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)?

  • 5
    $\begingroup$ This is the Binomial formula. $\endgroup$ – Stefan Hansen Nov 13 '12 at 19:53
  • $\begingroup$ 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 :\ $\endgroup$ – TopGunCpp Nov 13 '12 at 21:02
  • $\begingroup$ It all makes sense now, it "is" a syntactically simplified way to write the Binomial Theorem. Thanks for the clarification $\endgroup$ – TopGunCpp Nov 13 '12 at 21:07
  • $\begingroup$ 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. $\endgroup$ – Stefan Hansen Nov 13 '12 at 21:07
  • 1
    $\begingroup$ You are missing an $e^{tx}$ in the first line. $\endgroup$ – robjohn Aug 26 '14 at 23:32

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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