A random variable $ X $ is called (discretely) uniformly distributed on $ \{1, \ldots, n\}, n \in \mathbb{N} $ if it holds for its probability mass function (also called count density) $ \left\{p_{j}\right\} $ that $ p_{j}=P(X=j)=\frac{1}{n}, \forall j=1, \ldots, n $.
Let the random variables $ X $ and $ Y $ be independent and uniformly distributed on $ \{1, \ldots, n\} $ and let $ Z=X+Y $.
Determine the count density of $ Z $, i.e. the individual probabilities $ P(Z=j) $ for all $ j $ from the image of $ Z $.
Given solution:
$Z$ is a random variable, as a linear combination of random variables X and Y.
The image of Z is equal to $\{2,3, ..., 2n\}$.
If $i \in \{2,...,n\}$, then $\forall j \in \{1,...,i-1\}, j \in \{1,...,n\} \text{ and } i-j \in \{1,...,n\}$, thus:
$P(X = j, Y = i-j) = \frac{1}{n^2}$ and therefore $P(Z=i) \geq \sum_{j=1}^{i-1} \frac{1}{n^2} = \frac{i-1}{n^2}$ (#).
Suppose there exists a $j > i-1$ such that $P(X=j, Y=i-j) > 0$.
$\Rightarrow P(Y=i-j) > 0$ is a contradiction, since $i-j \leq 0$.
$\Rightarrow $ equality in (#).
Now let $i \in \{n+1, ..., 2n\}$ i.e. $i = n+k, k > 0$. Then $\forall j \in \{k,...,n\}$ the following statement applies:
$j \in \{1,...,n\} \text{ and } i-j \in \{1,...,n\}$, so $P(X = j, Y = i-j) = \frac{1}{n^2}$ and therefore:
$P(Z=i) \geq \sum_{j=k}^n \frac{1}{n^2} = \frac{n-k+1}{n^2} = \frac{n-(i-n)+1}{n^2} = \frac{2n-i+1}{n^2}$.
Suppose there exists $j \in {1,...,k-1}$ such that $P(X = j , Y = i-j) > 0$, so in particular $P(Y=i-j) > 0$ is a contradiction, since $i-j > n$.
Unfortunately, my instructor only wrote this solution down quickly on the board at the very end of the exercise, so I don't know if everything was written correctly. I also don't really understand what was actually done in this proof. It's not really clear to me what the aim of this proof is and how they tried to prove it. Can someone try to explain to me the steps in this proof, what was considered?
It would be enough for me if someone could describe more clearly what is done and shown in the individual steps in order to reach the goal. Because when I try to read through it, I just don't understand why these steps are necessary to show the goal and how to get to these specific steps, so I just don't see a clear pattern behind it.
Here is a more systematic, bruteforce way that you may like better.
Fix $m\in \{2,\ldots2n \}$ and note that
$$\begin{aligned} P(X+Y=m) &= \sum_{k=1}^n P(X=k) P(Y=m-k) \\&= \frac 1{n^2} \sum_{k=1}^n 1_{m-k\in \{1,\ldots ,n\}} \\&= \frac 1{n^2} |I|, \end{aligned}$$ where $I$ denotes the integer interval $[\max(1,m-n),\min(n,m-1)]\cap \mathbb N$, hence $|I|=\min(n,m-1) - \max(1,m-n) +1 $ and $$\begin{aligned} P(X+Y=m) &= \frac 1{n^2} \big(\min(n,m-1) - \max(1,m-n) +1\big) \\&= \frac 1{n^2} \big[\frac 12(n+m-1-|n-m+1|) - \frac 12(1+m-n-|m-n-1|) +1\big] \\&= \frac 1{n^2} (n-|n-m+1|). \end{aligned}$$
Thus the pmf of $X+Y$ is $\frac 1{n^2} (n-|n-m+1|)$ for each $m\in \{2,\ldots2n \}$.