Independence of $\frac1n\sum X_i$ and $\max X_i-\min X_i$ where $X_i$'s are i.i.d $N(0,1)$

Question

I'm trying to solve a question which requires proving independence of one $RV$ of $\max X_i-\min X_i$ RV and I don't know what to assume regarding $\max X_i$ and $\min X_i$ so that I can say that if A independent of B, then A independent of $\max X_i$. Could you help me with this question?

Question:

Let $(X_i)_{0<i<n+1}$ ($n>1$) be independent and identically distributed r.v of Gaussian law $N(0,1)$. Prove that the r.v. (a) $^*X_n=\frac1n\sum{X_i}$ and (b) $(\max_{0<i<n+1}X_i - \min_{0<i<n+1}X_i)$ are independent.

Hint: Consider the vector $(^*X_n,X_1-{^*X_n},X_2-{^*X_n},...,X_n-{^*X_n})^t$.

[My Way of Proof]

For the vector in the hint the 1st row of the Covariance matrix is all zeroes - I proved it and I think that the 1st row zeroes leads also independence since possibly it's a random vector. But I don't know how to move on from here? What is the connection of $\min/\max$ to $X_i-{^*X_n}$? Also, $^*{X_n}$ is Gaussian due to CLT (Central Limit theorem). But why this vector is a Gaussian vector? I need it for my independence to happen.

Can someone please help me solve this question?

Answer 1

I will use simpler notations: $\bar{X} = \frac{1}{n} \sum_{i=1}^n X_i$, $\min{\{X_i\}} = \min_{0 < i < n+1}X_i$.

First, we have $E[\bar{X}] = 0, E[X_i - \bar{X}] = 0$.

$E[\bar{X}(X_i - \bar{X})] = E[\bar{X}X_i]-E[\bar{X}^2] = \frac{1}{n} - \frac{1}{n^2}n = 0$.

Therefore $\bar{X}$ and $X_i - \bar{X}$ are uncorrelated. Because the two RVs are jointly Gaussian, they are independent. (correlation and independence)

Note that $\max{\{ X_i \}} - \min{\{ X_i \}} = \max{\{ X_i - \bar{X} \}} - \min{\{ X_i- \bar{X} \}}$. $\bar{X}$ is independent of RHS and thus LHS of the equation.

[EDIT] For completeness we also need to show that in the last paragraph in RHS, the min max are also independent. Alas: $E[(X_j-\bar{X})(X_i - \bar{X})] = 0$. But this equals to $\frac 1 n$.

Hence we can show that $E[\bar{X}(X_i-X_j)]={\frac 1 n}-{\frac 1 n}=0$. Since it holds for any $i,j$ it also holds for $max{X_i}-min{X_i}$. And as sum of Gaussian i.i.ds is also Gaussian, this uncorolatence grabs independence. & This is All We Had to Proove.

Note that the hint in the original question here is misleading https://math.unice.fr/~diel/TD0stocal.pdf

Answer 2

Let $X\equiv[X_1,\ldots,X_n]^{\top}$, $Z\equiv[\bar{X}_n,X_1-\bar{X}_n,\ldots,X_n-\bar{X}_n]^{\top}$, and $$ B=\begin{bmatrix} \phantom{-}1/n & \phantom{-}1/n & \cdots &\phantom{-}1/n \\ 1-1/n & -1/n & \cdots & -1/n \\ \vdots & \vdots & \ddots & \vdots \\ -1/n & -1/n & \cdots &1-1/n \end{bmatrix}. $$ Then $Z=BX$ and, therefore, $Z\sim N(0, BB^{\top})$. Since $B_1 B_i^{\top}=0$ for $i>1$ ($B_i$ is the $i$-th row of $B$), $\bar{X}_n$ is independent of $W\equiv\{X_i-\bar{X}_n:1\le i\le n\}$ and any measurable function of $W$.

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Alternatively, considering the following statistical model - $(\mathbb{R}^n,\mathcal{B}(\mathbb{R}^n),\{P_{\mu}^n:\mu\in\mathbb{R}\})$, where $P_{\mu}$ is a normal distribution with mean $\mu$ and unit variance, the result follows from Basu's theorem because $\bar{X}_n$ is a complete sufficient statistics for $\mu$ and the range ($X_{(n)}-X_{(1)}$) is ancillary to $\mu$.