HW6 - Mathematics Department UCLA T Richthammer spring 09...

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Mathematics Department, UCLA T. Richthammer spring 09, sheet 6 May 1, 2009 Homework assignments: Math 170B Probability, Sec. 1 52. Show that E ( X ) = E ( E ( X | Y )) in the case that X, Y have a joint PDF. Answer: Let f be the joint PDF, then f X ( x | Y = y ) = f X,Y ( x,y ) f Y ( y ) , so E ( X | Y = y ) = R dx f X,Y ( x,y ) f Y ( y ) x , so E ( E ( X | Y )) = R dyf Y ( y ) R dx f X,Y ( x,y ) f Y ( y ) x = R dydxf X,Y ( x, y ) x = E ( X ). 53. Conditional covariance formula. (a) Show that Cov ( X, Y ) = E (( Cov ( X, Y | Z )) + Cov ( E ( X | Z ) , E ( Y | Z )). (b) Let X 1 , X 2 be chosen completely at random in the interval [0 , x ], where X = x is randomly chosen with respect to a uniform distribution on [0 , 1]. Calculate E ( X i ), V ( X i ) and Cov ( X 1 , X 2 ) by conditioning on X . Answer: (a) E (( Cov ( X, Y | Z )) = E ( E ( XY | Z ) - E ( X | Z ) E ( Y | Z )) = E ( E ( XY | Z )) - E ( E ( X | Z ) E ( Y | Z )) and Cov ( E ( X | Z ) , E ( Y | Z )) = E ( E ( X | Z ) E ( Y | Z )) - E ( E ( X | Z )) E ( E ( Y | Z )). Using the power property we thus get E (( Cov ( X, Y | Z )) + Cov ( E ( X | Z ) , E ( Y | Z )) = E ( XY ) - E ( X ) E ( Y ) = Cov ( X, Y ). (b) We are given that f X 1 ,X 2 ( x 1 , x 2 | X = x ) = 1 x 1 [0 ,x ] ( x 1 ) 1 x 1 [0 ,x ] ( x 2 ) (independent uni- form distribution on an interval of length x ). Recall from 170A that a RV Y that is uniformly distributed on [0 , a ] has E ( Y ) = a 2 and V ( Y ) = a 2 12 . So E ( X i | X ) = X 2 , V ( X i | X ) = X 2 12 and Cov ( X 1 , X 2 | X ) = 0. So E ( X i ) = E ( E ( X i | X )) = E ( X 2 ) = 1 2 E ( X ) = 1 4 , V ( X i ) = E ( V ( X i | X )) + V ( E ( X i | X )) = E ( X 2 12 ) + V ( X 2 ) = 1 12 1 3 + 1 4 1 12 = 7 144 . Cov ( X 1 , X 2 ) = E (( Cov ( X 1 , X 2 | X )) + Cov ( E ( X 1 | X ) , E ( X 2 | X )) = 0 + Cov ( X 2 , X 2 ) = 1 4 V ( X ) = 1 48 . 54. A bin contains N = N 1 + N 2 lightbulbs of 2 types ( N i of type i ). The type i light bulbs function for a random amount of time (with mean μ i and standard deviation σ i ). Let X denote the lifetime of a bulb that is chosen at random. Calculate E ( X ) and V ( X ). Answer: Let Y denote the type of the chosen lightbulb. We are given that E ( X | Y = i ) = μ i and V ( X | Y = i ) = σ 2 i . Using the tower property for the expectation we get E ( X ) = i E ( X | Y = i ) P ( Y = i ) = μ 1 N 1 N + μ 2 N 2 N = μ 1 N 1 + μ 2 N 2 N . For the variance we have to cal- culate E ( V ( X | Y )) = i σ 2 i N i N = σ 2 1 N 1 + σ 2 2 N 2 N and V ( E ( X | Y )) = i μ 2 i N i N - ( μ 1 N 1 + μ 2 N 2 N ) 2 = ( μ 1 - μ 2 ) 2 N 1 N 2 N 2 , where we have used that E ( X | Y ) has the PMF p ( μ i ) = N i N . Using the conditional variance formula we get V ( X ) = σ 2 1 N 1 + σ 2 2 N 2 N + ( μ 1 - μ 2 ) 2 N 1 N 2 N 2 . 55. Conditional expectation, correlation and independence.
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This note was uploaded on 07/07/2010 for the course MATH 170B 170B taught by Professor Richthammer during the Winter '10 term at UCLA.

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HW6 - Mathematics Department UCLA T Richthammer spring 09...

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