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 random sample of size \(n\) from a Poisson distribution, i.e \(T(x) = \sum_{i=1}^n x_i\), is a sufficient statistic for \(\theta\), the parameter of the Poisson distribution.

Step-by-step solution

Define the Random Sample and obtain the joint probability mass function

Let \(X = (X_1, X_2, ..., X_n)\) be a random sample of size \(n\) from a Poisson distribution with parameter \(\theta\). The joint probability mass function of \(X\) is given by: \[ f(x_1,x_2,...,x_n|\theta) = e^{-n\theta} \frac{\theta^{\sum_{i=1}^n x_i}}{\prod_{i=1}^n x_i!} \] where \(x_i\) is the \(i^{th}\) observation of the sample.

Express the joint pmf as a product

The joint probability mass function can be expressed as a product of two factors: one involving \(\theta\) and the other involving the sample \(X\). This yields: \[ f(x_1,x_2,...,x_n|\theta) = h(x)g(T(x)|\theta) = \frac{e^{-n\theta}\theta^{\sum_{i=1}^n x_i}}{\prod_{i=1}^n x_i!} = \frac{1}{\prod_{i=1}^n x_i!} e^{-n\theta}\theta^{\sum_{i=1}^n x_i}\]

Apply the Factorization Theorem of Neyman

According to Neyman's Factorization Theorem, a statistic \(T(X)\) is sufficient for the parameter \(\theta\) if the joint density of \(X\) can be factored into a product of two factors: one involving \(\theta\) and \(T(X)\) and the other depending only on the data \(x_1, x_2, ..., x_n\) and not on \(\theta\). From the formula in Step 2, it's clear that the joint pmf can be factored as such with \(T(x) = \sum_{i=1}^n x_i\), which is the sum of the sample observations.

Conclusion

Since the function \(T(x) = \sum_{i=1}^n x_i\) satisfies the factorization theorem of Neyman, it can be concluded that this function is a sufficient statistic for the parameter \(\theta\) of a Poisson distribution.