Lorena Becker

2022-11-22

Consider a random sample of size n$n$ that follows a density probability function given by:
$f\left(x,\theta \right)=\frac{1}{\theta }{x}^{-\frac{\theta +1}{\theta }}{\mathbb{1}}_{\left(1,+\mathrm{\infty }\right)},\phantom{\rule{mediummathspace}{0ex}}\phantom{\rule{mediummathspace}{0ex}}\theta >0$
where $\theta$ is unknown.

Averi Shaffer

Expert

Indeed you can use logarithms if you want to, and you don’t need to consider $\mathrm{log}f$ instead of $f$ in order to do so. Since
$f\left(x\mid \theta \right)=\prod _{i=1}^{n}f\left({x}_{i},\theta \right)=\frac{1}{{\theta }^{n}}\prod _{i=1}^{n}{x}_{i}^{-\frac{\theta +1}{\theta }}=\frac{1}{{\theta }^{n}}\mathrm{exp}\left(-\frac{\theta +1}{\theta }\sum _{i=1}^{n}\mathrm{ln}{x}_{i}\right)\phantom{\rule{thickmathspace}{0ex}},$
we have
$f\left(x\mid \theta \right)=g\left(T\left(x\right)\mid \theta \right)h\left(x\right)\phantom{\rule{thickmathspace}{0ex}},$
with
$T\left(x\right)=\sum _{i=1}^{n}\mathrm{ln}x\phantom{\rule{thickmathspace}{0ex}}.$
Thus $T\left(x\right)$ is a sufficient statistic.
But you don’t have to use logarithms. You could also just write
$f\left(x\mid \theta \right)=\prod _{i=1}^{n}f\left({x}_{i},\theta \right)=\frac{1}{{\theta }^{n}}\prod _{i=1}^{n}{x}_{i}^{-\frac{\theta +1}{\theta }}=\frac{1}{{\theta }^{n}}{\left(\prod _{i=1}^{n}{x}_{i}\right)}^{-\frac{\theta +1}{\theta }}\phantom{\rule{thickmathspace}{0ex}},$
which factorizes with $S\left(x\right)=\prod _{i=1}^{n}{x}_{i}$. That $S\left(x\right)$ and $T\left(x\right)$ are likewise sufficient statistics is not surprising, since one is an injective function of the other$S\left(x\right)=\mathrm{exp}\left(T\left(x\right)\right)$ and $T\left(x\right)=\mathrm{log}\left(S\left(x\right)\right)$, and an injective function of a sufficient statistic is again a sufficient statistic.

Do you have a similar question?