# one last joint distribution

• Oct 27th 2009, 08:02 PM
billym
one last joint distribution
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?
• Oct 27th 2009, 10:17 PM
matheagle
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
• Oct 28th 2009, 06:11 AM
billym
When it comes to finding the marginal densities, would I then use 1-x and 1-y as the upper limits?
• Oct 28th 2009, 06:34 AM
matheagle
no, review calc 3
IF you had variables on that outer integral HOW can you end up with a constant?
• Oct 28th 2009, 06:56 AM
billym
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?
• Oct 28th 2009, 08:42 AM
matheagle
These look ok

Quote:

Originally Posted by billym
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?