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Suppose $X,Y$ are uniformly distributed independent random variable on $\{1,...,N\}$ , compute the density of $X+Y$.

So the density of $X$ or $Y$ is $f_X (x) = \frac{1}{N}$ (so if we sum the terms, we get $1$). I believe that I have to use the convolution but not sure how to do this in the discrete case, I have:

$f_{X+Y} (z) = \sum_x f_X (x) f_Y(z-x) = \sum_x \frac{1}{N} f_Y(z-x)$, but now I don't know what to do. Any help is greatly appreciated, thank you.

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    Hint: Instead of using fancy formulas and confusing yourself, begin by noting that $X+Y$ can take on values $2, 3, \ldots, 2N$, and then try computing $P\{X+Y=2\}$. What possible values of $X$ and $Y$ will result in $X+Y=2$? What is the probability of this event? Now try $P\{X+Y=3\}$. There are two _mutually exclusive_ cases to be considered here $\{X=1,Y=2\}$ and the other you should discover for yourself. Continue in this vein till inspiration strikes and you shout "Hey Ma! I see a pattern". Then start at the other end and figure out $P\{X+Y=2N\}$ to see a different pattern emerge.2012-12-12
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    Thanks this works, but unfortunately I'm trying to understand the fancy formula because I have a final exam on thursday. But thanks none the less2012-12-12

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I suspect that the reason you don't know what to do about the last expression is that you haven't thought carefully about the implied range of the sum. The summand in the convolution only exists for $1\le x\le N$ and $1\le z-x\le N$. On the other hand, if it does exist, both factors are $1/N$. Thus you just have to count the values of $x$ that fulfill those conditions and divide by $N^2$. That's basically also what Dilip suggested to do in the comment.