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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 $n<m$. 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?

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

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.

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  • $\begingroup$ Is there an intuitive way to understand how we choose the boundaries for different cases? $\endgroup$ – dxdydz Nov 28 '18 at 5:53
  • $\begingroup$ I added that to the answer. $\endgroup$ – Alejandro Nasif Salum Nov 28 '18 at 6:01

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