# Finding c in a joint PDF

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1. Nov 8, 2015

### whitejac

1. The problem statement, all variables and given/known data
E = { (x,y) | |x| + |y| ≤ 1}

fx,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.
2. Relevant 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.

3. 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∫00,1cdxdy? 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.

2. Nov 8, 2015

### 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.

3. Nov 8, 2015

### 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?

4. Nov 8, 2015

### andrewkirk

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.

5. Nov 8, 2015

### 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?

6. Nov 8, 2015

### 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.