# Finding the Moment Generating function of a Binomial Distribution

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)?

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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: youtube.com/watch?v=XEm3lzquu5c – 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.

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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 )

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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

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