Sampling distribution of normalized sample in normal family Suppose $X_1,...,X_n \sim iid N(\mu,\sigma^2)$, $\mu$ and $\sigma^2$ are unknown. 
I'm interested in the distribution of normalized sample $Z=\frac{X_1-\bar{X}}{S}$, where $S=\sqrt{\frac{1}{n-1}\sum_{i=1}^n(X_i-\bar{X})^2}$. 
I know that in normal family, $\bar{X}$, $S^2$ and $Z$ are mutually independent, but how can I find the distribution of $Z$? The difficult is that $X_1-\bar{X}$ and $S^2$ are not independent anymore.
 A: $\newcommand{\one}{\mathbf 1}\newcommand{\R}{\mathbf R}\newcommand{\Q}{\mathbf Q}$First, note that $S$ doesn't depend on $\mu$ and $X_1 - \bar X = (X_1 - \mu) - (\bar X - \mu)$, and we can also divide both numerator and denominator by $\sigma$ without changing anything, so without loss of generality we can take $\mu=0$ and $\sigma=1$.
Let $\R = I - \frac 1n \one\one^T$ so $X - \bar X \one = \R X$. Also
$$
X^T\R X = X^TX - n (\bar X)^2 = \sum_{i=1}^n (X_i -\bar X)^2
$$
so we have your quantity as the first coordinate of
$$
\sqrt{n-1} \frac{\R X}{\sqrt{X^T \R X}}.
$$
$\R$ is the projection onto the space orthogonal to $\one$, so it has $n-1$ eigenvalues with a value of $1$ and a single eigenvalue of $0$. The eigenspace of $1$ is the part of $\mathbb R^{n}$ orthogonal to $\one$, so we can factor $\R$ as $\R = \Q\Q^T$ with $\Q \in \mathbb R^{n\times n-1}$ having orthonormal columns and $\Q^T\one = \mathbf 0$. This means $\R X \stackrel{\text d}= \Q Z$ where $Z \sim \mathcal N_{n-1}(\mathbf 0, I)$, and we now have
$$
\sqrt{n-1} \frac{\R X}{\sqrt{X^T \R X}} = \sqrt{n-1} \Q \frac{Z}{\sqrt{Z^T \Q^T \Q Z}} = \sqrt{n-1} \Q \frac{Z}{\|Z\|}.
$$
It is a standard result that $\frac{Z}{\|Z\|}$ is uniform on the sphere in $n-1$ dimensions, so this tells us what's going on here:

*

*we have a uniform distribution on the unit sphere in $\mathbb R^{n-1}$

*we lift that sphere into the hyperplane in $\mathbb R^n$ with normal vector $\one$ via $\Q$, still centered at the origin

*we scale by $\sqrt{n-1}$
A few immediate consequences:

*

*$\text E\left[\sqrt{n-1} \frac{\R X}{\sqrt{X^T \R X}}\right] = \sqrt{n-1} \Q \operatorname E[Z / \|Z\|] = \mathbf 0$ since the sphere is symmetric and centered at the origin.


*From here we see that $\operatorname{Var}[Z / \|Z\|] = \frac{1}{n-1}I$ so $\operatorname{Var}\left[\sqrt{n-1} \frac{\R X}{\sqrt{X^T \R X}}\right] = \Q\Q^T = \R$, so each coordinate has a variance of $1 - \frac 1n$. This fits the fact that $\frac{X_i - \bar X_n}{S_n} \stackrel{\text d}\to \mathcal N(0,1)$ as $n\to\infty$.
We can then follow up with the Beta distribution comments as is discussed in the comments on the answer in the second link, and as @Glen_b has written on your question.
