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I would like to prove that Skewness and Kurtosis are sufficient statistics for gaussian distribution.

Later on I will try to prove on loglogistic distribution.

Do you have any idea how to do that?

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    $\begingroup$ You are going to fail --- you cannot prove what is not true $\endgroup$ May 26, 2017 at 12:00
  • $\begingroup$ Can you give some detail? What is the reason? $\endgroup$
    – Cerenimo
    May 26, 2017 at 12:08
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    $\begingroup$ You obtain absolutely no information about the location of a Gaussian from the skewness or kurtosis. Proof: they are the same for a Normal$(\mu,\sigma)$ and Normal$(\mu^\prime,\sigma)$ distribution, for any $\mu$ or $\mu^\prime$. Indeed, skewness is worthless, because it's always zero. $\endgroup$
    – whuber
    May 26, 2017 at 12:25
  • $\begingroup$ Thanks a lot. That's right. How can I prove these two are sufficient statistic on Loglogistic distribution? Any idea? $\endgroup$
    – Cerenimo
    May 26, 2017 at 12:45
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    $\begingroup$ Why would you think they are? What theorems exist that relate to sufficiency? $\endgroup$
    – Glen_b
    May 26, 2017 at 12:50

1 Answer 1

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You are going to fail --- you cannot prove what is not true!

More details: You obtain absolutely no information about the location of a Gaussian from the skewness or kurtosis. Proof: they are the same for a $\text{Normal}(\mu,\sigma)$ and $\text{Normal}(\mu′,\sigma)$ distribution, for any μ or μ′. Indeed, skewness is worthless, because it's always zero. (stated in comments by whuber).

As for the same question for loglogistic distribution, you would have to tell us why you expect skewness and kurtosis to be sufficient in that case (it is not).

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  • $\begingroup$ This answer is presently flawed. Sample skewness is not always zero. In fact, it is non-zero with probability one (though it has a distribution that does not depend on $\mu$). $\endgroup$
    – Ben
    Sep 1, 2019 at 0:21

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