Derivatives of PGF of Poisson Distribution
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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
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.
$\blacksquare$