Let $X \sim U(0,\theta_1) \text{ and } Y \sim U(0,\theta_2)$ be independent RV's. We're interested in $\mathbb{P}[X-Y\le z]$.
I've done the convolution (or double integrals), denote $Z=X-Y$
\begin{align}
\mathbb{P}[Z\le z] &= \int_{[x-y<z]} dF_X dF_Y \\
&=\int_0^{\theta_2} \int_0^{z+y} \frac{1}{\theta_1} dx \frac{1}{\theta_2} dy\\
&=\frac{z}{\theta_1}+\frac{\theta_2}{2\theta_1}
\end{align}
is this correct?
for $\theta_1=\theta_2 = 1$ we're getting $\mathbb{P}[X-Y\le z] = z+\frac12 $, which intuitively makes sense.
I'm also curious about the joint pdf. We get the pdf by differentiating the cdf w.r.t. $z$. So $f_Z(z) = \frac{1}{\theta_1}$. Had we chosen the convolution the other way around $$\int_0^{\theta_1} \int_0^{z+x} \frac{1}{\theta_2} dy \frac{1}{\theta_2} dx $$ we would have obtained the joint pdf $f_Z(z) = \frac{1}{\theta_2}$. Isnt' that weird?
Best Answer
Assuming that $X$ and $Y$ are independent, the joint distribution is
$f(x,y) = \frac{1}{\theta_1 \theta_2}$ if $0 \le x \le \theta_1, 0 \le y \le \theta_2$ and $0$ otherwise.
The correct way to do this is to integrate $f(x,y)$ over the region $x-y\le z$. The easiest way to compute this integral is to do it geometrically. Compute the area of the intersection of the region $x-y\le z$ and the rectangle $[0, \theta_1] \times [0, \theta_2]$ and divide it by $\theta_1 \theta_2$.
You can also use the convolution to compute the distribution of $Z = X - Y$ and find $P\{Z \le z\}$ by integrating $\int_0^z f_Z(s) ds$.