Question on determining the variance of joint probability distribution $(X,Y)$ has joint distribution $f_{X,Y}(x,y)=\frac{1}{2x}e^{-x}$, for $x>0, -x\le y\le x$, so determine the $Var[X], Var[Y]$ and correlation coefficient
But I don't know how to get one of marginal functions. $h(y)=\int_{0}^{\infty}\frac{1}{2x}e^{-x}dx$, so it's hard to continue to find $\sigma$
 A: Michael Hardy answered the question you seemingly asked: "I don't know how to get one of marginal functions" but as you discovered here, the marginal density of $Y$ seems to be a special function.  On the other hand, computing the moments of $Y$ (which is what really you want to do) does not require that you first find the marginal density of $Y$.  We have
$$
\begin{align*}
E[Y^n] &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}
y^n f(x,y)\ \mathrm dy\ \mathrm dx\
&= \int_{x = 0}^{\infty}\int_{y = -x}^{y = x} y^n \ \mathrm dy \frac{1}{2x} \exp(-x)\mathrm dx\
&= \int_{x = 0}^{\infty} \left . \frac{y^{n+1}}{n+1}\right\vert_{-x}^x
\frac{1}{2x} \exp(-x)\mathrm dx\
&= \begin{cases}
\int_{x = 0}^{\infty}  \frac{1}{n+1}
x^n \exp(-x)\mathrm dx = \frac{\Gamma(n+1)}{n+1}, & n ~\text{even},\
\quad & \
\quad & \
0, & n ~\text{odd},
\end{cases}\
\
\end{align*}
$$
from which you can get the variance of $Y$ and the covariance of $X$ and $Y$.
A: $h(y)$ is not $\int_0^\infty \frac{1}{2x} e^{-x}\;dx$.  It is
$$
\int_{|y|}^\infty \frac{1}{2x} e^{-x}\;dx.
$$
That is because of the inequality $-x \le y \le x$, which is the same as $|y| \le x$.  In other words, the density is positive only for those values of $x$ that are $\ge |y|$.
The fact that no "$y$" appears in the expression you had, $\int_0^\infty \frac{1}{2x} e^{-x}\;dx$, should arouse suspicions.  It would mean that the marginal density does not depend on $y$, i.e. is constant.  Probability density functions on the whole line are never constant, since the integral of a constant over the whole line is either $0$ or $\infty$ or $-\infty$, never $1$.
