# The exact value of Welch's t test degrees of freedom

Wikipedia defines an equation to approximate the degrees of freedom in Welch's t test and does not state anything about the exact value. Is there a reason why we could not evaluate the exact df and have to use the approximation? Also, it says that:

The approximate degrees of freedom are rounded down to the nearest integer [citation needed]

An explanation of why we should be rounding down would be also appreciated.

• As a side note, all frequentist solutions to this problem are approximate. You might instead use an exact Bayesian solution as in John Kruschke's BEST procedure which is implemented in R. – Frank Harrell Feb 4 '19 at 13:12
• Some relevant discussion can be found here - stats.stackexchange.com/questions/124961/… – Glen_b Feb 4 '19 at 15:07

Short answer: There is no exact degrees-of-freedom because the variance estimator in this test does not follow an exact chi-squared distribution.

Longer answer: The Welch T-test gives an approximate solution to the Fisher-Behrens problem (comparing the means of two samples with different variances). It uses the studentised test statistic:

$$T = \frac{\bar{X}_1 - \bar{X}_2}{\sqrt{S_1^2/N_1 + S_2^2/N_2}}.$$

The denominator in this test statistic is the square-root of an estimator of the mean difference:

$$\hat{\mathbb{V}}(\bar{X}_1-\bar{X}_2) = \frac{S_1^2}{N_1} + \frac{S_2^2}{N_2} \sim \frac{\chi_{N_1-1}^2}{N_1} \cdot \sigma_1^2 + \frac{\chi_{N_2-1}^2}{N_2} \cdot \sigma_2^2.$$

This quantity is a weighted sum of independent chi-squared random variables. Its exact distribution is quite complicated (and is best represented through its moment generating function), but it is not an exact chi-squared distribution.

The test uses the Welch-Satterwaite approximation, which approximates the distribution of this quantity by a single scaled chi-squared distribution. In this approximation the degrees-of-freedom formula arises as the best approximation of the chi-squared distribution to the true distribution of this quantity. Without this approximation to the chi-squared distribution there is no single exact degrees-of-freedom. Instead, the exact distribution is of a weighted sum of chi-squared random variables with the above weights and degrees-of-freedom.

@Ben's answer is very clear about why an exact solution for the degrees of freedom isn't possible.

As for

The approximate degrees of freedom are rounded down to the nearest integer [citation needed]

This seems unusual. There is a section on the Talk page for the article that questions whether this is usual or not, and why one would round down rather than up.

This CV question discusses reporting non-integer degrees of freedom in much more detail, and gives a few sources saying it is "conventional" (mostly from the days when computations were done by hand and critical values of the $$t$$ distribution looked up in reference tables that gave values for integer df only).

• I second this (+1). With modern computing there is no reason to round the DF from the Welch-Satterwaite approximation. It is best to use the approximating value without rounding. – Ben Jun 27 '20 at 12:08