Let's say that we have a graph A->B->C, and its coditional independence representation is like follows:
| a0 | a1 |
|:-----|--------:|
| 0.6 | 0.4 |
| A | P(b0|A) | P(b1|A) |
|:-----|--------:|--------:|
| a0 | 0.30 | 0.70 |
| a1 | 0.20 | 0.80 |
| B | P(c0|B) | P(c1|B) |
|:-----|--------:|--------:|
| b0 | 0.20 | 0.80 |
| b1 | 0.90 | 0.10 |
This graph is valid since every probability is non-negative and the sum of the graph distribution is 1: $P(A, B, C)=P(A)P(B|A)P(C|B)=0.6*0.3*0.2+0.6*0.3*0.8+\cdots + 0.4*0.8*0.9+0.4*0.7*0.1=1$
But what if I change the C to A? The first two probability tables don't change:
| a0 | a1 |
|:-----|--------:|
| 0.6 | 0.4 |
| A | P(b0|A) | P(b1|A) |
|:-----|--------:|--------:|
| a0 | 0.30 | 0.70 |
| a1 | 0.20 | 0.80 |
But how to draw the third one?
| B | P(a0|B) | P(a1|B) |
|:-----|--------:|--------:|
| b0 | 0.60 | 0.40 |
| b1 | 0.60 | 0.40 |
I am wonder if it is an infinit loop and the probability tables are not possible for such graphs. Any suggestions are hight appreciated. Thanks.