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Using Basu's theorem, prove that $\sum\limits_{i = 1 }^n {(X_i - X_{(1)}) }$ and $X_{(1)}$ are independent for any $(\theta, \lambda)$. You may assume that $X_{(1)}$ is complete and sufficient for $θ$ for any fixed $λ$,

Since $X_{(1)}$ is complete and sufficient, it is independent of every ancillary statistics. For $\sum\limits_{i = 1 }^n {(X_i - X_{(1)}) }$, I'm not completely sure how to prove it is independent. I thought about getting the joint pdf of $(X_{(1)},X_{(2)},...,X_{(n)})$ but I was not able to identify any distributions and prove that it doesn't depend on $\theta$

Lstan14
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    The joint pdf of $(X_{(1)},X_{(2)},...,X_{(n)})$ does depend on $\theta$, but the conditional pdf of $(X_{(2)},...,X_{(n)} \mid X_{(1)} = k)$ does not depend on $\theta$; even more helpfully, the pdfs of $(X_{(2)} - X_{(1)},X_{(3)} - X_{(2)}, ...,X_{(n)}-X_{(n-1)}\mid X_{(1)} = k)$ and of $(X_{(2)} - X_{(1)},X_{(3)} - X_{(1)}, ...,X_{(n)}-X_{(1)}\mid X_{(1)} = k)$ do not depend on either $\theta$ or $k$ – Henry Apr 29 '19 at 15:13

1 Answers1

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You can rewrite $\sum\limits_{i=1}^n (X_i-X_{(1)})$ as $$\sum_{i=1}^n (X_i-X_{(1)})=\sum_{i=1}^n\left[ (X_i-\theta)-(X_{(1)}-\theta)\right]$$

Now argue that distributions of both $X_i-\theta$ and $X_{(1)}-\theta$ are independent of $\theta$ for fixed $\lambda$, so that $\sum\limits_{i=1}^n (X_i-X_{(1)})$ is ancillary for $\theta$ for a given $\lambda$.

Since $X_{(1)}$ is complete sufficient for $\theta$, by Basu's theorem it follows that $\sum\limits_{i=1}^n (X_i-X_{(1)})$ is independent of $X_{(1)}$ for any fixed $\lambda$. Since $\lambda$ is arbitrary, your desired result follows.

StubbornAtom
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  • I don't understand the last sentence. Do you mean if for all $\lambda$, $X\sim f(\lambda,\theta)$ independent with $Y\sim g(\lambda,\theta)$, then $X$ and $Y$ independent? – Tan Jan 26 '21 at 21:53
  • @Tan Yes $\hspace{1cm}$. – StubbornAtom Jan 27 '21 at 06:17