Property A: If z has distribution N(0, 1), u2 has distribution χ2(m) and z and u are independent, then

has distribution T(m).

Proof: Since z has distribution N(0, 1), any linear combination of z is also normal, and in particular y = z$\sqrt{m}$ has distribution N(0,$\sqrt{m}$). Let f the pdf for y. Therefore

Let x = u2. Thus u$\sqrt{x}$, and so using the change of variables technique (Theorem 2 of General Properties of Distributions), if the pdf of x is h, then the pdf g of u is

But since x = u2 has distribution χ2(m), we have

Since t = y/u is an increasing function of y, keeping u fixed, if k(u, y) is the joint frequency function of u and y, and h(u, t) is the joint frequency function of u and t, then by a corollary to the change of variables technique,

Since y and u are independently distributed and y = tu,

Thus,

It now follows that the pdf q(t) of t is given by

where

Let

Then

and so

Thus,

which completes the proof.

Theorem 1: If x has normal distribution N(μ, σ), then for samples of size n, the random variable

has distribution T(n – 1).

Proof: Since x has normal distribution, by the Central Limit Theorem the sample mean has normal distribution $\sigma/\!\sqrt{n}$, and so z has distribution N(0, 1) where

By Corollary 3 of Chi-square Distribution, for samples of size n the sample variance s2 has distribution

Now define the random variable  as follows:

It now follows that u has distribution χ2(n–1). Finally define the random variable t as follows

From Property A, since and s2 are independently distributed, it follows that t has distribution T(n–1). But also,