The joint density function of the random variables X and Y i

The joint density function of the random variables X and Y is given by f(x, y) = {6/3 (x^2 + y) 0, 0 Less then equal x Greater then equal 1, 0 Less then equal y Greater then equal 1 elsewhere Find the marginal densities of X and Y. Are X and Y independent Find E(X), the expected value of X. Find F1 (x/y) and use it to find P (X Less 0.5/Y = 1).

Solution

f( x,y) = ( x^2 + y) * 6 /5......0 <= x <=1 and , 0<= y <=1.....

marginal density of X=f(x) = integration from 0 to 1 [ ( x^2 + y) * 6* dy /5 ] = ( x^2 + 0.5)*6/5...

marginal density of Y=f(y) = integration from 0 to 1 [ ( x^2 + y) * 6* dx /5 ] = ( 0.333 + y )*6/5......


f(x) * f(y) = ( x^2 + 0.5)*6/5 * ( 0.333 + y )*6/5 which is not equals to the joint density f(x,y)..

so, they are not independent!..............

c) E(X) =  integration from 0 to 1 [ x * f(x) *dx ]
=  integration from 0 to 1 [ x * ( x^2 + 0.5)*6/5 * dx ] = 3/5....

d) f ( x | y) = f( x,y) / f(y) =  [ ( x^2 + y) * 6/5 ] / [  ( 0.333 + y )*6/5 ] = ( x^2 + y) / ( 0.333 + y)....

p [ x < 0.5 | y =1 ] = integration from 0 to 0.5 [ ( x^2 + 1) * 6* dx /5 ] / [ ( 0.333 + 1 )*6/5 ] .........
= ( 13 /20 ) / ( 8 /5 ) = 13 / 32.......