1

Let's say that I know the following:

  • $P(A|B)$ is the probability that a storm is coming given it's cloudy.
  • $P(A|C)$ is the probability that a storm is coming given that the dogs bark.
  • $P(B)$ and $P(C)$ are independent.

How do I compute the following?:

  • $P(A|B,C)$, the probability of a storm coming given that it's cloudy AND the dogs are barking.

In layman's terms, I know that there is some likelihood that a storm is coming if it's cloudy. And, I know that there is some likelihood that a storm is coming if the dogs are barking. Therefore, shouldn't I have more confidence that a storm is coming if it's cloudy AND the dogs are barking? How do compute this?

The reason that I ask this question is because I am trying to combine measurements from two different sensors that measure the same thing. If I combine the measurements, shouldn't I expect greater confidence in my measurement?

This post and this post are related to my question, but the answers fall short in that I do not know the general probabilities of $P(B)$ and $P(C)$ to compute $P(A|B,C)$.

William Grand
  • 123
  • 1
    Two concepts to familiarize yourself with are conditional independence and Bayes rule. Also know that you can sum over a joint distribution to get a marginal distribution: e.g. in loose notation $P(Y) = \sum_X P(XY)$. – Matthew Gunn Nov 03 '22 at 15:39
  • 1
    If the only inputs are the conditional probabilities $P(B|A)$ and $P(C|A)$ one cannot compute $P(A|B\cap C)$ [in general]. – Xi'an Nov 03 '22 at 16:01
  • Thanks @Xi'an. To put my question into perspective, I would liken it to a scenario where I seek a second opinion from a doctor to determine if I have cancer:

    Let's say I want to know if I have cancer ($P(A)$). Doctor $B$ says I have cancer and is 75% sure ($P(B|A)=0.75$). I get a second opinion and Doctor $C$ also says I have cancer and is 75% sure ($P(C|A)=0.75$). Shouldn't I have a greater confidence than 75% that I have a cancer? Surely it is worth seeking a second opinion.

     – William Grand Nov 03 '22 at 19:21

3 Answers3

3

Note: This answer is to the original version of the question, which asked whether $\mathbb{P}(B|A)$ and $\mathbb{P}(C|A)$ can be used with no other information to obtain $\mathbb{P}(A|B,C)$.

Consider two scenarios, both of which have $\mathbb{P}(B|A) = \mathbb{P}(C|A) = 0.5$:

Scenario 1:

\begin{align*} \mathbb{P}(A,B,\overline C) &= 0.2 \\ \mathbb{P}(A,\overline B, C) &= 0.2 \\ \mathbb{P}(\overline A, B, C) &= 0.2 \\ \mathbb{P}(\overline A, \overline B, \overline C) &= 0.4 \end{align*}

Scenario 2:

\begin{align*} \mathbb{P}(A,B,C) &= 0.1 \\ \mathbb{P}(A,B,\overline C) &= 0.1 \\ \mathbb{P}(A,\overline B, C) &= 0.1 \\ \mathbb{P}(A,\overline B,\overline C) &= 0.1 \\ \mathbb{P}(\overline A, \overline B, \overline C) &= 0.6 \end{align*}

In scenario 1, $\mathbb{P}(A|B,C) = 0$. In scenario 2, $\mathbb{P}(A|B,C) = 1$. Clearly, you'll need more than just $\mathbb{P}(B|A)$ and $\mathbb{P}(C|A)$ to compute $\mathbb{P}(A|B,C)$.

josliber
  • 4,367
  • 28
  • 43
  • I'm not sure I follow. Perhaps I phrased my question incorrectly. To put it into perspective, my question could be changed to a scenario that is more familiar to us, "How do I quantify the probability of having cancer if I got an opinion from two different doctors?" – William Grand Nov 03 '22 at 19:33
  • 1
    @WilliamGrand I think you're still going to need some additional information. If A indicates you having cancer and B and C are two different doctors' prognosis, then $\mathbb{P}[B|A] = \mathbb{P}[C|A] = 0.75$ could be very informative (for instance, if $\mathbb{P}[B|\overline A] = \mathbb{P}[C|\overline A] = 0$) or very uninformative (for instance, if $\mathbb{P}[B|\overline A] = \mathbb{P}[C|\overline A] = 0.75$ -- the doctors are just guessing!). – josliber Nov 03 '22 at 19:54
  • Thanks. I believe that is exactly how to phrase the question. What additional information are you suggesting that I need? – William Grand Nov 03 '22 at 20:32
1

The answer is the conflation of probabilities, explained here.

In the syntax that answers my question, the equation becomes:

$P(A|B,C)=\eta{P(A|B) P(A|C)}$

where $\eta$ is the normalization factor:

$\eta=\left({P(A|B)P(A|C) + P(\overline A|B)P(\overline A|C)}\right)^{-1}$

My question could be changed to a scenario that is more familiar to us, "How do I quantify the probability of having cancer if I got an opinion from two different doctors?" Surely getting a second opinion will give us more confidence about the prognosis! If $A$ indicates you having cancer and $B$ and $C$ are two different doctors' prognosis (thanks @josliber), where $P(A|B)=0.75$ and $P(A|C)=0.75$, then applying those values in the equation above gives us:

$P(A|B,C)=\frac{0.75 * 0.75}{0.75 * 0.75 + (1-0.75)*(1-0.75)}=0.9$

Two doctors were 75% confident about their prognosis. Combining those prognoses gives us 90% confidence that I have cancer.

William Grand
  • 123
  • By reading the linked answer, it looks like you're actually treating your data as $\mathbb{P}(A|B) = 0.75$ and $\mathbb{P}(A|C) = 0.75$, aka the probability that you have cancer given diagnosis by each of the two doctors. This is not what's written in the question or answer. Also, there's a very strong independence assumption being made, that's not acknowledged in your question or answer. – josliber Nov 04 '22 at 14:12
  • As a thought exercise, imagine if doctor C always copied whatever doctor B says. Then clearly you wouldn't get any more confidence in your diagnosis from collecting C in addition to B. This highlights your need for independence assumptions. – josliber Nov 04 '22 at 14:15
  • Ok, yes, $B$ and $C$ are independent. I suppose I should make that clear in the question. $P(A|B)$ vs. $P(B|A)$ almost sounds like a debate over semantics, i.e. what is the likelihood I have cancer given a positive prognosis vs. what is the likelihood I have a positive prognosis given I have cancer. It's difficult for me to identify which to use. – William Grand Nov 04 '22 at 14:30
  • $\mathbb{P}(A|B)$ and $\mathbb{P}(B|A)$ seem pretty different to me (in more than semantic ways!); luckily they're linked by Bayes' theorem as long as we know $\mathbb{P}(A)$ and $\mathbb{P}(B)$. Please add the independence assumption if you know it to be true! For cancer diagnosis by two differennnt doctors, I'd say it's rarely if ever true, but perhaps in your setting it could be. – josliber Nov 04 '22 at 14:46
  • @josliber, I appreciate your help. The English semantics confuse me, but I went ahead and flipped the variables in the question and in my answer. I hope that they make more sense now. – William Grand Nov 04 '22 at 17:01
  • OK, I guess it's a totally different question now, so my answer makes less sense =/. Glad you found the answer you were looking for, though. – josliber Nov 07 '22 at 14:46
  • 1
    Your equation makes little sense why should we have the value $P(A|B,C) = 0.9$? Counterexample: consider the case where P(A) = P(B) = P(C) = 0.75 and A,B,C are independent, then the same values are entered in your formula but we should have $P(A|B,C) = 0.75$. – Sextus Empiricus Nov 07 '22 at 15:57
0

given values of

P(A|B,C) P(A|B,!C) P(A|!B,C) P(A|!B,!C)

given

P(B), P(C) and P(A|B) and P(A|C)

given B and C are independent

Then:

P(A|B) = P(A|B,C) * P(C) + P(A|B,!C) * P(!C) P(A|C) = P(A|B,C) * P(B) + P(A|!B,C) * P(!B)

This only gives you two equations but three unknowns P(A|B,!C), P(A|!B,C) and P(A|B,C).

Sextus Empiricus
  • 77,915