Conditioning on a joint exponential distribution I'm working on a problem from Hogg (7.3.4) where there is a joint pdf. 
$$f(x,y) = \frac{2}{\theta^2}e^\frac{-(x+y)}{\theta}$$
Valid for $ 0 < x < y < \infty$
As part of the problem, I need to find the conditional distribution of $Y$ given $X=x$. I know that one possible way of doing this is to find the marginal of $X$ and divide the joint by the marginal. However, that leads to a pretty messy result, and I believe that there is an easier way of doing this considering that $f(x,y)$ looks like an exponential distribution and if you fix $X=x$, it still looks like an exponential.
 A: The approach using the ratio of densities is allright, provided one writes the joint density correctly. Considering $a=1/\theta$, you are told that
$$
f_{X,Y}(x,y)=2a^2\mathrm e^{-a(x+y)}\,\mathbf 1_{0\lt x\lt y},
$$
hence
$$
f_X(x)=\int f_{X,Y}(x,y)\mathrm dy=\mathbf 1_{x\gt0}\int_x^\infty2a^2\mathrm e^{-a(x+y)}\,\mathrm dy,
$$
that is,
$$
f_X(x)=2a\mathrm e^{-2ax}\mathbf 1_{x\gt0}.
$$
Thus,
$$
f_{Y\mid X}(y\mid x)=\frac{f_{X,Y}(x,y)}{f_X(x)}=a\mathrm e^{-a(y-x)}\,\mathbf 1_{0\lt x\lt y}.
$$
In words, conditionally on $X=x$, $Y$ is distributed as $x$ plus an exponential random variable with parameter $a$.
Sanity check: The random vector $(X,Y-X)$ is independent exponential with parameters $2a$ and $a$ respectively. Finally, $(X,Y)$ is distributed as an ordered sample $(X_{(1)},X_{(2)})$ from an i.i.d. sample with exponential distribution of parameter $a$.
A: The joint PDF of $X$ and $Y$ is
$$
f_{X,Y}(x,y) = \frac{2}{\theta^2}e^{-\dfrac{x+y}{\theta}}\quad\text{for }0<y<x<\infty,
$$
then the marginal PDF of $X$ is
$$
\begin{align}
f_{X}(x)&=\int_{y=0}^xf(x,y)\ dy\quad\text{the region of $Y$ is }\ 0<y<x\\
&=\int_{y=0}^x \frac{2}{\theta^2}e^{-\dfrac{x+y}{\theta}}\ dy\\
&=\frac{2}{\theta^2}e^{-\dfrac{x}{\theta}}\int_{y=0}^x e^{-\dfrac{y}{\theta}}\ dy\quad\text{let $u=\dfrac{y}{\theta}\;\Rightarrow\;y=\theta u$ then }dy=\theta\ du\\
&=\frac{2}{\theta^2}e^{-\dfrac{x}{\theta}}\left[-\theta e^{-\dfrac{y}{\theta}}\right]_{y=0}^x\\
&=\frac{2}{\theta^2}e^{-\dfrac{x}{\theta}}\cdot\theta\left(1- e^{-\dfrac{x}{\theta}}\right)\\
&=\frac{2}{\theta}e^{-\dfrac{x}{\theta}}\left(1- e^{-\dfrac{x}{\theta}}\right)\quad\text{for }\ 0<x<\infty.
\end{align}
$$
Thus, the conditional PDF of $Y$ given that $X=x$ is
$$
\begin{align}
f_{Y|X=x}(y|x)&=\frac{f_{X,Y}(x,y)}{f_{X}(x)}\\
&=\frac{\frac{2}{\theta^2}e^{-\dfrac{x+y}{\theta}}}{\frac{2}{\theta}e^{-\dfrac{x}{\theta}}\left( 1-e^{-\dfrac{x}{\theta}}\right)}\\
&=\color{blue}{\frac{1}{\theta}e^{-\dfrac{y}{\theta}}\left(1- e^{-\dfrac{x}{\theta}}\right)^{-1}}, \quad\text{for }0<y<x<\infty.
\end{align}
$$
Wolfram Alpha has confirmed that integration of $f_{Y|X=x}(y|x)$ for $\ 0<y<x$ is equals to $1$.
