Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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

share|improve this question
add comment

2 Answers

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.

share|improve this answer
    
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
add comment

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 $h$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.