# sum of two discrete uniform random variables

Let $$X$$ be an integer chosen uniformly at random from the set $$\{1,2,...,n\}$$ and $$Y$$ be an independent integer chosen uniformly at random from the set $$\{1,2,...,m\}$$. Find the probability mass function of $$X+Y$$.

My attempt: Without loss of generality suppose $$n. Possible values of $$X+Y: 2,3,...,m+n.$$ For $$2\leq k\leq n$$ using independence of $$X$$ and $$Y$$ we have $$P(X+Y=k)=\sum_{i=2}^{n}P(X=k)P(Y=n-k)=\frac{n-1}{mn}.$$

I'm stuck here and don't know how to proceed. What other cases are needed to be considered?

• Perform the simple combinatorics to find the number of ways (and hence relative probability) of obtaining each output, $1, \ldots, m+n$. – David G. Stork Nov 28 '18 at 5:08

You have confused a little your variables and indexes. The sum for the case $$2\le k\le n$$ should be $$\sum_{i=1}^{k-1}P(X=i)P(Y=k-i).$$
Also consider the cases $$n and $$m
To see this, think that $$i$$, the values that $$X$$ take, can be from $$1$$ to $$n$$, as long as the respective values for $$Y$$, that is $$k-i$$ are between $$1$$ and $$m$$. This implies: $$1\le k-i \le m \iff -m \le i-k \le -1 \iff k-m \le i \le k-1.$$
So we have $$i\ge 1$$ and also $$i\ge k-m$$. This means that if $$k\le m$$, $$i$$ can take the value $$1$$, but if $$k>m$$, then the minimum value for $$i$$ has to be $$k-m$$ (actually, this is also $$1$$ for $$k=m+1$$, but this does not contradicts what we said).
In the same way, $$i$$ has to be smaller or equal than $$n$$ and $$k-1$$, so if $$k\le n$$, the maximum value for $$i$$ is $$k-1$$, but when $$k>n$$, the maximum value for $$i$$ is $$n$$. Combining these results, we get the three cases mentioned above.