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Suppose we have p dimensional vector of $X =[X_1 \dots X_n]$ where X is Laplace distributed. What will be a sufficient statistics for estimating covariance of $X$?

Would it be the sample covariance, i.e.$$\frac{1}{N}\sum_{i=1}^{N} (X_i-\bar X)(X_i-\bar X)^T\,?$$

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    $\begingroup$ Can you give us the density of the multivariate Laplace distribution? $\endgroup$
    – Xi'an
    Jul 29, 2015 at 12:36
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    $\begingroup$ The point of the comment by @Xi'an is that there is more than one meaning to a (multivariate) Laplace distribution: see stats.stackexchange.com/questions/52276/…. $\endgroup$
    – whuber
    Jul 29, 2015 at 12:37
  • $\begingroup$ If we use the density given in the tagged link I get sufficient statistic as $\frac{1}{N}\sqrt{(X-\mu)(X-\mu)^T}^{2-p/2}$ but I am not sure if it is correct. Can we take sample covariance itself as sufficient statistic assuming one to one? $\endgroup$
    – undefined
    Jul 29, 2015 at 13:00
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    $\begingroup$ You did not read properly the definition of this distribution: the correct answer is that there is no non-trivial sufficient statistic. This is because the Laplace distribution is not an exponential family of distributions. $\endgroup$
    – Xi'an
    Jul 29, 2015 at 13:02

1 Answer 1

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Ìf by multivariate Laplace distribution you mean any distribution such that the marginals all are distributed from unidimensional Laplace distributions, with densities$$f(x_i|\mu_i,\sigma_1)=\dfrac{1}{2\sigma_i}\exp\left\{-|x_i-\mu_i|/\sigma_i\right\},$$then there cannot be a sufficient statistic of fixed dimension for the parameters $(\mathbf{\mu},\mathbf{\Sigma})$ of the joint distribution. This is because the distribution cannot belong to an exponential family, hence cannot have a sufficient statistic of fixed dimension by virtue of the (Darmois-)Pitman-Koopman lemma.

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    $\begingroup$ +1 It does leave one wondering, though, what a sufficient statistic for any specific $n$ would be. $\endgroup$
    – whuber
    Jul 29, 2015 at 19:29
  • $\begingroup$ I would think an order statistic is the only possible reduction. "An" because the vectors can be reordered according to any rule. $\endgroup$
    – Xi'an
    Jul 29, 2015 at 19:49
  • $\begingroup$ Sorry I didnt understood what do you mean by order statistics is possible reduction? $\endgroup$
    – undefined
    Jul 30, 2015 at 5:12
  • $\begingroup$ For distributions outside the exponential family, the order statistics is always sufficient. In dimension one. In larger dimensions, any reordering of the iid sample works as well, since it integrates out the random permutation of the indices, which carries no information about the parameter(s). $\endgroup$
    – Xi'an
    Aug 2, 2015 at 10:46

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