Finding a marginal PDF of a joint probability distribution I understand the idea of how to do it, but I'm currently getting a constant as my marginal PDF, which doesn't make sense.
The overall distribution is as follows: 
$f(x,y) = 5ye^{-xy}$ for $0 < x, 0.2 < y < 0.4$
I'm trying to find the probability that $0 < x < 2$ given that $y = 0.25$, which should be $\frac{f(0<x<2,y=0.25)}{f_y(0.25)}$. As a minor double-check, should the top be a single integral evaluated with $y=0.25$?
The bulk of my question is that I'm finding $f_y$ to be a constant, which doesn't make sense. What am I doing wrong?
$f_y = \int_{0}^{\infty} 5ye^{-xy} dx = 5[-e^{-xy}]_{0}^{\infty} = 5[0-(-1)]=5$
 A: 
What am I doing wrong?

Answer: you are forgetting the support of the density. 
A sure way to stop making this mistake is to write the densities rigorously, for example using indicator functions. In your case the density $f$ is the function defined on the whole set $\mathbb R^2$, by
$$
f(x,y)=5y\mathrm e^{-yx}\mathbf 1_{x\gt0}\mathbf 1_{0.2\lt y\lt0.4}.
$$
Hence, the second marginal of $f$ is the function $f_Y$ defined on the whole set $\mathbb R$, by
$$
f_Y(y)=\int_\mathbb R f(x,y)\mathrm dx=5\,\mathbf 1_{0.2\lt y\lt0.4}\int_0^\infty y\mathrm e^{-yx}\mathrm dx=5\,\mathbf 1_{0.2\lt y\lt0.4}.
$$
Thus, $f_Y$ is a constant on the interval $(0.2,0.4)$, which makes perfect sense. Once again, note that $f_Y(y)$ is well defined, for every $y$ in $\mathbb R$, for example $f_Y(3)=0$.
Edit: By definition, the conditional density of $X$ conditionally on $Y$ is defined for every $x$ and every $y$ such that $f_Y(y)\ne0$ by 
$$f_{X\mid Y}(x\mid y)=f(x,y)/f_Y(y).
$$ 
(If $f_Y(y)=0$, one can set $f_{X\mid Y}(x\mid y)=0$.) Then, for every $y$ such that $f_Y(y)\ne0$,
$$
P[0\lt X\lt 2\mid Y=y]=\int_0^2f_{X\mid Y}(x\mid y)\mathrm dx.
$$
Try this computation and you will get a result for $P[0\lt X\lt 2\mid Y=y]$ between $0$ and $1$, as it should.
