Derivatives of PGF of Poisson Distribution

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

Then the derivatives of the PGF of $X$ with respect to $s$ are:

$\dfrac {d^k} {\d s^k} \, \map {\Pi_X} s = \lambda^k e^{- \lambda \paren {1 - s} }$


The Probability Generating Function of Poisson Distribution is:

\(\ds \map {\Pi_X} s\) \(=\) \(\ds e^{-\lambda \paren {1 - s} }\)
\(\ds \) \(=\) \(\ds e^{-\lambda + \lambda s}\)
\(\ds \) \(=\) \(\ds e^{-\lambda} e^{\lambda s}\) Exponential of Sum

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

From Higher Derivatives of Exponential Function, we have that:

$\dfrac {\d^k} {\d s^k} \paren {e^{\lambda s} } = \lambda^k e^{\lambda s}$

Thus we have:

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

Hence the result.