Sum of two independent binomial variables

Solution 1:

Let $(B_k)_k$ be a sequence of iid Bernoulli distributed random variable with $P(B_k=1)=p$ for $k=1,2,\dots$

Then $$X:=B_1+\cdots+B_n$$ is binomially distributed with parameters $n,p$ and $$Y:=B_{n+1}+\cdots+B_{n+m}$$ is binomially distributed with parameters $m,p$. It is evident that $X$ and $Y$ are independent.

Now realize that $$X+Y=B_1+\cdots+B_{n+m}$$ is binomially distributed with parameters $n+m,p$.

This spares you any computations.

Solution 2:

Just compute. Suppose $X \sim \def\Bin{\mathord{\rm Bin}}\Bin(n,p)$, $Y \sim \Bin(m,p)$. Now let $0 \le k \le n+m$, then \begin{align*} \def\P{\mathbb P}\P(X+Y = k) &= \sum_{i=0}^k \P(X = i, Y = k-i)\\ &= \sum_{i=0}^k \P(X=i)\P(Y=k-i) & \text{by independence}\\ &= \sum_{i=0}^k \binom ni p^i (1-p)^{n-i} \binom m{k-i} p^{k-i} (1-p)^{m-k+i}\\ &= p^k(1-p)^{n+m-k}\sum_{i=0}^k \binom ni \binom m{k-i} \\ &= \binom {n+m}k p^k (1-p)^{n+m-k} \end{align*} Hence $X+Y \sim \Bin(n+m,p)$.