Let f(x,y) = 6(1-y) where 0

Question

Let f(x,y) = 6(1-y) where 0 < x < y < 1 be a possible joint distribution.

 Demonstrate that this is a viable distribution (show your work)

The 0 < x < y < 1 is throwing me off. What domains should I be using when integrating with respect to x and with respect to y?

anonymous · Answer

I've tried 0 to y for x and x to 1 for y, but I am unsure of this answer. To prove a viable distribution, I need to show that the volume of f(x,y) with respect to x and y is equal to one. I need a definite integral to equal 1.

irishboy123 · Answer

the volume of f(x,y) with respect to x and y is the double integral of f(x,y).  and your f(x,y) is the plane $0x + 6y +z = 6$

your limit $0<x<y<1$ seems to be looking at the $x = 1, \ y= 1$ square  but above the y = x line itself.  

so you could try $\int_{y=0}^{1} \ \int_{x=0}^{y} (6 - 6y) \ dx \ dy$

irishboy123 · Answer

notice the order of integration.

anonymous · Answer

Wow. You are the best. I've integrated and received the correct answer of 1. I'm ecstatic. Thanks!