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<x_1,\cdots,x_n<0} \\&=\frac{1}{(\theta)^n}\mathbf1_{(x_{(1)}\ge-\theta)}\mathbf1_{(x_{(n)}\le0)} \end{align}

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 $ ?


1 Answer 1


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).

  • $\begingroup$ 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. $\endgroup$
    – Elisa
    Commented Feb 9, 2023 at 1:38
  • 1
    $\begingroup$ @Elisa: notice the introduction "As a function of $\theta$". The $X_i$'s are fixed and therefore negative. $\endgroup$
    – Xi'an
    Commented Feb 9, 2023 at 12:08

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