Derivatives of PGF of Geometric Distribution

Theorem
Let $$X$$ be a discrete random variable with the geometric distribution with parameter $p$.

Then the derivatives of the PGF of $$X$$ w.r.t. $$s$$ are:


 * $$\frac {d^n} {ds^n} \Pi_X \left({s}\right) = \frac {q p^n n!} {\left({1 - ps}\right)^{n+1}}$$

where $$q = 1 - p$$.

Proof
The Probability Generating Function of Geometric Distribution is:
 * $$\Pi_X \left({s}\right) = \frac {q} {1 - ps}$$

where $$q = 1 - p$$.

From Derivatives of Function of ax + b, we have that:
 * $$\frac {d^n} {ds^n} \left({f \left({1 - ps}\right)}\right) = \left({-p}\right)^n \frac {d^n} {dz^n} \left({f \left({z}\right)}\right)$$

where $$z = 1 - ps$$.

Here we have that $$f \left({z}\right) = \frac q z$$.

From Nth Derivative of Reciprocal of Mth Power:
 * $$\frac {d^n} {dz^n} \frac q z = q \frac {d^n} {dz^n} \frac 1 z = \frac {\left({-1}\right)^n n!} {z^{n + 1}}$$

where $$n!$$ denotes $n$ factorial.

So putting it together:
 * $$\frac {d^n} {ds^n} \Pi_X \left({s}\right) = q \left({-p}\right)^n \frac {\left({-1}\right)^n n!} {\left({1 - ps}\right)^{n + 1}}$$

whence (after algebra):
 * $$\frac {d^n} {ds^n} \Pi_X \left({s}\right) = \frac {q p^n n!} {\left({1 - ps}\right)^{n+1}}$$