Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Would someone please provide me an example of where we take a p.g.f and use it to derive the p.m.f. ?

I understand that you were have to take the derivatives of the pmf, which is understandable because the derivatives tell us the probabilities, but I would just like to see an example of where it all comes together.

share|improve this question

1 Answer 1

up vote 1 down vote accepted

Here's an example:

The probability mass function for a binomial random variable is given by

$$\mathrm{Pr}(X=i)=f(i; n, p) = \begin{pmatrix} n \\ i\end{pmatrix}p^i(1-p)^{n-i}.$$

Expand the generating function $G(z) = [(1-p)+pz]^n$ using the binomial theorem:

$$G(z) = \sum_{k=0}^n \begin{pmatrix} n \\ k\end{pmatrix} (pz)^k(1-p)^{n-k}.$$

Taking the $i$th derivative with respect to $z$, we must see that all the terms with index less than $i$ are annihilated, leaving,

$$G^{(i)}(z) = \sum_{k=i}^n \begin{pmatrix} n \\ k\end{pmatrix} (1-p)^{n-k} \frac{d^i}{dz^i} (pz)^k.$$

Equally obviously, any terms with index greater than $i$ have a power of $z$ that does not annihilate. By computing $G^{(i)}(0)$, these terms vanish, leaving only the $i$th term:

$$G^{(i)}(0) = \begin{pmatrix} n \\ i\end{pmatrix} (1-p)^{n-i} p^i \left.\frac{d^i}{dz^i} (z)^i\right|_{z=0}.$$

Of course, we know that $\frac{d^m}{dx^m} x^m = m!$, so this gives us

$$G^{(i)}(0) = \begin{pmatrix} n \\ i\end{pmatrix} (1-p)^{n-i} p^i i! = i!\ \mathrm{Pr}(X=i)$$

share|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.