Compute marginal probabilities for a given substitution I am trying to compute the marginal probability for this:
$$f_{XY}(x,y) =  \begin{cases}
      2e^{-y} & \text{if  } -y< x < y \ \text{ and }\ 0 < y < \infty ,\\
      0 & \text{otherwise.}   \end{cases} $$
Is my following solution correct?
$$f_X(x) = \int_{y}^{-y}\,2e^{-y}\,\mathrm{d}y = -2e^{-y} + 2e^y$$
and
$$f_Y(y) = \int_{0}^{\infty}\,2e^{-y}\,\mathrm{d}x = \infty$$
 A: Your joint pdf is not correctly normalized:
$$
\int_0^\infty dy\int_{-y}^y dx\ 2 e^{-y}=4
$$
instead of $1$. So, assuming that your jpdf is instead $(1/2)e^{-y}$ over the same $(x,y)$-domain, the marginal pdf of $X$ is obtained by integrating over $y$ (be careful about the integration range, though!)
$$
f_X(x)=\Theta(-x)\int_{-x}^\infty dy (1/2)e^{-y}+\Theta(x)\int_{x}^\infty dy (1/2)e^{-y}=\Theta(-x) e^{x}/2+\Theta(x)e^{-x}/2\ ,
$$ 
where $\Theta(x)$ is the Heaviside step function. The integration limits are obtained by resolving the constraints $-y<x<y$ together with $y>0$ for $y$. Note that this marginal pdf is correctly normalized over the support $x\in (-\infty,\infty)$.
For the other marginal
$$
f_Y(y)=\Theta(y)\int_{-y}^y dx (1/2)e^{-y}=y e^{-y}\Theta(y)\ ,
$$
correctly normalized over $y\in (0,\infty)$.
A: Nope your answer is not correct.
The fact you need the pdf of X shoul prompt you to integrate y.which you have done but the limits of integration are incorrect..limits should be from $ 0 to  \infty $
For your second part the limit should be
$ -y to y $.
