Lorena Becker

Answered

2022-11-22

Consider a random sample of size n$n$ that follows a density probability function given by:

$f(x,\theta )=\frac{1}{\theta}{x}^{-\frac{\theta +1}{\theta}}{\mathbb{1}}_{(1,+\mathrm{\infty})},\phantom{\rule{mediummathspace}{0ex}}\phantom{\rule{mediummathspace}{0ex}}\theta >0$

where $\theta $ is unknown.

$f(x,\theta )=\frac{1}{\theta}{x}^{-\frac{\theta +1}{\theta}}{\mathbb{1}}_{(1,+\mathrm{\infty})},\phantom{\rule{mediummathspace}{0ex}}\phantom{\rule{mediummathspace}{0ex}}\theta >0$

where $\theta $ is unknown.

Answer & Explanation

Averi Shaffer

Expert

2022-11-23Added 10 answers

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

$f(x\mid \theta )=\prod _{i=1}^{n}f({x}_{i},\theta )=\frac{1}{{\theta}^{n}}\prod _{i=1}^{n}{x}_{i}^{-\frac{\theta +1}{\theta}}=\frac{1}{{\theta}^{n}}\mathrm{exp}(-\frac{\theta +1}{\theta}\sum _{i=1}^{n}\mathrm{ln}{x}_{i})\phantom{\rule{thickmathspace}{0ex}},$

we have

$f(x\mid \theta )=g(T(x)\mid \theta )h(x)\phantom{\rule{thickmathspace}{0ex}},$

with

$T(x)=\sum _{i=1}^{n}\mathrm{ln}x\phantom{\rule{thickmathspace}{0ex}}.$

Thus $T(x)$ is a sufficient statistic.

But you don’t have to use logarithms. You could also just write

$f(x\mid \theta )=\prod _{i=1}^{n}f({x}_{i},\theta )=\frac{1}{{\theta}^{n}}\prod _{i=1}^{n}{x}_{i}^{-\frac{\theta +1}{\theta}}=\frac{1}{{\theta}^{n}}{\left(\prod _{i=1}^{n}{x}_{i}\right)}^{-\frac{\theta +1}{\theta}}\phantom{\rule{thickmathspace}{0ex}},$

which factorizes with $S(x)=\prod _{i=1}^{n}{x}_{i}$. That $S(x)$ and $T(x)$ are likewise sufficient statistics is not surprising, since one is an injective function of the other$S(x)=\mathrm{exp}(T(x))$ and $T(x)=\mathrm{log}(S(x))$, and an injective function of a sufficient statistic is again a sufficient statistic.

$f(x\mid \theta )=\prod _{i=1}^{n}f({x}_{i},\theta )=\frac{1}{{\theta}^{n}}\prod _{i=1}^{n}{x}_{i}^{-\frac{\theta +1}{\theta}}=\frac{1}{{\theta}^{n}}\mathrm{exp}(-\frac{\theta +1}{\theta}\sum _{i=1}^{n}\mathrm{ln}{x}_{i})\phantom{\rule{thickmathspace}{0ex}},$

we have

$f(x\mid \theta )=g(T(x)\mid \theta )h(x)\phantom{\rule{thickmathspace}{0ex}},$

with

$T(x)=\sum _{i=1}^{n}\mathrm{ln}x\phantom{\rule{thickmathspace}{0ex}}.$

Thus $T(x)$ is a sufficient statistic.

But you don’t have to use logarithms. You could also just write

$f(x\mid \theta )=\prod _{i=1}^{n}f({x}_{i},\theta )=\frac{1}{{\theta}^{n}}\prod _{i=1}^{n}{x}_{i}^{-\frac{\theta +1}{\theta}}=\frac{1}{{\theta}^{n}}{\left(\prod _{i=1}^{n}{x}_{i}\right)}^{-\frac{\theta +1}{\theta}}\phantom{\rule{thickmathspace}{0ex}},$

which factorizes with $S(x)=\prod _{i=1}^{n}{x}_{i}$. That $S(x)$ and $T(x)$ are likewise sufficient statistics is not surprising, since one is an injective function of the other$S(x)=\mathrm{exp}(T(x))$ and $T(x)=\mathrm{log}(S(x))$, and an injective function of a sufficient statistic is again a sufficient statistic.

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