# Multiple sufficient statistics and the factorization theorem

Suppose you are using the factorization theorem to find a sufficient statistic. Let us say that we have a negative sign in front of $T(x)$. How do you know whether or not to "absorb" a negative sign into a sufficient statistic or any other sign? So instead of $T(x)$ we could have $$\widetilde T(x):=−T(x) \>.$$

For example, suppose you have $$g(T(x)| \theta) = e^{-T(x)}$$ in the factorization theorem. How do you know whether or not to include the negative sign in $T(x)$?

• They are both sufficient and we can say the same thing in a more general context: If $T$ is a sufficient statistic for some parameter $\theta$ than any almost-surely invertible transformation $U = h(T)$ is also sufficient for $\theta$. Commented Apr 8, 2012 at 20:39
• @cardinal: So if $\bar{X}$ is sufficient for $\mu$ then how can $-\bar{X}$ be sufficient for $\mu$? If $\mu$ is known to be positive then how can $-\bar{X}$ be sufficient for $\mu$?
– ross
Commented Apr 9, 2012 at 4:18
• Hi ross, you should be able to click the "Add comment" link to respond to the comments you see. It might help to think about two things: (1) The statement of the Factorization Theorem itself and (2) The definition of sufficiency vis a vis conditioning on a sufficient statistic. I can post an answer with some more details, if that would help. Commented Apr 9, 2012 at 14:24

Let $$\mathcal{P}=\{B_k\}_{k \in K}$$ be a partition of the sample space for some index set $$K$$. We say that the partition $$\mathcal{P}$$ is sufficient for $$\theta$$ if the density $$f(x | X \in B_k)$$ does not depend on the parameter $$\theta$$ for any $$k \in K$$. A nice way to define a partition is through level sets of a statistic $$T(x)$$. Since the level sets $$\{\{X : T(X) = k\}\}_{k \in K} = \{\{X : -T(X) = k\}\}_{k \in -K}$$ are the same, this shows that $$-T$$ is sufficient if and only if $$T$$ is sufficient. With this perspective, we do not distinguish between $$T$$ and $$-T$$, or any other invertible function of $$T$$.