2
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ 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, 2017 at 12:23
  • 1
    $\begingroup$ @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, 2017 at 14:45

2 Answers 2

3
$\begingroup$

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).

$\endgroup$
-1
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ 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, 2017 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, 2017 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, 2017 at 10:25
  • 1
    $\begingroup$ I am afraid this proof is meaningless and the result is wrong. I suggest you remove the answer. $\endgroup$
    – Xi'an
    Apr 18, 2018 at 10:44
  • 1
    $\begingroup$ "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 $\endgroup$ Apr 18, 2018 at 11:37

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.