# How to find P(A and B and C) given all the marginal probabilities and two conditional probabilities w/r/t to same random variable A?

Given marginal probabilities $$P(A)$$, $$P(B)$$, and $$P(C)$$ and conditional probabilities $$P(A|B)$$, $$P(B|A)$$, $$P(A|C)$$, and $$P(C|A)$$ is it possible to find $$P(A\cap B \cap C)$$ ?

Attempt: $$P(A \cap B \cap C) = P(A \cap B)\cap P(A \cap C)= P(B|A)\cdot P(A)\cap P(C|A)\cdot P(A)= P(A) [P(B|A) \cap P(C|A)] = P(A) \cdot [P(B|A | C|A)] \cdot P(C|A) = P(C) \cdot [P(B|A | C|A)]$$

Can I distribute the conditioning variable? Are there useful properties I should know about? Any help is appreciated. My intuition tells me it should be possible to calculate $$P(A\cap B \cap C)$$. Any help appreciated.

• Your notation does not make sense to me: after writing $P(A \cap B \cap C)$ you take the intersection of two real numbers: $P(A \cap B) \cap P(A \cap C)$. What exactly are you trying to do? – Maurits M Sep 25 '19 at 7:52

This is actually asking if we can find the joint with marginals and $$P(A|C),P(A|B)$$, i.e. knowing $$P(C|A)$$ and $$P(B|A)$$ doesn't provide additional information since they can be found via Bayes rule. First of all, we know nothing about the dependence relation between $$B$$ and $$C$$, and also we don't know what happens when we condition on two of the events, which should give you a heads up already. We can't do it.

Let's find an example. Assume event $$A$$ has nothing to do with the other events both jointly and mutually, i.e. $$P(A|B)=P(A),P(A|C)=P(A),P(A|B,C)=P(A)$$. (e.g. $$A$$ can represent the event of raining in Amazons today, where $$B,C$$ represents the probability of heads of the same (maybe biased) coin in different tosses). Then, $$P(A,B,C)=P(A)P(B,C)$$, and we can't calculate $$P(B,C)$$ with only $$P(B)$$ and $$P(C)$$.

Also, your solution attempt is not meaningful because of @MauritusM 's comment.

Let's say for simplicity that each variable $$A$$, $$B$$ and $$C$$ has only two states such that they jointly have $$8$$ states. As described in the answers to What is the number of parameters needed for a joint probability distribution? you will need $$8-1=7$$ independent parameters to describe this distribution.

You only have $$5$$ parameters $$P(A)$$, $$P(B)$$, $$P(C)$$, $$P(B \vert A)$$ and $$P(C \vert A)$$. Note that $$P(C \vert A) = P(A \vert C) P(C)/P(A)$$ and $$P(B \vert A) = P(A \vert B) P(B)/P(A)$$ are not independent from the other $$5$$ parameters.

### Counter-example

Let the probabilities $$P(A=a \cap B=b \cap C=c)$$ be expressed by the numbers $$n_{abc}$$ with subscripts $$a$$, $$b$$, $$c$$ taking the values $$0$$ or $$1$$ depending on whether $$A$$, $$B$$, $$C$$ are true or not. Then we have $$8$$ numbers, but only $$6$$ equations to define them.

$$\begin{array}{rcl} n_{000}+n_{001}+n_{010}+n_{011}+n_{100}+n_{101}+n_{110}+n_{111} &=&1 \\ n_{100}+n_{101}+n_{110}+n_{111} &=& P(A=1)\\ n_{010}+n_{011}+n_{110}+n_{111} &=& P(B=1)\\ n_{001}+n_{011}+n_{101}+n_{111} &=& P(C=1)\\ (n_{110}+n_{111})/(n_{100}+n_{101}+n_{110}+n_{111}) &=& P(B=1 \vert A=1) \\ (n_{101}+n_{111})/(n_{100}+n_{101}+n_{110}+n_{111}) &=& P(C=1 \vert A=1) \\ \end{array}$$

So this will have multiple solutions that can be expressed by two additional numbers. Let $$p_a = 1-q_a = P(A=1)$$ (and similar for $$B$$ and $$C$$)

$$\begin{array}{rcl} n_{000}=q_a q_b q_c + r + s + t + u \\ n_{001}=q_a q_b p_c + r - s - t - u \\ n_{010}=q_a p_b q_c - r + s - t - u \\ n_{011}=q_a p_b p_c - r - s + t + u \\ n_{100}=p_a q_b q_c - r - s + t - u \\ n_{101}=p_a q_b p_c - r + s - t + u \\ n_{110}=p_a p_b q_c + r - s - t + u \\ n_{111}=p_a p_b p_c + r + s + t - u \\ \end{array}$$

In these equations you can see the expression of the numbers $$n_{abc}$$ as independent (the product of terms $$q$$ and $$p$$) plus extra terms that make the numbers deviate from independence. Those terms are $$r$$, $$s$$, $$t$$ and $$u$$ which will leave probabilities like $$P(A=1)$$ invariant (you can see it has the higher dimensional version of this case: https://stats.stackexchange.com/a/363777/164061 ). When we put these $$8$$ solutions into the before mentioned 6 equations you get:

$$\begin{array}{rcl} 1 &=&1 \\ p_a &=& P(A=1)\\ p_b &=& P(B=1)\\ p_c &=& P(C=1)\\ (p_b p_a + 2 r)/(p_a) &=& P(B=1 \vert A=1) \\ (p_c p_a + 2 s )/(p_a) &=& P(C=1 \vert A=1) \\ \end{array}$$

So you can find counter examples by using different values for the parameters $$t$$ and $$u$$ which will render different solutions that still satisfy your conditions $$P(A)$$, $$P(B)$$, $$P(C)$$, $$P(B \vert A)$$ and $$P(C \vert A)$$.

For example the two cases below:

$$n_{000}=n_{001}=n_{010}=n_{011}=n_{100}=n_{101}=n_{110}=n_{111} =0.125$$ or $$\begin{array}{rcl} n_{000}=n_{011}=n_{101}=n_{110}&=&0.1 \\ n_{001}=n_{010}=n_{100}=n_{111} &=&0.15 \end{array}$$

are an example of two different distributions with the same $$P(A) = P(B) = P(C) = P(B \vert A) = P(C \vert A) = 0.5$$ and in this particular case also $$P(C \vert B) = 0.5$$.

### Graphical

The image below sketches a cube which is how I intuitively imagine these variables $$r$$, $$s$$, $$t$$, as a crossed pattern that changes the distribution while leaving the marginal distributions (including some conditional) invariant. And the variable $$u$$ will leave all marginal distributions invariant (when you take the sum of $$n$$ allong any rib then you get both a $$+u$$ and $$-u$$ term). 