# How to prove $(\hat{X}-\mu)/(\hat{S}/\sqrt{n})$ is student t with $n-1$ degrees of freedom if $X_i$ are iid $N(\mu, \sigma)$?

It is commonly stated that if $$X_i$$ are iid $$N(\mu, \sigma)$$, then with $$\hat{X}$$ the sample mean, and $$\hat{S}$$ the sample error (sample standard deviation), then $$\frac{ \hat{X}-\mu}{\hat{S}/\sqrt{n}}$$ follows a t distribution with $$n-1$$ degrees of freedom.

e.g., wikipedia link states it, and another Wikipedia link states it as the natural way a t distribution arises, but nothing proves this fact.

How do we know/prove this?

The definition of a t distributed variable is $$\frac{U}{\sqrt{W/(n-1)}}$$ with $$U$$~$$N(0,1)$$, $$W$$~$$\chi^2(n-1)$$.

So we can write $$\frac{ \hat{X}-\mu}{\hat{S}/\sqrt{n}} = \frac{ \frac{\hat{X}-\mu}{\sigma/\sqrt{n}}}{ \frac{\hat{S}}{\sigma}} = \frac{U}{D}$$.

$$U$$ is $$N(0,1)$$ b/c $$X_i$$ are, meaning $$\hat{X}\sim N(\mu, \sigma/\sqrt n)$$.

We need to show that the bottom, $$D = \sqrt{W/(n-1)}$$ with $$W = \frac{(n-1)S^2}{\sigma^2} \sim \chi^2(n-1)$$. Looking at other post here, it seems the proof is not quite complete, but it gives a good start. Basically, the proof is: (1.) define auxiliary variable $$V=\sum_{i=1}^n(\frac{X_i-\mu}{\sigma})^2$$, clearly $$\chi^2(n)$$. (2.) Algebra shows that $$V= W + Y$$, where $$W = \frac{(n-1)\hat{S}^2}{\sigma^2 }$$ (quantity of interest) and $$Y = (\frac{n(\bar{X}-\mu)}{\sigma})^2$$ which is $$\chi^2(1)$$. (3.) Conclude that this implies $$W$$ is $$\chi^2(n-1)$$.

So I can follow (1.) and (2.). I do not think (3.) follows directly from (2.) without justification. (e.g., if $$W$$ and $$Y$$ were independent, I think the relationship above shows $$W$$ is $$\chi^2(n+1)$$!

Overall, how do we prove (3.)? OR Please provide another proof of this whole thing. I just want to understand this rigorously.

For $$\zeta\sim \mathcal N(0,1),~ \chi^2\sim\chi^2_{n},$$ both being independent to each other, \begin{align}\textrm{Fisher's t} := ~ \frac{\zeta}{\sqrt{\frac{\chi^2}n}}\sim t_n.\tag I\end{align}

Now take $$\zeta:= \frac{\bar x-\mu}{\sigma/\sqrt n}\sim \mathcal N(0, 1), \chi^2 :=\frac{\sum(x_i-\bar x) ^2}{\sigma^2}=\frac{ns^2}{\sigma^2}\sim \chi^2_{n-1},$$ both being independent (to be proved)

\begin{align} \text{Student's t}&:= \frac{\sqrt n (\bar x-\mu)}{\sigma}\times \frac\sigma{\sqrt{\sum(x_i-\bar x) ^2/(n-1)}}\\&= \frac{\bar x-\mu}{S/\sqrt n} \sim t_{n-1}.\tag{II} \end{align}

Fisher's lemma: If $$X_i\mapsto Y_i$$ via an orthogonal linear transformation, where $$X_i, ~i\in\{1, 2,\ldots, n\}$$ are independent $$\mathcal N(0, \sigma^2),$$ then $$Y_i\sim \mathcal N(0, \sigma^2)$$ and independent.

The proof is easy. Let $$\mathbf A$$ be the matrix corresponding to the transformation i.e. $$\bf Y = AX;$$ since $$\bf A$$ is orthogonal, $$\bf Y^\mathsf TY = X^\mathsf TX$$ and $$\bf A^\mathsf TA= I.$$ Now one can evaluate the joint density of $$Y_i$$ using the relation stated; the jacobian would yield $$1$$ and the rest would follow.

Now, take $$\bf A$$ such that

$$a_{11} = a_{12} =\ldots= a_{1n}= \frac1{\sqrt n};$$ this would imply $$y_1 = \sqrt n \bar x.$$

Now, as the transformation is orthogonal,

\begin{align} \sum y_i^2 &= \sum x_i^2\\ &= \sum (x_i-\bar x) ^2 +\underbrace{ n\bar x^2}_{=y_1^2}\\\implies \sum_{i= 2}^n y_i^2 &= \sum_{i=1}^n (x_i -\bar x) ^2; \end{align}

moreover

$$\sum_{i=1}^n (x_i -\mu) ^2= \sum_{i= 2}^n y_i^2 + n(\bar x - \mu)^2.$$

Therefore, following Fisher's lemma and the results found,

\begin{align}\mathrm dG(y_1, y_2, \ldots,,y_n) &= \frac1{\sigma^n(2\pi)^{n/2} }\exp\left [-\frac1{2\sigma^2}\left\{ \sum_{i= 2}^n y_i^2 + n(\bar x - \mu)^2 \right\}\right]|\mathcal J|\prod \mathrm dy_i\\ &= \left[\frac{1}{\sqrt{2\pi}(\sigma/\sqrt n) }\exp\left\{-\frac{ n}{2\sigma^2}(\bar x -\mu)^2\right\}~\mathrm d\bar x\right]\times \left[\frac{1}{\left(\sigma{\sqrt {2\pi}}\right)^{n-1} }\exp\left\{-\sum_{i=2}^n\frac{ y_i^2}{2\sigma^2}\right\}~\prod \mathrm dy_i\right]; \end{align}

proving the independence of $$\bar X$$ and $$\sum_{i=2}^n Y_i^2 = ns^2.$$

Now, since $$Y_i \sim \mathcal N(0, \sigma^2), ~i\in\{2, 3,\ldots, n\},$$

$$\sum_{i=2}^n \frac{Y_i^2}{\sigma^2} = \sum_{i=1}^n \left(\frac{X_i -\bar X}{\sigma}\right)^2\sim \chi^2_{n-1}.$$

## Reference:

Fundamentals of Mathematical Statistics, S. C. Gupta, V. K. Kapoor, Sultan Chand & Sons, 2014.

• So if you reduce to the case \mu = 0, \sigma =1 initially (WLOG), then the proof required is completed as soon as you derive \sum_{i>1} Y_i^2 = sum_{i=1}^n (X_i - \barX)! very slick!. As for the independence (which doesn't seem to be needed but we get for free from Fisher's Lemma): I think it is easy to show that if random variables Y_1, ..., Y_n are all mutually independent, then for any i, j, Y_i is independent of Y_j^2 and Y_1 independent of \sum_{i>1} Y_i. This means Y_1^2 is independent of sum_{i>1} Y_i^2. Commented Aug 25, 2022 at 18:08