ho7_1 - CS237 Problem Set 7 Expectation Ungraded Please...

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CS237. Problem Set 7: Expectation Ungraded. Please complete by Thursday March 10 March 8, 2011 Reading. L&L Expectation I and II. M&U 2.1-2.4. Exercise 1. We independently roll two dice. Let X 1 be the number that comes up on the first die, X 2 be the number on the second die, and let Y = X 1 + X 2 . 1. What is E [ X | X 1 even]? 2. What is E [ X | X 1 = X 2 ]? 3. What is E [ X 1 | X = 8]? Solution: For 1. If X 1 is even, then we can have the following outcomes: X 1 = x and X 2 = y where ( x, y ) ∈ S = { (2 , 1) , (2 , 2) , ..., (2 , 6) , (1) (4 , 1) , (4 , 2) , ... (4 , 6) , (2) (6 , 1) , (6 , 2) , ... (6 , 6) } (3) Notice that there are 3 · 6 = 18 possible outcomes. Given that X 1 is even, each of these outcomes happens with with probablity 1 18 . So we have E [ X | X 1 even ] = X ( x,y ) ∈S ( x + y ) 1 18 (4) = X x ∈{ 2 , 4 , 6 } X y ∈{ 1 , 2 , 3 , 4 , 5 , 6 } ( x + y ) · 1 18 (5) = 1 18 X x ∈{ 2 , 4 , 6 } X y ∈{ 1 , 2 , 3 , 4 , 5 , 6 } x + X x ∈{ 2 , 4 , 6 } X y ∈{ 1 , 2 , 3 , 4 , 5 , 6 } y (6) = 1 18 X x ∈{ 2 , 4 , 6 } 6 x + X y ∈{ 1 , 2 , 3 , 4 , 5 , 6 } 3 y (7) = 1 18 ((2 + 4 + 6) · 6 + (1 + 2 + 3 + 4 + 5 + 6) · 3) (8) = 1 18 (72 + 126) = 10 (9) For 2, we condition on X 1 = X 2 and we have the following outcomes: (1 , 1) , (2 , 2) , (3 , 3) , (4 , 4) , (5 , 5) , (6 , 6) . Given that X 1 = X 2 , each one of these outcome occurs with probability 1 6 so we have: E [ X | X 1 = X 2 ] = X x ∈{ 1 , 2 , 3 , 4 , 5 , 6 } ( x + x ) · 1 6 (10) = 2 · 6 · 7 2 1 6 = 7 (11) For 3, the following outcomes exist given X = 8 : (2 , 6) , (3 , 5) , (4 , 4) , (5 , 3) , (6 , 2) , so we have E [ X 1 | X = 8] = 2 · 1 6 + 3 · 1 6 + 4 · 1 6 + 5 · 1 6 + 6 · 1 6 (12) = 20 6 3 . 3 (13) 1
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Exercise 2. A student delays laundry for a few days. All random values below are mutually independent.
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