ST 558 Homework 8

# ST 558 Homework 8 - Kelly Harrison ST 558 PROBLEM 1 a) P(...

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Unformatted text preview: Kelly Harrison ST 558 PROBLEM 1 a) P( x, y) = 1 x =0 y =0 * " ... # B C 2 4 x +y x!y! =C -4 x ! y ! x =0 y =0 2 2 x y =C x ! y ! x =0 y =0 2 x 2 y = Ce e = Ce = 1 2 2 4 C =1/ e = e x 0 n NOTE : n! = e x b) P( x) = P( y) = e y =0 -4 2 2 x y x!y! -4 = e = e -4 2 2 y ! x! y =0 x 2 2 y = e = e -4 2 2 x x! y -4 e = e 2 -2 2 2 x x! y -2 e x =0 2 2 x y -4 x!y! x ! y! x =0 y x y! e = e 2 y! c) P( x, y) = e -4 2 x +y x!y! -2 2 x -2 2 y -4 2 x + y = e y ! e x ! y ! x! Since P(x,y) = P(x)P(y), then yes, x and y are independent. P( x)P( y) = e PROBLEM 2 a) 1 1 x 2 3 1/12 0 1/18 5/36 x P(x) y 2 1/6 1/9 1/4 19/36 y 3 0 1/5 2/15 1/3 P(y) 1/4 14/45 79/180 1 1 2 3 1/4 14/45 79/180 1 2 3 5/36 19/36 1/3 b) P(x 2) = + 14/45 = 101/180 c) P(y 2) = 5/36 + 19/36 = 2/3 PROBLEM 3 a) f( x) = 3 0 1 1 2 ( x + 1)dy = 2 3 ( x + 1) dy = 0 1 2 3 ( x + 1) ( y | ) = 0 1 2 3 ( x + 1) (1 - 0) = 2 3 ( x + 1) f( y) = 2 3 ( x + 1)dx = 0 1 2 x 2 ( + x) | = 0 3 2 3 2 12 2 - = 2 + 1 ( 0 + 0 ) 1 b) f( x | y) = f( x, y) f ( y) 2 = 3 ( x + 1) 1 = 2 3 ( x + 1) c) P( x + y 2 1 1)= 2 3 ( x+ 0 0 1 1-x 2 1)dydx= 2 3 (x + 0 1 1) 1-x dydx = 0 2 3 (x + 0 1 1) ( y | )dx= 0 1-x 2 3 (x + 0 1 1) (1- x)dx 1 2 x3 2 1 2 4 = (1 - x )dx = - x | = - = 3 1 3 = 3 9 0 3 0 3 3 3 PROBLEM 4 2 a) E( x y) = 2 x y( 4( x - xy) )dxdy 2 0 0 1 1 = 4 x dx y(1 - y)dy = 4 3 0 0 1 1 x 4 1 4 |( 0 y 2 2 - y 3 3 ) | = 1 (1 / 2 - 1 / 3) = 4 0 1 1 6 b) E( x - y) = 1 0 0 1 1 ( x - y) ( x - xy)dxdy = (1 - y) ( x - yx)dxdy = (1 - y) 0 2 1 0 1 2 0 1 2 1 x x 3 -y | dy 0 3 2 = (1 dy ( - y) - 2 = 3 3 0 0 1 y 1 1 - y 6 - y 2 )dy = ( y 3 - y 2 12 - y 3 6 )| = 0 1 1 12 c) ? PROBLEM 5 E( xy) = 1 x y 2 51 2 3 4 | = 2 dy = y | = y 0 3 0 5 5 0 0 0 0 0 0 3 1 y 1 y 1 1 x y 2 41 1 2 3 x = xdxdy = 6 dxdy = 6 = 2 dy = y | = x6 x y |0 0 3 4 2 0 0 0 0 0 0 1 y xy6 xdxdy = 6 y x dxdy = 6 y 2 1 y 1 1 x y 3 31 2 2 y = 6xdxdy == 6 = 3 dy = y | = 1 | y 0 0 2 3 0 0 0 0 2 1 1 Cov( x, y) = E( xy) - x y = - (1) = - 5 2 10 1 y 1 b) f( x) = 1 6xdy = 6xy | = 6 x(1 - x) x 2 3 1 x 1 1 x 1 3 x 4 - 1 1 E( x) = 6( x - x )dx = 6 - | =6 = 4 2 0 3 4 3 0 1 1 1 x 4 x 5 3 2 2 2 3 4 E( x ) = 6( x - x )dx = 6 - x )dx = 6 - x (x | = 0 4 5 10 0 0 V( x ) = 1 1 - = 10 20 2 y 3 2 f ( y) = E( y ) = 2 6xdx = 4 6 2 x | = 3y 2 0 y 2 0 1 3y dy = 2 3 5 0 V( y) = 3 5 - 1 = -2 / 5 PROBLEM 6 a) E( xy) = xyP( xy) = xy 5xy 4 = 5 4 = 4( 4 / 5) = 16 / 5 ...
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## This note was uploaded on 05/06/2010 for the course ST 558 taught by Professor Staff during the Spring '08 term at NMT.

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