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:


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

where $$q = 1 - p$$.

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

where $$q = 1 - p$$.

First we need to obtain the first derivative:

$$ $$ $$ $$ $$

Now we can use

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

where $$z = 1 - qs$$.

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

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

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

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