= Solution
For each normal sample, its <sample mean> is independent of its <sample variance>, and the scaled residual sum of squares is chi-squared. <Independence> of the samples also makes their two residual sums independent. Adding their chi-squared distributions gives
$$
\frac{(m-1)S_X^2+(n-1)S_Y^2}{\sigma^2}\sim\chi^2_{m+n-2}.
$$
Consequently
$$
\boxed{d=(m+n-2)/\sigma^2,\qquad dS^2\sim\chi^2_{m+n-2}}.
$$
The pooled <variance> is also independent of the difference of sample means.
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