Derivatives of Probability Generating Function at One

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Theorem

Let $X$ be a discrete random variable whose probability generating function is $\map {\Pi_X} s$.


Then the $n$th derivative of $\map {\Pi_X} s$ at $s = 1$ is given by:

$\dfrac {\d^n} {\d s^n} \map {\Pi_X} 1 = \expect {X \paren {X - 1} \cdots \paren {X - n + 1} }$

for $n = 1, 2, \ldots$


Proof

Proof by induction:

For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:

$\dfrac {\d^n} {\d s^n} \map {\Pi_X} 1 = \expect {X \paren {X - 1} \cdots \paren {X - n + 1} }$


Basis for the Induction

$\map P 1$ is the case:

$\dfrac \d {\d s} \map {\Pi_X} 1 = \expect X$

This is demonstrated in Expectation of Discrete Random Variable from PGF.

This is our basis for the induction.


Induction Hypothesis

Now we need to show that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1})$ is true.


So this is our induction hypothesis:

$\dfrac {\d^k} {\d s^k} \map {\Pi_X} 1 = \expect {X \paren {X - 1} \cdots \paren {X - k + 1} }$


Then we need to show:

$\dfrac {\d^{k + 1} } {\d s^{k + 1} } \map {\Pi_X} 1 = \expect {X \paren {X - 1} \cdots \paren {X - k} }$


Induction Step

This is our induction step:


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\(\ds \) \(=\) \(\ds \)
\(\ds \leadsto \ \ \) \(\ds \) \(=\) \(\ds \)

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \N_{>0}: \dfrac {\d^n} {\d s^n} \map {\Pi_X} 1 = \expect {X \paren {X - 1} \cdots \paren {X - n + 1} }$


Sources

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