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 $T(X) =\sum_{i=1}^n \log(x_i)$?
Best Answer
Both works.
$$T_1(X) = \prod_{i=1}^nx_i$$ $$T_2(X) =\sum_{i=1}^n \log(x_i)$$
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.