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hwk4_soln

# hwk4_soln - IEOR E4702 Statistical Inference for Financial...

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IEOR E4702 Statistical Inference for Financial Engineering HWK Solution 4 1. a. Denote X i ’s the samples, i = 1 , . . . , n , and O i ’s frequencies O i , i = 1 , . . . , 12 . With the assumption of geometric distribution and statistical independence of samples, the likelihood is L ( p ) = n Y i =1 p X i 1 (1 p ) = p P n i =1 X i n (1 p ) n , and ln L ( p ) = ( n X i =1 X i n ) ln p + n ln(1 p ) . Taking derivative with respect to p and letting it to zero gives P n i =1 X i n p n 1 p = 0 . Therefore, the estimator for p is ˆ p = P n i =1 X i n P n i =1 X i = 363 130 363 = 0 . 6419 . b. There are 12 cells in the test. Rigorously speaking, we cannot directly apply the chi- square test, as some of the cells have very low frequencies. However, for this hwk problem, we shall ignore this issue. Let O i be the observed cell counts and E i be the expected cell counts under H 0 , where under the geometric distribution E i = n ˆ p i 1 (1 ˆ p ) , and E 1 = 46 . 5565 , E 2 = 29 . 8834 , E 3 = 19 . 1813 , E 4 = 12 . 3120 , E 5 = 7 . 9027 , E 6 = 5 . 0726 , E 7 = 3 . 2559 , E 8 = 2 . 0899 , E 9 = 1 . 3414 , E 10 = 0 . 8610 , E 11 = 0 . 5527 , E 12 = 0 . 3547 . The likelihood ratio Chi-square test statistic is 2 ln Λ = 2 12 X i =1 O i ln μ O i E i = 6 . 1875 . 1

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The Pearson’s Chi-square test statistic is 12 X i =1 ( O i E i ) 2 E i = 6 . 7186 . The d.f. is 12 1 1 = 10 , which gives the critical value χ 2 0 . 05 , 10 = 18
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hwk4_soln - IEOR E4702 Statistical Inference for Financial...

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