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I recently started reading about sufficient statistics. I have the following questions:

1) Is sample mean a sufficient statistic for Laplace Distribution (aka Double Exponential) if we already know the scale parameter?

2) If not, is this also related to the fact that it is not an efficient estimator of the mean/location?

EDIT

For example sample median with known scale parameter is sufficient and efficient.

Thanks

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    $\begingroup$ I noticed you have deleted a few of your recent questions that were quite interesting. I will not vote to undelete, I will leave that for your discretion (at least for now). But would you care to at least post a comment under the OP or an answer before deleting? That way at least the users with sufficient reputation would see what solution you found or why you thought the question was no longer relevant. Thank you! $\endgroup$
    – Richard Hardy
    Commented Jan 4, 2017 at 15:06
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    $\begingroup$ The one you deleted an hour ago and another recent one (I forgot what exactly it was about). If I were you, I would keep them a bit longer. By deleting them early you also discourage slow users who perhaps plan to answer when they find time. And the questions are quite good, IMHO (even if they do not get many upvotes). $\endgroup$
    – Richard Hardy
    Commented Jan 4, 2017 at 15:15
  • $\begingroup$ I am not sure how to do that. But mine was a general comment applicable to future posts as well. Good luck with your statistical problems! $\endgroup$
    – Richard Hardy
    Commented Jan 4, 2017 at 15:22

2 Answers 2

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For on observation, the Laplace pdf is $$f_X(x) = \dfrac 1 {2b} \exp(-\dfrac {|x-\mu|} b)$$ For multiple iid observations, the pdf is $$f_\boldsymbol X(\boldsymbol x) = \dfrac 1 {(2b)^n} \exp(- \dfrac 1 b \sum_{i=1}^n {|x_i-\mu|})$$ The easiest way to determine what statistics are sufficient for $\boldsymbol X$ is to try to use the Factorization Theorem (https://en.wikipedia.org/wiki/Sufficient_statistic#Fisher.E2.80.93Neyman_factorization_theorem). However, if you start to work with this expression, you'll see that the absolute values in the sum make it impossible to do any simplification/factorization.

To answer your first question, the sample mean is not a sufficient statistic (event if $b$ is known). However, if $\mu$ is known, then $\sum_{i=1}^n |x_i-\mu|$ is a sufficient statistic for $b$. But $\mu$ will almost never be known unless it's assumed to be zero.

As for your second question, I don't believe there are any theorems which directly state for some conditions, inefficiency implies insufficiency or vice-versa. However, there are theorems which connect sufficient statistics to maximum likelihood estimators and MLEs are asymptotically efficient under certain regularity conditions. So in that sense, I suppose you could view the insufficiency and inefficiency of the sample mean as related results.

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  • $\begingroup$ +1 for the first part. For the second question, I believe sample median is a sufficient statistic and it is efficient. That hints me of some connection. $\endgroup$
    – Cagdas Ozgenc
    Commented Dec 19, 2016 at 14:00
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    $\begingroup$ Hm. Heuristically, that doesn't make sense to me. Suppose you have two distinct samples, x = {-1, 0, 1} and y = {-100, 0, 1}. If all you know is the sample median (0 in both cases), you certainly don't have a sufficient amount of info to make inference about b. Given x, we'd expect b to be much smaller than if we were given y. In both cases, you'd need the whole sample to make the best possible inference. $\endgroup$
    – jjet
    Commented Dec 19, 2016 at 14:06
  • $\begingroup$ Sorry about that. You did mention that you were assuming b was fixed and you wanted to make inference about $\mu$. Disregard my last comment. $\endgroup$
    – jjet
    Commented Dec 19, 2016 at 15:28
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    $\begingroup$ (+1) For independent observations from a Laplace distribution the minimal sufficient statistic for the location parameter (scale being known) is the order statistic - there's indeed no further reduction possible. You can separate this into the median (the maximum-likelihood estimate) & an ancillary complement, the configuration statistic. $\endgroup$
    – Scortchi ♦
    Commented Dec 19, 2016 at 16:49
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    $\begingroup$ It's a special situation owing to the Gaussian distribution's being in the exponential family - see the Pitman-Koopman-Darmois theorem, $\endgroup$
    – Scortchi ♦
    Commented Dec 19, 2016 at 20:39
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The sample median is the maximum likelihood estimator of the mean if the scale is known, but it is not the sufficient statistics. I agree that the order statistics is the sufficient statistics. This can be seen using the factorization theorem and examining the dependence of the RN derivative, upon the examination of the absolute values therein, on the data across all mu values.

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