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Sufficient Statistics and Maximum Likelihood Estimators of the Poisson Distribution 📂Probability Distribution

Sufficient Statistics and Maximum Likelihood Estimators of the Poisson Distribution

Theorem

Given a random sample X:=(X1,⋯ ,Xn)∼Poi(λ)\mathbf{X} := \left( X_{1} , \cdots , X_{n} \right) \sim \text{Poi} \left( \lambda \right) following a Poisson distribution,

the sufficient statistic TT and the maximum likelihood estimator λ^\hat{\lambda} for λ\lambda are as follows: T=∑k=1nXkλ^=1n∑k=1nXk \begin{align*} T =& \sum_{k=1}^{n} X_{k} \\ \hat{\lambda} =& {{ 1 } \over { n }} \sum_{k=1}^{n} X_{k} \end{align*}

Proof

Sufficient Statistic

f(x;λ)=∏k=1nf(xk;λ)=∏k=1ne−λλxkxk!=e−nλλ∑kxk∏kxk!=e−nλλ∑kxk⋅1∏kxk! \begin{align*} f \left( \mathbf{x} ; \lambda \right) =& \prod_{k=1}^{n} f \left( x_{k} ; \lambda \right) \\ =& \prod_{k=1}^{n} {{ e^{-\lambda} \lambda^{x_{k}} } \over { x_{k} ! }} \\ =& {{ e^{-n \lambda} \lambda^{ \sum_{k} x_{k}} } \over { \prod_{k} x_{k} ! }} \\ =& e^{-n \lambda} \lambda^{ \sum_{k} x_{k}} \cdot {{ 1 } \over { \prod_{k} x_{k} ! }} \end{align*}

Neyman Factorization Theorem: Let a random sample X1,⋯ ,XnX_{1} , \cdots , X_{n} have the same probability mass/density function f(x;θ)f \left( x ; \theta \right) for parameter θ∈Θ\theta \in \Theta. A statistic Y=u1(X1,⋯ ,Xn)Y = u_{1} \left( X_{1} , \cdots , X_{n} \right) is a sufficient statistic for θ\theta if there exist two non-negative functions k1,k2≥0k_{1} , k_{2} \ge 0 satisfying the following condition: f(x1;θ)⋯f(xn;θ)=k1[u1(x1,⋯ ,xn);θ]k2(x1,⋯ ,xn) 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) Note, k2k_{2} must not depend on θ\theta.

According to the Neyman Factorization Theorem, T:=∑kXkT := \sum_{k} X_{k} is the sufficient statistic for λ\lambda.

Maximum Likelihood Estimator

log⁡L(λ;x)=log⁡f(x;λ)=log⁡e−nλλ∑kxk∏kxk!=−nλ+∑k=1nxklog⁡λ−log⁡∏kxk! \begin{align*} \log L \left( \lambda ; \mathbf{x} \right) =& \log f \left( \mathbf{x} ; \lambda \right) \\ =& \log {{ e^{-n \lambda} \lambda^{ \sum_{k} x_{k}} } \over { \prod_{k} x_{k} ! }} \\ =& -n\lambda + \sum_{k=1}^{n} x_{k} \log \lambda - \log \prod_{k} x_{k} ! \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 λ\lambda must be 00, therefore 0=−n+∑k=1nxk1λ  ⟹  λ=1n∑k=1nxk \begin{align*} & 0 = - n + \sum_{k=1}^{n} x_{k} {{ 1 } \over { \lambda }} \\ \implies & \lambda = {{ 1 } \over { n }} \sum_{k=1}^{n} x_{k} \end{align*}

Consequently, the maximum likelihood estimator λ^\hat{\lambda} for λ\lambda is as follows: λ^=1n∑k=1nXk \hat{\lambda} = {{ 1 } \over { n }} \sum_{k=1}^{n} X_{k}

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