# Joint density function

1. May 16, 2010

### squenshl

I have a region R = {(x,y): |x| + |y| <= 1}
I have sketched the graph on the x-y plane and shaded in the region R (I got a diamond).
How do I find the area of R and hence find the joint density function f(x,y) and how do I specify the region for which f(x,y) is non-zero.

2. May 16, 2010

### EnumaElish

A diamond is 4 triangles: Area(R) = 4 times the area of one triangle.

Are x and y random variables?

3. May 16, 2010

### squenshl

They are just random points chosen uniformly on the region

4. May 17, 2010

### squenshl

So the area is 0.5 x 4 = 2, how does that relate to finding the joint density function.
So f(x,y) = 2 for -1 < x < 1 and -1 < y < 1, 0 otherwise

Last edited: May 17, 2010
5. May 17, 2010

### squenshl

No. Wouldn't it be f(x,y) = x + y for 0 < x < 1 and 0 < y < 1 and 0 otherwise
because when I do the double integral, the integral equals 1 which it should.

6. May 17, 2010

### LCKurtz

You said your region is R = {(x,y): |x| + |y| <= 1}. That is not the same as you are describing as -1 < x < 1 and -1 < y < 1. And a uniform distribution gives a constant density, not a variable like x + y. You need to have f(x,y) be a constant value on R such that the integral of f(x,y) over R is 1.

7. May 17, 2010

### squenshl

f(x,y) = 1/2 for 0 < x < 1 - |x|, but I don't know the limits of y.

8. May 17, 2010

### LCKurtz

No. Can't x be negative?

Didn't you say you had drawn a picture of your region R and it was diamond shaped?

Use the description of R itself to say where f = 1/2.