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hw7-sol

# hw7-sol - AMS 311(Fall 2009 Joe Mitchell PROBABILITY THEORY...

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AMS 311 (Fall, 2009) Joe Mitchell PROBABILITY THEORY Homework Set # 7 – Solution Notes (1). (15 points) If E (3 X ) = 1 and var (2 X ) = 5 , find (a). E [(2 + X ) 2 ] and (b). var (4 + 3 X ) . If E (3 X ) = 1 and var (2 X ) = 5, then we know that E ( X ) = 1 / 3 and 5 = var (2 X ) = 4 var ( X ) = 4[ E ( X 2 ) - [ E ( X )] 2 ] = 4[ E ( X 2 ) - (1 / 3) 2 ], so E ( X 2 ) = (5 / 4) + (1 / 9) = 49 / 36. Thus, E [(2 + X ) 2 ] = E [4 + 4 X + X 2 ] = 4 + 4 E ( X ) + E ( X 2 ) = 4 + 4 · (1 / 3) + 49 / 36 = 241 / 36 and var (4 + 3 X ) = 3 2 var ( X ) = 9 · (5 / 4) = 45 / 4 (2). (15 points) The random variables X and Y have a joint density function given by f ( x, y ) = 2 e - 2 x /x if 0 x < , 0 y x 0 otherwise Compute cov ( X, Y ) . cov ( X, Y ) = E ( XY ) - E ( X ) E ( Y ) = 0 x 0 xy · 2 e - 2 x /xdydx - 0 x 0 x · 2 e - 2 x /xdydx · 0 x 0 y · 2 e - 2 x /xdydx (3). (30 points) [See examples 5a, 5b of Ross, Chapter 6.] Let X and Y be continuous random variables with joint probability density function given by f ( x, y ) = 1 25 if 0 x < 5 and - x y x 0 otherwise Compute the following quantities. (a). (6 points) The marginal density, f Y ( y ) , of Y . (Be explicit about all cases.) (b). (6 points) The conditional density, f X | Y ( x | 2) , of X given Y = 2 . Be explicit about all cases!

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hw7-sol - AMS 311(Fall 2009 Joe Mitchell PROBABILITY THEORY...

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