Joint PDF Help?

Question

Let X and Y have the joint probability density function given by f(x,y) = k(1-y) for 0 ≤ x ≤ y ≤ 1. Find the following:

a. The value of k that makes this a probability density function.

For this I got k = 6

b. P(X ≤ ¾, Y ≥ ½)

I don't know how to go about doing this.

Answer 1

a. ∫(0 to 1) ∫(x to 1) k(1-y) dy dx
= ∫(0 to 1) (k[y - y^2/2][x to 1]) dx
= ∫(0 to 1) (k(1/2 - x + x^2/2)) dx
= k[x/2 - x^2/2 + x^3/6][0 to 1]
= k (1/6 - 0)
= 1 => k = 6.

b. Here x ranges from 0 to 3/4; y ranges from 1/2 to 1 but must also be greater than or equal to x. So we split the integral in two (where y = x intersects y = 1/2, i.e. at x = 1/2) to get
P(X ≤ 3/4, Y ≥ 1/2)
= ∫(0 to 1/2) ∫(1/2 to 1) 6(1-y) dy dx + ∫(1/2 to 3/4) ∫(x to 1) 6(1-y) dy dx
= ∫(0 to 1/2) ([6y-3y^2][1/2 to 1]) dx + ∫(1/2 to 3/4) ([6y-3y^2][x to 1]) dy dx
= ∫(0 to 1/2) (3/4) dx + ∫(1/2 to 3/4) (3 - 6x + 3x^2) dx
= 3/8 + [3x - 3x^2 + x^3][1/2 to 3/4]
= 3/8 + 9/4 - 27/16 + 27/64 - 3/2 + 3/4 - 1/8
= 31/64.