Theorem 7.3.5. Geometric Statistics Theorem.

For the geometric distribution,
\begin{equation*} \mu = 1/p \end{equation*}
\begin{equation*} \sigma^2 = \frac{1-p}{p^2} \end{equation*}
\begin{equation*} \gamma_1 = \frac{2-p}{\sqrt{1-p}} \end{equation*}
\begin{equation*} \gamma_2 = \frac{p^2-6p+6}{1-p} + 3 \end{equation*}

Proof.

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