Sufficient Statistic I have a question:

Does a sufficient statistic have to be one to one? For example, can $T(x) = x^2$ or $T(x) = |x|$ be sufficient statistics? I know that one to one functions of sufficient statistics are sufficient.

 A: No, it doesn't have to be one-to-one.  Consider a distribution family with pmf
$$f(x \mid \theta) =
\begin{cases}
0 & \text{with probability 0.5 if } \theta=0 \\
1 & \text{with probability 0.5 if } \theta=0 \\
2 & \text{with probability 0.5 if } \theta=1 \\
3 & \text{with probability 0.5 if } \theta=1 \\
\end{cases}$$
Then, the sufficient statistic for $x$ is $\left\lfloor\frac x2\right\rfloor$.
The way I think of sufficient statistics is that they tell you what you learned from $x$ to identify $\theta$.
When you don't use all of the information in a single observation $x$, that is, if there exists two outcomes $\omega_1, \omega_2$ such that $T(\omega_1) = T(\omega_2)$, then there is no way to alter the parameters $\theta$ of the distribution to change the relative probabilities $f(\omega_1 \mid \theta) / f(\omega_2 \mid \theta)$ of the two outcomes.
A: I like Neil G's example. I wondered if we could find some simple conditions that would prohibit a sufficient statistics to be one-to-one. Here is my take on this. 
(Following the comment by cardinal, in all that follows, suppose that $n\geq 2$.)
Suppose that we have the usual setup: the random variables $X_1,\dots,X_n$ are conditionaly i.i.d., given $\Theta=\theta$, with some density $f_{X_i\mid\Theta}(x_i\mid\theta)$.
As pointed out by cardinal, the identity map $\mathrm{id}:\mathbb{R}^n\to\mathbb{R}^n$ is always sufficient for the parameter $\Theta,$ and it is one-to-one. It is also easy to see that the order statistics $U:\mathbb{R}^n\to\mathbb{R}^n$ defined by $U(x_1,\dots,x_n)=(x_{(1)},\dots,x_{(n)})$ is also sufficient, but it is not one-to-one. The intuitive idea is that the value of both carry the same information about $\Theta$ as the original sample $(x_1,\dots,x_n).$
We generally want our sufficient statistic $T$ to reduce the dimension of the original sample, and this reduction will, in general, as we can see from many of the usual examples, imply that $T$ is not one-to-one.
Now, remember that $T$ is $\textit{minimal}$ sufficient if it is sufficient, and for $\textit{any}$ other sufficient statistic $S,$ if $S(x)=S(y)$, then $T(x)=T(y)$, which means that $T$ is a function of every other sufficient statistic $S$. This definition captures the idea that, if $T$ is minimal sufficient, then you can't find another sufficient statistic that provides more data reduction than $T$ does.
Proposition. If $T$ is minimal sufficient, then $T$ is not one-to-one.
Proof. Suppose that $T$ is minimal sufficient and one-to-one. Let $U$ be the order statistics as defined above. Take a sample point $x=(x_1,\dots,x_n)$, a permutation $\pi:\mathbb{R}^n\to\mathbb{R}^n$, and $y=(x_{\pi(1)},\dots,x_{\pi(n)}),$ such that $x\neq y$. It is clear that $U(x)=U(y)$ and, since $U$ is sufficient, we must have, by definition, that $T(x)=T(y)$, which contradicts the fact that $T$ is one-to-one.
