Odds ratio of the intesections of 3 events with dependence

Question

Let's consider 3 events A,B and C. On a Venn diagram representing these three events, the different intersections are noted as follows:

$$abc = P(A\cap B\cap C)\;;\; ab = P(A\cap B \cap \bar C)\;;\; ac = P(A\cap \bar B \cap C)\;;\; a = P(A\cap \bar B \cap \bar C)$$

Venn diagram

If the three events A, B and C are mutually independent then :

$$ OR = \frac{a/ac}{ab/abc} = 1$$

I know that if event B and C are not independent then the odds ratio define above will no longer be equal to one.

Now here's my question, what happens to this odds ratio when event B and C are independent but events A and B (or A and C) are not independent ?

Thank you in advance for your clarifications on this subject !

Answer 1

I'll stick with the traditional $\mathbb P(\cdot)$ notation (rather than your $a, b, c$ notation) in order to discuss some conditional probability concepts.

First, without any assumptions of independence at all, by the definition of conditional probability, your original odds ratio can be rewritten as:

\begin{align*} & \frac{\mathbb P(A \mid \overline B \cap \overline C) \mathbb P(\overline B \cap \overline C) \bigg/ \left[\mathbb P(A \mid \overline B \cap C) \mathbb P(\overline B \cap C) \right] } {\mathbb P(A \mid B \cap \overline C) \mathbb P(B \cap \overline C) \bigg/ \left[ \mathbb P(A \mid B \cap C) \mathbb P(B \cap C)\right]} \\ = \ & \frac{\mathbb P(A \mid \overline B \cap \overline C) \mathbb P(A \mid B \cap C) \mathbb P(\overline B \cap \overline C) \mathbb P(B \cap C)} {\mathbb P(A \mid \overline B \cap C) \mathbb P(A \mid B \cap \overline C) \mathbb P(\overline B \cap C) \mathbb P( B \cap \overline C)} \end{align*}

The latter expression sheds some light on what's going on in the independence cases you've asked about. First, if all three are independent, then:

• $\mathbb P(A \mid \cdot) = \mathbb P(A)$, whence those four conditional expectation terms immediately cancel
• the terms that look like $\mathbb P(B \cap C)$ become $\mathbb P(B) \mathbb P(C)$; each of $\mathbb P(B), \mathbb P(\overline B), \mathbb P(C), \mathbb P(\overline C)$ appear exactly once in the numerator and the denominator, so they all cancel out as well

which is why you're left with 1 when the dust settles. This formulation also suggests how you might proceed when only $B, C$ are independent of one another; we get the same cancellation of the four terms on the right side, but the conditional probability terms remain uncancellable and your original odds ratio is equal to $$\frac{\mathbb P(A \mid \overline B \cap \overline C) \mathbb P(A \mid B \cap C)}{\mathbb P(A \mid \overline B \cap C) \mathbb P(A \mid B \cap \overline C)}.$$

Intuitively, these four terms encode how $A$ depends on $B, C$; specifically, they fully encode the probabilities of $A$ against all four of the possible configurations of the other two events.

I don't see a way to get an expression that's much cleaner than this, since those four conditional probabilities are basically free variables -- but it's possible I'm just missing something right now. In particular, it cannot be guaranteed that the ratio is 1 under these conditions.

For a quick example of why this is thorny, consider a case where $A$ and $B \cap C$ are mutually exclusive (i.e. $\mathbb P(A \mid B \cap C) = 0$); then this odds ratio collapses to $0$.