HW5Solutions (dragged) 2

# HW5Solutions (dragged) 2 - Cumulative distribution p value...

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Problem 4.35 prob (same color) = 2 ( 5 2 ) ( 10 2 ) = 4 / 9, prob (di erent color) = 1 - prob (same color) = 5 / 9. E [ value ] = 1 . 1 4 / 9 + ( - 1) 5 / 9 = - 1 15 V ar ( value ) = E [ value 2 ] - ( E [ value ]) 2 = (1 . 1 2 4 / 9 + ( - 1) 2 5 / 9) - ( - 1 15 ) 2 Problem 4.38 a. We know that for a random variable Y , V ar ( Y ) = E [ Y 2 ] - ( E [ Y ]) 2 Let Y = 2 + X , then 5 = V ar ( X ) = V ar (2 + X ) = E [(2 + X ) 2 ] - ( E [2 + X ]) 2 = E [(2 + X ) 2 ] - (2 + E [ X ]) 2 = E [(2 + X ) 2 ] - (2 + 1) 2 Thus E [(2 + X ) 2 ]=14 Problem 4.38 b. V ar (4 + 3 X ) = 3 2 V ar ( X ) = 9 5 = 45 Extra problem 1 By enumeration. Sequence Prize selected With probability 1,2,3 2 1/6 1,3,2 3 1/6 2,1,3 3 1/6 2,3,1 3 1/6 3,1,2 2 1/6 3,2,1 1 1/6 Thus, the selected prize will have value 1 with prob 1/6, value 2 with prob 2/6 and value 3 with prob 3/6. E [ value ] = 1 1 / 6 + 2 2 / 6 + 3 3 / 6 = 14 / 6, V ar ( value ) = E [ value 2 ] - ( E [ value ]) 2 = 1 1 / 6 + 4 2 / 6 + 9 3 / 6 - (14
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Unformatted text preview: Cumulative distribution: p ( value 1) = 1 / 6 , p ( value 2) = 3 / 6 , p ( value 3) = 1 Extra problem 2 a. E [ R ] = E [( X 1 + X 2 ...X n ) /n ] = 1 n E [ X 1 + X 2 ... + X n ] = 1 n ( E [ X 1 ]+ E [ X 2 ] ... + E [ X n ]) = 0 . 04 V ar ( R ) = V ar [( X 1 + X 2 ...X n ) /n ] = 1 n 2 V ar [ X 1 + X 2 ... + X n ] = 1 n 2 ( V ar [ X 1 ]+ V ar [ X 2 ] ... + V ar [ X n ]) = V ar ( X 1 ) n std ( R ) = std ( X 1 ) p n = . 03 p n Extra problem 2 b. Reduce risk while maintain the same expected return. Extra problem 3 E [ x ] = 3 , V ar ( x ) = 1 3...
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