one last joint distribution

billym · #1 $Old$ October 27th, 2009, 10:02 PM

joint density function of X and Y:

$f(x,y)=\left\{\begin{array}{lr}k(x+y)&0\le x+y\le 1; x,y\ge 0\\0&otherwise\end{array}\right.$

If I was looking for k, would $\int_0^{1-y} \int_0^{1-x} k(x+y) dy dx=1$ be the right way to start?

matheagle · #2 $Old$ October 28th, 2009, 12:17 AM

NO
The outer integral cannot contain variables...
see my change below

$\int_0^{1} \int_0^{1-x} k(x+y) dy dx=1$ is the right way to start

billym · #3 $Old$ October 28th, 2009, 08:11 AM

When it comes to finding the marginal densities, would I then use 1-x and 1-y as the upper limits?

matheagle · #4 $Old$ October 28th, 2009, 08:34 AM

no, review calc 3
IF you had variables on that outer integral HOW can you end up with a constant?

billym · #5 $Old$ October 28th, 2009, 08:56 AM

I understand that about finding the constant. What I mean is, once i have found the constant (3 I think), and want to find the marginal densities, would I then use:

$f_Y(y)=\int_0^{1-y} f(x,y) dx$

and

$f_X(x)=\int_0^{1-x} f(x,y) dy$

or do I have to stick with the initial values?

matheagle · #6 $Old$ October 28th, 2009, 10:42 AM

These look ok

Quote:

Originally Posted by billym $View Post$

I understand that about finding the constant. What I mean is, once i have found the constant (3 I think), and want to find the marginal densities, would I then use:

$f_Y(y)=\int_0^{1-y} f(x,y) dx$

and

$f_X(x)=\int_0^{1-x} f(x,y) dy$

or do I have to stick with the initial values?