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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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