Suppose that female and male students apply to schools A and B. Given that $p>q$ and $r>s$ where $p$ is the ratio of female students accepted to A, $q$ is the ratio of males accepted to A, $r$ is the ratio of females accepted to B and $s$ is the ratio of males accepted to B, prove that Simpson's paradox occur if and only if $q>r$ or $s>p$. Also, how can I generalize this result to more institutions?
Showing the only if part is easy but when I'm trying to show that paradox occurs if $q>r$ I get an inequality that leads to nowhere:
Let $p=p_1/p_2$, $q=q_1/q_2$, $s=s_1/s_2$, $r=r_1/r_2$. If $q>r$, $p>q$ and $r>s$ we have $q_1r_2+p_1q_2+r_1s_2+p_1s_2>q_2r_1+p_2q_1+r_2s_1+p_2s_1$. Now if I show that $q_1r_2$ is so big that the paradox occurs I'm done, but I can't. Is there something I'm missing?
You may have difficulty proving the if direction, as it not always true. Indeed in a sense it is usually not true
Suppose there are $40$ students in total and the number of applicants of each gender to each college is $10$, and consider $p=\frac{8}{10}$, $q=\frac{6}{10}$, $r=\frac{4}{10}$ and $s=\frac{2}{10}$, implying $p > q > r > s$
Then the overall ratio of females accepted is $\frac{12}{20}$, greater than the overall ratio of males accepted of $\frac{8}{20}$, and there is no paradox