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

[alias]

Homework Statement

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.

Homework 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

The Attempt at a Solution

I need quite a bit of direction with this question, I have no idea where to start.

 

[alias]

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

 

[Mark44]

Homework Statement

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.

Homework 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

The Attempt at a Solution

I need quite a bit of direction with this question, I have no idea where to start.

This is the right place for this question.

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

 

[alias]

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

 

[Mark44]

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

 

[alias]

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

 

[alias]

got it, thanks.