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How to find the real valued sufficient statistic of $\theta$ for a random sample from a distribution with the probability density function written below?

$$f(x;\theta) = \theta ax^{a−1} \exp(−\theta x^a), x > 0, \theta > 0, a > 0$$

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up vote 3 down vote accepted

$$f(x,\theta) = \theta a x^{a-1} \exp(-\theta x^{a}) = a x^{a-1} \exp(-\theta x^a + \log(\theta))$$

$$\displaystyle \prod_{i=1}^n f(x_i;\theta) = a^n \left( \prod_{i=1}^n x_i^{(a-1)} \right) \exp(-\theta \sum_{i=1}^n x_i^a + n \log(\theta))$$

Given $a$, define $$h(x_1,x_2,\ldots,x_n) = a^n \left( \prod_{i=1}^n x_i^{(a-1)} \right)$$ $$T(x_1,x_2,\ldots,x_n) = \sum_{i=1}^n x_i^a$$ $$g(\theta, T(x_1,x_2,\ldots,x_n)) = \exp(-\theta T + n \log(\theta))$$

Hence, by Fisher factorization theorem $\displaystyle T(x_1,x_2,\ldots,x_n) = \sum_{i=1}^n x_i^a$ is a sufficient statistic.

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You made it look simple. I knew I had to use factorization theorem but was struggling to use it. Had I known to put $\theta$ in exponential it would have been simpler for me. Thanks. – Sunil Feb 22 '11 at 1:33

Use the factorization theorem. In particular we have the following:

Theorem. $T$ is a sufficient statistic for $\theta$ if the liklihood factorizes in the following form: $$L(x_1, \dots, x_n| \theta) = g(\theta, T(x_1, \dots, x_n)) \cdot h(x_1, \dots, x_n)$$ for some functions $g$ and $h$.

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