For some events $A$, $B$, $C$
I know the occurrence probabilities $P(A),\: P(B),\: P(C)$
I also know the pairwise co-occurance probabilities $P(A,B),\: P(A,C),\: P(B,C)$

I want to approximate the triple co-occurance probability: $P(A,B,C)$ By making some (incorrect) assumptions (eg about independence.)

I know that if I assume all events are mutually independent then I can make the approximation: $P(A,B,C)\approxeq P(A)\;P(B)\;P(C)$

But I have more information than just the individual marginal probability since I also have the pairwise information, which is not used in the approximation above. So a better estimate should be possible

  • $\begingroup$ Do you know the probability$P(A\cup B\cup C)$? then you can exactly calculate the $P(A\cap B \cap C)$ $\endgroup$
    – Deep North
    Oct 2, 2015 at 6:05
  • $\begingroup$ No. I only know pairwise statistics. $\endgroup$ Oct 2, 2015 at 6:11

1 Answer 1



$$ P(A \cup B \cup C) = P(A) + P(B) + P(C) - P(A \cap B) - P(B \cap C) - P(C \cap A) + P(A \cap B \cap C) $$ $\implies$

$$ P(A \cap B \cap C) = P(A \cup B \cup C) - \big(P(A) + P(B) + P(C)\big) + \big(P(A \cap B) + P(B \cap C) + P(C \cap A)\big) $$


$$ 0 \leq P(A \cup B \cup C) \leq 1$$

$\implies$ $$ - \big(P(A) + P(B) + P(C)\big) + \big(P(A \cap B) + P(B \cap C) + P(C \cap A)\big) \leq P(A \cap B \cap C) \leq 1- \big(P(A) + P(B) + P(C)\big) + \big(P(A \cap B) + P(B \cap C) + P(C \cap A)\big) $$


$$ \begin{align} P(A \cap B \cap C) &= P\big((A \cap B) \cap C\big)\\ &= P(C|(A \cap B))P(A \cap B) \tag{1}\\ &or\ P(B|(A \cap C))P(A \cap C) \tag{2}\\ &or\ P(A|(B \cap C))P(B \cap C) \tag{3} \end{align} $$ Which assuming conditional independence is respectively equal to:

$$ \begin{align} P(A \cap B \cap C) &= P(C)P(A \cap B) \tag{1a}\\ &or\ P(B)P(A \cap C) \tag{2a}\\ &or\ P(A)P(B \cap C) \tag{3a} \end{align} $$

  • $\begingroup$ Thanks. Still reading over this a few times, but I don't understand the part of your answer below "Approximating". 1) What is "or" 2) How is $P(C|(A \cup B)$ calculated from only the pairwise/individual probabilities? $\endgroup$ Oct 2, 2015 at 9:09
  • $\begingroup$ Ahh I see how 2) works it $P(C|(A \cup B)=P(C)$ if $C$ is conditional independent of $A\capB$ And 1.) the "or" is depending on which I dependancy I take. Could I take the average of the 3 ors to get a better approximation? $\endgroup$ Oct 2, 2015 at 9:13
  • $\begingroup$ @rightskewed are you approximating the value if you give three possibilities? I think that you would want to aggregate these three values in some way based on the underlying problem. As Oxinabox suggests, this could be an average, but if this is in some application, then you might rather want to go for the $\min$ or the $\max$. $\endgroup$
    – Gumeo
    Oct 2, 2015 at 9:46
  • $\begingroup$ I can't think of a strict inequality right now that will motivate the use of $max$ or $min$ $\endgroup$ Oct 2, 2015 at 19:07

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