2

Suppose $$\prod_{i=1}^n[(N-x_i)!]=\prod_{i=1}^n[(N-y_i)!]$$ for all $N\in\mathbb{Z}^+$.

Why does it imply $x_{(i)}=y_{(i)}$, for all $i\in\{1,\cdots,n\}$? Here $X_{(i)}$ is the order statistic.

I am trying to find a minimal sufficient statistic for $\{X_i\}_{i=1}^n\sim\text{Binomial}(N,\frac{1}{2})$. The last step is showing $\prod_{i=1}^n[(N-x_i)!]=\prod_{i=1}^n[(N-y_i)!] \implies x_{(i)}=y_{(i)}$.

mxdxzxyjzx
  • 296

1 Answers1

1

\begin{equation} \prod_{i=1}^n[(N-x_i)!]=\prod_{i=1}^n[(N-y_i)!]\iff\prod_{i=1}^n[(N-x_{(i)})!]=\prod_{i=1}^n[(N-y_{(i)})!] \end{equation} $(N-x_{(i)})!$ is a polynomial of $N$ to the degree of $N-x_{(i)}$.

$\prod_{i=1}^n[(N-x_{(i)})!]$ is a polynomial of $N$ to the degree of $nN-\sum_{i=1}^nx_{(i)}$. Its powers include $nN-\sum_{i=1}^{n} x_{(i)}$, $(n-1)N-\sum_{i=1}^{n-1} x_{(i)}$, $(n-2)N-\sum_{i=1}^{n-2} x_{(i)},\cdots$.

This means $$\sum_{i=1}^m x_{(i)}=\sum_{i=1}^m y_{(i)}, \forall m\in\{1,\cdots,n\}\iff x_{(i)}=y_{(i)},\forall i\in\{1,\cdots,n\}$$

--- Update:

$(N-x_{(i)})!$ is like $N^{N-x_{(i)}}+c_{i,1}N^{N-x_{(i)}-1}+c_{i,2}N^{N-x_{(i)}-2}+\cdots$

$\prod_{i=1}^n[(N-x_{(i)})!]$ is like $(N^{N-x_{(n)}}+c_{n,1}N^{N-x_{(n)}-1}+c_{n,2}N^{N-x_{(n)}-2}+\cdots)\times(N^{N-x_{(n-1)}}+c_{n-1,1}N^{N-x_{(n-1)}-1}+c_{n-1,2}N^{N-x_{(n-1)}-2}+\cdots)\times(N^{N-x_{(n-2)}}+c_{n-2,1}N^{N-x_{(n-2)}-1}+c_{n-2,2}N^{N-x_{(n-2)}-2}+\cdots)\underset{\text{n terms product}}{\cdots}$

Thus the largest degree term of $\prod_{i=1}^n[(N-x_{(i)})!]$ is $N^{nN+\sum_{i=1}^nx_{(i)}}$. Also it has terms like $N^{(n-1)N+\sum_{i=1}^{n-1}x_{(i)}}$, $N^{(n-2)N+\sum_{i=1}^{n-2}x_{(i)}}$, $\cdots$

If $\prod_{i=1}^n[(N-x_{(i)})!]=\prod_{i=1}^n[(N-y_{(i)})!],\forall N\in\{1,2,\dots\}$, their terms must match. Thus this implies $$\sum_{i=1}^mx_{(i)}=\sum_{i=1}^my_{(i)},\forall m\in\{1,2,\dots\,n\}$$

Tan
  • 621
  • How did you get the summation to $m$? I believe that you just have the summation to $n$. Note that we could have $x_i$ be a permutation of $y_i$ (unless they are ordered), so I'm surprised you could make such a strong conclusion. – Calvin Lin Jan 04 '21 at 00:54
  • Also, note that OP defines $x_{(i)} $ to be the order statistic, as opposed to being $x_i$. – Calvin Lin Jan 04 '21 at 00:55
  • @CalvinLin I think the powers of the polynomial of $N$ is associated with the $\sum_i^m$ terms, right? – mxdxzxyjzx Jan 04 '21 at 01:52
  • @CalvinLin updated. Does it answer your first question? I do not quite understand your second question. – Tan Jan 04 '21 at 02:25
  • No to first. The polynomial has degree $ \sum_{i=1}^n (N-x_i) $, and you're comparing 2 polynomials of the same degree, so that only gives you one equation. Alternatively, can you explain how you concluded that $x_1 = y_1$? Perhaps it came from me not understanding the phrase "Its powers include ... ". – Calvin Lin Jan 04 '21 at 02:28
  • @CalvinLin Updated. How about now? – Tan Jan 04 '21 at 02:47
  • I did not conclude $x_1=y_1$. I concluded $x_{(1)}=y_{(1)}$. – Tan Jan 04 '21 at 03:10