How to get the Joint probability of two dependent continuous random variables I would like to calculate the following probability of :
$$
\begin{equation}
\begin{array}{l}
\displaystyle P_u = \Pr\{x<a_1 , x \geq a_2 \,\,y + a_3\} 
\end{array}
\end{equation}
$$
where $x$ and $y$ are independent exponential random variables  with a unit mean and $$a_1 , a_2, a_3 $$ are positive constants.
I tried to use 
$$
\begin{equation}
\begin{array}{l}
\displaystyle P_u = \Pr\{x<a ,\,\, x -a_2 \,\,y \geq  a_3\}\\
\displaystyle  \quad \,\,=  \Pr\{x<a , z \geq  a_3\}\\
\end{array}
\end{equation}
$$
where $$z = x-a_2 \, y$$ 
and 
$$
\begin{equation}
\begin{array}{l}
\displaystyle P_u = \int_{x=0}^{a_1}  \int_{z=a_3}^{\infty} f_{xz}(x,z)\, dx\, dz
\end{array}
\end{equation}
$$
I got PDFs of x and z, but since they are not independent , I can not assume :
$$
\begin{equation}
\begin{array}{l}
\displaystyle f_{xz}(x,z)=f_{x}(x)\, f_{z}(z)
\end{array}
\end{equation}
$$
How to complete from here or another alternative at all?
 A: The probability will take value $0$ is $\mathsf{P}\left(a_{2}y+a_{3}\geq a_{1}\right)=1$.
So you must take care of suitable conditions on the $a_{i}$.
Let $\left[x<a_{1},x\geq a_{2}y+a_{3}\right]$ denote the function
$\mathbb{R}^{2}\to\mathbb{R}$ that takes value $1$ if $x<a_{1}$
and $x\geq a_{2}y+a_{3}$ are both satisfied and takes value $0$ otherwise. 
Then:
$\begin{aligned}\mathsf{P}\left(x<a_{1},x\geq a_{2}y+a_{3}\right) & =\mathsf{E}\left[x<a_{1},x\geq a_{2}y+a_{3}\right]\\
 & =\int\int\left[x<a_{1},x\geq a_{2}y+a_{3}\right]f_{X}\left(x\right)f_{Y}\left(y\right)dxdy\\
 & =\int f_{Y}\left(y\right)\left[a_{2}y+a_{3}<a_{1}\right]\int_{a_{2}y+a_{3}}^{a_{1}}f_{X}\left(x\right)dxdy
\end{aligned}
$
Where $[a_2y+a_3<a_1]$ denotes the function $\mathbb R\to\mathbb R$ that takes value $1$ if $a_2y+a_3<a_1$ is satisfied and takes value $0$ otherwise.
This can further be worked out discerning cases $a_{2}<0$, $a_{2}=0$
and $a_{2}>0$.
In e.g. the last case (i.e. $a_2>0$) we get $\int_{-\infty}^{\frac{a_{1}-a_{3}}{a_{2}}}f_{Y}\left(y\right)\int_{a_{2}y+a_{3}}^{a_{1}}f_{X}\left(x\right)dxdy$.
Note that it will take value $0$ if $a_1\leq a_3$ (then $\mathsf{P}\left(a_{2}y+a_{3}\geq a_{1}\right)=1$)  and simplifies to $\int_{0}^{\frac{a_{1}-a_{3}}{a_{2}}}f_{Y}\left(y\right)\int_{a_{2}y+a_{3}}^{a_{1}}f_{X}\left(x\right)dxdy$ otherwise.
A: If $a_1 \leq 0$ then $P_u = 0$ so, we assume that $a_1 > 0$. For $a_2 = 0$, $P_u$ is straightforward: $ Pu = 0$ if $a_3 < a_1$ else $P_u = \int_{a_1}^{a_3}f_X(x)dx$.
For $a_2 \neq 0$ we can use change of variables
$$\begin{align*}
U &= X \\
V &= X - a_2 Y
\end{align*}$$
The inverse tranformation is
$$\begin{align*}
X &= U \\
Y &= \frac{U-V}{a_2}
\end{align*}$$
The joint pdf of U and V will be
$$f_{U,V}(u,v) = \left. \frac{f_{X,Y}(x,y)}{\left | \frac{\partial(U,V) }{\partial(X,V)} \right | } \right|{}_{x = u \\ y= \frac{u-v}{a_2} }  =
\frac{1}{|a_2|}f_X(u)f_Y \left (\frac{u-v}{a_2} \right)$$
Then
$$ P_u = Pr\{U < a_1, V \geq a_3 \} =
\frac{1}{|a_2|} \iint_{D} f_X(u) f_Y \left (\frac{u-v}{a_2} \right)dvdu $$
Where $D$ is the domain
$$D = \left \{(x,y) \in \Re: 0<u<a_1, v>a_3, \frac{u-v}{a_2}>0 \right \}$$
Now, as @drhab mentions as well in his solution, we have to take two different cases into account:
1) If $a_2>0$ then $a_3 < a_1$ (otherwise the domain $D$ will be empty) and $P_u$ simplifies to
$$ P_u = \frac{1}{a_2} \int_{\max\{0,a_3\}}^{a_1}\int_{a_3}^{u}f_X(u) f_Y \left (\frac{u-v}{a_2} \right)dvdu $$
2) If $a_2<0$ then $P_u$ simplifies to
$$ P_u = -\frac{1}{a_2} \int_{0}^{a_1}\int_{\max\{0,a_3\}}^{\infty}f_X(u) f_Y \left (\frac{u-v}{a_2} \right)dvdu $$
Performing the relevant integration, will give the required result.
