Confidence interval and t-student distribution? Corollary:
Under the assumptions of Theorem, a $(1-\alpha)100\%$ confidence interval for $\mu_x-\mu_y$ is $(\overline X -\overline Y)\pm t_{n+m-2}(\alpha/2)S_{\overline X -\overline Y}$.
Theorem:

How can I prove the Corollary using the Theorem? The difference in the Corollary is that $\sigma^2_x\neq\sigma^2_y$.
Thank you in advance for your help.
 A: Since the Corollary says "under the assumptions of the Theorem," the two variances would still be equal in the Corollary.
So you have
\begin{align}
& -t < \frac{(\overline X - \overline Y) - (\mu_X-\mu_Y)}{s_p\sqrt{\frac 1 n + \frac 1 m}} < t \tag 1 \\[10pt]
& -ts_p \sqrt{\frac 1 n + \frac 1 m} < (\overline X - \overline Y) - (\mu_X-\mu_Y) < ts_p\sqrt{\frac 1 n + \frac 1 m} \\[10pt]
& \text{Then, multiplying all three expressions by $-1$, we get:} \\
&ts_p\sqrt{\frac 1 n + \frac 1 m} > (\mu_X-\mu_Y) - (\overline X - \overline Y) > - ts_p\sqrt{\frac 1 n + \frac 1 m} \\[10pt]
& (\overline X-\overline Y) + ts_p\sqrt{\frac 1 n + \frac 1 m} > \mu_X-\mu_Y > -ts_p\sqrt{\frac 1 n + \frac 1 m}. \tag 2
\end{align}
In the middle we have only the unobservable quantity to be estimated, and on the left and right we have only observable quantities. Since the probability of the event in line $(1)$ is $1-\alpha,$ so is the probability of the event in line $(2).$
If we take $S_{\overline X - \overline Y}$ to mean $s_p\sqrt{\frac 1 n + \frac 1 m},$ then that answers the question.
