Calculating convolution of binomial distribution using moment generating function

Question

I have two independent random variables $X_{1}, X_{2}$ on the same probability space. $X_{1}$ is bin bin(n, p) and $X_{2}$ ís bin (m, p) with n, m natural numbers and p in the interval [0,1]. I need to prove now, that the distribution of $X_{1}+X_{2}$ is bin (n+m, p), using the moment generating function $$\mathbb{E}[e^{tX}]$$. I have no idea.... Thanks for any help!

Equorne

Edit: Sorry, that I don't write any "hi" or anything at the beginning; the editor somehow always deletes it.

Answer 1

The moment generating function (MGF) of $X_1$ is $$M_{X_1}(t) = (1-p+pe^t)^n.$$ For $X_2$ it is $$M_{X_2}(t) = (1-p+pe^t)^m.$$ The MGF of the sum of two independent random variables is the product of both MGFs, i.e. $$ \begin{align} M_{X_1+X_2}(t) &= (1-p+pe^t)^n (1-p+pe^t)^m\\ &= (1-p+pe^t)^{n+m} \end{align} $$ which is again a binomial distribution with parameters $n+m$ and $p$.