Proving additivity of Bernoulli binomial distribution
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}
$\endgroup$ 32 Answers
$\begingroup$This is the 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)
$\endgroup$ $\begingroup$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.
$\endgroup$
