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Let $X \sim Bi(n,p), Y \sim Bi(m,p)$. “Visual arguments” suggest that $X+Y \sim Bi(m+n,p)$. However, I am unable to prove that.

Using the definition I can reduce the problem to $$\sum_{i=0}^k \binom{m}{i} \binom{n}{k-i} = \binom{n+m}{k},$$ which is – according to Wolfram Alpha – correct. But I have no idea for a proof here as well. (I only tried to use the definition of the binomial coefficients, but that did not help me.)

So: How to prove the statement above?

Thanks in advance, Keba.

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  • $\begingroup$ 31 minutes. $ $ $\endgroup$
    – Did
    Commented Jul 16, 2014 at 16:25
  • $\begingroup$ This assumes that X and Y are independent. $\endgroup$
    – Did
    Commented Jul 16, 2014 at 16:26

1 Answer 1

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This is known as Vandermonde's identity.

A combinatorial interpretation of the identity follows: Suppose you have $n$ boys and $m$ girls. The RHS counts the number of ways to choose $k$ children out of the total $n+m$ children regardless of sex; the LHS counts the number of ways to choose $i$ girls and $k-i$ boys for each $i = 0, 1, 2, \ldots, k$. Obviously, the two are equal.

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  • $\begingroup$ Thanks for your answer. Although this makes sense (and is probably fine enough for the test of my stochastic course), but I am still interested in a more mathematical answer. $\endgroup$
    – Keba
    Commented Jul 16, 2014 at 16:04
  • $\begingroup$ Oh, your edit links to a nice Wikipedia page, that‘s helpful. :) $\endgroup$
    – Keba
    Commented Jul 16, 2014 at 16:06

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