Let Y1 and Y2 have the joint probability density function given by:

f (y1, y2) = k (1 - y2), 0 ≤ y1 ≤ y2 ≤ 1, 0, elsewhere.

(a) Find the value of k that makes this a probability density function.

(b) Find P (Y1 ≤ 3/4, Y2 ≥ 1/2).

a) k=6

b) P (Y1 ≤ 3/4, Y2 ≥ 1/2) = 9/16

Step-by-step explanation:

a) if

f (y1, y2) = k (1 - y2), 0 ≤ y1 ≤ y2 ≤ 1, 0, elsewhere

for f to be a probability density function, has to comply with the requirement that the sum of the probability of all the posible states is 1, then

P (all possible values) = ∫∫f (y1, y2) dy1*dy2 = 1

then integrated between

y1 ≤ y2 ≤ 1 and 0 ≤ y1 ≤ 1

∫∫f (y1, y2) dy1*dy2 = ∫∫k (1 - y2) dy1*dy2 = k ∫ [ (1-1²/2) - (y1-y1²/2) ] dy1 = k ∫ (1/2-y1+y1²/2) dy1) = k[ (1/2 * 1 - 1²/2 + 1/2*1³/3) - (1/2 * 0 - 0²/2 + 1/2*0³/3) ] = k * (1/6)

then

k/6 = 1 → k=6

b)

P (Y1 ≤ 3/4, Y2 ≥ 1/2) = P (0 ≤Y1 ≤ 3/4, 1/2 ≤Y2 ≤ 1) = p

then

p = ∫∫f (y1, y2) dy1*dy2 = 6*∫∫ (1 - y2) dy1*dy2 = 6*∫ (1 - y2) * dy2 ∫dy1 =

6*[ (1-1²/2) - ((1/2) - (1/2) ²/2) ]*[3/4-0] = 6 * (1/8) * (3/4) = 9/16

therefore

P (Y1 ≤ 3/4, Y2 ≥ 1/2) = 9/16