Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose $X$ has a $Binomial(n,p)$ distribution. Then its moment generating function is

$$ M(t) = \sum_{x=0}^x {n \choose x}p^x(1-p)^{n-x} \\ =\sum_{x=0}^{n} {n \choose x}(pe^t)^x(1-p)^{n-x}\\ =(pe^t+1-p)^n $$

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

share|cite|improve this question
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
this video explains how to find the mgf of a binomial distribution: – user103477 Oct 27 '13 at 2:04

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.

share|cite|improve this answer

The Moment Generating Function of the Binomial Distribution Consider the binomial function (1)

b(x;n,p) = (n!/x!(n−x)!)p^x q^(n−x) with q = 1 − p. Then the moment generating function is given by (2)

Mx(t) = nX ext n! x=0

x!(n − x)!pxqn−x

nX (pet)x n! x=0 x!(n − x)!qn−x = (q + pet)n, where the final equality is understood by recognising that it represents the expansion of binomial.

(Copy answer to word and save as .doc or pdf for comprhension )

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.