Is it always possible to partition data in such a way that Simpson's paradox is achieved?

Question

$A,B,C,D $ are natural numbers such that: $$\frac AB ~>~\frac CD,~A\le B,~C\le D$$

Is it always possible to find natural numbers $a_1,a_2,b_1,b_2,c_1,c_2,d_1,d_2$ that satisfy: $$\text{1) }~~ a_1+a_2 = A,~b_1+b_2 = B,c_1+c_2 = C,d_1+d_2 = D~; $$ $$\text{2) }~~ \frac{a_1}{b_1} <~\frac{c_1}{d_1}~,\frac{a_2}{b_2} <\frac{c_2}{d_2}$$

Trivial cases that I exclude are if: $$A=B, A=0, C=0,B=1,D=1$$

Intuitively I expect the answer to be no. Because in the examples of Simpson's paradox percentages A/B and C/D differ slightly.

Answer 1

There are some additional obvious cases to exclude. For example, if $C=1$ we cannot construct the paradox, because either $c_1/d_1$ or $c_2/d_2$ is zero and not greater than any ratio of natural numbers.

If you allow $a_1 = 0$ and $c_1 \geq d_1,$ it seems possible to find the numbers in almost all other cases. For example, for $A=1,$ $B=2,$ $C=2,$ $D=5,$ we have $$ \frac12 > \frac25,$$ but $$ \frac01 < \frac14, \quad \frac11 < \frac21. $$

The case where $c_1 > d_1$ (which cannot arise in the natural context of the paradox) seems objectionable, however. If we exclude it, then I think we also exclude any cases where $B=2,$ since then the only possible values of $a_1/b_1$ and $a_2/b_2$ are $0$ and $1,$ as shown above.

There may be other small values of $A,B,C,D$ for which it is impossible to construct the paradox. I do not claim to have fully explored them.

The intuition that $A/B$ and $C/D$ must only "differ slightly," however, does not follow through if you are willing to consider large enough values of $A,B,C,D.$ That is, for large enough numbers, it seems possible to construct the paradox even when $A/B$ is much greater than $C/D.$ For example,

$$ \frac{1 + 899}{11+989} = \frac{9}{10} > \frac{1}{10} = \frac{90+10}{989+11}, $$

but

$$ \frac{1}{11} < \frac{90}{989}, \quad \frac{899}{989} < \frac{10}{11}. $$

So I think the limitation on finding instances of the paradox is associated only with the absolute sizes of $A,B,C,D,$ not with their relative sizes.

Answer 2

I assume (as antkam suggests in the comments) that you require $a_1\leq b_1$, etc, so that all of these fractions are probabilities.

Then there are a couple of potential barriers:

• If $C\leq 1$ then one of the $\frac{c_i}{d_i}$ must be $0$, and so can't be larger than $\frac{a_i}{b_i}$.
• If $B-A\leq 1$ then one of the $\frac{a_i}{b_i}$ must be $1$, since otherwise we would have $a_1+a_2\leq(b_1-1)+(b_2-1)<A$, and so it can't be smaller than $\frac{c_i}{d_i}$.

In all other cases we have the valid solution $$\frac{0}{1}<\frac{C-1}{D-1}\quad\text{and}\quad\frac{A}{B-1}<\frac11.$$