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Suppose $X_1,X_2,...,X_n$ is a i.i.d random sample with probability mass function $p(x_i,\theta)$ where $x_i \in \{\theta,\theta+1,\theta+2,...\}$ and $\theta \in \mathbb{R}$. I claim that minimal sufficient statistic is $(X_{(1)},...,X_{(n)})$. Am I right?

Special case

If it is hard to find a minimal sufficient statistic for this problem, you can consider a special case where $p(x_i,\theta)=p(x_i)$ which means p does not depend on $\theta$.

General case

I also believe that my claim is true when $\theta \in A$ where $card(A) \ge 2$.

My Approach

If you try to use factorization theorem: $$P(\textbf{X}=\textbf{x}) = {\displaystyle \prod_{i=1}^{n} p(x_i,\theta) I_{ \{ \theta,\theta+1,...\} }(x_i) }$$where $I_{ \{ \theta,\theta+1,...\} }(x_i)$ is the indicator function. This is where we can't do anything because we can't factor the product of indicator functions, so I guess the minimal sufficient statistic is $(X_{(1)},...,X_{(n)}) $.

My Interpretation

I think ${\displaystyle \prod_{i=1}^{n} I_{ \{ \theta,\theta+1,...\} }(x_i) }$ means "in order to find out $x_i$ is a valid observation, we need to know $\theta$ "

My Question

Is $(X_{(1)},...,X_{(n)})$ the minimal sufficient statistic?

I also asked this question in math stackexchange but I got no answer.

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    $\begingroup$ What is a minimal sufficient statistic is going to depend on the form of $p(x_i,\theta)$. Since the sample is i.i.d., $(X_{(1)},...,X_{(n)})$ will be a sufficient statistic but may in some cases not be minimal. $\endgroup$
    – Henry
    Commented May 30 at 20:11
  • $\begingroup$ @Henry Try to find a counterexample. Assume that $p(x_i,\theta)=(\dfrac{1}{2})^{x_i-\theta+1}$. Now find the minimal sufficient statistic. $\endgroup$
    – ArshakParsa
    Commented May 30 at 20:41
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    $\begingroup$ Just out of curiosity, what is a "cursed support"? Do you actually mean "curved support"? $\endgroup$
    – Zhanxiong
    Commented May 30 at 20:59
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    $\begingroup$ In your example of $p(x_i,\theta)=\left(\frac{1}{2}\right)^{x_i-\theta+1} I_{[\theta\le x_i]}$, the minimal sufficient statistic is $X_{(1)}$. If instead $p(x_i, \theta)=\frac{1}{\theta}I_{[\theta\le x_i \lt 2\theta]}$ then the minimal sufficient statistic is $(X_{(1)},X_{(n)})$. In both cases $(X_{(1)},\ldots,X_{(n)})$ is sufficient but not minimal. $\endgroup$
    – Henry
    Commented May 30 at 22:38
  • $\begingroup$ @Zhanxiong: in Modern Online English, 'cursed' means roughly 'very unusual, in an unpleasant and scary way, as if affected by a curse'. I think it came from Reddit $\endgroup$
    – Thomas Lumley
    Commented May 30 at 22:47

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