(Sheldon Ross) Proving the independence of sample mean and sample variance Please refer to the proof given in : Sheldon Rose, A first course in probability :
The author says that since $\{Y, X_i − \overline X, i = 1, \cdots , n\}$ and $\{\overline X, (X_i − \overline X), i = 1, \cdots, n\}$ have the same joint distribution, thus showing that $\overline X$ is independent of the sequence of
deviations $X_i − \overline X, i = 1, \cdots n.$
Somehow, the statement in bold doesn't look too obvious. Could someone please clarify? Am I missing something?

 A: It says in the previous sentence or two that $Y$ is independent of the $X_i$, so it is independent of $X_i-\bar X$ (bc $\bar X$ is a function of $X_i$). Since $Y, X_i -\bar X$ has the same distribution as $\bar X, X_i-\bar X$, this means that $\bar X$ is independent of the $X_i-\bar X$.
A: It is obvious. Duh.
Just kidding. So to add a bit on what the author said, there's a few things you have to put together to understand.

*

*First of all, $X_i - \overline X$ for $i=1,\cdots,n$ and $\overline X$ together are a linear combination of the $X_1, \cdots, X_n$, and so therefore have a joint multivariate normal distribution. If you were to write the matrix, it would look something like
$$
\begin{bmatrix} 
1-\frac1n           & -\frac 1n & \cdots     & -\frac 1n  \\
 - \frac 1n & 1-\frac1n         & \ddots     & \vdots \\
 \vdots     & \ddots    & \ddots     & - \frac 1n \\ 
 - \frac 1n & \cdots    & - \frac 1n & 1-\frac1n \\
\frac 1n    & \cdots    & \cdots     & \frac 1n
\end{bmatrix}
$$
with the $i^{\text{th}}$ row corresponding to $X_i$ and the $(n+1)^{\text{th}}$ row to $\overline X$. This is true regardless of the matrix, I just computed it for convenience so that you know what we're talking about.


*If you let $Y$ be a normal random variable with mean $\mu$ and variance $\sigma^2/n$, then $\{Y, X_1 - \overline X,\cdots,X_n - \overline X \}$ has the same joint multivariate distribution as $\{ \overline X, X_1 - \overline X, \cdots, X_n -  \overline X \}$. This is because the multivariate distribution of $\{Y,X_1 -  \overline X,\cdots,X_n -  \overline X\}$ is determined by its expectations and the covariance matrix (regardless of $Y$, as long as the joint distribution is multivariate normal). In the case of $Y$, the covariance matrix looks like this:
$$
\begin{bmatrix} 
\mathrm{Var}(X_1) & 0      & \cdots            & 0      \\
0                 & \ddots & \ddots            & \vdots \\
\vdots            & \ddots & \mathrm{Var}(X_n) & 0      \\
0                 & \cdots & 0                 & \mathrm{Var}(Y)
\end{bmatrix}
$$
The entries off the diagonal are 0 because the $X_i$ are independent of each other, thus so are the $X_i - \overline X$; the zeros on the last row and last column are there because $Y$ is independent of all the $X_i$, hence also of $X_i - \overline X$, which is just a linear combination of the $X_i$.
The situation is slightly different for $\{\overline X, X_1 - \overline X, \cdots, X_n - \overline X\}$; we have the following matrix:
$$
\begin{bmatrix} 
\mathrm{Var}(X_1) & 0      & \cdots            & 0      \\
0                 & \ddots & \ddots            & \vdots \\
\vdots            & \ddots & \mathrm{Var}(X_n) & 0      \\
0                 & \cdots & 0                 & \mathrm{Var}(\overline X)
\end{bmatrix}
$$
In the last row, we have $\mathrm{Cov}(X_i - \overline X, \overline X)$ (and similarly in the last column). As observed in the book, note that all these values are zero.

*

*Therefore, since $\mathrm{Var}(\overline X) = \frac{\sigma^2}n = \mathrm{Var}(Y)$, and $Y$ is independent of the $X_i$, we conclude that $\overline X$ is independent of the $X_i$ because the joint distributions of $\{Y,X_1 - \overline X, \cdots, X_n - \overline X\}$ and $\{\overline X, X_1 - \overline X, \cdots, X_n - \overline X\}$ are the same. This combines together a bunch of things:

*

*Multivariate normal distributions are determined by their expectation and covariance matrix.

*If $Y$ is independent of the $X_i$, then it is independent of the $X_i - \overline X$.

*$\{Y, X_1 - \overline X, \cdots, X_n - \overline X\}$ and $\{\overline X, X_1 - \overline X, \cdots, X_n - \overline X\}$ have the same multivariate normal distribution (look at their covariance matrix and use the first fact in this list).

*The joint multivariate distribution of two multivariate variables $\{X_1,\cdots,X_n\}$ and $\{Y_1,\cdots,Y_m\}$ determines their independence (if you can factor the distribution as a product
$$
F_{X_1,\cdots,X_n,Y_1,\cdots,Y_m}(x_1,\cdots,x_n,y_1,\cdots,y_m) = F_{X_1,\cdots,X_n}(x_1,\cdots,x_n) F_{Y_1,\cdots,Y_m}(y_1,\cdots,y_m)
$$
which is equivalent to the definition of independence).



Combine all those things together, and you can see that $\overline X$ is independent of $X_1 - \overline X, \cdots, X_n - \overline X$.
Fact worth nothing: if you look at the linear transformation that sends $(x_1,\cdots,x_n)$ to $(x_1 - \overline x, \cdots, x_n - \overline x)$, this maps $\mathbb R^n$ into an $(n-1)$-dimensional subspace of $\mathbb R^n$, the subspace of those vectors for which the sum of the coordinates is zero. There's another linear transformation which sends $(x_1,\cdots,x_n)$ to $(\overline x, \cdots, \overline x)$, i.e. the one-dimensional subspace spanned by the average (if it were non-zero of course, but that's the space in which the average lands in, spanned by the vector $(1,\cdots,1)$). Because of the properties of the multivariate distribution, the orthogonality of those two vectors implies that they are uncorrelated (because of the way covariance is calculated), and therefore independent (by what we discussed above). (This is another proof of your concern, taking $(Y_1,\cdots,Y_m) = (\overline X, \cdots, \overline X)$ with $n=m$.)
Hope that helps,
A: An alternative  proof is the following: in a gaussian model $\overline{X}_n$ is CSS (Complete and Sufficient Statistic) for $\mu$ while $\frac{(n-1)S_n^2}{\sigma ^2}\sim \chi_{(n-1)}^2$ thus the sample variance is ancillary for $\mu$.
At this point we can invoke Basu's Theorem concluding that
$$\overline{X}_n \perp\!\!\!\perp S_n ^2$$
