Thread: Figuring out limits of integration when doing joint densities

1. Figuring out limits of integration when doing joint densities

The problem in question:

f(x,y)=24xy for 0<= x<=1, 0<=y<=1, 0<=x+y<=1, otherwise x=0

The goal is to show that f(x,y) is a joint probability fxn and I believe we do that by doubly integrating f(x,y) and showing the result equals 1.

I believe that the answer is: $\int_0^1\int_0^{1-y} \! 24xy \, \mathrm{d}x{d}y$, however I'm having trouble understanding the limits of integration. Why is it that we integrate from 0 to 1-y on x and from 0 to 1 on y?

Here's a similar problem I had earlier:

f(x,y)=2 0<x<y, 0<y<1

The problem asks the student to determine whether X,Y are independent. So, I set out to find the marginal distribution of x:

$\int_x^1 \! 2 \, \mathrm{d}y$ Why do we integrated from x to 1
$\int_0^y \! 2 \, \mathrm{d}x$ Why do we integrate from 0 to y

Could we switch the limits of integration as long as we adjust them for both X and Y? For example, could we integrate from 0 to 1 instead of x to 1 if we change the limits of integration on Y?

Thanks a bunch in advance...I hate probability

2. you need to review calculus three
To get started you need to draw the regions and that will explain the bounds
AND by the way I LOVE probability.
And this seems to be a problem from Walpole, I'm the first reviewer of Walpole in the introduction.