Conditional independence of events that result in a definite outcome

Question

Two different diseases cause a certain weird symptom; anyone who has either or both of these diseases will experience the symptom. Let $D_{1}$ be the event of having the first disease, $D_{2}$ be the event of having the second disease, and $W$ be the event of having the weird symptom. Suppose $D_{1}$ and $D_{2}$ are independent with $P(D_{j}) =p_{j}$, and that a person with neither of these diseases will have the weird symptom with probability $w_{0}$. Let $q_{j} = 1 - p_{j}$, and assume that $0 < p_{j} < 1$.

a) Find $P(W)$

Let $H$ be the event that a person doesn't have either of the diseases. Then

$$P(W) = P(D_1) + P(D_2) - P(D_1, D_2) + P(H)P(W\mid H) = p_1 + p_2 - p_1 p_2 + q_1 q_2 w_0.$$

b) Find $P(D_{1}|W), P(D_{2}|W), P(D_{1}, D_{2}|W)$

$$P(D_{1}\mid W) = \frac{P(D_{1})P(W|D_{1})}{p_{1} + p_{2} + q_{1}q_{2}w_{0}} = \frac{p_{1}}{p_{1} + p_{2} - p_{1}p_{2} + q_{1}q_{2}w_{0}}.$$

$$P(D_{2}\mid W) = \frac{P(D_{2})P(W\mid D_{2})}{p_{1} + p_{2} + q_{1}q_{2}w_{0}} = \frac{p_{2}}{p_{1} + p_{2} - p_{1}p_{2} + q_{1}q_{2}w_{0}}.$$

$$P(D_{1}, D_{2}\mid W) = \frac{P(D_{1}, D_{2})}{P(W)} = \frac{p_{1}p_{2}}{p_{1} + p_{2} - p_{1}p_{2} + q_{1}q_{2}w_{0}}.$$

c) Determine algebraically whether or not $D_{1}$ and $D_{2}$ are conditionally independent given $W$.

Since $P(D_{1}\mid W)P(D_{2}\mid W) \neq P(D_{1},D_{2}\mid W)$, they are not independent.

d) Suppose for this part only that $w_{0} = 0.$ Give a clear, convincing intuitive explanation in words of whether $D_{1}$ and $D_{2}$ are conditionally independent given $W$.

Even if $q_{1}q_{2}w_{0} = 0,$ $P(D_{1}\mid W)P(D_{2}\mid W) \neq P(D_{1},D_{2}\mid W)$ so the events are still not conditionally independent.

The way part $d$ is phrased makes me think the events are supposed to be independent, but my algebra shows otherwise, and I am struggling to come up with a clear, intuitive explanations for why they would be independent. Any suggestions?

Answer 1

When $w_0=0$ they are not conditionally independent, and you have an example of Berkson's paradox. They are only conditionally independent when $w_0=1$, in which case everyone is weird.

I would say you may not have given an "explanation in words". Perhaps you could say something like

"When $w_0=0$ everyone with the weird symptom has at least one of the diseases. So anyone with the weird symptom but not $D_1$ must have $D_2$ while only a proportion $p_2$ of those with the weird symptom and $D_1$ are also expected to have $D_2$, and thus $D_1$ and $D_2$ are not conditionally independent given the weird symptom."

Answer 2

This should be a comment, but not enough reputation here.

The entire reasoning depends on $P(D_1 \cup D_2) < 1$, an assumption which is not stated. There must be at least one healthy person.

If $P(D_1 \cup D_2) = 1$, the definition of $w_0$ is meaningless:

$$ \begin{aligned} w_0 &= P(W \mid \neg(D_1 \cup D_2))\\ &= \frac{P(W \cap \neg(D_1 \cup D_2))}{P(\neg(D_1 \cup D_2))} \end{aligned} $$

but $P(\neg(D_1 \cup D_2)) = 0$.