TITLE: Sum of two independent binomial variables
QUESTION [40 upvotes]: How can I formally prove that the sum of two independent binomial variables X and Y with same parameter p is also a binomial ?

REPLY [2 votes]: 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 $(n_1+n_2,p)$