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Unformatted text preview: SIEO 3600 (IEOR Majors) Solution to Assignment #6 Introduction to Probability and Statistics March 4, 2010 Solution to Assignment #6 ( Expectation and variance of random variables ) 1. (continued with Problem 2 in HW5) X and Y are two continuous random variables with joint probability density function f X,Y ( x,y ) = 10 xy 2 , < x < y < 1; , otherwise . (a) Compute E [ X ] and E [ Y ]. ( Hint : You have calculated the marginal PDF of X and Y in HW5) (b) Compute E [ X 2 ] and E [ Y 2 ]. (c) Use (a) and (b) to compute var( X ) and var( Y ). Solution: (a) Recall that from HW5, we have the marginal PDFs of X and Y f X ( x ) = 10 3 x 10 3 x 4 , < x < 1 , , otherwise. f Y ( y ) = 5 y 4 , < y < 1 , , otherwise. Hence, we have E ( X ) = Z 1 x 10 3 x 10 3 x 4 dx = 5 9 , E ( Y ) = Z 1 y 5 y 4 dy = 5 6 . (b) We have the second moments of X and Y E ( X 2 ) = Z 1 x 2 10 3 x 10 3 x 4 dx = 5 14 , E ( Y 2 ) = Z 1 y 2 5 y 4 dy = 5 7 . (c) From (a) and (b), we have the variances of X and Y var( X ) = E ( X 2 ) ( E [ X ]) 2 = 5 14 5 9 2 , var( Y ) = E ( Y 2 ) ( E [ Y ]) 2 = 5 7 5 6 2 . 2. Suppose that X and Y are independent continuous random variables. Show that (a) P ( X + Y a ) = R  F X ( a y ) f Y ( y ) dy (b) P ( X Y ) = R  F X ( y ) f Y ( y ) dy where f Y is the density function of Y , and F X is the distribution function of X . 2 SIEO 3600, Solution to Assignment #6 Solution: (a) Independence of X and Y implies that f X,Y ( x,y ) = f X ( x ) f Y ( y ). Therefore, P ( X + Y a ) = Z Z x + y a f X,Y ( x,y ) dxdy = Z Z x + y a f X ( x ) f Y ( y ) dxdy = Z  f Y ( y ) Z a y f X ( x ) dx dy = Z  f Y ( y ) P ( X a y ) dy = Z  f Y ( y ) F X ( a y ) dy. (b) Again, by independence, we have P ( X Y ) = Z Z x y f X,Y ( x,y ) dxdy = Z Z x y f X ( x ) f Y ( y ) dxdy = Z  f Y ( y ) Z y f X ( x ) dx dy = Z  f Y ( y ) P ( X y ) dy = Z  f Y ( y ) F X ( y ) dy....
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This note was uploaded on 10/03/2011 for the course SIEO W3600 taught by Professor Yunanliu during the Spring '10 term at Columbia.
 Spring '10
 YUNANLIU

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