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Let $X_{1},\ldots {, X_{n}}$ be iid random variables with $X_{1} ∼ N(µ,\sigma ^{2}).$Let $\bar{X}= \sum_{i=1}^{n} \frac{X_{i}}{n}$, $R=max_{1\le i \le n} \{X_{i}\}$-$min_{1\le i \le n}\{X_{i}\}$.Show that $\bar{X} $and $ R$ are independently distributed.

First Method:- I do this problem firstly assuming that $\sigma ^{2}$ is fixed constant. So, by taking $\sigma ^{2} = \sigma^{2}_{0}$(fixed constant) $R=max_{1\le i \le n} \{X_{i}\}$-$min_{1\le i \le n}\{X_{i}\}$$=X_{(n)}-X_{(1)}$ $=(X_{(n)}- \mu) -(X_{(1)}- \mu)$= $max_{1 \le i \le n}\{X_{i}-\mu\}-min_{1\le i \le n}\{X_{1}-\mu\}$. Now, By$ X_{i}-\mu\stackrel{iid}{\sim} N(0,\sigma^{2}_{0}) $ $\forall i=1(1)n. $ So, We see that the distribution of $max_{1 \le i \le n}\{X_{i}-\mu\}-min_{1\le i \le n}\{X_{1}-\mu\} $is independent of parameter $\mu .$So, $R=max_{1\le i \le n} \{X_{i}\}$-$min_{1\le i \le n}\{X_{i}\}$$=X_{(n)}-X_{(1)}$ is an ancillary statistics. Also,Sample mean $\bar{X}$ is Complete sufficient statistic for $\mu.$ by.So, by Basu's theorem Sample mean , $ \bar{X}$ and Sample Range, $R=max_{1\le i \le n} \{X_{i}\}$-$min_{1\le i \le n}\{X_{i}\}$ are independently distributed.

Second Method:- Secondly, For unknown $\sigma$ , I assume Without Loss Of Generality, Let $X_{1} \le X_{2}\le X_{3}\ldots{\le X_{n}}.$ So $Covariance(\bar{X}, X_{(n)}-X_{(1)})=Covariance(\bar{X},X_{n}-X_{1})=0 $,as $X_{1},\ldots{,X_{n}}$ are iid.

But I am not sure that whether these methods are valid or not for this problem. If possible, Please give another proof.

Debarghya Jana
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    See https://stats.stackexchange.com/questions/151096/independence-of-sample-mean-and-sample-range-of-normal-distribution?rq=1 – kjetil b halvorsen Mar 14 '22 at 03:38
  • Your second method is not valid (or at least it must be incomplete); 0 covariance doesn't imply independence in general. – Glen_b Mar 16 '22 at 01:47
  • $ ( \bar X, X_{n}-X_{1})$ jointly follows Bivariate normal distribution. So, if they are uncorrelated , then they will be independent. – Debarghya Jana Mar 16 '22 at 05:54
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    @DebarghyaJana, the range does not follow a normal distribution (e.g. it can not be negative), and it is not $X_n-X_1$ but rather $X_{(n)}-X_{(1)}$. – Matt F. Apr 14 '22 at 18:31

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