So I want to prove that any one-to-one function of minimal sufficient statistic is also minimal sufficient. Here is my proof:

Let $T$ be a minimal sufficient statistic and $f$ is a one-to-one function such that $S=f(T)$. Since $T$ is minimal sufficient, then for any other sufficient statistic $T'$, $T=h(T')$ for some function $h$.

$\Rightarrow S=f(T)=f(h(T'))=g(T')$, where $g=f \circ h$. So $S$ is a function of $T'$, therefore $S$ is a minimal sufficient.

I'm not sure if this proof is correct because I'm not using the fact that $f$ is a one-to-one function. Can anyone give me some explanation or figure out which part is wrong?

  • 1
    $\begingroup$ You have not shown that S is also sufficient. $\endgroup$
    – frank
    Oct 10, 2022 at 4:56
  • $\begingroup$ Oops...I think I need to use $f$ is one-to-one to show $S$ is sufficient? Since $T=f^{-1}(S)$ is sufficient, by factorization theorem $F(x|\theta)=k_1(T|\theta)k_2(x)=k_1(f^{-1}(S)|\theta)k_2(x)=L(S|\theta)k_2(x)$, where $L=k_1 \circ f^{-1}$, so $S$ is also sufficient. Is this correct? $\endgroup$
    – Alex He
    Oct 10, 2022 at 6:28

2 Answers 2


By the measure-theoretic definition of sufficiency and minimal sufficiency, this statement is quite obvious. While the proof itself is just a one-liner, to comprehend underlying definitions needs a solid understanding of measure-theoretic conditional probability. The definitions below are quoted from the Sufficient Subfields subsection in Section 34 of Probability and Measure (2nd edition) by Patrick Billingsley.


Suppose that for each $\theta$ in an index set $\Theta$, $P_\theta$ is a probability measure on $(\Omega, \mathscr{F}$). A $\sigma$-field $\mathscr{G}$ in $\mathscr{F}$ is sufficient for the family $[P_\theta: \theta \in \Theta]$ if versions $P_\theta[A\|\mathscr{G}]$ can be chosen that are independent of $\theta$ -- that is, if there exists a function $p(A, \omega)$ of $A \in \mathscr{F}$ and $\omega \in \Omega$ such that, for each $A \in \mathscr{F}$, $p(A, \cdot)$ is a version of $P_\theta[A\|\mathscr{G}]$. A sufficient statistic is a random variable or random vector $T$ such that $\sigma(T)$ is a sufficient subfield.

A sub-$\sigma$-field $\mathscr{G}_0$ sufficient with respect to $[P_\theta: \theta \in \Theta]$ is minimal if, for each $\mathscr{G}$, $\mathscr{G}_0$ is essentially contained in $\mathscr{G}$ in the sense that for each $A \in \mathscr{G}_0$ there is a $B$ in $\mathscr{G}$ such that $P_\theta(A\triangle B) = 0$ for all $\theta$ in $\Theta$.


In view of the above two definitions, if one can show that $\sigma(T) = \sigma(f(T))$, then the proof is complete. To this end, recall that for a random variable $X$, by definition $\sigma(X)$ is the smallest $\sigma$-field with respect to which it is measurable. Now since $f(T)$ is a (measurable, technically this assumption needs to be added) function of $T$, it is $\sigma(T)$-measurable, whence $\sigma(f(T)) \subset \sigma(T)$ -- note that this relation holds for general measurable function $f$, one-to-one is not needed. Conversely, writing $T = f^{-1}(f(T))$ (here we used the one-to-one condition), then the similar argument easily yields $\sigma(T) \subset \sigma(f(T))$. In conclusion, $\sigma(T) = \sigma(f(T))$.

  • $\begingroup$ I have always preferred the treatment with sufficient sigma fields to be more intuitive to via kernels although they are both explaining the same thing. +1. $\endgroup$ Jul 4, 2023 at 13:23

It follows from the more general result:

Observation $1.$ Any one-to-one function of a sufficient statistic is a sufficient statistic. (cf. $\rm [I], $ section $6.2, $ p. $280$).

Let $T$ be a sufficient statistic and $f$ be a one-to-one function. Let $S(\mathbf x):=f(T(\mathbf x)). $ Now $f$ is onto its range, so $f^{\leftarrow}$ exists. Therefore by Neyman Factorization Theorem, for a certain $g, ~h, $

$$\begin{align}p(\mathbf x|\boldsymbol\theta) &=g(T(\mathbf x) |\boldsymbol\theta) h(\mathbf x) \\ &=g(f^{\leftarrow} (S(\mathbf x))|\boldsymbol\theta) h(\mathbf x)\\ &= k(S(\mathbf x))|\boldsymbol\theta) h(\mathbf x).\tag 1\end{align}$$



Observation $2.$ If $S$ and $T$ are both minimal sufficient, then $T = w(S) $ and $S = l(T).$ $w,~ l$ are inverse functions which are one-to-one and onto.

This is evident but it does tell that $S, ~T$ are equivalent. (cf. $[\rm II], $ section $4.2, $ p. $111$).


$\rm [I]$ Statistical Inference, George Casella, Roger L. Berger, Wadsworth, $2002.$

$\rm [II]$ Statistical Theory and Inference, David J. Olive, Springer International, $2014.$

  • 1
    $\begingroup$ +1 very neat, straight to the point! $\endgroup$
    – utobi
    Jul 4, 2023 at 13:40

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