Why doesn't the Welch statistic follow the t-distribution Basing to my previous (still unanswered) question, I'm trying to find what happens when we constructing confidence interval for means difference using Student's t-test (for one or two samples), assuming that variances $\sigma_x^2$ and $\sigma_y^2$ for two populations $X \sim \mathcal{N}(\mu_x, \sigma_x^2)$ and $Y\sim \mathcal{N}(\mu_y, \sigma_y^2)$ are equal, when it really is not.
I know that in this case we simply use Welch t-test and in the bag, but I want to understand reasons of that.
Say we know our differences are equal and want to find confidence interval for means difference. Than with assuming null hypothesis $H_0: \mu_x = \mu_y$ we can just merge two samples into one and find $[\bar{X}_n - t_{\alpha, n-1}\frac{S_n}{\sqrt{n}}, \bar{X}_n + t_{\alpha, n-1}\frac{S_n}{\sqrt{n}}]$, where $S_n$ -- sample variance estimation. However, remember the fact that variances are not equal, so it means we can not perform such a test and need to perform Welch t-test with this statistic:
$$t = \frac{\bar{X}_n - \bar{Y}_m}{\sqrt{\frac{\sigma_x^2}{n} +  \frac{\sigma_y^2}{m} }}.$$
Also I believe we could use Student's t-test for two samples instead of one.
Question is: what problems causes using Student's tests instead of Welch's with $t$-statistic described above? Why exaclty do we need use it?
 A: 
What problems causes using Student's tests instead of Welch's with
t-statistic described above?

Let's back up a bit and rephrase. By "using Student's tests" what you are really saying is that in the case in which the variances $\sigma_x^2$ and $\sigma_y^2$  are unknown but can be assumed to be equal, a certain statistic $\frac{\bar{X}-\bar{Y}-(\mu_X-\mu_Y)}{S_p \sqrt{\frac{1}{n}+\frac{1}{m}}}$ follows the t-distribution with n+m-2 degrees of freedom ($S_p^2$=$\frac{(n-1)S_X^2+(m-1)S^2_Y}{n+m-2}$ is the pooled variance). That means we know the sampling distribution of this statistic and that is why we can contextualize probabilistically the single realized value of this statistic in our sample ($\bar{X}-\bar{Y}$) and draw an inference from it to the population parameter of interest ($\mu_x-\mu_Y$). Why does $\frac{\bar{X}-\bar{Y}-(\mu_X-\mu_Y)}{S_p \sqrt{\frac{1}{n}+\frac{1}{m}}}$ follow a t-distribution?
$\frac{\bar{X}-\bar{Y}-(\mu_X-\mu_Y)}{S_p \sqrt{\frac{1}{n}+\frac{1}{m}}}$=$\frac{\frac{\bar{X}-\bar{Y}-(\mu_X-\mu_Y)}{\sigma\sqrt{\frac{1}{n}+\frac{1}{m}}}}{{\sqrt{\frac{S_p^2}{\sigma^2}}}}$
The numerator has a standard normal distribution, while the denominator can be shown to have a $\chi^2$ distribution with $n+m-2$ degrees of freedom. Because the numerator and denominator are independent, we conclude (from the definition of the t-distribution) that their ratio is t distributed with $n+m-2$ degrees of freedom. Why does the numerator have the standard normal distribution? Because $\bar{X}-\bar{Y}$ is Normally distributed with mean $\mu_X-\mu_Y$ and variance $\frac{\sigma^2}{n}+\frac{\sigma^2}{m}=\sigma^2(\frac{1}{n}+\frac{1}{m})$. This last equality would not hold if $\sigma_x^2$ and $\sigma_y^2$ are not equal, which means that the $\sigma's$ in the above rightmost fraction do not cancel out; we cannot get from the right expression to the left one (which does not have any $\sigma's$).

So what does that mean?
If the variances $\sigma_x^2$ and $\sigma_y^2$  are unknown and cannot be assumed to be equal, the statistic $\frac{\bar{X}-\bar{Y}-(\mu_X-\mu_Y)}{S_p \sqrt{\frac{1}{n}+\frac{1}{m}}}$ does not follows the t-distribution. Can we find another statistic that has a well-behaved sampling distribution, maybe even one we've already catalogued? We've been unsuccessful at it thus far and not for the lack of trying. It is known as the Behrens-Fisher problem. B.L. Welch came up with an approximate solution: the Welch statistic. The Welch statistic is approximately distributed as a $t$ random variable. Why approximately distributed and not exactly distributed? Because the Welch statistic is the ratio of a standard normal random variable and a variable that is not exactly but only approximately $\chi^2$ distributed.
A: At the very end of your long question, you get around to the point, asking why not go ahead and use the two-sample pooled test, even when variances are not known to be equal, instead of using the Welch test. Here's why not:
Suppose you have a sample of size 10 from a normal distribution with SD $\sigma_1 = 4$
and a sample of size 40 from a normal distribution with $\sigma_2 = 1.$ Also suppose the two
sample means are equal: $\mu_1=\mu_2 = 50$
(so, $H_0$ is true) and that we want a test at the 5% level.
The following simulation in R shows the very bad behavior of
the pooled 2-sample t test. (Note parameter var.eq=T to get the pooled test.)
set.seed(2022)
pv = replicate(10^5, t.test(rnorm(10,50,4),
             rnorm(40,50,1), var.eq=T)$p.val)
mean(pv <= 0.05)
[1] 0.2948

Instead of the expected 5% significance level (Type I Error), we have significance level about 30%.
That is a massive false discovery rate.
By contrast, the significance level using the default Welch test in the t.test procedure,
we get very nearly the nominal 5% significance level. Although it uses an approximation, the Welch
test takes the unequal variances into account to give very nearly the correct results.
set.seed(2022)
pv = replicate(10^5, t.test(rnorm(10,50,4),
             rnorm(40,50,1))$p.val)
mean(pv <= 0.05)
[1] 0.05084

It is as simple as that. The pooled test gives
incorrect and misleading answers and the Welch
test gives correct answers.
