Question

Prove that the sum of the observations of a random sample of size \(n\) from a Poisson distribution having parameter \(\theta, 0<\theta<\infty\), is a sufficient statistic for \(\theta\).

Short answer

The sum of the observations of a Poisson distribution is a sufficient statistic for \(\theta\), because when the joint pmf of the Poisson sample is factorized (following the factorization theorem), the sum of the observations appears in the factor that includes \(\theta\), while the remaining parts of the sample appear in the factor that does not include \(\theta\).

Step-by-step solution

Define a Random Sample and Joint Probability Function

A random sample \(X_1, X_2,...,X_n\) from a Poisson distribution has the probability mass function:\(P(X = k) = \frac{{e^{-\theta}\theta^k}}{{k!}}\)\nThe joint probability function for \(n\) independent observations from this distribution is: \(f(x_1,x_2,...,x_n; \theta) = \prod_{i=1}^{n} \frac{{e^{-\theta}\theta^{x_i}}}{{x_i!}}\)

Simplify the Equation

Observe that the joint pmf can be rewritten as \(f(x_1,x_2,...,x_n; \theta) = e^{-n\theta}(\theta^{\sum_{i=1}^{n}{x_i}})/\prod_{i=1}^{n}{x_i!}\)

Apply Factorization Theorem

The factorization theorem states that \(T(X)\) is a sufficient statistic for \(\theta\) if the joint pmf/pmf can be factored into two parts: one part that is a function of \(T(X)\) and \(\theta\), and another part only of \(X\). Compare this with the previous expression of the joint pmf and write it in this factorized form:\(f(x_1,x_2,...,x_n; \theta) = e^{-n\theta}(\theta^{\sum_{i=1}^{n}{x_i}})g(x_1,x_2,...,x_n)\) where \(g(x_1,x_2,...,x_n) = 1/\prod_{i=1}^{n}{x_i!}\)

Identify the Sufficient Statistic

Clearly from the factorized form, the statistic \(T(X) = \sum_{i=1}^{n}{X_i}\) is a sufficient statistic for \(\theta\) given that it appeares in the factor that includes \(\theta\), while the remaining parts of the sample \(X_1, X_2, ..., X_n\) appear in the function that does not include \(\theta\).