$X$ and $Y$ are distributed according to the joint PDF
$$ f_{X,Y} (x,y) = \{
\begin{array}{lr}
\frac{3}{7}x & : 1 \leq x \leq 2, 0 \leq y \leq x\\
0 & : otherwise
\end{array}
$$
The random variable $Z$ is defined by $Z = Y – X$. Determine the PDF $f_{Z}(z)$.
The solution to this problem first determines the CDF then differentiates to determine the PDF. We are given that
$$
f_{X,Y} (x,y) = \{
\begin{array}{lr}\\
\int_{x = -z}^{x=0} \int_{y=0}^{y=x+z} \frac{3}{7}x dy dx & : -2 \leq z \leq -1\\
\int_{x = -z}^{x=1} \int_{y=0}^{y=x+z} \frac{3}{7}x dy dx & : -1 < z \leq 0
\end{array}
$$
I have a few questions.
-
Why is my integration bound not y=0 to y=1. I suspect this has to do with the values of z and also some bound of integration concept. I should know this, but could someone provide some intuition as to why this is?
-
I understand intuitively that $Z$ is dependent on values $X,Y$ which is the reason we want to consider the joint PDF $f_{X,Y}(x,y)$. I can see why we want to use the CDF then differentiate to find PDF approach. However, I'm so lost as to how we get our bounds for our integrals. Could I bother someone to step through the thought process of how I should attack this problem? Thanks.
Geez, this problem has seriously been bothering me for days!
Best Answer
You will need to draw at least one picture. First we identify the part $L$ of the plane where the joint density function of $X$ and $Y$ "lives."
Draw the vertical lines $x=1$ and $x=2$. Draw the line $y=x$. The joint density lives on the part $L$ of the first quadrant that is between $x=1$ and $x=2$, and below $y=x$. This region $L$ is a trapezoid with corners $(1,0)$, $(2,0)$, $(2,2)$, and $(1,1)$.
Let $Z=Y-X$. We want the cdf $F_Z(z)$ of $Z$, that is, $\Pr(Z\le z)$. So we want $\Pr(Y\le X+z)$. We want the probability that $(X,Y)$ lands in the part $D$ of $L$ that lies below the line $y=x+w$.
For fixed $z$, we will draw the line $y=x+z$. The only interesting values of $z$ are between $-2$ and $0$. Note that the line $y=x+z$ is always parallel to the line $y=x$.
The geometry is different for small negative $z$ than for large negative $z$. For if $z$ is small negative, the region $D$ below $y=x+z$ but in $L$ is a trapezoid. At the value of $z$ for which the line $y=x+z$ passes through $(1,0)$, the part $D$ of $L$ below $y=x+z$ turns into a triangle. This happens when $z=-1$.
Now we can calculate. For $-1\lt z\le 0$, we integrate first with respect to $y$, $y=0$ to $y=x+z$, and then with respect to $x$, from $1$ to $2$. So we have $$F_Z(z)=\int_{x=1}^2\left(\int_{y=0}^{x+z}\frac{3}{7}x\,dy \right)\,dx$$ when $-1\lt z\le 0$.
For $-2\le z\le -1$, we are integrating over a triangle. Note that the triangle has corners $(-z,0)$, $(2,0)$, and $(2,2+z)$.
So as before, in the inner integration, $y$ will go from $0$ to $x+z$. Then for the outer integration, $x$ goes from $-z$ to $2$. So $$F_Z(z)=\int_{x=-z}^2\left(\int_{y=0}^{x+z}\frac{3}{7}x\,dy \right)\,dx$$ when $-2\le z\le -1$.
The rest is mechanical. Compute the integrals, and differentiate to find $f_Z(z)$.
Remark: There are other ways to solve the problem. One can find the density function of $Z$ directly, by using a variant of the usual convolution procedure. Doing this still requires careful attention to the geometry, to get the limits right.