Hey folks,

I've got a problem with a joint probability density function question.

Given f(x, y) = 6(x^2)y for 0 <= x <= y, x + y <= 2 and 0 elsewhere, I need to verify that f is a valid joint PDF.

The fact that f(x, y) >= 0 for all x, y is obvious and not difficult to prove. The part I'm having trouble with is showing that:

∫ ∫ f(x,y)*dx*dy = 1

...where both integrals are from -∞ to ∞.

Here's my thinking: based on the bounds given in the function definition, x must be bounded from 0 to 1, since if x is greater than 1 then x + y is greater than 2 and f goes to 0. y must be bounded from 0 to 2.

However, simply integrating with those bounds doesn't work; the answer I get is 4. The problem is that the integral using those bounds includes some area which should not be included according to the definition of the PDF since it would include, for example, x = 1 and y = 2.

However, I don't know what to do to get around this. I think the bounds need to be changed, but I don't know what they should be. I also toyed with the idea of subtracting a different integral, but I think that would invalidate the proof anyway.

Thanks for any help.

Edit: So I'm able to get the correct answer (1) if I subtract the integral ∫ ∫ f(x,y)*dx*dy bounded from x = 0 to 1 and y = 1 to 2, since this is equal to the area that's currently in my integral but shouldn't be. This is the answer I'll go with for now, but can anyone tell me if doing this subtraction is a valid way of showing that the density function is equal to 1 when integrated from -∞ to ∞, or if I need to go about this another way?

x+y<2

so y<2-x

and x<y

so x<y<2-x

=> 0<x<1

so integrate from 0 to 1 with x and x to 2-x with y