**Property A**: If* z* has distribution *N*(0, 1), *u*^{2} has distribution *χ*^{2}(*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* has distribution *N*(0,). Let *f* the pdf for y. Therefore

Let *x* = *u*^{2}. Thus *u* = , 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* = *u*^{2} has distribution *χ*^{2}(*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*,

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

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 , and so *z* has distribution *N*(0, 1) where

By Corollary 3 of Chi-square Distribution, for samples of size *n* the sample variance *s*^{2} has distribution

Now define the random variable as follows:

It now follows that *u* has distribution *χ*^{2}(*n*–1). Finally define the random variable *t* as follows

From Property A, since *x̄* and *s*^{2} are independently distributed, it follows that *t* has distribution *T*(*n*–1). But also,

Charles, I believe you have an error above, “Let x = u^2. Thus u = z * sqrt{x}, and so using the change of variables…”

If x=u^2 then that means u = sqrt(x), not z * sqrt(x).

regards,

Cristian

Cristian,

Thanks for catching this typo. I have just corrected the webpage per your suggestion. I appreciate your help in making the website better.

Charles