Question

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample of size \(n\) from a geometric distribution that has pmf \(f(x ; \theta)=(1-\theta)^{x} \theta, x=0,1,2, \ldots, 0<\theta<1\), zero elsewhere. Show that \(\sum_{1}^{n} X_{i}\) is a sufficient statistic for \(\theta\).

Short answer

Yes, by applying the factorization theorem, we show that \( \sum_{i=1}^{n} X_{i} \) is a sufficient statistic for \( \theta \) in a geometric distribution.

Step-by-step solution

Write down the joint pmf of the sample

As the geometric distribution is memoryless and each \( X_{i} \) is independent of the others, the joint pmf of the sample of size \( n \) is given by \( f(x_{1}, x_{2}, ..., x_{n} ; \theta) = \prod_{i=1}^{n} f(x_{i} ; \theta) = \prod_{i=1}^{n} (1-\theta)^{x_{i}} \theta \)

Rearrange the joint pmf using algebra

We can break down the joint pmf into two parts: one that contains the statistic \( \sum_{i=1}^{n} x_{i} \) and another that is free of it. Doing the algebra, we have \( f(x_{1}, x_{2}, ..., x_{n} ; \theta) = \theta^{n} (1-\theta)^{\sum_{i=1}^{n} x_{i}} \)

Apply Factorization Theorem

According to the factorization theorem, the joint pmf can be expressed as a function of \( \theta \) and the sufficient statistic times a function of the data that does not contain \( \theta \). Here, it can be written as \( \theta^{n} (1-\theta)^{\sum_{i=1}^{n} x_{i}} = g(\sum_{i=1}^{n} x_{i}, \theta) h(x_{1}, x_{2}, ..., x_{n}) \), where \( g(t, \theta) = \theta^{n} (1-\theta)^{t} \) and \( h(x_{1}, x_{2}, ..., x_{n}) = 1 \), which does not involve \( \theta \)