# prove that maximum and minimum is sufficient statistic for uniform($\theta$ , $\theta$+1)

I'm trying to understand the proof that $$T(X_1,..., X_n)=(max\{X_1,..., X_n\}, min\{X_1,..., X_n\})$$ , is a sufficient statistic for $$\theta$$, given that $$f_{\theta}=Uniform(\theta, \theta+1)$$

• from Elements of Information Theory by Cover & Thomas, page 37.

I know that the joint pdf is: $$f_{\theta}(X_1=x_1,...,X_n=x_n, max X_i = X(n),min X_i = X(1))=\prod_{i=1}^{n}f_\theta(X_i)=\prod_{i=1}^{n}[1 \boldsymbol{1}({\theta.

However, I do not sure that $$f_{\theta}(max X_i = X(n), min X_i=X(1))=n (n-1)\boldsymbol{1}({X(n)-1<\theta and thus $$f_\theta(X_1=x_1,..., X_n=x_n|max X_i = X(n), min X_i=X(1))=\frac{1}{n(n-1)}$$

in summary, I need some help to prove that this $$T(X)$$ is a sufficient statistic.

Suposse we are given a random sample $$X=(X_1, \ldots, X_n)$$ whose distribution is $$f_\theta = Uniform(\theta, \theta + 1)$$. Therefore $$f_\theta (x) = \frac{1}{\theta + 1 - \theta}\boldsymbol 1_{(\theta, \theta + 1)}$$ As the sample is random $$f_\theta (X_1,\ldots,X_n) = \prod_{i=1}^nf_\theta(X_i) = \left(\frac{1}{\theta + 1 - \theta} \right)^n \prod_{i=1}^n \boldsymbol 1_{(\theta, \theta + 1)}(X_i)$$ $$= \boldsymbol 1_{(\theta, \theta + 1)}\left(\min_{1\leq i \leq n} X_i\right)\boldsymbol 1_{(\theta, \theta + 1)}\left(\max_{1\leq i \leq n} X_i\right)$$ The Fisher-Neyman theorem for sufficient statistics states that

An statistic $$T$$ is sufficient if and only if $$f(X;\theta) = h(X)g(T(X);\theta)$$ for two nonnegative functions $$h, g$$.

Considering the statistic $$T(X)= (\max\{X_1,\ldots,X_n\}, \min\{X_1,\ldots,X_n\}) = (T_1(X), T_2(X))$$, as well as the fuctions

• $$h(X) = 1$$
• $$g(T(X);\theta) = \boldsymbol 1_{(\theta, \theta + 1)}\left(T_1(X)\right)\boldsymbol 1_{(\theta, \theta + 1)}\left(T_2(X)\right)$$

The theorem ensures as that $$T(X)$$ is indeed a sufficient statistic :)

• thank you, however I think $$f_\theta (x) = \frac{1}{\theta + 1 - \theta}\boldsymbol 1_{(\theta, \theta + 1)}=\boldsymbol 1_{(\theta, \theta + 1)}$$ and thus $\theta$ does not appeared any more. Aug 4, 2023 at 15:30
• You are totally right, I missed that sorry. Now it is fixed :) Aug 4, 2023 at 15:31
• thank you, in the cover book, he said that "one can show that the distribution of the data is independent of the parameter given the statistic T", how can I see this observation? Aug 12, 2023 at 6:32