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Assume I have three events A, B, and C, and I know the following probabilities:

Scenario 1:

  • $P(A)$ and $P(B)$
  • $P(C|A)$ and $P(C|B)$

Scenario 2:

  • I additionally know $P(C)$.

I am looking for $P(C|A\cap B)$.

I think it is clear that I cannot calculate this conditional probability in both scenarios because no information is available about the three-fold interaction of A, B, and C; in particular, $P(A\cap B \cap C)$ is unknown.

But what would be the best approximation for $P(C|A\cap B)$?

$P(C|A)$ and $P(C|B)$, and even $P(C)$ might be candidates, but it is clear that none of them can generally be the best approximation because each of them "ignores" available information (about B, A, and both, resp.).

Maybe someone could also tell me terms or even sources where I should lookup this kind of question in literature; "information fusion" seems to be related, but I could not find the answer there, and it seems to be a basic question actually.

Edit: Changed "estimate" by "approximate" as suggested in the comments.

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  • $\begingroup$ Hi: I don't know the answer to your question but you already know P(C) in scenario 1 because P(C) = P(C|A)P(A) = P(C|B) P(B). So, you can just stick to scenario 1. Also, do you happen to know if A and B are independent ? It is an interesting question. I thought it would be easy but it doesn't seem to be. I don't know if there's a "field" associated with this problem. It seems more like a "it can be done or it can't be done" type of question. $\endgroup$
    – mlofton
    Commented Aug 1, 2023 at 6:05
  • $\begingroup$ @mlofton: I don't think this is true because $P(C|A) \cdot P(A) = P(C \cap A)$, not $P(C)$. In general, A and B are not independent, but that is of course an interesting special case. Estimation can surely be done ($P(C|B)$, e.g., is an example of an estimator), but the question is, what is the best one? $\endgroup$
    – Remirror
    Commented Aug 1, 2023 at 6:22
  • $\begingroup$ Do you have any further information? For example, is it possible that C occurs but none of A or B is true? Does this question have a practical background? $\endgroup$
    – Ute
    Commented Aug 1, 2023 at 8:57
  • $\begingroup$ @Ute: No further information in the general case. Yes, I think the question has large practical background, but not a specific one. A common usecase should be that you have data about C in two different datasets, where one dataset additionally contains data about A but not B and vice versa for the other dataset. $\endgroup$
    – Remirror
    Commented Aug 1, 2023 at 9:19
  • $\begingroup$ Ah, you got an answer: insufficient information. That is why I was asking ;-) $\endgroup$
    – Ute
    Commented Aug 1, 2023 at 9:21

1 Answer 1

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It can't be done, you have insufficient information. What you need to know is

$$ P(C|A \cap B) = \frac{P(A \cap B \cap C)}{P(A \cap B)} $$

As you can see, for this you need to know the joint probabilities for $A \cap B$, which you don't. You said that you know that it cannot be calculated and you are looking for the "best estimator", but there is really nothing good. You would need to know the relation between $A$ and $B$. It is a completely different story if $A$ happens always when $B$ happens, then if $A$ happens never when $B$ happens, or if they are independent.

For a simple example, imagine that you are trying to guess if a patient will get cured $C$ using the medicine $A$ or $B$. You know that both medicines are highly effective, hence $P(C|A)$ and $P(C|B)$ are high, what you don't know is what is going to happen if the patient uses both drugs. It may be the case that using the two drugs together would be not better than any of the drugs alone, it also might be the case that they together are twice as good, on another hand, there is a risk that the drugs used together come into a chemical reaction that not only makes them not successful, but they lead to possibly deadly side effects. As you can see, it can be anything, and the individual probabilities don't tell you much about the joint probability.

You could assume that $C$ is independent of $A$ and $B$, then $P(C|A\cap B) = P(C)$, or assume conditional independence and use one of the conditional probabilities as a replacement, but I doubt that this is the answer you were looking for. Neither of the solutions can be considered "best" unless you know something about their joint distribution, but if you knew it, you would not have the problem in the first place.

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  • $\begingroup$ True, but I just do not believe that there is no approach that, at the very least, is "on average" (i.e. accross all possible cases) better than others. For instance, I assume that using $P(C|A)$ as estimator should be better on average than using $P(C)$. $\endgroup$
    – Remirror
    Commented Aug 1, 2023 at 9:22
  • $\begingroup$ @Remirror why? By the law of total probability $P(C) = \sum_i \sum_j P(A_i \cap B_j \cap C)$, so $P(C)$ is the "on average" answer itself. No reason why it would be better or worse than the conditional probabilities. $\endgroup$
    – Tim
    Commented Aug 1, 2023 at 9:27
  • $\begingroup$ For my understanding, $P(C|A)$ should be a better estimator than $P(C)$ iff C is not independent of A. In other words, your sum averages accross all potential values of A, but that is inefficient if we know the true value of A. $\endgroup$
    – Remirror
    Commented Aug 1, 2023 at 10:30
  • $\begingroup$ @Remirror Yes, if you assume conditional independence on $B$. Imagine a trivial counterexample where $P(A \cap B \cap C) = 0$ and all the other probabilities are non-zero. Then, $P(C | A \cap B) = 0$, and the other probabilities (marginal and conditional) are some values that are non-zero. The other probabilities tell you nothing about this case. $\endgroup$
    – Tim
    Commented Aug 1, 2023 at 10:54
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    $\begingroup$ @Remirror to convince yourself, create a simple $8 \times 3$ table in Excel for states (yes/no) of $A$, $B$, $C$, and with the fourth column for counts, then you can calculate probabilities like $P(A \cap B \cap \neg C)$ by summing appropriate counts and dividing by total. You would easily see that by manipulating the counts you can make the conditional and marginal probabilities whatever you want and the conditionals can easily be made very different from each other. $\endgroup$
    – Tim
    Commented Aug 1, 2023 at 11:04

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