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The joint density function of
Y_1and
Y_2is given by


f(y_1,y_1)=\left \{ {{30y_1y_2^2} \atop {0}} \right. \left ,y_1-1\leq y_2\leq 1, 0\leq y_2 \leq 1 \atop{, elsewhere}

a. Find F(1/2,1/2).

b. Find F(1/2,2).

c. Find P(
Y_1>
Y_2).

1 Answer

5 votes

The joint density function of Y and Y is only non-zero for values between 0 and 1, and where Y1 is less than or equal to Y2. The probability that both Y1 and Y2 are less than or equal to 1/2 is 1/4, while the probability that Y1 is less than or equal to 1/2 and Y2 is less than or equal to 2 is 0. Finally, the probability that Y1 is greater than Y2 is 1/2.

The joint density function of Y and Y is given by the following equation:

f(y1, y2) = 1, if 0 <= y1 <= 1 and 0 <= y2 <= 1 and y1 <= y2

f(y1, y2) = 0, otherwise

This means that the density function is only non-zero for values of y1 and y2 between 0 and 1, and where y1 is less than or equal to y2.

To find F(1/2, 1/2), we need to calculate the probability that Y1 and Y2 are both less than or equal to 1/2. Because the density function is 1 for this range, the probability is simply the area of the region in the y1-y2 plane where both y1 and y2 are less than or equal to 1/2. This is a right triangle with base and height of 1/2, so the area is (1/2)*(1/2) = 1/4.

To find F(1/2, 2), we need to calculate the probability that Y1 is less than or equal to 1/2 and Y2 is less than or equal to 2. However, the density function is only non-zero when Y1 <= Y2, and in this case, Y2 cannot be less than or equal to 2 while Y1 is less than or equal to 1/2. Therefore, the probability F(1/2, 2) is 0.

To find P(Y1 > Y2), we need to calculate the probability that Y1 is greater than Y2. The area where this occurs is the triangle in the y1-y2 plane above the line y1 = y2. This triangle has base and height of 1, so the area is 1/2.

Therefore, the answers are:

a. F(1/2, 1/2) = 1/4

b. F(1/2, 2) = 0

c. P(Y1 > Y2) = 1/2

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