# Can the marginal distributions of A,C and B,C be used to build joint distribution of A and B?

There are three random variables $$A$$, $$B$$ and $$C$$. If the variables $$A$$ and $$B$$ were independent, their marginal joint distribution would be given by $$P(A,B) = P(A)P(B)$$

For example, given the discrete probability distributions of $$A = \{ A_1, A_2 \}$$:

A1 0.6
A2 0.4


and of $$B = \{B_1, B_2 \}$$:

B1 0.4
B2 0.6


The joint probability distribution would then be $$P(A,B)=P(A)P(B)$$:

A1 B1 0.24
A1 B2 0.36
A2 B1 0.16
A2 B2 0.24


However, If the variables are not independent, and we know the marginal joint distributions of $$P(A,C)$$ and $$P(B,C)$$, e.g.

A1 C1  0.4       B1 C1  0.1
A1 C2  0.2       B1 C2  0.3
A2 C1  0.2       B2 C1  0.5
A2 C2  0.2       B2 C2  0.1


Is that enough to build the joint distribution of $$P(A,B,C)$$? And, additionally, the marginal joint distribution of $$P(A,B)$$?

• You already have the joint distribution of P(AB)=P(A)P(B). I don't understand why are you trying to use C to obtain P(AB) when you already have that from your first step? Also, P(AB) is not equal to P(A)P(B|A). – StatsStudent Nov 7 '18 at 18:08
• Please explain to us how $C$ "gives us more information:" what does it represent? What do the entries in your last table mean? – whuber Nov 7 '18 at 18:21
• The P(AB)=P(A)P(B) only in the case of the events being independent. Their respective joint distributions with C show this not to be the case. – Sakari Cajanus Nov 7 '18 at 19:12
• I reopened the question despite the fact the criteria in your example are inconsistent. For instance, you are asserting that the probability of C1 is $0.4+0.1=0.5$ while also asserting it equals $0.1+0.5=0.6.$ The question might be clearer to some readers if the example actually had a solution. That might help everyone see that in general there are many solutions. – whuber Nov 7 '18 at 20:09
• Thanks, I updated the probabilities so they should be consistent. – Sakari Cajanus Nov 7 '18 at 20:16

In the general case, no. The marginals of $$P(A,C)$$ and $$P(B,C)$$ are enough to determine the full joint distribution of $$P(A,B,C)$$ only in the case of conditional independence: $$P(A \mid B,C) = P(A\mid C)$$ and equivalently: $$P(B \mid A,C) = P(B\mid C).$$

Starting with the full joint probability $$P(A,B,C)$$. Given the formula for joint probability $$P(X,Y)=P(X\mid Y)P(Y),$$ we get:

$$P(A,B,C)=P(A,C)P(B\mid A,C).$$

If the events $$A$$ and $$B$$ are not conditionally independent given $$C$$, now we would need to know the distribution of $$P(B \mid A,C)$$, which would be enough to form the full joint distribution.

If we assume the probability of B given C be independent of A, the formula simplifies to: $$P(A,B,C)=P(A,C)P(B\mid C).$$

Using the example probabilities, e.g. $$P(A_1,B_1,C_1) = P(A_1,C_1)P(B_1\mid C_1) = 0.4 \cdot \frac{0.1}{0.1+0.5} = \frac{2}{30},$$

where $$P(B_1 \mid C_1)$$ is the conditional probability of observing $$B_1$$ after observing $$C_1$$.

The full table of probabilities is then:

A1 B1 C1    4/60
A1 B1 C2    9/60
A1 B2 C1   20/60
A1 B2 C2    3/60
A2 B1 C1    2/60
A2 B1 C2    9/60
A2 B2 C1   10/60
A2 B2 C2    3/60


And the joint distribution of A, B (summing over C):

A1 B1   13/60   0.2166...
A1 B2   23/60   0.3833...
A2 B1   11/60   0.1833...
A2 B2   13/60   0.2166...


We can see that $$A$$ and $$B$$ are not independent -- but they are still conditionally independent given $$C$$! If this is not true, the full joint distribution cannot be built from marginal joint distributions like this.

• So that readers don't leave with the wrong impression that this kind of information determines the joint distribution, you really ought to point out how unusual your set of conditions is: it permits a unique answer, whereas in general there will be a two-parameter family of solutions. – whuber Nov 7 '18 at 23:11