I am having trouble finding documentation showing a proof, or at least some outline for it, illustrating how to derive the distribution and degrees of freedom of the test statistic for a two-sample t-test when the population variances are assumed to be different, $\sigma_1^2\ne \sigma_2^2$. If $ \overline{X}=\frac{1}{n_1}\sum\limits_{i=1}^{n_1}X_i$ where $X_1, \ldots, X_{n_1}$ are i.i.d. $\text{N}\left(\mu_1, \sigma_1^2 \right)$, $ \overline{Y}=\frac{1}{n_2}\sum\limits_{i=1}^{n_2}Y_i$ where $Y_1, \ldots, Y_{n_2}$ are i.i.d. $\text{N}\left(\mu_2, \sigma_2^2 \right)$, $S_1^2$ and $S_2^2$ are the unbiased sample variances for $X_i$ and $Y_i$ respectively, the t-statistic is: $$\frac{(\overline{X}-\overline{Y})-(\mu_1-\mu_2)}{\sqrt{\frac{S_1^2}{n_1}+\frac{S_2^2}{n_2}}}\sim t(\nu) $$ which has a t-distribution with $\nu$ degrees of freedom equal to: $$\Large \nu=\frac{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2} \right)^2}{\frac{\left(\frac{s_1^2}{n_1} \right)^2}{n_1-1}+\frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}} $$ Actually, I read that this statistic only approximates a t-distribution. Regardless, is there any resource where I can read more about this? I only see resources citing this result with no references giving/outlining a proof for this.


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