# Sufficient Statistic of Uniform $(-\theta,0)$

Let $$X_1, ... , X_n$$ be i.i.d random variables Uniform $$(-\theta,0)$$ , with $$\theta > 0$$ parameter

\begin{align}f_{\theta}(x_1,x_2,\cdots,x_n)&=\prod_{i=1}^nf(x_i;\theta) \\&=\frac{1}{(\theta)^n}\mathbf1_{-\theta

So can we conclude that the sufficient statistic for $$-\theta$$ so for $$\theta$$ too is $$X_{(1)}$$.

Also that the sufficient statistic for $$\theta$$ is $$-X_{(1)}$$. ??

Thus, both $$X_{(1)}$$ and $$-X_{(1)}$$ are sufficient statistics for $$\theta$$ ?

As a function of $$\theta$$ $$\frac{1}{\theta^n}\mathbf1_{x_{(1)}\ge-\theta}\mathbf1_{x_{(n)}\le0}= \frac{1}{\theta^n}\mathbf1_{x_{(1)}\ge-\theta}$$ Hence the likelihood function only depends on $$X_{(1)}$$, which makes it a sufficient statistic for $$\theta$$, $$-\theta$$, $$\sin(3\theta)$$ and any other function of $$\theta$$ (as sufficiency is not to be confused with unbiasedness or any other estimation property).
• I don't understand that why we have $$\frac{1}{\theta^n}\mathbf1_{x_{(1)}\ge-\theta}\mathbf1_{x_{(n)}\le0}= \frac{1}{\theta^n}\mathbf1_{x_{(1)}\ge-\theta} .$$ We have no restriction that $X_{i}(1 \le i \le n)$ are non-positive i.i.d random variables. Commented Feb 9, 2023 at 1:38
• @Elisa: notice the introduction "As a function of $\theta$". The $X_i$'s are fixed and therefore negative. Commented Feb 9, 2023 at 12:08