Let $X \sim P_{\theta} \in \mathcal{P} = \{P_{\theta}: \theta \in \Theta,\; P_{\theta} \; \text{symmetric around} \; 0\}$, and absolutely continuous w.r.t. the Lebesgue measure. If $T(X)=|X|$, then $T$ is sufficient?
Let i.i.d. $X_1,...,X_n \sim \text{Uniform}[a,b]$ and $-\infty<a<b<\infty$. If we denote the corresponding order statistics by $X_{(1)},...,X_{(n)}$ s.t. $X_{(1)} \le ... \le X_{(n)}$, then $T:\mathbb{R}^n \to \mathbb{R}^2$ with $T(X_1,...,X_n) = (X_{(1)},X_{(n)})$ is sufficient?
If $a$ is known in Q2, then $T:\mathbb{R}^n \to \mathbb{R}$ with $T(X_1,...,X_n) = X_{(n)}$ is sufficient?
I want to learn how to show them sufficient or not.
In the first case note that $f_{\theta}$, the density of $P_{\theta}$ needs to be an even function. So $f_{\theta}(-x)=f_{\theta}(x)=f_{\theta}(|x|)$.
Also given the order statistics $(X_{(1)},X_{(2)},\cdots,X_{(n)})$, $(X_1,X_2,\cdots,X_n)$ can take only $n!$ distinct values. Thus $f_{\theta}(x_1,x_2,\cdots,x_n)=n!\times f_{\theta}^{*}(x_{(1)},x_{(2)},\cdots,x_{(n)})$, where $f_{\theta}^{*}$ is the pdf of the order statistics.