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

Can someone please explain how the sum is obtained from lines (2) to (3)?

• This is the Binomial formula. – Stefan Hansen Nov 13 '12 at 19:53
• It makes sense to me that the Binomial Theorem would be applied to this, I'm just having a hard time working out how they get to the final result using it :\ – TopGunCpp Nov 13 '12 at 21:02
• It all makes sense now, it "is" a syntactically simplified way to write the Binomial Theorem. Thanks for the clarification – TopGunCpp Nov 13 '12 at 21:07
• Call $l=pe^t$ and $j=1-p$, then the second line is $\sum_{x=0}^n {n \choose x} l^x j^{n-x} = (l+j)^n$ by the binomial formula. – Stefan Hansen Nov 13 '12 at 21:07
• You are missing an $e^{tx}$ in the first line. – robjohn Aug 26 '14 at 23:32

## 2 Answers

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) &=\mathrm{E}(e^{tk})\\ &=\sum_{k=0}^n\binom{n}{k}p^k(1-p)^{n-k}e^{tk}\\ &=\sum_{k=0}^n\binom{n}{k}\left(pe^t\right)^k(1-p)^{n-k}\\ &=\left(pe^t+(1-p)\right)^n \end{align} The last step is simply an application of the binomial theorem.

φ(t) = E(e^(tX)) =>E(e^(t.(Σx))) =>E(e^(tx1).e^(tx2).e^(tx3)...e^(txn)) =>E(e^(tx1)).E(e^(tx2)).E(e^(tx3))...E(e^(txn)) ; Since all individual events are independant => [e^t + (1-p)].[e^t + (1-p)].[e^t + (1-p)]...[e^t + (1-p)] ; n times, since all n random variables are bernoulli random variables => [e^t + (1-p)]^n