I have two stochastic variable $X\sim B(n,p)$ and $Y\sim B(m,p)$, I need to prove that $Z=X+Y \sim B(n+m,p)$.
Why is $$\sum_{k=0}^{u} \binom{n}{k} \binom{m}{u-k} = \binom{n+m}{u}$$
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Vandermonde's Identity. Look it up. – Rushabh Mehta Feb 24 '21 at 21:01
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This is exactly the Vandermonde convolution/identity. https://en.m.wikipedia.org/wiki/Vandermonde%27s_identity – Sean Roberson Feb 24 '21 at 21:02
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If you are selecting $(n+m)$ items, $u$ at a time, then, for each selection, you will end up with $k$ of the items, coming from the first $n$, where $k$ must be in $\{0,1,\cdots, u\}$. Of course, I am assuming that $u \leq n$ and $u \leq m$.
user2661923
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It is unnecessary to assume $u\leq n$ or $u\leq m$, when conveniently $k>h$ or $k<0$ implies $\tbinom hk=0$ – Graham Kemp Feb 24 '21 at 23:18