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I came cross an interesting comment saying

If the dimension of a sufficient statistic $T(X)$ is the same as that of the parameter space, then $T(X)$ is minimal sufficient.

Is this is true? I examined some examples e.g. full-rank exponential families, curved exponential families e.g. $N(\theta, \theta^2)$, and some uniform distributions, e.g. $U(-\theta,\theta)$, $U(0,\theta)$, $U(\theta,\theta+a)$ and found their sufficient statistic has the same dimension with the parameter space and indeed they are minimal sufficient.

Is this quoted statement correct? Can you please provide a reference of a proof, or a counter example?

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    $\begingroup$ Hint: for a Normal$(\mu,\sigma^2)$ family, let $\bar x$ and $s^2$ be the sample mean and variance, respectively, and let $x_1$ be the first observation in sample. It's well-known that $(\bar x, \sqrt{s^2})$ is minimal sufficient. Define $T=(\bar x, \operatorname{sgn}(x_1)\sqrt{s^2}).$ (1) What is the dimension of $T$? (2) Is $T$ minimal sufficient? $\endgroup$
    – whuber
    Jan 7, 2021 at 19:08
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    $\begingroup$ @whuber (1) 2; (2) No. $T$ is a two-dimensional sufficient statistic for the two-dimensional parameter $(\mu,\sigma^2)$. But $T$ is not minimal sufficient statistic as $(\bar{x},\text{sgn}(x_1)\sqrt{s^2})$ and $(\bar{x},\sqrt{s^2})$ is not one-to-one corresponded. $\endgroup$
    – Tan
    Jan 7, 2021 at 19:19
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    $\begingroup$ @StubbornAtom for $N(\theta,\theta^2)$ sufficient statistic $(\sum X_i, \sum X_i^2)$ is two dimensional and $(\theta,\theta^2)$ lives in $\mathbb{R}^2$, and it happens that $(\sum X_i, \sum X_i^2)$ is also minimal sufficient statistic. So it satisfied the quoted statement. For $U(\theta,\theta+1)$, sufficient statistic is two-dim $(X_{(1)},X_{(n)})$ but $\theta$ is one-dim. Thus $U(\theta,\theta+1)$ is not a suitable example for the quoted statement in OP. $\endgroup$
    – Tan
    Jan 7, 2021 at 20:05
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    $\begingroup$ @Tan: In the $N(θ,θ^2)$ model, the parameter is one-dimensional even though this is a two-dimension (curved) exponential family. $\endgroup$
    – Xi'an
    Jan 8, 2021 at 8:33
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    $\begingroup$ @Tan what is the parameter space of $N(\theta,\theta^2)$? Don't you have a single parameter $\theta \subset \mathbb{R}^1$? $\endgroup$ Feb 3, 2021 at 14:42

2 Answers 2

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The construction by Whuber in the comments gives an example to create a sufficient statistic that is not minimal sufficient.

The trick of that counterexample is to design a sufficient statistic that has a larger space, but still the same dimension. For a normal distribution with fixed $\mu$ and parameterized by only $\sigma$, the parameter is the set of positive numbers $\sigma \subset \mathbb{R_{>0}}$ and the minimal sufficient statistic has the same space $s \subset \mathbb{R_{>0}}$. This leaves room for the creation of a statistic $s^\prime \subset \mathbb{R}$ by adding the negative numbers (in the example by Whuber this is done my multiplying with the sign of a variable $x_1$), which has a non-injective surjective relationship with the minimal sufficient statistic $s$ such that the inverse function does not exist.

d

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  • $\begingroup$ I need to improve this answer. After one year re-reading it I had troubles understanding it. I have some ideas to do these improvements but am not gonna do it right away. If anybody reads this comment after a month (Feb 2022 or further) then I have forgotten it and please remind me to edit the question. $\endgroup$ Jan 12, 2022 at 17:58
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The quoted statement is incorrect. The post Minimal sufficient statistic whose dimension is less than dimension of parameter has one example, @whuber another one in a comment.

This: Minimum dimension of sufficient statistics is also relevant.

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  • $\begingroup$ The first link's sufficient statistic's dimension does not equal to the parameter space dimension. Thus it is not a counterexample of the statement. $\endgroup$
    – Tan
    Jan 29, 2021 at 17:43
  • $\begingroup$ @Tan: Did you look at the answer in that post? It is an example. $\endgroup$ Jan 29, 2021 at 17:46
  • $\begingroup$ I think so.. It says when sample size is one then sufficient statistic $X$ is a minimal sufficient statistic for $\text{Beta}(\alpha,\beta)$. So $X$ is one dimensional and $(\alpha,\beta)$ is two dimensional. My question was asking when sufficient statistic dimension is same (not less) with parameter space dimension, would the sufficient be minimal? $\endgroup$
    – Tan
    Jan 29, 2021 at 22:50

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