find $c$ in $f(x; y) = c(x^2 − y^2)e^{−x}$; $x > 0$; $−x < y < x$

Question

find $c$ in $f(x; y) = c(x^2 − y^2)e^{−x}$; $x > 0$; $−x < y < x$

I know I have to take the integral such that :

$$ \int \int f(x; y) = c(x^2 − y^2)e^{−x} dx dx = 1 $$

but I have troubled to define the boundaries.

Since $x > 0$ and $−x < y < x$, can i say that $0 < y < x$?

I tried solving the following two cases :

1.

$$ \int_0^\infty \int_{-x}^x c(x^2 − y^2)e^{−x} dx dy = c \int_0^\infty \int_{-x}^x exp{(−x \times 2log(x))} − exp{(−x \times 2log(y))} dx dy $$

2. $$ \int_0^\infty \int_{0}^x c(x^2 − y^2)e^{−x} dx dy $$

but i get $0$ in the integration for both cases.

Are my boundaries correct?

Is this the right approach to find c?

How do the integral boundaries change when I want to find marginal densities?

marginal

$$ f(y) = \int_{0}^\infty c(x^2 − y^2)e^{−x} dx $$

$$ f(x) = \int_{-x}^x c(x^2 − y^2)e^{−x} dy $$

Answer 1

Let's find the marginal densities first.

The marginal of $X$ is

\begin{align} f_X(x)&=\int f(x,y)\,dy \\&=\int_{-x}^x c(x^2-y^2)e^{-x}\,dy\,\mathbf1_{x>0} \\&=ce^{-x}\left[x^2\int_{-x}^x \,dy-\int_{-x}^xy^2\,dy \right]\mathbf1_{x>0} \\&=ce^{-x}\left[2x^3-\frac{2x^3}{3}\right]\mathbf1_{x>0} \\&=\frac{4c}{3}x^3e^{-x}\,\mathbf1_{x>0} \end{align}

Now,

\begin{align} \int_0^\infty f_X(x)\,dx&=\frac{4c}{3}\Gamma(4)=8c \end{align}

So it must be that $$c=\frac{1}{8}$$

But note that $$-x<y<x\implies x>\max(-y,y)=|y|$$

, so that the marginal of $Y$ for all $y\in\mathbb R$ must be given by

$$f_Y(y)=\int_{|y|}^\infty f(x,y)\,dx$$