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In parametric statistical inference, a statistic $T$ of some variable $X$ (thought of as experimental or observational data) is sufficient for the parameter $\theta$ if is captures the essential information in the data $X$ about the parameter $\theta$ (there are several equivalent definitions for this). It is implicitly understood that the random variable $X$ has a probability distribution $f(x;\theta)$, where the parameter $\theta$ is of course unknown.

Related is the concept of a minimal sufficient statistic, which captures nothing more than the essential.

  1. Is it true that any random variable $X$ as above has a sufficient statistic?
  2. Suppose $X$ has a sufficient statistic $T$. Must it also have a minimal sufficient statistic?

I'd be glad to have either a concise proof or a simple counter-example for each of these two.

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    $\begingroup$ en.wikipedia.org/wiki/Sufficient_statistic. Wikipedia article on sufficient estimators: A case in which there is no minimal sufficient statistic was shown by Bahadur, 1954. However, under mild conditions, a minimal sufficient statistic does always exist. In particular, in Euclidean space, these conditions always hold if the random variables (associated with $P_{\theta }$) are all discrete or are all continuous. $\endgroup$ – Marc Nov 23 '16 at 15:13
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  1. In a degenerate manner your sample $(X_1,...,X_n)$ is always a sufficient statistic from Fisher's information perspective. $T(X)$ is sufficient statistic for $\theta$ iff $\mathcal{I}_{\theta}(T(X))=\mathcal{I}_{\theta}(X_1,...,X_n)$.Thus for every parametric structure you can take the data points themselves as the sufficient statistic.
  2. No. You have a minimial sufficient statistic iff your parametric distribution $f(X;\theta)$ can be factorized into $h(X)g(\theta; T(X))$, then the $T(X)$ is the minimal sufficient statistic or can be reduced to one. Moreover, it is pretty common to fail to satisfy this condition, e.g., Weibull distribution which density function is given by $$ f(x;\theta)=\frac{\theta x^{\theta-1}}{\lambda^\theta}\exp\{-x^\theta/\lambda^\theta\}, x\ge 0, $$
    where $\theta$ is unknown shape parameter, cannot be factorized w.r.t $\theta$.
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  • $\begingroup$ Thanks @a-e. Is your answer to (2) not in contradiction to Marc's quote from Wikipedia, "under mild conditions, a minimal sufficient statistic does always exist"? $\endgroup$ – Shlomi A Nov 25 '16 at 21:23
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    $\begingroup$ Don't know. I guess "mild" it's a relativistic statement. From my humble experience, although the existence of a minimal sufficient statistic is more common in parametric structures, yet a non-existence is not that rare phenomenon. $\endgroup$ – V. Vancak Nov 25 '16 at 21:30
  • $\begingroup$ Alright. Thanks @a-e! As for (1), if I'm interested in non-trivial sufficient statistic, then are there necessary and/or sufficient conditions for their existence? $\endgroup$ – Shlomi A Nov 26 '16 at 11:32
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    $\begingroup$ If by non-trivial you mean not $(X_1,...,X_n)$ then yes - the factorization theorem is iff statement. I.e., inter alia it provides a sufficient condition for the existence of a non-trivial sufficient statistic. $\endgroup$ – V. Vancak Nov 26 '16 at 11:38
  • $\begingroup$ @V.Vancak Why the sample itself is a statistic? I thought a statistic must be a real-valued function of the sample. $\endgroup$ – Tony B Nov 16 '19 at 18:59

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