Question

the joint density function of y1 and y2 is given by f(y1, y2) = 30y1y22, y1 − 1 ≤ y2 ≤ 1 − y1, 0 ≤ y1 ≤ 1, 0, elsewhere. (a) find f 1 2 , 1 2 .

Answer 1

The joint density function is given by f(y1, y2) = 30y1y22. To find f(1/2, 1/2), substitute y1 = 1/2 and y2 = 1/2 into the joint density function.

Step-by-step explanation:

The joint density function of y1 and y2 is given by:

f(y1, y2) = 30y1y22, y1 - 1 ≤ y2 ≤ 1 - y1, 0 ≤ y1 ≤ 1, 0 elsewhere.

(a) To find f(1/2, 1/2), we substitute y1 = 1/2 and y2 = 1/2 into the joint density function:

f(1/2, 1/2) = 30(1/2)(1/2)2 = 30(1/2)(1/4) = 30(1/8) = 3.75

Answer 2

To find the value of f(1/2, 1/2) for the given joint density function, substitute y1 and y2 with 1/2 resulting in f(1/2, 1/2) = 3.75.

Step-by-step explanation:

The student asks to find the value of the joint density function f(y1, y2) at the point (1/2, 1/2). The provided joint density function is f(y1, y2) = 30y1y2^2 for { y1 − 1 ≤ y2 ≤ 1 − y1, 0 ≤ y1 ≤ 1, 0}, and zero elsewhere. To find f(1/2, 1/2), we simply substitute y1 and y2 with 1/2 into the function, assuming (1/2, 1/2) is within the defined range.

So, we calculate it as follows:

f(1/2, 1/2) = 30 * (1/2) * (1/2)^2 = 30 * 1/2 * 1/4 = 30/8 = 3.75