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Suppose $1\le x \le k, 1\le y \le k$, and we need to find a explicit formula for the the number of integers solution for the equation such that $$x+y=z$$ where $z \in\{2,...,2k\}$


I'm doing this because I am trying to find the probability mass function of the distribution $$Z=X+Y$$ where $X$ and $Y$ are two discrete uniform distribution random variables on $\{1,...,k\}$.

And while I was trying to find the pmf of $Z$, it occured to me that $$f_Z(z) = \dfrac{\text{ The number of solutions for the linear Diophantine equation above}}{k^2}$$

Am I overthinking this? Or am I approaching the problem in a wrong way?

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By "$z =\{2,...,2k\}$", do you mean $z \in\{2,...,2k\}$? –  joriki Nov 18 '11 at 20:17
    
@joriki Yes, it was a typo.Thanks for pointing it out –  geraldgreen Nov 18 '11 at 20:20
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up vote 2 down vote accepted

Note that $1 \leq x \leq z$.

Case 1: $2 \leq z \leq k+1$. Then any $1 \leq x \leq z-1 $ together with $ y= z-x$ works. Note that in this case $1 \leq y \leq k+1-1$, is in the right range.. In this case there are $z-1$ solutions.

Case 2: $k+1 < z$. Since $k+1< z =x+y \leq x+k$ we need $x >1$. Actually $z \leq x+k$ gives us the right lower bound:

$$x \geq z-k \,.$$

Now for each $z-k \leq x \leq k$ we can set $y=z-x$. $y $ is in the right range, since $z-k \leq x$ means $y \leq k$, while $x+1 \leq k+1 < z$ means $y \geq 1$.

In this case there are $k-(z-k)+1=2k-z+1$ solutions.

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You are right. I might have been overthinking this problem. –  geraldgreen Nov 19 '11 at 23:28
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