(Bayesian probability) Show that $P(H|E) = \frac{h (c +a \overline c)}{hc+a\overline c}$

Question

The question, briefly

How does the calculation $P(H|E) = \frac{h (c +a \overline c)}{hc+a\overline c}$ work?

Some background

I'm trying to work through the proof of the following theorem in Heinzelmann and Hartmann (2022):

Theorem 1. $P(H | E, \neg O ) > P(H|E) > P(H)$.

I don't quite follow the proof. I quote Heinzelmann and Hartmann's (2022) proof below, up to the step I find puzzling. Specifically, my problem is with the calculation $P(H|E) = \frac{h (c +a \overline c)}{hc+a\overline c}$.

We consider the Bayesian network in Fig. 1 and use the machinery of Bayesian networks (..). With this, it is easy to see that $P(H) = op + \overline o q =: h$, where $\overline x := 1 - x$. Note that the above expression for $P(H)$ and condition (3) imply that $p < h <q$. **Next we use the product rule and calculate $P(H|E) = \frac{h (c +a \overline c)}{hc+a\overline c}$.

Here is the Directed Acyclical Graph (DAG) relevant for the calculation.

And here's some notation and probabilities they use in the proof:

• $o := P(O)$
• $p := P(H|O)$
• $q := P(H|\neg O)$
• $c := P(c)$
• $a := P(E|H, \neg C) = P(E|\neg H, \neg C)$
• $P(E|H,C)=1$
• $P(E|\neg H,C)= 0$
• $P(E|H,\neg C)=a$
• $P(E|\neg H, \neg C)=a$
• Condition (3): $p < q$

Here is my understanding of how the proof goes. Firstly, they apply the Law of Total Probability to express the marginal probability P(H). So far, so good.

Next, they state they use the product rule it in calculating $P(H|E) = \frac{h (c +a \overline c)}{hc+a\overline c}$. This is where I don't follow. My understanding is that the product rule (i.e., the chain rule) applies to joint probabilities, not conditional probabilities:

Chain rule. $P(A \cap B) = P(B|A)\cdot P(A)$

I didn't figure out where the product rule was used. So I tried (unsuccessfully) doing the calculation in detail myself.

My attempt at showing $P(H|E) = \frac{h (c +a \overline c)}{hc+a\overline c}$.

My unsuccessfull attempt follows.

Attempt. By definition of conditional probability, $$P(H|E) = \frac{P(E \cap H)}{P(E)}.$$ Then, by applying the law of total probability in the numerator we'd get $$\frac{P(E|H,C)\cdot P(H,C)+P(E| H\neg C)\cdot P(H,\neg C)}{P(E)}. $$ Now, we see that H and C are d-separated in the DAG, so $P(H, C) = P(H)\cdot P(C)$. We use this to get $$\frac{P(E|H,C)\cdot P(H)\cdot P(C)+P(E| H\neg C)\cdot P(H)\cdot P(\neg C)}{P(E)}. $$ Then we move the $P(H)$ outside the brackets like this: $$ \frac{(P(E|H,C)\cdot P(C)+P(E| H\neg C)\cdot P(\neg C))\cdot P(H)}{P(E)}. $$ Now we plug in the values, as specified above, which results in $$ \frac{(1 \cdot c+a\cdot \overline c)\cdot h}{P(E)} = \frac{h(c+a\overline c)}{P(E)}, $$ as desired. So, the numerator is done.

Next, I think I should expand $P(E)$ to $hc+a\overline c$. Namely, I should show $P(E) = P(H)\cdot P(C) + P(E| H, \neg C) \cdot P(\neg C)$. Perhaps the law of total probability (LOTP) was used at some point in the Heinzelmann and Hartmann calculation? Applying the LOTP to $P(E)$ gets us: $$\frac{h(c+a \overline c)}{hc + ah \overline c + a \overline h \overline c},$$ which is not quite there yet, and I don't see how to proceed. End of attempt.

What am I missing here? Could someone perhaps give me some pointers on how to proceed, or comments on where I've gone wrong? I'd very much appreciate it.

Answer 1

$\overline h= 1-h$ by definition, so: $$ah\overline c+a\overline h\overline c ~{= a(h+\overline h)\overline c\\=a\overline c}$$