ) = 1 (1 + x ) 4 and so x F * ( x ) F n ( x ) F n ( x ) max difference . 2 . 517 . . 2 . 517 . 7 . 880 . 2 . 4 . 680 . 9 . 923 . 4 . 6 . 523 1 . 1 . 949 . 6 . 8 . 349 1 . 3 . 964 . 8 1 . . 164 The max in the right hand column is our test statistic of 0 . 680. For our critical values, we plug into the formula they gave us. The 0.025 critical value is 0.66, which is less than our test statistic, and the 0.010 critical value is 0.72, which is c 2010 The Infinite Actuary, LLC C.4 Solutions above out test statistic, so we reject at the 0.025 level and do not at the 0.010 level, giving an answer of D 9 . We have truncation at 10. To find....
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This note was uploaded on 07/16/2011 for the course STATS 141 taught by Professor Rahul during the Spring '11 term at Drake University .
 Spring '11
 Rahul

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