Question

Show that the function defined by f (x, y, z,u) = 24/(1+x+y+z+u)5 , x > 0, y > 0, z > 0,u > 0 and 0 elsewhere is a joint density function. (a) Find P(X > Y < Z > U). (b) Find P(X+Y +Z+U ≥ 1).

Answer 1

To show that the function defined by f(x, y, z, u) = 24/(1+x+y+z+u)^5 is a joint density function, we need to check the non-negativity of the function and its integral over the entire domain. We also need to find the probabilities for the given conditions.

Step-by-step explanation:

To show that the function defined by f(x, y, z, u) = 24/(1+x+y+z+u)^5 is a joint density function, we need to check the following:

1. The function is non-negative for all values of x, y, z, and u.

2. The integral of the function over its entire domain is equal to 1.

To find P(X > Y < Z > U), we need to find the region in the function's domain that satisfies the given condition and calculate the probability.

To find P(X+Y+Z+U ≥ 1), we need to find the region in the function's domain that satisfies the given condition and calculate the probability.