probability of a set of events from the unconditional probabilities and the probabilities of every pair of events

Question

Obviously, for two events $A$ and $B$, you can not infer $P(A \cap B)$ from $P(A)$ and $P(B)$ if you don't know $P(A|B)$ or $P(B|A)$.

I have three events, $A,B$ and $C$, and I know $P(A), P(B), P(C), P(A \cap B), P(A \cap C), P(B \cap C)$. I suppose, from this, there is no way to compute $P(A \cap B \cap C)$.

Is this right and is there a way to prove it? And how does this generalise to more Events: If you only know the unconditional probabilities of every event, and the probabilities of every pair of events $(P(A \cap B), P(A \cap C), P(A \cap D), P(B \cap C), ...)$, can you show that there is no way of computing the probability of all events together $(P(A \cap B \cap C \cap D ...))$?

Answer 1

The events $A,B,C$ induce a pseudo-partition of $\Omega$ that contains $8$ events.

With a pseudo-partition (unofficial and personal terminology) I mean a collection of disjoint sets that cover $\Omega$ but do not necessarily satisfy the condition that they are not empty.

Elements are the events of the form $E\cap F\cap G$ where $E\in\{A,A^\complement\}$, $F\in\{B,B^\complement\}$ and $G\in\{C,C^\complement\}$.

If $(U_1,U_2,\dots,U_8)$ is a tuple containing these $8$ sets then at first hand we have the equality:$$\sum_{i=1}^8P(U_i)=1$$

To get a situation where the $P(U_i)$ are determined we need $7$ extra conditions.

The knowledge of e.g. $P(A\cap B)$ can be looked at as such a condition.

But knowing $P(A), P(B), P(C), P(A \cap B), P(A \cap C), P(B \cap C)$ is not enough because that only gives us $6$ conditions.

The same approach works if we are dealing more generally with $n$ events.

In that case we need $2^n-1$ extra conditions.

Answer 2

You need only one counterexample for a disproof.

One way is to suppose a probability space containing several disjoint events $E_1,E_2,E_3,\ldots.$ Assign a probability to each $E_i$ and define each of the events $A_1,B_1,C_1,A_2,B_2,C_2$ as a union of some collection of the events $E_1,E_2,E_3,\ldots$ in such a way that

\begin{align} P(A_1) &= P(A_2), \\ P(B_1) &= P(B_2), \\ P(C_1) &= P(C_2), \\ P(A_1 \cap B_1) &= P(A_2 \cap B_2), \\ P(A_1 \cap C_1) &= P(A_2 \cap C_2), \\ P(B_1 \cap C_1) &= P(B_2 \cap C_2), \\ P(A_1 \cap B_1 \cap C_1) &\neq P(A_2 \cap B_2 \cap C_2). \\ \end{align}

Then given only these same probabilities for the first six of the corresponding combinations of $A,$ $B,$ and $C$ it is impossible to know whether the three sets are $A_1,B_1,C_1$ or $A_2,B_2,C_2$ and therefore impossible to determine the value of $P(A \cap B \cap C).$

Of course for a counterexample you must provide the assignments of $A_i,B_i,C_i$ to $E_j$ and the probabilities of the $E_j.$ Some choices of these assignments and probabilities yield cases where $P(A \cap B \cap C)$ is determined, so you need to choose carefully.