Answer:
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
