Another continuous joint distribution

Suppose that X and Y are random variables with joint probability distribution f(x,y) = cx^2y^3 over D=[0,2]\times[0,1].

• Find c such that f is a good probability distribution over D
• Evaluate E(X) and E(Y).
• Show that X and Y are independent.

• To solve for c you must take the integral of f(x,y) with respect to x on the domain of [0,2] and with respect to y on the domain [0,1] and set it equal to 1. This got me the c value of 3/2.
• E(X) is the sum of f(x,y)*x with respect to x and y on the given domains, which gives the value of 3/2
  E(Y) is the sum of f(x,y)*y with respect to x and y on the given domains, which gives the value of 4/5
• To check for independence you must check that E(XY) == E(X) * E(Y) and in this case it does (1.2 == 1.2)

After doing the double integral of cx^2y^3, over the given bounds, found c=1.5;

E(X) = 1.5, E(Y) = 4/5, E(X,Y) = 1.2

E(XY) == E(X)*E(Y), 1.2 =1.2 so yes are independent