[proofplan]
The $t_{n-1}$ distribution is defined as the law of $Z/\sqrt{W/(n-1)}$, where $Z \sim N(0,1)$ and $W \sim \chi^2_{n-1}$ are independent. Our strategy is to algebraically massage the studentised statistic $T = \sqrt{n}(\bar{X} - \mu)/\tilde{\sigma}$ into exactly this canonical ratio. Two ingredients do all the work: the standardisation $Z := \sqrt{n}(\bar{X} - \mu)/\sigma \sim N(0,1)$ comes from the distribution of $\bar{X}$, and the chi-squared variable $W := S_{XX}/\sigma^2 \sim \chi^2_{n-1}$ together with its independence from $\bar{X}$ comes from the sampling theorem for $\bar{X}$ and $S^2$. A single divison inside the square root — factoring $\tilde{\sigma}$ into $\sigma$ times $\sqrt{W/(n-1)}$ — identifies $T$ with the canonical ratio and concludes the argument.
[/proofplan]
[step:Set up the standardised sample mean $Z$]
Let $X_1, \ldots, X_n$ be i.i.d. $N(\mu, \sigma^2)$ with $n \geq 2$, and let $\bar{X}$, $S_{XX}$, $\tilde{\sigma}^2$ denote the sample mean, sum of squared deviations, and sample variance
\begin{align*}
\bar{X} = \tfrac{1}{n}\sum_{i=1}^n X_i, \qquad S_{XX} = \sum_{i=1}^n (X_i - \bar{X})^2, \qquad \tilde{\sigma}^2 = \tfrac{S_{XX}}{n-1}.
\end{align*}
Define
\begin{align*}
Z: \Omega &\to \mathbb{R} \\
\omega &\mapsto \tfrac{\sqrt{n}(\bar{X}(\omega) - \mu)}{\sigma}.
\end{align*}
By the [Independence of $\bar{X}$ and $S^2$ for Normal Samples](/theorems/1435) theorem (part 1), $\bar{X} \sim N(\mu, \sigma^2/n)$. Standardising, $Z = \sqrt{n}(\bar{X} - \mu)/\sigma \sim N(0, 1)$.
[/step]
[step:Set up the chi-squared variable $W$ and invoke its independence from $Z$]
Define
\begin{align*}
W: \Omega &\to \mathbb{R} \\
\omega &\mapsto \tfrac{S_{XX}(\omega)}{\sigma^2}.
\end{align*}
By the same sampling theorem (part 2), $W \sim \chi^2_{n-1}$ and $S_{XX}$ is independent of $\bar{X}$. Since $Z$ is a Borel-measurable function of $\bar{X}$ alone and $W$ is a Borel-measurable function of $S_{XX}$ alone, $Z$ and $W$ are independent.
[guided]
We unpack two facts from [Independence of $\bar{X}$ and $S^2$ for Normal Samples](/theorems/1435):
1. $S_{XX}/\sigma^2 \sim \chi^2_{n-1}$. So if we define $W := S_{XX}/\sigma^2$, then $W \sim \chi^2_{n-1}$ as desired.
2. $\bar{X}$ and $S_{XX}$ are independent.
We need independence of $Z$ and $W$, not $\bar{X}$ and $S_{XX}$. But $Z = \sqrt{n}(\bar{X} - \mu)/\sigma$ is a (deterministic, measurable) function of $\bar{X}$, and $W = S_{XX}/\sigma^2$ is a measurable function of $S_{XX}$. Measurable functions of independent random variables are independent (this follows from the factorisation of the joint distribution, or equivalently from the fact that $\sigma(f(X)) \subseteq \sigma(X)$ for any measurable $f$). Therefore $Z$ and $W$ are independent.
This independence is essential: the definition of the $t$-distribution requires the numerator $Z$ and the chi-squared inside the denominator to be independent. It is the non-trivial input that distinguishes the $t$-distribution from a ratio of dependent normal and chi-squared variables.
[/guided]
[/step]
[step:Rewrite $T$ as the canonical ratio $Z/\sqrt{W/(n-1)}$]
We compute directly. Multiply numerator and denominator of $T = \sqrt{n}(\bar{X} - \mu)/\tilde{\sigma}$ by $1/\sigma$:
\begin{align*}
T = \frac{\sqrt{n}(\bar{X} - \mu)}{\tilde{\sigma}} = \frac{\sqrt{n}(\bar{X} - \mu)/\sigma}{\tilde{\sigma}/\sigma} = \frac{Z}{\tilde{\sigma}/\sigma}.
\end{align*}
The denominator $\tilde{\sigma}/\sigma$ is a positive quantity (take the positive square root of $\tilde{\sigma}^2/\sigma^2$):
\begin{align*}
\tfrac{\tilde{\sigma}}{\sigma} = \sqrt{\tfrac{\tilde{\sigma}^2}{\sigma^2}} = \sqrt{\tfrac{S_{XX}/(n-1)}{\sigma^2}} = \sqrt{\tfrac{S_{XX}/\sigma^2}{n-1}} = \sqrt{\tfrac{W}{n-1}}.
\end{align*}
Substituting,
\begin{align*}
T = \frac{Z}{\sqrt{W/(n-1)}}.
\end{align*}
[guided]
The goal of this step is purely algebraic: start from the definition of $T$ and rearrange until it looks like $Z/\sqrt{W/(n-1)}$, the canonical form of a $t_{n-1}$ variable.
**Numerator.** We already have $Z = \sqrt{n}(\bar{X} - \mu)/\sigma$ from Step 1, so $\sqrt{n}(\bar{X} - \mu) = \sigma Z$. Substituting,
\begin{align*}
T = \frac{\sqrt{n}(\bar{X} - \mu)}{\tilde{\sigma}} = \frac{\sigma Z}{\tilde{\sigma}} = \frac{Z}{\tilde{\sigma}/\sigma}.
\end{align*}
**Denominator.** We need to rewrite $\tilde{\sigma}/\sigma$ in terms of $W$. Start from the definition $\tilde{\sigma}^2 = S_{XX}/(n-1)$ and divide by $\sigma^2$:
\begin{align*}
\tfrac{\tilde{\sigma}^2}{\sigma^2} = \tfrac{S_{XX}}{(n-1)\sigma^2} = \tfrac{1}{n-1}\cdot\tfrac{S_{XX}}{\sigma^2} = \tfrac{W}{n-1}.
\end{align*}
Both $\tilde{\sigma}$ and $\sigma$ are positive, so $\tilde{\sigma}/\sigma = \sqrt{W/(n-1)}$ (positive square root). Substituting,
\begin{align*}
T = \frac{Z}{\sqrt{W/(n-1)}}.
\end{align*}
Notice what happened: the unknown parameter $\sigma$ vanished from the formula. That is the whole point of studentising — $T$ depends on the data only through $Z$ and $W$, both of which have distributions not depending on $\sigma$. The statistic $T$ is therefore **pivotal** for $\mu$.
[/guided]
[/step]
[step:Conclude $T \sim t_{n-1}$ from the definition of the $t$-distribution]
The $t$-distribution with $k$ degrees of freedom is defined as the law of $U/\sqrt{V/k}$, where $U \sim N(0, 1)$ and $V \sim \chi^2_k$ are independent. We verify all three conditions with $k = n-1$, $U = Z$, $V = W$:
(i) $Z \sim N(0,1)$ — established in Step 1.
(ii) $W \sim \chi^2_{n-1}$ — established in Step 2.
(iii) $Z$ and $W$ are independent — established in Step 2.
The representation $T = Z/\sqrt{W/(n-1)}$ from Step 3 now matches the definition exactly, so $T \sim t_{n-1}$. This completes the proof.
[/step]
