C are x and y independent 3 Suppose the joint pdf of X Y giv

3. Suppose the joint pdf of (X, Y) given by Find: (a) The marginal density function fx(x), and the conditional density function f y|x(y|x)

Solution

f(x)=marginal density fn of X= integration from 0 to 1 ( 6/7 * (x+y)^2 )dy
   = ( 6/7 * (x^2*y +y^3/3 + x* y^2)) over 0 to 1
   =6/7[ x^2 + 1/3 + x ] (the limit in action)

conditional density function
f (y|x) = P(Y = y|X = x) = P(X = x and Y = y) / P(X = x) = f(x, y) / f(x)
= (6/7 (x+y)^2 ) / ( 6/7 * ( x^2 + 1/3 + x)) = (x+y)^2 / (x^2 +x+1/3)

(b) E(x)= int from 0 to 1( x*f(x) dx)= int from 0 to 1 ( x * 6/7(x^2+x+1/3)) =6/7 ( x^4/4 + x^3/3 + x^2 /6) over 0 to 1

= 6/7 * ( 1/4 + 1/3 + 1/6 ) (the limits in action) = 9/14
p(x<= 1/2 ) = int from 0 to 1/2 ( f(x)dx) = int from 0 to 1/2 ( 6/7 (x^2 + x + 1/3))
= 6/7 ( x^3 /3 + x^2/2 + x/3) over 0 to 1/2 = 6/7 * ( 1/24 + 1/8 + 1/6 ) = 2/7

(C) No,x and y are not independent as you can see the condition distribution of Y|X is not free from \'x\' values.
For independence, f(x,y)=f(x)*f(y)

Thus,f(y|x) = f( x,y)/ f(x) = f(x)* f(y) /f(x) = f(y) ; which is not the case here,so they are not independent in this problem!