Suppose $X_1,\cdots,X_n$ are $i.i.d$ from a distribution with p.d.f $$\delta_{(\theta,c)}(x)=\frac{1}{c}\mathbb{1}_{(x\in[\theta,\theta+c])},$$ where $\theta\in\mathbb{R}$ and $c\in\mathbb{R}^+$ unknown.

Find a minimal sufficient statistic for $(\theta,c)$.

From the range of $x_i$, i.e. $x_i\in[\theta,\theta+c]$, we can determine that $\theta\leq x_i$ and $\theta+c\geq x_i$, $\forall i\in\{1,\cdots,n\}$. This implies $$\theta\leq x_{(1)}\text{ and }\theta+c\geq x_{(n)},$$ where $x_{(1)}=\underset{i\in\{1,\cdots,n\}}{\min}x_i$ and $x_{(n)}=\underset{i\in\{1,\cdots,n\}}{\max}x_i$.

The plots shows the area of $\theta\leq x_{(1)}\text{ and }\theta+c\geq x_{(n)}.$ It look like that $(x_{(1)},x_{(n)})$ is a minimal sufficient statistic for $(\theta,c)$ because $(x_{(1)},x_{(n)})$ uniquely determines the shape of the log-likelihood function. Is this correct?


  • 1
    $\begingroup$ Sufficient statistics are for the entire parameter vector, not for components of the parameter vector. Cfr the definition of sufficiency. $\endgroup$
    – Xi'an
    Commented Jan 5, 2021 at 17:36
  • 2
    $\begingroup$ This shows the pair $(X_{(1)},X_{(n)})$ is sufficient. To show it is minimal, you have to demonstrate that two different realisations of $(X_{(1)},X_{(n)})$ lead to two different likelihood functions. $\endgroup$
    – Xi'an
    Commented Jan 5, 2021 at 20:56
  • $\begingroup$ @Xi'an. Thanks. This is a good reminder. $t(x)=t(y)\iff l_x(\theta)= l_y(\theta)+const$ here $l$ is log-likelihood and $\theta$ is the parameter. I've updated my question. I still have a little bit confusion on how to actually step-by-step show that $(X_{(1)},X_{(n)})$ is a minimal sufficient statistics. $\endgroup$
    – Tan
    Commented Jan 5, 2021 at 22:03

1 Answer 1


(Nobody answers so I post this answer which I am not sure is correct)

We show $(Y_{(1)},Y_{(n)})$ is a sufficient complete statistic, which implies $(Y_{(1)},Y_{(n)})$ is minimal sufficient. We first give the p.d.f of $(Y_{(1)},Y_{(n)})$ $$f(y_1, y_n,\theta,c)=\frac{n(n-1)}{c^n}(y_n-y_1)^{n-1},\quad\forall \theta\leq y_1\leq y_n\leq \theta+c\text{ and } (\theta,c)\in\mathbb{R}\times\mathbb{R}^+$$

Then, for any function $g(x_1,x_n)$ so that $\mathbb{E}_{\theta,c}\left[g(y_1,y_n)\right]=0,\forall (\theta,c)\in\mathbb{R}\times\mathbb{R}^+$, we have $$0=\frac{n(n-1)}{c^n}\int_{\theta}^{\theta+c}\int_{\theta}^{y_n}g(y_1,y_n)(y_n-y_1)^{n-2} dy_1dy_n,\forall (\theta,c)\in\mathbb{R}\times\mathbb{R}^+$$ The integral area of $(y_1, y_n)$ is a triangle with vertices $(\theta,\theta)$, $(\theta,\theta+c)$ and $(\theta+c,\theta+c)$. With varying $\theta\in\mathbb{R}$ $c\in\mathbb{R}^+$, these triangles generate the Beral $\sigma$-algebra of $\mathcal{B}=\{(x,z)\in\mathbb{R}^2:x\leq z\}$. Thus $$0=\frac{n(n-1)}{c^n}\int_{A}g(y_1,y_n)(y_n-y_1)^{n-2} d(y_1,y_n),\text{ for any Borel set }A\subset\mathcal{B}.$$ This means $$g(y_1,y_n)(y_n-y_1)^{n-2}\equiv0,a.e.\iff g\equiv 0,a.e.$$ Thus we conclude $(Y_{(1)},Y_{(n)})$ is a complete statistic for $(\theta,c).$

Since $(Y_{(1)},Y_{(n)})$ is sufficient by factorization theorem, we conclude $(Y_{(1)},Y_{(n)})$ is minimal sufficient.


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