ST 558 Homework 8

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

This note was uploaded on 05/06/2010 for the course ST 558 taught by Professor Staff during the Spring '08 term at NMT.

Ask a homework question - tutors are online