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