# Marginal density function for dependent support

Given the joint density function of $(X,Y)$ is $$f(x,y)=\begin{cases} \frac{1}{8}(y^2-x^2)e^{-y} \quad \text{ if } -y<x<y \text{ , } 0<y<\infty \\ 0 \quad \quad \quad \quad \quad \quad \text{ otherwise } \end{cases}$$

What is the marginal density function of $X$?

I can compute $f_X(x)=\frac{1}{4}(x+1)e^{-x}$ by integrating $dy$ from $0$ to $\infty$, but the answer states we have to consider for both $x>0$ and $x<0$ by symmetry. As I am new to joint distribution, I don't understand it very well. Thank you in advance!

I would like to advise you first to plot the region where your Joint PDF is non zero, this will ease the way for you to compute the integrals when finding the marginal PDFs.

The region where your joint PDF can be seen in the yellow color in the following image:

Thus in order to compute the marginal PDF of $X$, you must integrate over $Y$, so we have to find how $Y$ is varying wrt $X$ in the yellow region.

• When $x \in ]-\infty , 0[$, we have $-x < y< + \infty$.
• When $x \in ]0, +\infty [$, we have $x < y< + \infty$.

Thus $$f_X(x)=\begin{cases} \int_{-x}^{\infty } \frac{1}{8}(y^2-x^2)e^{-y} dy \quad \text{ if } -\infty <x<0 \\ \\ \int_{x}^{\infty } \frac{1}{8}(y^2-x^2)e^{-y} dy \quad \text{ if } 0<x< \infty \\ \\ 0 \qquad \qquad \qquad \qquad \qquad \qquad \text{ if } x=0 \end{cases}$$

Hope this helps you.

Well, if you consider the region given $-y<x<y,0<y<\infty$, you want the inside part of the two inequalities, $y>x$ AND $-y<x$. There is not restriction that says $x$ has to be positive, so you have to consider the right side of y-axis and the left side of y-axis.