# How to prove any one-to-one function of minimal sufficient statistic is minimal sufficient?

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?

• You have not shown that S is also sufficient. Commented Oct 10, 2022 at 4:56
• 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? Commented Oct 10, 2022 at 6:28

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 (3rd edition) by Patrick Billingsley.

### Definitions

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 $$P_\theta[A\parallel\mathscr{G}]$$ 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 $$P_\theta[A\parallel\mathscr{G}]$$. 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 $$\Theta$$.

### Proof

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 $$f$$ is a (measurable, technically this assumption needs to be added) real-valued function, $$f(T)$$ is $$\sigma(T)$$-measurable, whence $$\sigma(f(T)) \subset \sigma(T)$$ -- note that this relation holds for any general measurable function $$f$$, the one-to-one condition of $$f$$ is not needed. Conversely, that $$f$$ is one-to-one enables us to write $$T = f^{-1}(f(T))$$, then the same argument as before immediately yields $$\sigma(T) \subset \sigma(f(T))$$ (if $$f$$ is measurable and invertible, then its inverse $$f^{-1}$$ must also be measurable). In conclusion, $$\sigma(T) = \sigma(f(T))$$.

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}

$$\square$$

Also,

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. $$111$$).

## References:

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

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

• +1 very neat, straight to the point! Commented Jul 4, 2023 at 13:40

So $$S$$ is a function of $$T'$$, therefore $$S$$ is a minimal sufficient.

It is not true that every function of a sufficient statistic is a sufficient statistic. For example, suppose $$X_1,\ldots,X_n\sim\text{i.i.d.}\operatorname N(\mu,\sigma^2).$$ The pair $$(X_1+\cdots+X_n,\, X_1^2+\cdots+X_n^2)$$ is a sufficient statistic for this family of distributions. The sum $$X_1+\cdots+X_n$$ is a function of the pair $$(X_1+\cdots+X_n,\, X_1^2 + \cdots + X_n^2),$$ but that sum is not a sufficient statistic.

To say that $$T$$ is a sufficient statistic for some family of distributions of $$X$$ means that the conditional probability distribution of $$X$$ given $$T$$ is the same for all distributions in that family. For example, the conditional distribution of $$(X_1,\ldots,X_n)$$ given $$(X_1+\cdots + X_n,\, X_1^2+\cdots + X_n^2)$$ is the same for all values of $$(\mu,\sigma^2).$$

The function whose input is $$(X_1+\cdots+X_n,\, X_1^2 + \cdots + X_n^2)$$ and whose output is $$X_1+\cdots+X_n$$ is not one-to-one.

Suppose $$S=f(T)$$ and $$f$$ is one-to-one. Then the event $$S=s$$ is the same as the event $$T=f^{-1}(s).$$ The conditional distribution of $$X$$ given $$S=s$$ is the conditional distribution of $$X$$ given $$T = f^{-1}(X),$$ and that is the same for all distributions in the family. Therefore $$S$$ is a sufficient statistic for that family.