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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 sum of the observations of a random sample from a gamma distribution with a given pdf is indeed a sufficient statistic for \(\theta\) because it fulfills the Factorization Theorem's requirements.

Step by step solution

01

Write the Joint PDF

The first step is to write the joint pdf of \(x_1, x_2, ..., x_n\), which is the product of the individual pdfs, each of which is \(f(x_i; \theta) = \frac{1}{\theta} e^{\frac{-x_i}{\theta}}\) for \(i = 1, 2, ..., n\). The joint pdf becomes \(f(x_1, x_2,...,x_n; \theta) = \frac{1}{\theta^n} e^{\frac{-\sum_{i=1}^{n}x_i}{\theta}}\) since the observations are independent.
02

Apply the Factorization Theorem

The Factorization Theorem states that a statistic T(X) is sufficient for \(\theta\) iff the joint pdf can be factorized into two factors, one which is a function of T only and the other which is a function of observations only. We can now express the joint pdf: \(f(x_1, x_2,...,x_n; \theta) = (1 / \theta^n) e^{\frac{-T(X)}{\theta}} \cdot (1)\) where \(T(X) = \sum_{i=1}^{n}x_i\). The first part is only a function of the sufficient statistic and the second part is only a function of the observations which satisfies the Factorization theorem.
03

Conclusion: Prove Sufficiency

Since we have successfully factorised the function such that one factor is a function of \(\theta\) and T(X) and the other is a function of observations only, we can say that T(X) is a sufficient statistic for \(\theta\).

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