# Finding a joint probability density function given marginal probability density functions

X and Y are independent random variables with the following PDFs:

$f_{x}(x) = \begin{cases} (1/3)e^{-x/3}, & x \geq\text{ 0} \\ 0, & \text{otherwise} \end{cases}$

$f_{y}(y) = \begin{cases} (1/2)e^{-y/2}, & y \geq\text{ 0} \\ 0, & \text{otherwise} \end{cases}$

a) What is $P[X > Y]$?
b) What is $E[XY]$?
c) What is $Cov[XY]$?

Attempt at Solution:
I know the formulas by which to find both the expected value and covariance for $XY$, but the problem lies in finding the joint PDF of X and Y. I understand that to find the marginal PDFs, which have been given, the integral from negative infinity to positive infinity of the joint PDF with respect to the opposite variable must be taken. However, this is the opposite case. My initial instinct was to take the derivative of each marginal PDF with respect to the opposite variable, but in order for that to work, both derivatives would have to be equivalent, which is not the case.

Other than this issue, I am not sure how to solve for $P[X > Y]$.

• By independence, the joint pdf is the product of the individual pdf. You do not need the joint pdf to answer the last two questions, but you do need it for $\Pr(X\gt Y)$. Aug 3, 2014 at 18:54
• $\int_{y=0}^\infty\left(\int_{x=y}^\infty (\text{joint density)}\,dx\right)\,dy$. Or else integrate first with respect to $y$, $y=0$ to $x$. The first is somewhat more pleasant. Aug 3, 2014 at 18:59
• @AndréNicolas what if they are not independent? Can you still find the joint pdf? Oct 9, 2018 at 12:35

Hints:

$P\left\{ X>Y\right\} =\int f_{Y}\left(y\right)P\left\{ X>Y\mid Y=y\right\} dy=\int f_{Y}\left(y\right)P\left\{ X>y\right\} dy$

Alternative:

$\int_{0}^{\infty}\int_{y}^{\infty}f_{\left(X,Y\right)}\left(x,y\right)dxdy=\int_{0}^{\infty}\int_{y}^{\infty}f_{X}\left(x\right)f_{Y}\left(y\right)dxdy=\int_{0}^{\infty}f_{Y}\left(y\right)\int_{y}^{\infty}f_{X}\left(x\right)dxdy$

Note that integrand $\int_{y}^{\infty}f_{X}\left(x\right)dx$ can be recognized as: $P\left\{ X>y\right\}$

Since $X$ and $Y$ are independent the joint PDF of $(X,Y)$ is the product of the PDFs of $X$ and $Y$.

• Since you know the PDF of $X$ you must be able to find $P(X>y)$ right? Also have a look here: math.stackexchange.com/q/882579/75923 Aug 3, 2014 at 19:05
• what if they are not independent? Can you still find the joint pdf? Oct 9, 2018 at 12:35
• Not purely on base of knowledge of the marginals. You need more information for that. Oct 9, 2018 at 12:53
• It is even possible that no joint pdf exists. This for instance if $X=Y$. Oct 9, 2018 at 13:00
• Thank you for answering! What kind of information would I need? Oct 9, 2018 at 13:17