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I'm trying to prove that for a random sample $X_1,\ldots,X_n$ that depends on $\theta$, if $T$ is a sufficient statistic for $\theta$, then so is $T'=f\circ T$, for any injective function $f$.

My attempt:

Since $f$ is injective, if we restrict the target space of $f$ to $\operatorname{range}(f)$, then we can consider $f$ as bijective. Now $T'\in \operatorname{range}(f)$ since $T'=f\circ T$, so $T=f^{-1}\circ T'$. Define $\phi'(u(\vec{x}),\theta)=\phi(f^{-1}\circ u(\vec{x}),\theta)$. Since $T$ is sufficient, we have that $f(\vec{x};\theta)=\phi(T(\vec{x};\theta))h(\vec{x})=\phi(f^{-1}\circ T'(\vec{x};\theta))h(\vec{x})=\phi'(T'(\vec{x};\theta))h(\vec{x})$. Hence, $T'$ is also sufficient.

Does this look alright?

Michael Hardy
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user153582
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1 Answers1

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I believe your argument works, although it's a bit more on the notation-intensive side than what I am comfortable with.

But here's how I would do it:

It is enough to prove that the conditional probability distribution of $X_1,\ldots,X_n$ given $f(T(X_1,\ldots,X_n))$ does not depend on $\theta$. To do that, it suffices to show that the conditional probability distribution of $X_1,\ldots,X_n$ given $f(T(X_1,\ldots,X_n))$ is the same as the conditional probability distribution of $X_1,\ldots,X_n$ given $T(X_1,\ldots,X_n)$. And that follows from the fact that the event that $f(T(X_1,\ldots,X_n))=\text{some particular number or vector or whatever}$ is the same as the event that $T(X_1,\ldots,X_n)=\text{some particular number}\ldots\text{(the inverse-image)}\ {}$. Or, if you want to talk about conditioning on a sigma-algebra, etc., then you can say that the event that $f(T(X_1,\ldots,X_n))\in\text{some particular set}$ is the same as the event that $T(X_1,\ldots,X_n)\in\text{some particular set}$.

(Rather than "a random sample $X_1,\ldots,X_n$ that depends on $\theta$", it would be better to refer to "a random sample $X_1,\ldots,X_n$ whose probability distribution depends on $\theta$".)

Michael Hardy
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