Finding the Moment Generating function of a Binomial Distribution Suppose $X$ has a $\rm{Binomial}(n,p)$ distribution. Then its moment generating function is


  \begin{align}
M(t)
&amp;= \sum_{x=0}^x e^{xt}{n \choose x}p^x(1-p)^{n-x} \\
&amp;=\sum_{x=0}^{n} {n \choose x}(pe^t)^x(1-p)^{n-x} \\
&amp;=(pe^t+1-p)^n
\end{align}


Can someone please explain how the sum is obtained from lines (2) to (3)?
The moment generating function for the binomial distribution $B_{n,p}$, whose discrete density is $\binom{n}{k}p^k(1-p)^{n-k}$, is defined as
$$
\begin{align}
M_{B_{n,p}}(t)
&amp;=\mathrm{E}(e^{tk})\\
&amp;=\sum_{k=0}^n\binom{n}{k}p^k(1-p)^{n-k}e^{tk}\\
&amp;=\sum_{k=0}^n\binom{n}{k}\left(pe^t\right)^k(1-p)^{n-k}\\
&amp;=\left(pe^t+(1-p)\right)^n
\end{align}
$$
The last step is simply an application of the binomial theorem.
$\phi(t) = \mathbb{E}[e^{tX}]
       \Rightarrow \mathbb{E}[e^{t\cdot(\Sigma x)}]
       \Rightarrow \mathbb{E}[e^{tx_1}\cdot e^{tx_2}\cdot...\cdot e^{tx_n}]
       \Rightarrow \mathbb{E}[e^{tx_1}]\cdot \mathbb{E}[e^{tx_2}]\cdot...\cdot \mathbb{E}[e^{tx_n}]$
Since all individual events are independent,
$\Rightarrow [pe^t + (1-p)].[pe^t + (1-p)].[pe^t + (1-p)]...[pe^t + (1-p)]$ n times, since all n random variables are Bernoulli random variables
$\Rightarrow[pe^t + (1-p)]^n=[pe^t + q]^n$, where $q=1-p$.