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$$ f(x|\theta) = e^{x-\theta}\exp\left(-e^{x-\theta}\right),\;\;\; -\infty < x < \infty,\;\;\; -\infty < \theta < \infty.$$

Find acompliete sufficient statistics, or show that one does not exists.

What I have found was that this given $f$ is not an exponential family, and my solution says as followed;

There is no complete sufficient statistics for that. In detail, the solution said the order statistics are minimal sufficient, and this is location family. Thus, the range $R = X_{(n)} - X_{(1)}$ is ancilliary, and expectation does not depend on $\theta$. So this sufficient statistics is not complete.

I don't understand two points. First, the solution could be the proof for no-existency of complete statistics, because this is only the proof for the case of $R = X_{(n)} - X_{(1)}$,

and second, The fact that the expectation does not depend on $\theta$ could imply that this sufficient statistics is not complete.

Could anybody help me to understand this?

Davide Giraudo
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Block Jeong
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1 Answers1

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Contrary to your solution that says "There is no complete sufficient statistic...", I claim the following:

Claim: Given a random sample $\mathbf{X} = (X_1,\ldots,X_n)$ from the pdf $$f(x|\theta) = e^{x-\theta}\exp\left(-e^{x-\theta}\right),\;\;\; -\infty < x < \infty,\;\;\; -\infty < \theta < \infty\,, $$ the statistic defined by $T(\mathbf{X}) = \sum\limits_{i=1}^{n} \exp{(X_i)}$ is a complete sufficient statistic for $\theta$.

Proof:

Sufficiency of $T(\mathbf{X})$. Rewriting the pdf, $$ f(x|\theta) = e^{-\theta}\left[\exp(-e^{-\theta})\right]^{e^{x}}\cdot e^{x} $$ resulting to the likelihood $$ f(\mathbf{x}|\theta) = e^{-n\theta}\left[\exp(-e^{-\theta})\right]^{T(\mathbf{x})}\cdot \exp\left(\sum\limits_{i=1}^{n}x_i\right)\,. $$ Thus, by the Factorization Theorem, $T(\mathbf{X})$ is sufficient.

Completeness of $T(\mathbf{X})$. Showing that $T(\mathbf{X})$ is a complete statistic is equivalent to showing that the family of its pdf is complete. By simple transformation, it easy to see that $\exp(X_i)\sim \text{exponential}(\beta)$ with $\beta = e^{\theta}$ for every $i=1,\ldots,n$, implying $$T(\mathbf{X}) = \sum_{i=1}^{n} \exp{(X_i)} \sim \text{Gamma}(n,e^{\theta})$$ where we note that $\exp(X_i)$'s are also i.i.d. The conclusion follows immediately from the fact that the family of gamma distributions is complete as a result of the uniqueness of Laplace transforms.

venrey
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  • No, that's incorrect. If a minimal sufficient statistics is not complete, then there is no complete sufficient statistics. Let R be a minimal sufficient statistic, suppose there is a complete minimal sufficient statistic,T, for contradiction. Then there exist a 1-1 function, g, s.t. g(R) = T. Since g is 1-1, you can define an inverse, h. If E(f(R)) = 0, then E(f(h(T)) = 0. Since h is non-zero, f must be 0 almost everywhere. So R is complete. – William Ambrose Nov 03 '20 at 14:45
  • @WilliamAmbrose, good day. Thanks for the comment. I am confused. If I did not misunderstood your point, what you are trying to say is that way I prove completeness is incorrect? Or the my entire proof is incorrect? Which is it? – venrey Dec 16 '20 at 22:30
  • I am not exactly sure which part of your proof went wrong, but I believe there are some flaws. OP mentioned that the range statistic is minimal sufficient, and not complete. If my statement about showing one minimal sufficient statistic being not complete is equivalent to showing the non-existence of complete sufficient statistic, then whatever you have constructed must have been flawed. – William Ambrose Dec 17 '20 at 13:06