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Chapter 7: Problem 5

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

Expert verified
The sufficient statistic for the parameter \(\theta\) in the gamma distribution, whose PDF is given as \(f(x ; \theta)=(1 / \theta) e^{-x / \theta}, 0<x<\infty, 0<\theta<\infty\), and zero elsewhere, is the sum of observations of the random sample. This is established through the successful application of Factorisation Theorem on the joint pdf of the sample.

Step by step solution

01

Joint pdf of a Sample

First, establish the joint pdf of the random sample (X1, X2,... Xn). This can be found by taking the product of the individual pdfs as the observations are independent. The joint pdf is \(g(x_1, x_2, ..., x_n ; \theta) = \prod_{i=1}^{n}(1 / \theta) e^{-x_i / \theta} = (1 / \theta^n) e^{-\sum_{i=1}^n x_i / \theta}\).
02

Apply Factorization Theorem

The Factorization Theorem states that a statistic T(X) is a Sufficient Statistic for \(\theta\) if and only if the joint pdf of the sample can be factorized into two factors. One factor is dependent on \(\theta\) and T(X) and the other factor depends only on the X's and not on \(\theta\). Next, express the joint pdf in the form h(x)g(T(x);\(\theta\)). Here, it can be factored as: \(h(x) = 1 \quad and \quad g(T(x);\(\theta\)) = (1 / \theta^n) e^{-T / \theta}\). Here, T(x) is \(\sum_{i=1}^{n}x_i\), the sum of observations.
03

Proclaim Sufficiency

Since a factorization was found such that one of the factors depends only on the X's and not on \(\theta\) (h(x)) and the other factor depends on \(\theta\) and the function T(X) (where T(X) is the sum of observations), we must conclude that T(X) is our sufficient statistic for \(\theta\).

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