Sufficient Statistics and Maximum Likelihood Estimators for the Geometric Distribution
Theorem
Given a random sample $\mathbf{X} := \left( X_{1} , \cdots , X_{n} \right) \sim \text{Geo} \left( p \right)$ that follows a geometric distribution, the sufficient statistic $T$ and the maximum likelihood estimator $\hat{p}$ for $p$ are as follows. $$ \begin{align*} T =& \sum_{k=1}^{n} X_{k} \\ \hat{p} =& {{ n } \over { \sum_{k=1}^{n} X_{k} }} \end{align*} $$
Proof
Sufficient Statistic
$$ \begin{align*} f \left( \mathbf{x} ; p \right) =& \prod_{k=1}^{n} f \left( x_{k} ; p \right) \\ =& \prod_{k=1}^{n} p \left( 1 - p \right)^{x_{k} - 1} \\ =& p^{n} \left( 1 - p \right)^{\sum_{k} x_{k} - n} \\ =& p^{n} \left( 1 - p \right)^{\sum_{k} x_{k} - n} \cdot 1 \end{align*} $$
Neyman factorization theorem: Suppose a random sample $X_{1} , \cdots , X_{n}$ has the same probability mass/density function $f \left( x ; \theta \right)$ with respect to the parameter $\theta \in \Theta$. A statistic $Y = u_{1} \left( X_{1} , \cdots , X_{n} \right)$ being a sufficient statistic for $\theta$ implies the existence of two non-negative functions $k_{1} , k_{2} \ge 0$. $$ f \left( x_{1} ; \theta \right) \cdots f \left( x_{n} ; \theta \right) = k_{1} \left[ u_{1} \left( x_{1} , \cdots , x_{n} \right) ; \theta \right] k_{2} \left( x_{1} , \cdots , x_{n} \right) $$ However, $k_{2}$ must not depend on $\theta$.
According to the Neyman factorization theorem, $T := \sum_{k} X_{k}$ is a sufficient statistic for $p$.
Maximum Likelihood Estimator
$$ \begin{align*} \log L \left( p ; \mathbf{x} \right) =& \log f \left( \mathbf{x} ; p \right) \\ =& \log p^{n} \left( 1 - p \right)^{\sum_{k} x_{k} - n} \\ =& n \log p + \sum_{k=1}^{n} x_{k} \log \left( 1 - p \right) \end{align*} $$
The log-likelihood function of the random sample is as above, and for the likelihood function to be maximized, the partial derivative with respect to $p$ needs to be $0$. Therefore, $$ \begin{align*} & 0 = n {{ 1 } \over { p }} - {{ 1 } \over { 1 - p }} \left( \sum_{k=1}^{n} x_{k} - n \right) \\ \implies & {{ n } \over { p }} + {{ n } \over { 1 - p }} = {{ 1 } \over { 1 - p }} \sum_{k=1}^{n} x_{k} \\ \implies & {{ n } \over { p(1-p) }} = {{ 1 } \over { 1 - p }} \sum_{k=1}^{n} x_{k} \\ \implies & {{ 1 } \over { p }} = {{ 1 } \over { n }} \sum_{k=1}^{n} x_{k} \end{align*} $$
Thus, the maximum likelihood estimator $\hat{p}$ for $p$ is as follows. $$ \hat{p} = {{ n } \over { \sum_{k=1}^{n} X_{k} }} $$
■