Sum of two independent binomial variables How can I formally prove that the sum of two independent binomial variables X and Y with same parameter p is also a binomial ?
 A: 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.
A: 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 $X+Y \sim \Bin(n+m,p)$.
A: We can prove this using Moment generating function as follows if someone is not comfortable with characterstic functions as answered above:
Let $X \sim B(n_1,p_1)$ and $Y \sim B(n_2,p_2)$ be independent random variables.
We know the MGF of the Binomial distribution is as follows:
$M_X(t)=(q_1+p_1e^t)^{n_{1}},M_Y(t)=(q_2+p_2e^t)^{n_{2}} $
Since X and Y are independent
$M_{X+Y}(t)=M_X(t) \cdot M_y(t) =(q_1+p_1e^t)^{n_{1}} \cdot (q_2+p_2e^t)^{n_{2}}$
We see that we cannot express it in the form $(q+pe^t)^{n}$ and thus by uniqueness property of MGF $X+Y$ is not a binomial variate.
However if we take $p_1=p_2=p$ then we have:
$M_{X+Y}(t)=M_X(t) \cdot M_y(t) =(q+pe^t)^{n_{1}} \cdot (q+pe^t)^{n_{2}}$
$=(q+pe^t)^{n_{1}+n_{2}} $
which is MGF of binomial variate with parameters $(n_1+n_2,p)$
A: Another way:
Suppose $X\sim$ Bin$(n, p)$ and $Y\sim$ Bin$(m, p)$.
The characteristic function of $X$ is then
$$\varphi_X(t) = E[e^{itX}]=\sum_{k=0}^ne^{itk}{n\choose k}p^k(1-p)^{n-k}=\sum_{k=0}^n{n\choose k} (pe^{it})^k(1-p)^{n-k}=(1-p+pe^{it})^n.$$
Since $X, Y$ independent, 
$$\varphi_{X+Y}(t)=\varphi_{X}(t)\varphi_Y(t)=(1-p+pe^{it})^n(1-p+pe^{it})^m=(1-p+pe^{it})^{n+m}.$$
By uniqueness, we get $X+Y\sim$ Bin$(n+m, p)$.
