Finding the marginal PDF with dependent variable bounds

Question

Consider the following joint PDF: $$ f_{X,Y}(x,y)= \begin{cases} \frac{1}{8}(x-y) & -x<y<x, \; 0<x<2 \\ 0 & \text{else} \end{cases} $$ We know that $f_{Y}(y)=\int_{-\infty}^{+\infty} f_{X,Y}(x,y)\;dx$. but since the variables are dependent ($-x<y<x$), merely integrating over the interval $(0,2)$ would not suffice. What is the approach to find the marginal PDF in such cases?

Answer 1

The inequality $-x < y < x$ is the same as $x > |y|$. Also $-x < y < x$ along with bounds $0 < x < 2$ suggests $-2 < y < 2$ as the bounds for the support of $Y$. So pdf of $Y$ is \begin{align*} f_Y(y) = \int_{-\infty}^{+\infty} f_{X,Y}(x,y)\;dx &= \int_{-\infty}^{|y|} f_{X,Y}(x,y)\;dx + \int_{|y|}^{2} f_{X,Y}(x,y)\;dx + \int_{2}^{+\infty} f_{X,Y}(x,y)\;dx \\ &= 0 + \int_{|y|}^{2} \frac{1}{8}(x - y)\;dx + 0, \quad -2 < y < 2 \end{align*} and $0$ otherwise. You can easily work out the integral.

Answer 2

If you sketch the region bounded by the inequalities $-x<y<x, 0<x<2$, it's the interior of the triangle with vertices $(0,0), (2,-2), (2,2)$.

Algebraically, by changing the inequality $-x<y$ to $-y < x$, the lower bounds of $x$ given $y$ has to satisfy all $3$ of

$$-y<x,\quad y<x,\quad0<x$$

i.e. $|y|<x$. Combining with the upper bound of $x$,

$$|y|<x<2$$

The range of $y$ that has some corresponding $x$ is

$$|y|<2\\ -2<y<2$$