The forward direction is actually quite a deep result, known as the Hammersley-Clifford Theorem. The counterexample for non-positive distributions was found by Moussouris, and you can see it on page 12 here, and the explanation that follows on page 13. The given network is not factorable.

In case the link goes stale, consider a distribution 4 nodes that form a square, with vertices labeled $(a,b,c,d)$ in clockwise order. Then define a uniform distribution on 8 of the following configurations (with the other 8 having zero probability):

a----b
|    |
|    |
d----c

$$(1,1,1,1)$$ $$(1,1,1,0)$$ $$(0,1,1,0)$$ $$(0,0,1,0)$$ $$(0,0,0,0)$$ $$(0,0,0,1)$$ $$(1,0,1,0)$$ $$(1,0,0,1)$$ $$(1,1,0,1)$$

The proof is to then assume the distribution factorizes. An exhaustive search on the above configurations will show all four factors are positive. This is a contradiction since the remaining 8 states have 0 probability.

The forward direction is actually quite a deep result, known as the Hammersley-Clifford Theorem. The counterexample for non-positive distributions was found by Moussouris, and you can see it on page 12 here, and the explanation that follows on page 13. The given network is not factorable.

In case the link goes stale, consider a distribution 4 nodes that form a square, with vertices labeled $(a,b,c,d)$ in clockwise order. Then define a uniform distribution on 8 of the following configurations (with the other 8 having zero probability):

1----1    1----1
|    |    |    |
|    |    |    |
1----1    0----1

1----0    0----1
|    |    |    |
|    |    |    |
1----0    0----1

1----0    0----0
|    |    |    |
|    |    |    |
0----1    0----1

0----0    0----0
|    |    |    |
|    |    |    |
1----0    0----0

The proof is to then assume the distribution factorizes. An exhaustive search on the above configurations will show all four factors are positive. This is a contradiction since the remaining 8 states have 0 probability.

The forward direction is actually quite a deep result, known as the Hammersley-Clifford Theorem. The counterexample for non-positive distributions was found by Moussouris, and you can see it on page 12 here, and the explanation that follows on page 13. The given network is not factorable.

The forward direction is actually quite a deep result, known as the Hammersley-Clifford Theorem. The counterexample for non-positive distributions was found by Moussouris, and you can see it on page 12 here, and the explanation that follows on page 13. The given network is not factorable.

In case the link goes stale, consider a distribution 4 nodes that form a square, with vertices labeled $(a,b,c,d)$ in clockwise order. Then define a uniform distribution on 8 of the following configurations (with the other 8 having zero probability):

a----b
|    |
|    |
d----c

$$(1,1,1,1)$$ $$(1,1,1,0)$$ $$(0,1,1,0)$$ $$(0,0,1,0)$$ $$(0,0,0,0)$$ $$(0,0,0,1)$$ $$(1,0,1,0)$$ $$(1,0,0,1)$$ $$(1,1,0,1)$$

The proof is to then assume the distribution factorizes. An exhaustive search on the above configurations will show all four factors are positive. This is a contradiction since the remaining 8 states have 0 probability.