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We know that sufficient statistic is not necessarily minimal. For example, consider a random sample $X_1,\cdots,X_n \sim \text{Bernoulli}(\theta) $. It can be shown easily that both $T'(X_1,\cdots,X_n) = (X_1,\cdots,X_n)$ and $T(X_1,\cdots,X_n) = \sum_{i=1}^n X_i$ are sufficient statistic. However, $T'$ is not minimal. This is not surprising because after all, $T'$ has "more dimensions" than $T$ , and thus achieve less data reduction. More specifically, range of $T'$ is in $\mathbb{R}^n$ and range of $T$ is in $\mathbb{R}$.

In general, we consider a random sample $X_1,\cdots,X_n \sim f_X (x\mid\theta)​$. Does there exist a sufficient statistic $S$ for $\theta$ such that it is sufficient, its range is in $\mathbb{R}$, but it is not minimal?

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  • $\begingroup$ Just to be clear, are you stipulating that $T(X_1,...,X_n) \in \mathbb{R}$, and not something like $\mathbb{R}^n$? You mention that $T$ has range $\mathbb{R}$ but then go on to rule out a case in which $T$ is not in $\mathbb{R}$ (the trivial case), so wasn't sure. $\endgroup$ – Taimur May 6 '17 at 12:23
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    $\begingroup$ @Taimur What I meant to say was this. I know that $T = \text{id}$ is a sufficient statistic. In many cases it will not be a MSS, and a MSS will have lower dimensions. So this is an "easy example". But I am interested in an example where $T$ has dimension one, is sufficient, but not minimal. $\endgroup$ – 3x89g2 May 6 '17 at 14:45
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Consider a Normal sample $(x_1,\ldots,x_n)$ from ${\cal N}(0,\sigma^2)$. Then it is a standard result that the statistic$$S(x_1,\ldots,x_n)=\sum_{i=1}^n x_i^2$$is minimal sufficient. If I now define the statistic $$Z(x_1,\ldots,x_n)=\text{sign}(x_1)\sum_{i=1}^n x_i^2$$where$$\text{sign}(x_1)=\begin{cases} 1&\text{if }x_1\ge 0\\ -1&\text{if }x_1< 0\\\end{cases}$$ then $Z$ is also sufficient but not minimal sufficient (since the signs of the observations are ancillary).

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Suppose $T$ and $T'$ are both sufficient statistics of dimension 1 in $\mathbb{R}$, and suppose without loss of generality, that $T$ is minimal sufficient but $T'$ is not.

Then there exists a function $g$ such that $T = g(T')$ but there is no function $f$ such that $T' = f(T)$ (otherwise $T'$ would be minimal sufficient). It follows that $g$ isn't invertible, so $g(x)$ is not unique for some $x \in \mathbb{R}$. Suppose now that $T'(X_1,...,X_n) = x$. Then $T = g(T')$ takes more than one value, and so (after a bit of hand-waving) $T$ cannot be a sufficient statistic. This is a contradiction, and so both $T$ and $T'$ are minimal.

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  • $\begingroup$ I don't see why the first sentence of the second paragraph is true? Should it be the other way around? Namely, since $T$ is minimal, $T'$ is a function of $T$. Since $T'$ is not minimal, it is not a function of $T$? $\endgroup$ – 3x89g2 May 7 '17 at 3:11
  • $\begingroup$ @Misakov you're right - changed it to the other way round. The rest of the proof works pretty much the same I think. $\endgroup$ – Taimur May 7 '17 at 10:02
  • $\begingroup$ What does "$g(x)$ is not unique for some $x$" mean? Are you saying that for a given $x$, $g(x)$ can take on different values? $\endgroup$ – 3x89g2 May 7 '17 at 10:25
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    $\begingroup$ I am afraid this proof is meaningless and the result is wrong. I suggest you remove the answer. $\endgroup$ – Xi'an Apr 18 '18 at 10:44
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    $\begingroup$ "Then T=g(T′) takes more than one value, and so (after a bit of hand-waving) T cannot be a sufficient statistic." this part is unclear $\endgroup$ – Sextus Empiricus Apr 18 '18 at 11:37

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