Comparison of means with different sample variances: why using smallest sample size as degrees of freedom? Suppose you want to determine whether the means of two populations can reasonably be considered equal. You draw one sample from each population, and those samples have different bias-adjusted  variance. Then you're computing the following test statistic:
$$
T=\frac{\bar{X_1}-\bar{X_2}-(\mu_1-\mu_2)}{\sqrt{\frac{\sigma^2_1}{n_1}+\frac{\sigma^2_2}{n_2}}}
$$
Where $\bar{X_1}$ and $\bar{X_2}$ are observed sample means; $\mu_1$and $\mu_2$ are unknown population means; $\sigma^2_1$ and $\sigma^2_2$ are Bessel-corrected sample variances; and $n_1$ and $n_2$ are sample sizes.
Why is it common practice to use $(min(n_1,n_2)-1)$ degrees of freedom for the test statistic? This is in particular what's taught to students in introductory statistics classes.
As far as I know, you should be running a Welch's t-test with a more complex approximation of the degrees of freedom.
I get the intuition that the amount of information you get is bottlenecked by the smaller sample, and I see that the resulting expression mirrors the degrees of freedom for computing a confidence interval on a single population mean, but I failed to find out where this $(min(n_1,n_2)-1)$ actually comes from.
Thanks in advance for your explanations.
ANSWER
Page 3 of the following document. $(min(n_1,n_2)-1)$ is a lower bound of the Welch–Satterthwaite approximation for the degrees of freedom. Thus, using it leads to sometimes underestimating the degrees of freedom but never overestimating them, so at least you're not going to underestimate the fatness of tails and attribute false power to your test.
 A: The Welch 2-sample t test uses (approximate) degrees of freedom $\nu^\prime$ based on the two sample sizes $n_1, n_2$ and sample variances $S_1^2, S_2^2,$ respectively.
Degrees of freedom in a pooled t test are simply $\nu = n_1 + n_2 - 2,$
but the formula for Welch's $\nu^\prime$ is a little messier.
One can show that
$$\min{(n_1-1, n_2 -1)} \le \nu^\prime \le n_1 + n_2 - 2,$$
As a simplification, David Moore has observed in some of his elementary
books (authored and co-authored) that if $n_1 > 30$ and $n_2 > 30,$ then $\nu^\prime \ge 30.$ [See one reference below.]
Thus, if both samples are of size greater than 30 and the absolute value of the Welch t statistic exceeds $c = 2.04,$ one can reject $H_0$ against a two-sided alternative at the 5% level.
qt(.975, 30)
[1] 2.042272

I have never seen use of $\nu^\prime = \min(n_1-1, n_2-1),$
in all cases, recommended as a 'general practice'. If you are using statistical
software, procedures for the Welch 2-sample t test routinely base the P-value on the exact value of $\nu^\prime.$
In R, the following example uses $\nu^\prime = 14.492.$ Some other statistical
software programs round $\nu^\prime$ down to the nearest integer.
set.seed(2021)
x1 = rnorm(10, 50, 7)
x2 = rnorm(10, 65, 9)
t.test(x1, x2)  # Welch t is default

        Welch Two Sample t-test

data:  x1 and x2
t = -4.1494, df = 14.492, p-value = 0.0009169
alternative hypothesis: true difference in means is not equal to 0
95 percent confidence interval:
 -26.120658  -8.356365
sample estimates:
 mean of x mean of y 
 52.12437  69.36288 

Directly in R, using the (slightly rounded) t statistic, we have about the
same P-value reported in the output above for the Welch t procedure.
2*pt(-4.1492, 14.492)
[1] 0.0009172653

Ref: DS Moore & GP McCabe, Basic Practice of Statistics, 3e (1999), Freeman, p541.
