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Unformatted text preview: 18.05. Test 2. (1) Let X be the players fortune after one play. Then 1 c 1 P ( X = 2 c ) = and P ( X = ) = 2 2 2 and the expected value is 1 c 1 5 EX = 2 c = c. × 2 + 2 × 2 4 Repeating this n times we get the expected values after n plays (5 / 4) n c. S (2) Let X i , i = 1 , . . . , n = 1000 be the indicators of getting heads. Then n = X 1 + . . . + X n is the total number of heads. We want to find k such that P (440 S n k ) . 5 . Since µ = EX i = . 5 and β 2 = Var( X i ) = . 25 by central limit theorem, S n − nµ S n − 500 Z = = ≈ nβ ≈ 250 is approximately standard normal, i.e. k − 500 P (440 S n k ) = P ( 440 − 500 = − 3 . 79 Z ) ≈ 250 ≈ 250 ( k − 500 ) − ( − 3 . 79) = . 5 . ≈ 250 From the table we find that ( − 3 . 79) = . 0001 and therefore ( k − 500 ) = . 4999 . ≈ 250 500 Using the table once again we get k − and k 500 . 250 (3) The likelihood function is α n e n ( α ) = X i ) +1 ( and the loglikelihood is Y log ( α ) = n log α + nα −...
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This note was uploaded on 10/28/2008 for the course MATH 18.05 taught by Professor Panchenko during the Spring '05 term at MIT.
 Spring '05
 Panchenko
 Statistics, Probability

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