How to get the Joint probability of two dependent continuous random variables

Question

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?

Answer 1

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.

Answer 2

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.