# Find the constant that makes f(x,y) a PDF

1. Oct 5, 2011

### alias

1. The problem statement, all variables and given/known data
Find the value of k that makes this a probability density function.
The question does not specify whether X and Y are independent or dependent. That is actually another part this question.

2. Relevant equations
Let X and Y have a joint density function given by
f(x,y) =e^(-kX), 0<=X<inf, 0<=Y<=1,
k is a constant, and 0 elsewhere

3. The attempt at a solution
I need quite a bit of direction with this question, I have no idea where to start.

2. Oct 5, 2011

### alias

should I be posting this question in the statistics section of pf?

3. Oct 5, 2011

### Staff: Mentor

This is the right place for this question.

How can you tell whether a function is a "probability density function"?

4. Oct 5, 2011

### alias

The integral of a joint PDF = 1:
f(x,y) dxdy = 1
Sorry I don't have a better response.

5. Oct 5, 2011

### Staff: Mentor

That's the direction I was looking for.

For a PDF, the integral should be 1, right?

What does k need to be so that this integral is 1?
$$\int_{y = 0}^1 \int_{x = 0}^{\infty} e^{-kx} dx~dy = 1$$

6. Oct 5, 2011

### alias

I end up with k=1, still not sure if I'm right...

7. Oct 6, 2011

### alias

got it, thanks.