What should the bounds of integration be for calculating the marginal distribution function from a joint distribution function?

Question

I calculated $c=\dfrac{1}{8}$. To find $f_y(y)$, I integrated $f(x,y)$ from $-y$ to $y$. What should the bounds be to find $f_x(x)$?

I'm confused because x depends on y. Should the bounds just be $x$ to $\infty$?

Answer 1

A figure indicating the range of $(X, Y)$ often helps to determine the integral bounds and solve problems like this.

In this problem, the range of $(X, Y)$ is given by $$-y \leq x \leq y, \ 0 < y < \infty,$$ which corresponds to the highlighted area (in light blue).

As such, the marginal density of $Y$ is given by $$f_Y(y) = \int_{-y}^y f(x,y) dx = \int_{-y}^y c(y^2 - x^2)e^{-y} dx;$$ and the marginal density of $X$ is given by $$f_X(x) = \begin{cases}\int_x^\infty f(x,y) dy = \int_x^\infty c(y^2 - x^2)e^{-y} dy & \text{for } x \geq 0;\\ \int_{-x}^\infty f(x,y) dy = \int_{-x}^\infty c(y^2 - x^2)e^{-y} dy & \text{for } x < 0.\end{cases}$$ Remark 1: The integral bounds to get $f_X(x)$ depend on the value of $x$ (being positive or negative), which can be illustrated straightforwardly from the figure.

Remark 2: One may notice that the resulting $f_X(x)$ could also be summarized into a "one-liner" formula, given by $$f_X(x) = \int_{\left|x\right|}^\infty f(x,y) dy = \int_{\left|x\right|}^\infty c(y^2 - x^2)e^{-y} dy \quad \text{for } x \in \mathbb{R}.$$ These integral bounds are implied by the fact that $$-y \leq x \leq y \Longleftrightarrow y \geq x \text{ and } y \geq -x \Longleftrightarrow y \geq \left|x\right|.$$ The resulting $f_X(x)$ would then be a function of $\left|x\right|$.