Question

The joint density function of two continuous random variable X and Y is

f(x, y) = (cx² + xy/3 , : if 0 ⩽ x ⩽ 1, 0 ⩽ y ⩽ 2

( 0, : otherwise

(a) Find c.

(b) Find the marginal pdf of X and Y .

(c) Are X and Y independent (justify)? 

Answer 1

a.

Since it is a pdf then,

$\int$_y$\int$_x f(x, y)dxdy=1

$\intop$₀²$\int$₀¹ (cx²+xy/3) dxdy=1

$\intop$₀²(( cx³/3+x²y/6)|₀¹)dy=1

$\int$₀² (c/3+y/6)dy =1

(cy/3+y²/12)|₀²=1

=2c/3+4/12=1

2c/3=2/3

c=1

therefore, the joint density function is,

f(x, y) = (x² + x*y/3 , : if 0 ⩽ x ⩽ 1, 0 ⩽ y ⩽ 2

( 0, : otherwise

b.

Let g(x) be the marginal distribution for random variable X, then its marginal distribution is defined by,

g(x)= $\int$₀² f(x, y)dy

=$\int$₀² (x² + x*y/3)dy

=(x²y+xy²/6)|₀²

=2x²+4x/6

therefore,

g(x) = (2x²+4x/6, : if 0⩽ x ⩽ 1,

( 0, : otherwise

Let h(y) be the marginal distribution for random variable Y, then its marginal distribution is defined by,

h(y)= $\int$₀¹ f(x, y)dx

= $\int$₀¹ (x² + x*y/3)dx

=(x³/3+x²y/6)|₀¹

=1/3+y/6

therefore,

h(y)= (1/3+y/6, : if 0 ⩽ y ⩽ 2

( 0, : otherwise

c.

To test for independence then the condition, E(XY)=E(X)*E(Y) must hold.

E(XY) = $\int$_y $\int$_x (x*y)* (f(x, y))dxdy

=$\int$₀² $\int$₀¹ x*y*(x² + x*y/3)dxdy

=$\int$₀² $\int$₀¹(x³y+(x*y)²/3)dxdy

=$\int$₀²(x⁴y/4+x³y²/9)|₀¹dy

=$\int$₀²(y/4+y²/9)dy

=(y²/8+y³/27)|₀²

=4/8+8/27

=172/216

=43/54

E(X)= $\int$₀¹ x*g(x)dx

=$\int$₀¹ x(2x²+4x/6)dx

=$\int$₀¹(2x³+4x²/6)dx

=(2x⁴/4+4x³/18)|₀¹

=2/4+4/18

=26/36

=13/18

and

E(Y)=$\int$₀² y*h(y)dy

=$\int$₀² y(1/3+y/6)dy

=$\int$₀²(y/3+y²/6)dy

=(y²/6+y³/18)|₀²

=4/6+8/18

=20/18

=10/9

Let us know verify if the condition hold,

E(XY)= 43/54 and E(X)*E(Y)=13/18*10/9=65/81

Since E(XY) is not equal to E(X)*E(Y), then random variables X and Y are not independent.