Sorry in advance for the confusing title. I'm not quite sure on the proper way to describe this question.

Suppose I have $n$ events: $A, B, C, ...., N$ (I'm using individual letters instead of proper indexing to make it easier to see what I would like to do). For these events I know the following probabilities

$$P(A), P(A | B), P(A | C) , ... P(A | N)$$ $$P(B), P(B | A), P(B | C) , ... P(B | N)$$ $$:$$ $$P(N), P(N | A), P(N | B) , ... P(N | M)$$

What I'm interested in calculating is $$P(A|B \cap C \cap ... \cap N)$$

Is this possible? I feel as though this question is utterly trivial but I have not been able to figure out how to determine $P(A|B \cap C \cap ... \cap N)$ or show that it can't be calculated with the information I have. If it is possible, how would I do it for $n = 3$? I suspect that if I can do it for $n = 3$ a recursive function could be used to calculate for whatever $n$ I want. Ultimately I would like to be able to solve the above probability for around $n = 50$ and I would like to know whether the information I have is sufficient (sans computational complexity)

  • $\begingroup$ Unless you have some kind of conditional inpendence assumptions, it's difficult to figure out $P(A\cap B\cap C\cdots)$, when you are just given single conditionals. $\endgroup$
    – Alex R.
    Commented Feb 21, 2018 at 19:57

2 Answers 2


I've been working on this problem a few years ago. It is far from trivial.

Basically, you don't have enough information to answer. Actually, assume you have three events $A$, $B$, $C$. What you known is:

  • the probability of each individual event : $P(A), P(B), P(C)$ (1)
  • the probability of each pariwise conjunctions: $P(A\cap B), P(A\cap C), P(B\cap C)$ (2)

If you consider the sigma (=boolean) algebra generated by $A,B,C$, you know everything except the missing information $P(A\cap B\cap C)$. With this missing information, you could calculate absolutely everything involving $A,B$ and $C$ like for example $P(A|B\cap C)$

Since you have some missing information, you need to estimate it. The solution I've found (that works very well on many problems) is to use the principle of maximum entropy on the sigma algebra generated by $A,B,C$:

  • Call $P$ the probability measure on this sigma algebra
  • Assume (1) and (2) as constraints
  • Find the value of $P(A\cap B\cap C)$ that maximizes the entropy of the probability measure

The problem can be extended to $n>3$ events just the same, except that you have a high dimensional vector of missing information.

Again, the solution to this problem is far from trivial. But luckily maximum entropy has been intensively studied, and we know how it leads to exponential families solutions. The problem is very similar to logistic regression. You can read literature about fitting maximum entropy. Actually, the most efficient solution is to use standard convex optimization algorithms once you have transformed the problem into an unconstrained optimization problem. A good resource : https://web.cs.elte.hu/blobs/diplomamunkak/msc_mat/2013/biszak_elod.pdf


This is what is known.

  1. Marginal probabilities: $P(A) ... P(N)$
  2. Marginal complements: $P(A^c) ... P(N^c)$
  3. Simple conditionals: $P(A|B) ... P(N|M)$

By 1. know $P(B)$

By 3 know $P(A|B)$


$P(A|B) = P(A \cap B)/P(B)$

$\implies P(A \cap B)$

B. 1. Need to know $P(C|A \cap B) $ but it is not known (Mistake is here).

$P(C|A \cap B) = P(C \cap A \cap B)/P(A \cap B)$

$ \implies P(A \cap B \cap C)$

Eventually we will get $P(A \cap B \cap C ... \cap N)$


Then we will get $P(B \cap C ... \cap N)$

That would get the answer.

  • $\begingroup$ How do you calculate $P(C|A \cap B)$? In your second line there are two unknowns as I see it, $P(C|A \cap B)$ and $P(A \cap B \cap ... \cap N)$ $\endgroup$
    – HXSP1947
    Commented Feb 22, 2018 at 8:08
  • $\begingroup$ @HXSP1947 You are right. I assumed something $P(C|A \cap B)$ that was not given. Maybe knowing $P(A \cup B .. \cup N) = 1$ would be enough. $\endgroup$ Commented Feb 22, 2018 at 13:15

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