How is it possible to have $P(A|B \cup C)$ lower than both $P(A|B)$ and $P(A|C)$?

Question

Let's say we investigate disease probability given two symptoms. Thus we have 3 variables:

• A) has disease
• B) has symptom 1
• C) has symptom 2

We have following data:

Now we want to see which symptom makes probability of disease higher.

First we calculate probability of disease when having symptom 1. We can easily calculate that conditional probability P(A|B) = 2/3 as there are three cases of symptom 1 and two cases of disease when having symptom 1.

We do the same for symptom 2. As there are three cases of symptom 2 and two cases of disease when having symptom 2 the conditional probability P(A|C) = 2/3

Now we want to get some reference value, which would tell us what is overall probability of disease when having at least one symptom. So we calculate P(A|B∪C) = 1/2 as there are 4 cases with at least one symptom, and in 2 of them disease is also present.

However, how can we explain the fact, that having symptom 1 without consideration of symptom 2 (and vice versa) leads to higher probability of disease, than having at least one these symptoms?

If there would be 50 symptoms in dataset and we need to tell which of them makes the probability of disease lower than case has random set of symptoms we could not do that, as the reference value (which I would intuitively expect to be ~average) is actually the lowest one?

How should I define the reference general probability that random combinations of symptoms leads to disease? One way that I tried would be to calculate this reference, would be to multiply each row where disease is present with number of present symptoms in that row and then divide sum of this by total sum of symptoms, followingly:

Thus I get 4/6 = 2/3. Is this approach correct? If yes, is it called somehow?

Answer 1

Let's elucidate this issue using a Venn diagram for the three sets at issue. Let $A$ be the orange circle, let $B$ be the violet circle and let $C$ be the green circle, and denote the probabilities of the intersection-areas in the diagram by $P_1,...,P_7$ (ignoring the area outside the three sets).

From this diagram we can see that:

$$\begin{align} \mathbb{P}(C | A \cup B) &= \frac{P_4+P_5+P_7}{P_2+P_3+P_4+P_5+P_6+P_7}, \\[14pt] \mathbb{P}(C | A) &= \frac{P_4+P_7}{P_2+P_4+P_6+P_7}, \\[14pt] \mathbb{P}(C | B) &= \frac{P_5+P_7}{P_3+P_5+P_6+P_7}, \\[14pt] \max(\mathbb{P}(C | A) + \mathbb{P}(C | B)) &= \max \Bigg( \frac{P_4+P_7}{P_2+P_4+P_6+P_7}, \frac{P_5+P_7}{P_3+P_5+P_6+P_7} \Bigg). \\[12pt] \end{align}$$

From this result we can see that $\mathbb{P}(C | A \cup B)$ will tend to be lower than $\max(\mathbb{P}(C | A) + \mathbb{P}(C | B))$ when $P_7$ is large in comparison to $P_6$. To give a more specific example, suppose we take the symmetry conditions $P_2 = P_3$ and $P_4 = P_5$ for simplicity, which then reduces things to:

$$\begin{align} \mathbb{P}(C | A \cup B) &= \frac{2P_4+P_7}{2P_2+2P_4+P_6+P_7}, \\[14pt] \max(\mathbb{P}(C | A) + \mathbb{P}(C | B)) &= \frac{P_4+P_7}{P_2+P_4+P_6+P_7}, \\[12pt] \end{align}$$

Now, if we also take $P_2 P_7 > P_4 P_6$ then we have:

$$\begin{align} \max(\mathbb{P}(C | A) + &\mathbb{P}(C | B)) - \mathbb{P}(C | A \cup B) = \frac{P_4+P_7}{P_2+P_4+P_6+P_7} - \frac{2P_4+P_7}{2P_2+2P_4+P_6+P_7} \\[6pt] &= \frac{(P_4+P_7)(2P_2+2P_4+P_6+P_7) - (2P_4+P_7) (P_2+P_4+P_6+P_7)}{(P_2+P_4+P_6+P_7)(2P_2+2P_4+P_6+P_7)} \\[6pt] &= \frac{(P_4+P_7)(P_2+P_4) - P_4 (P_2+P_4+P_6+P_7)}{(P_2+P_4+P_6+P_7)(2P_2+2P_4+P_6+P_7)} \\[6pt] &= \frac{P_2 P_7 - P_4 P_6}{(P_2+P_4+P_6+P_7)(2P_2+2P_4+P_6+P_7)} > 0. \\[6pt] \end{align}$$

As you can see, this gives the inequality that has occurred in your example. So we can see that this is certainly possible, and it tends to occur when $P_7$ is large in comparison to $P_6$. In the context of your particular question, this occurs because the probability of having both symptoms and the disease is substantially larger then the probability of having both symptoms but no disease (the latter is zero in your analysis).