# 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.

• One thing you might find ... suggestive is to think about the distribution for $n=2$ and then to simulate the case for $n=4$, and then perhaps to think about $n=3$. Commented Oct 28, 2022 at 3:23
• It might also help to think about writing the sum of squares in the numerator of $S^2$ in terms of $(X_1-\bar{X})^2$ and the remaining squared deviations, and ponder $(X_1-\bar{X})^2/[(n-1)S^2]$ and then think about the connection between gamma and beta variates. Putting those thoughts together might be sufficient. Commented Oct 28, 2022 at 3:30
• stats.stackexchange.com/q/181964/119261 Commented Oct 28, 2022 at 5:30

$$\newcommand{\one}{\mathbf 1}\newcommand{\R}{\mathbf R}\newcommand{\Q}{\mathbf Q}$$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:

1. we have a uniform distribution on the unit sphere in $$\mathbb R^{n-1}$$
2. we lift that sphere into the hyperplane in $$\mathbb R^n$$ with normal vector $$\one$$ via $$\Q$$, still centered at the origin
3. we scale by $$\sqrt{n-1}$$

A few immediate consequences:

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

2. 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 $$n\to\infty$$.