Multiple Random Variables - find probability given joint pdf

[ksm100]

Homework Statement

Show that the function defined by f(x,y,z,u) = 24*(1+x+y+z+u)^(-5) for x,y,z,u>0 and f=0 elsewhere is a joint density function.

Find P(X>Y>Z>U) and P(X+Y+Z+U>=1).

Homework Equations

distribution function = quadruple integral from 0 to x (or y or z or u) here of the joint density.
P(X<=x, Y<= y, Z <= z, U <= u) = F(x,y,z,u) = multiple integral (each variable from 0 to infinity) of f(x,y,z,u)

The Attempt at a Solution

I've checked that f is a joint density function by integrating from 0 to infinity for all variables and getting 1 as an answer.

What I'm unsure of is how to find the probabilities requested. I've also found that the marginal density of X is f_X(x)=(1+x)^(-2) by integrating from 0 to infinity for the other three variables, so F_X(x)=1-1/(x+1).

I think this all has to do with the bounds of the integral of the joint pdf, but I'm having trouble figuring out what they are.
For the first probability, P(X<Y<Z<U), I know from the solutions it's equal to 1/24, and I'm tempted to think that since the marginal densities involve only one variable this can be reduced to essentially a counting problem where there are 4! orders and one that you want so the probability is 1/4! = 1/24. But I'm not sure if this is legitimate or if it's just a coincidence and the actual solution requires integration and such.

For the second probability, P(X+Y+Z+U>=1), I thought it should be the integral of the joint pdf for u,z,y from 0 to 1 (because if they are greater than 1 then P=1 which isn't interesting) and x from 0 to 1-y-z-u but this didn't work out (the solution says the answer should be 15/16).

If anyone could offer some advice I would greatly appreciate it!
Thanks so much!

 

[mighty2000]

Hello there! It's great to see another scientist on this forum. Let's take a look at the two probabilities you are trying to find.

For P(X>Y>Z>U), you can approach it as a counting problem, but it's important to make sure that the bounds of the integral reflect the given condition. In this case, the condition is all variables must be greater than 0, so the bounds should reflect that. You can also use the joint distribution function to find this probability, which would involve integrating from 0 to infinity for all variables.

For P(X+Y+Z+U>=1), your approach is on the right track, but the bounds of the integral are not quite correct. Remember that for this probability, the condition is that the sum of all variables must be greater than or equal to 1. So the bounds should reflect that, and you should also consider the order in which you integrate the variables. You can also use the joint distribution function to find this probability, which would involve integrating from 1 to infinity for the sum of all variables.

I hope this helps! Let me know if you have any further questions.