Let $X_1 , \dots, X_n$ be independent standard normal variables.
How do I show that $\hat{X}_n$ (average of $X_1 , \dots, X_n$) is independent with $\frac{1}{n}\sum_{i=1}^{n}|X_i-\hat{X}_n|$ and $(X_1-\hat{X}_n, \dots, X_n -\hat{X}_n)$?
My attempt for the second part:
I check whether $Cov(X_i-\hat{X}_n,\hat{X}_n)=0$ as follows:
$$\mathsf{Cov}(X_i-\hat{X}_n,\hat{X}_n)=\mathsf{Cov}(X_i,\hat{X}_n)-\mathsf{Cov}(\hat{X}_n,\hat{X}_n)=\mathsf{Cov}(X_i,\hat{X}_n)-\mathsf{Var}(\hat{X}_n)=$$$$\mathsf{Cov}(X_i,\hat{X}_n)-\frac{\sigma^2}{n}=\frac{1}{n}\sum \mathsf{Cov}(X_i,X_j)-\frac{\sigma^2}{n} $$
Since $X_i$ and $X_j$ are independent for $i \neq j$:
$$=\frac{1}{n}\sum \mathsf{Cov}(X_i,X_j)-\frac{\sigma^2}{n}=\frac{\sigma^2}{n}-\frac{\sigma^2}{n}=0 $$
Is this correct? And how to show the first part?