Confidence Interval on the Difference in Means, Variances Unknown and not Assumed Equal

Question

How can I prove that $$T=\dfrac{\overline{X}_1-\overline{X}_2-(\mu_1-\mu_2)}{\sqrt{\dfrac{S_1^2}{n_1}+\dfrac{S_2^2}{n_2}}}$$ has a $ t $ distribution?

Assumptions:

1. $ X_{11},\ldots,X_{1n_{1}} $ is a random sample from population 1.
2. $ X_{21},\ldots,X_{2n_{2}} $ is a random sample from population 2.
3. The two populations represented by $ X1 $ and $ X2 $ are independent.
4. $\sigma_{1}$, $\sigma_{2}$ unknown, and $ \sigma_1 \neq \sigma_2$.
5. Both populations are normal.

What I know:

• $ Z=\dfrac{\overline{X}_1-\overline{X}_2-(\mu_1-\mu_2)}{\sqrt{\dfrac{\sigma_1^2}{n_1}+\dfrac{\sigma_2^2}{n_2}}} \sim N(0,1)$.
• $ U=\dfrac{(n_1-1)S_1^2}{\sigma_1^2}+ \dfrac{(n_2-1)S_2^2}{\sigma_2^2} \sim \chi^{2}_{n_1+n_2-2}$
• If $ Z \sim N(0,1) $, $ U \sim \chi_{n_1+n_2-2}^{2} $ and $ Z $ and $ U $ are independent, then

$$ \dfrac{Z}{\sqrt{\dfrac{U}{n_1+n_2-2}}} \sim t_{n_1+n_2-2} $$

My attempt: I have formed the quotient

$$ \dfrac{\dfrac{\overline{X_{1}}-\overline{X_{2}}-(\mu_{1}-\mu_{2})}{\sqrt{\dfrac{\sigma_{1}^{2}}{n_{1}}+\dfrac{\sigma_{2}^{2}}{n_{2}}}}}{\sqrt{\left[\dfrac{(n_{1}-1)S_{1}^{2}}{\sigma_{1}^{2}}+ \dfrac{(n_{2}-1)S_{1}^{2}}{\sigma_{2}^{2}}\right]\dfrac{1}{n_{1}+n_{2}-2}}} \sim t_{n_1+n_2-2}$$

The problem with which I found is that unknown variances can not be eliminated as in the case $ \sigma_{1}^{2}=\sigma_{2}^{2}=\sigma^{2} $

Answer 1

I don't think the proposed theorem is true.

Google the term "Behrens–Fisher problem". The Wikipedia article on this is woefully incomplete, making it look as if one proposed way to model the Behrens–Fisher problem is the Behrens–Fisher problem. But use Google Scholar and Google Books.