# minimal sufficient statistic for $U(\theta, \theta+c)$. $(\theta,c)$ unknown

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?

• Sufficient statistics are for the entire parameter vector, not for components of the parameter vector. Cfr the definition of sufficiency. Jan 5, 2021 at 17:36
• 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. Jan 5, 2021 at 20:56
• @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.
– Tan
Jan 5, 2021 at 22:03

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.