Calculating C and Marginal PDFs for a Uniform Distribution on a Bounded Area

[whitejac]

Homework Statement

E = { (x,y) | |x| + |y| ≤ 1}

f_x,y (x,y) =
{
c (x,y) ∈ E
0 otherwise
}

Find C.
Find the Marginal PDFs
Find the conditional X given Y=y, where -1 ≤ y ≤ 1.
Are X and Y independent.

Homework Equations

I'm taking a guess here in the solution...
but F(x,y) = F(x)F(y)
and f(x,y) = f(x)f(y)
These will be used later, when I'm wishing to find the Marginals and the independence.

The Attempt at a Solution

So, this is a uniform distribution (if it's not stated in my pdf, it's stated in the problem's text.)
Considering that this an "area" I should just be able to integrate this with respect to the bondaries correct?
That would be ∫_0,1∫0_0,1cdxdy? Then c = 1, or do I base it off of E? Then it should be bounded from [-1,1]?
This is what I believe it to be, but I'm not entirely sure. My professor gave a solution that was probably more general where he found something else first, but i didn't quite get it because he was trying to rush it at the end of class.

After finding C, the marginals are the integrals with respect to y and x to give us the "trace" of the density function.

 

[andrewkirk]

Because it's a pdf, we must have $\int_{-\infty}^\infty\int_{-\infty}^\infty f_{X,Y} (x,y)dx dy =\int_E c dA=c\int_E dA=1$ where the $dA$ indicates integrating by area.

So just work out that last integral, which is the area of $E$, and then figure out $c$ by the requirement for the final equality to hold.

You'll find it easier to work out E if you first draw a picture.

 

[whitejac]

Okay, that's what I thought after reviewing a bit of what this meant.
Drawing E, we have what looks essentially like a diamond where each point is at y = 1, y = -1, x = 1, x = -1. This would bound the area from [-1,1] for dx and dy.
My question now is when evaluating the area, do we use E as the function of integration or do we use the PDF? I'm trying to grapple the idea of two things that are related by not the same. We have a distribution of probabilities... across a geometric area E?

 

[andrewkirk]

when evaluating the area, do we use E as the function of integration or do we use the PDF?

You are trying to get the cumulative probability of all (x,y) pairs, which requires integrating the pdf. So it's the latter. E is a set - a region in the number plane - not a function that can be integrated in this context. The relevance of E is that you know that the pdf is only nonzero inside E, so you can restrict your integration to inside E without changing the result.

 

[whitejac]

Oh okay, and within that set it has a uniform probability (0,1) where each point is smaller in probability by a factor of 1/4 because E takes the shape it does?

 

[LCKurtz]

Remember that $\iint_E c~dydx = c\iint_E 1~dydx = c\cdot \text{Area of }E$. You shouldn't need calculus and integrals to figure out that last expression.