Question

Show that the sum of the observations of a random sample of size \(n\) from a gamma distribution that has pdf \(f(x ; \theta)=(1 / \theta) e^{-x / \theta}, 0

Short answer

- T(x)=∑xi (from i=1 to n) is a sufficient statistic for θ.

Step-by-step solution

Understand the Probability Density Function (PDF)

First, we need to comprehend the given function \(f(x ; \theta)=(1 / \theta) e^{-x / \theta}\), for \(0<x<\infty\), \(0<\theta<\infty\), else zero. This function represents a gamma distribution. It describes the time lapse between events in a Poisson point process, i.e., a process in which events occur continuously and independently at a constant average rate. The parameter θ is the expected time lapse.

Compute Joint Probability Density

Take a random sample \(X_1,X_2,...,X_n\) from the gamma population and write down the joint pdf. It is simply the product of individual density functions since the observations are independent. This is given by: \[f(x_1,x_2,...,x_n ; \theta)= \prod_{i=1}^{n}(1 / \theta) e^{-x_i / \theta}= \frac{1}{\theta^n} e^{-\sum_{i=1}^{n}x_i / \theta}\]

Applying Factorization Theorem

Next, we apply the Factorization theorem. The theorem states that a statistic T(X) is a sufficient statistic for θ if and only if there exist functions g(t ; θ) and h(x) such that for all sample points and parameter points, the joint pdf or joint mass function of the sample observations can be expressed as the product of these functions. So, we need to factorise our joint pdf into two components, one that depends on the sample data through the sufficient statistic, and the other that depends on the sample data in a way that does not involve the parameter at all. Here, our joint pdf can be re-written as: \[f(x_1,x_2,...,x_n ; \theta)= \frac{1}{\theta^n} e^{-T(x) / \theta}e^{-n \bar{x}}\] Where, \( T(x)=\sum_{i=1}^{n}x_i \) and \( \bar{x}=\frac{\sum_{i=1}^{n}x_i}{n} \) Now compare this with factorization theorm expression, the function g(T,θ) is that part of f that contains θ. So here, \( g(T,θ)= \frac{1}{\theta^n}e^{-T / θ} \) and \( h(x)=e^{-n \bar{x}} \) So, \( T(x)=\sum_{i=1}^{n}x_i \) is a sufficient statistic for θ .