I am following this proof (actually, the "second" one) about the independence of sample mean and sample variance in the normal case.
I understand the computations but still a (big) element is missing in my understanding: why is the vector $\begin{pmatrix} \bar{X}, X_1 - \bar{X}, \dots, X_n - \bar{X}\end{pmatrix}^T$ jointly normally distributed ?
For lazy people who wouldn't read the link, I have an i.i.d. sample of a normal random variable $X$ and I am considering the random vector $\begin{pmatrix} \bar{X}, X_1 - \bar{X}, \dots, X_n - \bar{X}\end{pmatrix}^T$.
I am able to understand that $\bar{X}-X_j \sim N\left(0, \frac{n-1}{n} \sigma^2\right)$, this is because $\bar{X}-X_j$ can be written as a linear combination of independent (by i.i.d. assumption) variables:
$$\bar{X}-X_j =\frac{1}{n}\left(X_1+\cdots+X_{j-1}+X_{j+1}+\cdots+X_n\right)-\frac{n-1}{n} X_j .$$
I also know $\bar{X} \sim \mathcal{N}\left(\mu, \frac{\sigma}{\sqrt{n}}\right)$ but I don't understand why together, they are jointly normal ? In fact, even if you remove $\bar{X}$ from the vector and only focus on $\begin{pmatrix} X_1 - \bar{X}, \dots, X_n - \bar{X}\end{pmatrix}^T$, I don't get the joint normality since I don't think that these guys are independent.
I know that random variables are jointly normally distributed iff any linear combination of them follows a 1D normal distribution; should I explictly do this, or is there a more direct way to see the thing ?
If $X$ is a jointly Gaussian vector and $A$ is a matrix, then $AX$ is jointly Gaussian. Seeing that there exists a matrix $A$ such that $(\overline{X}, X_1 - \overline{X}, \ldots , X_n - \overline{X})^T = A (X_1, \ldots , X_n)^T$ we are done.