Derivatives of PGF of Binomial Distribution

Theorem
Let $$X$$ be a discrete random variable with the Binomial distribution with parameters $n$ and $p$.

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


 * $$\frac {d^k} {ds^k} \Pi_X \left({s}\right) = \begin{cases}

n^{\underline k} p^k \left({q + ps}\right)^{n-k} & : k \le n \\ 0 & : k > n \end{cases}$$ where $$n^{\underline k}$$ is the falling factorial.

Proof
Proof by induction:

For all $$k \in \N^*$$, let $$P \left({k}\right)$$ be the proposition:
 * $$\frac {d^k} {ds^k} \Pi_X \left({s}\right) = \begin{cases}

n^{\underline k} p^k \left({q + ps}\right)^{n-k} & : k \le n \\ 0 & : k > n \end{cases}$$

The Probability Generating Function of Binomial Distribution is:
 * $$\Pi_X \left({s}\right) = \left({q + ps}\right)^n$$

where $$q = 1 - p$$.

We have that for a given binomial distribution, $$p$$ and $$q$$ are constant.

Basis for the Induction
We have:

$$ $$

So $$P(1)$$ is true, as $$n^{\underline 1} = n$$.

This is our basis for the induction.

Induction Hypothesis
Now we need to show that, if $$P \left({r}\right)$$ is true, where $$r \ge 1$$, then it logically follows that $$P \left({r+1}\right)$$ is true.

So this is our induction hypothesis:
 * $$\frac {d^r} {ds^r} \Pi_X \left({s}\right) = \begin{cases}

n^{\underline r} p^r \left({q + ps}\right)^{n-r} & : r \le n \\ 0 & : r > n \end{cases}$$

Then we need to show:
 * $$\frac {d^{r+1}} {ds^{r+1}} \Pi_X \left({s}\right) = \begin{cases}

n^{\underline {r+1}} p^{r+1} \left({q + ps}\right)^{n-\left({r+1}\right)} & : r+1 \le n \\ 0 & : r+1 > n \end{cases}$$

Induction Step
First, let $$r < m$$. Then we have:

$$ $$ $$ $$ $$

At this stage, as $$r < n$$, we have that $$r+1 \le n$$. So far so good.

Now suppose that $$r = n$$. Then by the induction hypothesis:
 * $$\frac {d^r}{ds^r} \Pi_X \left({s}\right) = n^{\underline r} p^r \left({q + ps}\right)^0 = n! p^n$$

Then $$\frac {d^{r+1}}{ds^{r+1}} \Pi_X \left({s}\right) = \frac {d}{dx} n! p^n = 0$$ by Differentiation of a Constant.

At this stage, as $$r = n$$, we have that $$r+1 > n$$. Again, so far so good.

Finally, suppose that $$r > n$$. Then by the induction hypothesis:
 * $$\frac {d^r}{ds^r} \Pi_X \left({s}\right) = 0$$.

Then $$\frac {d^{r+1}}{ds^{r+1}} = 0$$ by Differentiation of a Constant.

At this stage, as $$r > n$$, we have that $$r+1 > n$$. Which is all we need.

So $$P \left({r}\right) \implies P \left({r+1}\right)$$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $$\frac {d^k} {ds^k} \Pi_X \left({s}\right) = \begin{cases}

n^{\underline k} p^k \left({q + ps}\right)^{n-k} & : k \le n \\ 0 & : k > n \end{cases}$$