Editing Student's t-test (section)

===Exact method for unequal variances and sample sizes===

The test<ref>{{cite arXiv | eprint=2210.16473 | last1=Wang | first1=Chang | last2=Jia | first2=Jinzhu | title=Te Test: A New Non-asymptotic T-test for Behrens-Fisher Problems | year=2022 | class=math.ST }}</ref> deals with the famous [[Behrens–Fisher problem]], i.e., comparing the difference between the means of two normally distributed populations when the variances of the two populations are not assumed to be equal, based on two independent samples.

The test is developed as an [[exact test]] that allows for '''unequal sample sizes''' and '''unequal variances''' of two populations. The exact property still holds even with '''extremely small''' and '''unbalanced sample sizes''' (e.g. <math>\ m \equiv n_\mathsf{X} = 50\ </math> vs. <math>\  n \equiv n_\mathsf{Y} = 5\ </math>).

The statistic to test whether the means are different can be calculated as follows:

Let <math>\ X = \left[\ X_1, X_2, \ldots, X_m\ \right]^\top\ </math> and <math>\ Y = \left[\ Y_1, Y_2, \ldots, Y_n\ \right]^\top\ </math> be the i.i.d. sample vectors (for <math>\ m \ge n\ </math>) from <math>\ \mathsf{Norm}\left(\ \mu_\mathsf{X},\ \sigma_\mathsf{X}^2\ \right)\ </math> and <math>\ \mathsf{Norm}\left(\ \mu_\mathsf{Y},\ \sigma_\mathsf{Y}^2\ \right)\ </math> separately.

Let <math>\ (P^\top)_{n\times n}\ </math> be an <math>n\times n</math> orthogonal matrix whose elements of the first row are all <math>\ \tfrac{ 1 }{ \sqrt{ n\ } }\ ,</math> similarly, let <math>\ (Q^\top)_{n\times m}\ </math> be the first <math>\ n\ </math> rows of an <math>\ m\times m\ </math> orthogonal matrix (whose elements of the first row are all <math>\ \tfrac{ 1 }{ \sqrt{ m\ } }\ </math>).

Then <math>\ Z \equiv \frac{\ \left( Q^\top \right)_{n\times m}\ X\ }{ \sqrt{ m\ } }\ -\ \frac{\ \left( P^\top \right)_{n\times n}\ Y\ }{ \sqrt{ n\ } }\ </math> is an {{mvar|n}}-dimensional normal random vector:

:<math> Z ~\sim~ \mathsf{Norm}\left(\ \left[\ \mu_\mathsf{X} - \mu_\mathsf{Y},\ 0,\ 0,\ \ldots,\ 0\ \right]^\top\ ,\ \left( \frac{\ \sigma_\mathsf{X}^2\ }{ m } + \frac{\ \sigma_\mathsf{Y}^2\ }{ n }\right)\ I_n\ \right) ~.</math>

From the above distribution we see that the first element of the vector {{mvar|Z}} is

:<math> Z_1 = \bar X - \bar Y = \frac{ 1 }{\ m\ } \sum_{i=1}^m\ X_i - \frac{ 1 }{\ n\ } \sum_{j=1}^n\ Y_j\ ,</math>

hence the first element is distributed as

:<math> Z_1 - \left( \mu_\mathsf{X} - \mu_\mathsf{Y} \right) ~\sim~ \mathsf{Norm}\left(\ 0,\ \frac{\ \sigma_\mathsf{X}^2\ }{ m } + \frac{\ \sigma_\mathsf{Y}^2\ }{ n }\ \right)\ ,</math>

and the squares of the remaining elements of {{mvar|Z}}  are [[chi-square distribution|chi-squared]] distributed

:<math> \frac{\ \sum_{i=2}^n Z^2_i\ }{\ n - 1\ } ~\sim~ \frac{\ \chi^2_{n - 1}\ }{\ n - 1\ } \times\left( \frac{\ \sigma_\mathsf{X}^2\ }{ m }+\frac{\ \sigma_\mathsf{Y}^2\ }{ n } \right) </math>

and by construction of the orthogonal matricies {{mvar|P}} and {{mvar|Q}} we have

:<math> Z_1 - \left( \mu_\mathsf{X} - \mu_\mathsf{Y} \right) \quad \perp \quad \sum_{i=2}^n Z^2_i\ ,</math>

so {{mvar|Z}}{{sub|1}}, the first element of {{mvar|Z}}, is statistically independent of the remaining elements by orthogonality.
Finally, take for the test statistic

:<math> T_\mathsf{e} ~\equiv~ \frac{\ Z_1 - \left( \mu_\mathsf{X} - \mu_\mathsf{Y} \right)\ }{\ \sqrt{ \left( \sum_{i=2}^{n} Z^2_i \right) /\left( n - 1 \right)\ }\ } ~\sim~ t_{n - 1} ~.</math>