Question

values of the pair (X, Y) are the points inside the triange ose vertices are at (0, 0), (0, 1), and (1, 0). A joint density for X and Y is f(x, y)- 60x2y on that triangle. a. Find the marginal density for Y. b. Are X and Y independent? Why? c. Find the CDF of Z = X + Y. ,

The answers are

Answer 1

given

f(x,y) =60x²y

a)

b)

$f(x)=\int_{0}^{x+y=1}f(x,y)dy =\int_{0}^{1-x}60x^2ydx =30x^2[y^2]_0^{1-x}=30x^2(1-x)^2$

now

f(x)*f(y) =600x²y(1-y)³(1-x)² which is not equal to f(x,y)

Hence X and Y are not independent

c)

Z=X+Y

so jacobian =1

y=z-x

so

0<z-x<1 this gives 0<z<1 and 1<z<2 but here x+y<1

also

this gives z>x>z-1 and also 0<x<1

so

now

f(x,z) =f(x,y(x,z)) =60x²(z-x)

so

$f(z)=\int_{0}^{z}f(x,z)dx =\int_{0}^{z}(60x^2z-60x^3)dx =[20x^3z-15x^4]^z_0=5z^4$

0<z<1

so

$F(z)=\int_{0}^{z}f(z)dz =\int_{0}^{z}5z^4 dz =[z^5]_0^z=z^5$ 0<z<1

=1 for z>1