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About LinkBacksI need some help understanding how to derive the moment generating function of a binomial distribution. I know that the solution is supposed to be:
so far I have:
 = \sum_{y=0}^{\infty} e^{ty}p(y) = \sum_{y=0}^{\infty} e^{ty} \binom{n}{y}p^y q^{n-y} = \sum_{y=0}^{\infty} \frac{n!}{y!(n-y)!} (pe^t)^y q^{n-y})
at which point I don't know how to carry on.
I need some help understanding how to derive the moment generating function of a binomial distribution. I know that the solution is supposed to be:
so far I have:
at which point I don't know how to carry on.
First the correction of some mistakes (the edit is in red). Do you understand why?Originally Posted by lllll
I need some help understanding how to derive the moment generating function of a binomial distribution. I know that the solution is supposed to be:
so far I have:
 = \sum_{y=0}^{\infty} e^{ty}p(y) = \sum_{y=0}^{{\color{red}n}} e^{ty} \binom{n}{y}p^y q^{n-y} = \sum_{y=0}^{{\color{red}n}} \frac{n!}{y!(n-y)!} (pe^t)^y q^{n-y})
at which point I don't know how to carry on.
One or two more steps and you have it!
The key observation is the binomial theorem:
.
In your caseand
. Capisce?
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