# Given the joint probability density function of $X$ and $Y$, find the pdf of each of them

The joint probability density function of X and Y is given by:

$$f(x,y) = ce^{-2x-3y},\qquad 0 < x < \infty,\ 0 < y < \infty$$

(a) Compute the probability density function of $X$, $f_X(x)$.
(b) Compute the probability density function of $Y$, $f_y(y)$.
(c) are random variables $X$ and $Y$ independent?

It's part of an assignment, one of the questions was to find the join probability density, so I did a double integral of the function and wound up with $C = 6$. The problem I'm having that I can't seem to understand is the parts of $A$ and $B$. I'm not entirely sure as to what I'm suppose to do.. I'm assuming I'm suppose to integrate once with respect to $x$ and the other with respect to $y$ but I'm not entirely sure how my answer is supposed to look... am I supposed to get an answer like "$6e^{-3y} {-e^{-2x}/2}$" or is that completely wrong ...

• What part of this question, which looks a lot like a homework assignment, is causing you difficulty? – Dilip Sarwate Sep 27 '13 at 22:15
• This is phrased just like a question on an assignment or a test. It makes it look as if you're just passing a question on to us that was written by someone other than you. You should tell us your own thoughts on this. – Michael Hardy Sep 27 '13 at 22:16
• Sorry my apologies, Yes it's part of an assignment, one of the questions was to find the join probability density, so I did a double integral of the function and wound up with C = 6. The problem I'm having that i can't seem to understand is the parts of A and B. I'm not entirely sure as to what i'm suppose to do.. im assuming i'm suppose to integrate once with respect to x and the other with respect to y but im not entirely sure how my answer is supposed to look.. am i supposed to get an answer like " 6e^(-3y)(-e^(-2x)/2) " or is that completely wrong ... – drocktapiff Sep 27 '13 at 22:24

To get the probability density function $f_X(x)$, we take the joint density and "integrate out" $y$. Specifically, $$f_X(x)=\int_{-\infty}^\infty f_{X,Y}(x,y)\,dy.$$ For $x\le 0$, we get $0$. For $0\lt x\lt\infty$, we need to find $$\int_0^\infty 6e^{-2x}e^{-3y}\,dy.$$ You should get $2e^{-2x}$.
To get the density function of $Y$, integrate out $x$.
If $f_{X,Y}(x,y)=f_X(x)f_Y(y)$, we have independence. In this case, we do.