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final_practice_sol

# final_practice_sol - ORIE 360/560 Practice Final Solutions...

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ORIE 360/560 Practice Final Solutions Fall, 2007 Problem 1. (a) EX = E [ E [ X | Y ]] = 1 2 E [ X | Y = 1] + 1 2 E [ X | Y = 2] = 3 4 . Var X = E [ X 2 ] - E 2 [ X ] = E [ E [ X 2 | Y ]] - 9 16 = 1 2 E [ X | Y = 1] + 1 2 E [ X | Y = 2] - 9 16 . Now, if X U (0 , a ), then EX 2 = Z a 0 1 a x 2 dx = a 2 3 , so Var X = 1 2 1 3 + 4 3 - 9 16 = 5 6 - 9 16 = 40 48 - 27 48 = 13 48 . (b) P [ X > 1] = P [ Y = 2 , X > 1] = P [ Y = 2] P [ X > 1 | Y = 2] = 1 2 · 1 2 = 1 4 . Problem 2. (a) For a given suit, there are ( 13 k )( 39 13 - k ) ways to have k cards of that suit (and hence 13 - k cards of the other three). Therefore, the probability of getting a hand with one suit of at least nine cards is 4 13 X k =9 ( 13 k )( 39 13 - k ) ( 52 13 ) . (b) For a given pair of suits, there are ( 26 13 ) - 2 ways to choose 13 cards from those two suits (not including the two hands consisting of only cards from one suit or only cards from the other). There are ( 4 2 ) ways to choose two suits. Hence, the answer is 4 2 ( 26 13 ) - 2 ( 52 13 ) . Problem 3. (a) We have EZ = E [4 X - 3 Y + 2] = 4 · 5 - 3 · ( - 2) + 2 = 28 . (b) We have Cov( X, Z ) = Cov( X, 4 X - 3 Y + 2) = Cov( X, 4 X ) + Cov( X, - 3 Y ) + Cov( X, 2) = 4 Var X + 0 + 0 = 4 . (c) Recall that when you condition on Y , Y behaves as a constant. Hence, Var[ Z | Y ] = Var[4 X - 3 Y + 2 | Y ] = Var[4 X | Y ] = Var 4 X (by independence of X and Y ) = 16 . Problem 4. (a) We have F X ( x ) = 0 if x 0, F X ( x ) 1 if x > 1, and for 0 < x < 1, F X ( x ) = Z x 0 θt θ - 1

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