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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.

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  • $\begingroup$ What do you mean by changing $C$ to $A$? Are you asking $P(A|B)$ when the network is still in the same form, i.e. $A\rightarrow B\rightarrow C$ $\endgroup$ – gunes Jan 24 at 12:44
  • $\begingroup$ @gunes I want to see what would happen. I want to make sure the network will be invalid if I substitute C by A(indicating the network has changed). If P(A|B) is just P(A) the sum would be larger than 1, I thought. $\endgroup$ – Lerner Zhang Jan 24 at 13:14
  • $\begingroup$ I think it would lead to an endless loop and the joint distribution would be summed larger than 1. $\endgroup$ – Lerner Zhang Jan 25 at 2:18
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When you construct a network in the form $A\rightarrow B\rightarrow C$, since every node depends on only its parents, by definition the joint probability is written as $P(A,B,C)=P(A)P(B|A)P(C|B)$. When we set $C=A$, i.e. form a cyclic relationship, this formula turns into $$P(A,B,A)=P(A,B)=P(A)P(B|A)P(A|B)$$ Here, $P(A)P(B|A)=P(A,B)$, then to satisfy the equation it should be $P(B|A)=1$, assuming non-zero probabilities. Conversely thinking, we'll have $P(A|B)=1$. If they're both $1$, then $P(A,B)=P(A)$, which means $B=A$ or $A$ and $B$ are deterministic, i.e. with probability $1$, e.g. The world is spinning.

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  • $\begingroup$ So is my head. LOL. $\endgroup$ – Lerner Zhang Jan 25 at 22:47

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