The joint probability density function of XY is given by fx

The joint probability density function of (X,Y ) is given by

f(x, y) =   c(x + y)^2 if 0<=x<=1 and 0<=y<=1

0, otherwise.

(a) Find the value of c. (b) Are X and Y independent. Justify your answer

Solution

(a)

f(x,y) = c(x + y)^2

integral (0->1) integral(0->1) c(x + y)^2 dxdy

= integral (0->1) integral(0->1) c(x^2 + y^2 + 2xy) dxdy

= integral (0->1) integral(0->1) c(x^3/3 + xy^2 + x^2y) dy

= integral (0->1) c(1/3 + y^2 + y) dy

= (0->1) c(y/3 + y^3/3 + y^2/2)

= c(1/3 + 1/3 + 1/2)

=7c/6

Now,

7c/6 = 1

=> c = 6/7 Answer

(b)

f(x,y) = x^2 +2xy + y^2

fX(x) = integral (x^2 +2xy + y^2)dy

= x^2y +xy^2 + y^3/3

fY(y) = integral (x^2 +2xy + y^2)dx

= x^3/3 +x^2y + xy^2

Now ,

f(x,y) is not equal to fX(x) . fY(y)

Therefore ,

X and Y are not independent.

The joint probability density function of (X,Y ) is given by f(x, y) = c(x + y)^2 if 0<=x<=1 and 0<=y<=1 0, otherwise. (a) Find the value of c. (b)

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