# 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 $X$ with respect to $s$ are:

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

## Proof

 $\ds \map {\Pi_X} s$ $=$ $\ds e^{-\lambda \paren {1 - s} }$ $\ds$ $=$ $\ds e^{-\lambda + \lambda s}$ $\ds$ $=$ $\ds e^{-\lambda} e^{\lambda s}$ 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.

$\blacksquare$