How to find the real valued sufficient statistic of $\theta$ for a random sample from a distribution with the probability density function written below?
$$f(x;\theta) = \theta ax^{a−1} \exp(−\theta x^a), x > 0, \theta > 0, a > 0$$
How to find the real valued sufficient statistic of $\theta$ for a random sample from a distribution with the probability density function written below?
$$f(x;\theta) = \theta ax^{a−1} \exp(−\theta x^a), x > 0, \theta > 0, a > 0$$
$$f(x,\theta) = \theta a x^{a-1} \exp(-\theta x^{a}) = a x^{a-1} \exp(-\theta x^a + \log(\theta))$$
$$\displaystyle \prod_{i=1}^n f(x_i;\theta) = a^n \left( \prod_{i=1}^n x_i^{(a-1)} \right) \exp(-\theta \sum_{i=1}^n x_i^a + n \log(\theta))$$
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.
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 $h$.