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}

  • $\begingroup$ ah ok, but how is that a Bernoulli distribution? $\endgroup$ – BeepBoop Dec 2 '18 at 3:13
  • 1
    $\begingroup$ Do you mean Binomial? $\endgroup$ – Thomas Shelby Dec 2 '18 at 6:23
  • $\begingroup$ Recall the uniqueness of moment generating functions. $\endgroup$ – Thomas Shelby Dec 2 '18 at 6:27

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