For the distribution $f_\theta (x) = \theta x^{\theta-1}$, what is the sufficient statistic corresponding to the Monotone Likelihood Ratio?

Suppose I have a sequence of iid random variables $X_1, \ldots, X_n$ following the pdf:

$$f_\theta (x) = \theta x^{\theta-1}$$

for $\theta >0$ and $0 <x<1$.

I would like to find a sufficient statistic $T(X)$, such that the family $f_\theta (x)$ has a monotone likelihood ratio (MLR) in $T(X)$.

I do this by having:

$$\frac{f(x|\theta_1)}{f(x|\theta_2)} = \frac{\prod_{i=1}^{n}\theta_1x_i^{\theta_1-1}}{\prod_{i=1}^{n}\theta_2x_i^{\theta_2-1}} = \left(\frac{\theta_1}{\theta_2}\right)^n \prod_{i=1}^n\left(x_i\right)^{\theta_1-\theta_2} = \left(\frac{\theta_1}{\theta_2}\right)^n \left(\prod_{i=1}^nx_i\right)^{\theta_1-\theta_2}$$

At this point, is the sufficient statistic corresponding to the MLR, $T(X) = \prod_{i=1}^nx_i$?

Or would it be:

$$\frac{f(x|\theta_1)}{f(x|\theta_2)} = \left(\frac{\theta_1}{\theta_2}\right)^n \left(\prod_{i=1}^nx_i\right)^{\theta_1-\theta_2} = \left(\frac{\theta_1}{\theta_2}\right)^n \left(e^{\sum_{i=1}^n \log(x_i)}\right)^{\theta_1-\theta_2}$$

and hence the sufficient statistic is $\sum_{i=1}^n \log(x_i)$? I understand that sufficient statistics are not unique, but which one of the above is the right answer?

Is it $T(X) = \prod_{i=1}^nx_i$ or $T(X) =\sum_{i=1}^n \log(x_i)$?

$$T_1(X) = \prod_{i=1}^nx_i$$ $$T_2(X) =\sum_{i=1}^n \log(x_i)$$
Notice that $$T_2(x)=\log(T_1(X))$$ and $$T_1(X)=\exp(T_2(X)),$$ injective function maps a sufficient statistics to another sufficient statistics.
• Given a set $X$ of independent identically distributed data conditioned on an unknown parameter $\theta$ , a sufficient statistic is a function $T(X)$ whose value contains all the information needed to compute any estimate of the parameter. It is a way to summarize the data. Both of them summarizes the same distribution. – Siong Thye Goh Dec 27 '16 at 19:11
• rather than "the" sufficient statistics, i think it should be "a" sufficient statistics. The distribution of $T_1(X)$ and $T_2(X)$ are different but they can be transformed from one to the other. – Siong Thye Goh Dec 27 '16 at 19:17