The joint density for XY is given by fxy xy x3y3 16 0 x 2

The joint density for (X,Y) is given by fxy (x,y) =x3y3 /16 ; 0 < x < 2 and 0 < y < 2 (a) Find the marginal densities for X and Y (b) Are X and Y independent? (c) Find P[X<1]. (d) If it is known that y = 1, what is P[X<1]? (Do not useany computation to answer this question!)

this question is mostly answered in another section the only difference is the C part is different in this one please answer

a)

Marginal density of X

fx(X) = (0,2)integral fxy(X,Y) dy = (0,2)integral (x^3*y^3)/16 dy = x^3/4

Marginal density of Y

fy(Y) = (0,2)integral fxy(X,Y) dx = (0,2)integral (x^3*y^3)/16 dx = y^3/4

b)

X and Y are independent if

fx(X)*fy(Y) = fxy(X,Y)

so,

LHS = fx(X)*fy(Y) = x^3/4*y^3/4 = (x^3*y^3)/16 =RHS

so,

Yes they are independent

c)

d)

Conditional Probability distribution of X is

fx(x|Y=y) = fxy(X,Y)/fy(Y) = ((x^3*y^3)/16)/(y^3/4) = x^3/4

so,

P(X<=1) = (0,1)integral fx(x|Y=y) dx = (0,1)integral x^3/4 dx = 1/16 = 0.0625

Solution

c)
fx(X) = (0,2)integral fxy(X,Y) dy = (0,2)integral (x^3*y^3)/16 dy = x^3/4
P[X<1] = (0,1)integral x^3/4 dx
= (0,1) (x^4/16)
= 1/16 Answer