The following is a homework problem. Please tell me if my solution is correct and if not please point out my mistakes.
Let $x_{1}, x_{2},...,x_{M}$ be i.i.d. samples from the absolute continuous distribution given by the p.d.f $$f(x| \theta) = \frac{ \theta}{(x+1)^{\theta +1}}$$
find a sufficient statistic for $\theta$
My answer is the following:
I assume the range of the function is $[0, \inf)$ otherwise the integral does not equal 1 so the function couldn't be a probability distribution.
$$f(x_{1}, x_{2},...,x_{n}) = \prod_{i=1}^n \frac{ \theta}{(x_{i}+1)^{\theta +1}}$$ $$= \prod_{i=1}^n \theta(x_{i}+1)^{-(\theta +1)} = \prod_{i=1}^nexp[ln(\theta)- ln (x_{i}+1)^{-(\theta +1)}]$$ $$= exp \hspace{1mm}[\hspace{1mm} \sum_{i=1}^n( ln(\theta)- (\theta +1)ln (x_{i}+1))]$$ $$= exp \hspace{1mm}[\hspace{1mm} n\hspace{1mm} ln(\theta)- (\theta +1) \sum_{i=1}^n ln (x_{i}+1)]$$ $ \therefore$ using Neyman - Fisher Factorization $$T(x) = \sum_{i = 1}^n ln(x_{i} +1)$$ $$g(T(x), \theta) = exp [ \hspace{2mm} n \hspace{1mm}ln \hspace{1mm}\theta - ( \theta + 1) T(x)]$$ $$h(x) = 1$$
and T(x) is the sufficient statistic
Could someone please tell me if my solution is correct and if not what is a better solution?
Also as the samples $x_{i}$ only appear once in the joint probability function would the entire joint p.d.f $\prod_{i=1}^n \frac{ \theta}{(x+1)^{\theta +1}}$ be a sufficient statistic?
What does "absolute continuous" mean? What is an absolute continuous distribution?
