# Showing the sum of binomial independent variables follows a binomial distribution using moment generating functions

So I'm trying to solve the following problem:

Show that if $$X_i$$ follows a binomial distribution with $$n_i$$ trials, and probability of $$p_i=p$$ for $$i = 1,2,3...n$$, and the $$X_i$$ are independent, then $$\sum_{i=1}^{n}{X_i}$$ follows a binomial distribution using moment generating functions.

Here's what I've tried so far:

$$M_{\sum_{i=1}^{n}}(t) = \prod_{i=0}^{n}M_{X_i}(t) = \prod_{i=0}^{n}(pe^t + 1 -p)^{n_i}$$

I'm not sure where to go from here, or if I've even done everything correctly until this point. Any help will be greatly appreciated. Thanks!

\begin{align} M_{\sum_{i=1}^{n}X_i}(t) &= \prod_{i=1}^{n}M_{X_i}(t)\\ & = \prod_{i=1}^{n}(pe^t + 1 -p)^{n_i}\\ &=(pe^t + 1 -p)^{\sum_{i=1}^{n} n_i} \end{align}

• ah ok, but how is that a Bernoulli distribution? – BeepBoop Dec 2 '18 at 3:13
• Do you mean Binomial? – Thomas Shelby Dec 2 '18 at 6:23
• Recall the uniqueness of moment generating functions. – Thomas Shelby Dec 2 '18 at 6:27