Convolution of discrete uniform distributions For two independent, discrete, uniformly distributed random variables $X$ and $Y$, I wish to obtain the distribution of the sum $Z=X+Y$. I have the densities:
$$f_X(x)=\left\{\begin{matrix} \dfrac{1}{m+1},\;x\in [0,m]\\0,\;\text{otherwise}\end{matrix}\right.$$
$$f_Y(y)=\left\{\begin{matrix}\dfrac{1}{n+1},y\in [0,n]\\0,\;\text{otherwise}\end{matrix}\right.$$
$$f_Z(z)=\sum_{x=-\infty}^\infty f_X(x)f_Y(z-x)$$
I know that $f_Z(z)$ should look like a trapezoidal distribution. However, without knowing this, is it possible to calculate that distribution in a convenient way symbolically? 
I'm finding it tedious to keep track of summation limits and such. For instance, separate cases for $m>n$ and $m<n$. I was wondering if there is a natural or intelligent notation for this, since this problem seems to be just so symmetric w.r.t $\,X$ and $Y$.
To make the question concrete: what is the most natural, symbolic/non-graphical way to solve for $f_Z(z)$? (I don't mind using another convolution formula if there is one, by the way)
Any help would be appreciated. Thanks.
 A: If $X$ and $Y$ are independent
integer-valued random variables uniformly distributed
on $[0,m]$ and $[0,n]$ respectively, then the probability mass function (pmf)
of $Z = X+Y$ has
a trapezoidal shape as you have already noted, and Khashaa has written down for
you.  The answer can be summarized as follows, but whether this is more
compact or appealing is perhaps a matter of taste.
$$P\{Z=k\} = \begin{cases}\displaystyle
\frac{k+1}{(m+1)(n+1)},& k \in [0, \min(m,n)-1],\\
 \\
\displaystyle\frac{1}{\max(m,n)+1},& k \in [\min(m,n), \max(m,n)],\\
 \\
\displaystyle\frac{(m+n)-(k-1)}{(m+1)(n+1)}, & k \in [\max(m,n)+1, m+n].\end{cases}$$
To my mind, the easiest way of solving this problem, and indeed a way that
works for dependent and non-uniformly distributed random variables as well,
is to write down the joint pmf of $(X,Y)$ as a rectangular array or matrix
of $m$ columns (numbered $0, 1, \ldots , m$ from left to right) and $n$ rows
(numbered $n, n-1, \ldots, 0$ from top to bottom.  Then, $P\{X+Y=k\}$
is the sum of the entries on the $k$-th diagonal of this array.  For the
case of constant entries, we get the nice trapezoidal shape that the 
OP has noticed.
