Derivatives of PGF of Shifted Geometric Distribution

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

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


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

where $q = 1 - p$.

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

where $q = 1 - p$.

First we need to obtain the first derivative:

From Derivatives of Function of $a x + b$:
 * $\dfrac {d^n} {ds^n} \left({f \left({1 - qs}\right)}\right) = \left({-q}\right)^n \dfrac {d^n} {dz^n} \left({f \left({z}\right)}\right)$

where $z = 1 - qs$.

Here we have that $f \left({z}\right) = p \dfrac 1 {z^2}$.

From Nth Derivative of Reciprocal of Mth Power:
 * $\dfrac {d^{n-1}}{dz^{n-1}} \dfrac 1 {z^2} = \dfrac {\left({-1}\right)^{n-1} 2^{\overline {n-1}}} {z^{\left({n-1}\right) + 2}}$

where $\overline {n-1}$ denotes the rising factorial.

Note that we consider the $n-1$th derivative because we've already taken the first one.

Also note that $2^{\overline {n-1}} = 1^{\overline {n-1}} = \left({n-1}\right)!$

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

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