Homework 3 Solution - MSPE PROGRAM ECON 506- FALL 2010...

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MSPE PROGRAM ECON 506- FALL 2010 SOLUTION TO HW3 (The problem numbers refer to numbers in 7 th edition) 3.148 Let denote q = 1-p. 2 ) 1 ( ) ( t t dt d qe pe t m - = . At t = 0, this is 1/ p = E ( Y ). 4 2 ) 1 ( ) )( 1 ( 2 ) 1 ( ) ( 2 2 t t t t t t dt d qe qe qe pe pe qe t m - - - - - = . At t = 0, this is (1+ q )/ p 2 . Thus, V ( Y ) = (1+ q )/ p 2 – (1/ p ) 2 = q / p 2 . 3.153 a. Binomial with n = 5, p = 1/3. b. If m ( t ) is multiplied top and bottom by ½, this is a geometric mgf with p = ½. c. Poisson with λ = 2. 3.155 Differentiate to find the necessary moments: a. E ( Y ) = 7/3. b. V ( Y ) = E ( Y 2 ) – [ E ( Y )] 2 = 6 – (7/3) 2 = 5/9. c. Since ). ( ) ( tY e E t m = Y can only take on values 1, 2, and 3 with probabilities 1/6, 2/6, and 3/6. 3.161 . ) ( ) ( ) ( ) ( ) ( 1 ) 1 ( * * t qe p t tY t Y t tY Y t m e e e E e E e E t m - - - - = = = = = 3.167 a. The value 6 lies (11–6)/3 = 5/3 standard deviations below the mean. Similarly, the value 16 lies (16–11)/3 = 5/3 standard deviations above the mean. By Tchebysheff’s theorem, at least 1 – 1/(5/3) 2 = 64% of the distribution lies in the interval 6 to 16. b. By Tchebysheff’s theorem, .09 = 1/ k 2 , so k = 10/3. Since σ = 3, k σ = (10/3) 3 = 10 = C .
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3.170 Similar to Ex. 3.167: the interval (.48, 52) represents two standard deviations about the
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Homework 3 Solution - MSPE PROGRAM ECON 506- FALL 2010...

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