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This is about proving the additivity of Bernoulli's binomial distribution. On another words,

$$X\sim\mathrm{Bin}(n,p),\quad Y\sim\mathrm{Bin}(m,p)$$

implies that $$ X+Y\sim\mathrm{Bin}(n+m,p).$$

meaning, if $X$ and $Y$ are in binomial distribution, with different population and the same success probability, the distribution of $X+Y$ is also binomial with $X+Y\sim\mathrm{Bin}(n+m,p)$.

And I can't understand the fourth line, where I marked '?'. What concept should I learn more to understand this?

\begin{align} \mathbb P(U+V=k) &= \sum_{j=0}^k \mathbb P(U=j,V=k-j)\\ &= \sum_{j=0}^k \mathbb P(U=j)\mathbb P(V=k-j)\\ &= \sum_{j=0}^k \binom nj p^j(1-p)^{n-j}\binom m{k-j}p^{k-j}(1-p)^{m-k+j}\\ &= p^k(1-p)^{n+m-k}\sum_{j=0}^k \binom nj \binom m{k-j}\quad \quad\mathbf ?\\ &= \binom {n+m}k p^k(1-p)^{n+m-k}. \end{align}

Michael Hoppe
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서영빈
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  • you can use moment generating functions to prove it. The proof you've shown us uses the combinatorial identity $\sum_{j=0}^k \begin{pmatrix} n\ j\end{pmatrix}\begin{pmatrix} m\ k-j\end{pmatrix} = \begin{pmatrix} n+m\ k\end{pmatrix}$ – fGDu94 Jan 26 '20 at 14:33
  • @fGDu94 Thank you! – 서영빈 Jan 26 '20 at 15:27
  • You need to assume that $X,Y$ are independent. – J. De Ro Jan 26 '20 at 19:12

2 Answers2

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This is the Chu-Vandermonde Identity (https://proofwiki.org/wiki/Chu-Vandermonde_Identity). In general the addition of two binomial distribution is a special case of the Poisson Binomial distribution (Addition of two Binomial Distribution)

Alessio K
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You should be able to do this without any calculation. It depends on definitions, but most would already know that that a binomial distribution, with parameters $n$ and $p$, is defined as the sum of $n$ iid Bernoulis each with parameter $p$. Your $X+Y$ then is the sum of $n':=n+m$ iid Bernoulis each with parameter $p$ and hence is binomially distributed.

user8675309
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