Derivatives of PGF of Poisson Distribution

Theorem

Let $X$ be a discrete random variable with the Poisson distribution with parameter $\lambda$.

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

$\dfrac {d^k} {ds^k} \Pi_X \left({s}\right) = \lambda^k e^{- \lambda \left({1-s}\right)}$

Proof

The Probability Generating Function of Poisson Distribution is:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \Pi_X \left({s}\right)\)	\(=\)	\(\displaystyle e^{-\lambda \left({1-s}\right)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle e^{-\lambda + \lambda s}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle e^{-\lambda} e^{\lambda s}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Exponent of Sum

We have that for a given Poisson distribution, $\lambda$ is constant.

From Higher Derivatives of Exponential Function, we have that:

$\dfrac {d^k}{ds^k} \left({e^{\lambda s}}\right) = \lambda^k e^{\lambda s}$

Thus we have:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac {d^k} {ds^k} \Pi_X \left({s}\right)\)	\(=\)	\(\displaystyle \frac {d^k}{ds^k} e^{-\lambda} \left({e^{\lambda s} }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle e^{-\lambda} \frac {d^k}{ds^k} \left({e^{\lambda s} }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Derivative of Constant Multiple
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle e^{-\lambda} \lambda^k e^{\lambda s}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	Higher Derivatives of Exponential Function

Hence the result.

$\blacksquare$