# Too many independences for decomposition of full joint probability?

Conditional independences given:

1. A ⊥ C | B ---------- A and C are independent given B
2. D ⊥ B | A, C
3. E ⊥ C, D | A
4. E ⊥ D | A

and I need to use this information for decomposing the full joint probability P(A, B, C, D, E).

My poor understanding leads me to

P(A, B, C, D, E) = P(B)P(A|B)P(C|B)P(E|A)P(D|A)

However, it does not seem to be correct, as 2. and 3. are just ignored by the answer.

Please kindly shed me some light on the question.

Thank you very much.

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Just to clarify; what does E ⊥ C, D | A mean? D is independent of what? – Stijn Apr 28 '11 at 11:15
@Stijn, I am not sure but I think it means E and C&D are independent given A. – user6023 Apr 28 '11 at 15:46

Some hints to help solving your problem:

I hope these hints bring you to a correct decomposition.

Best, S

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It seems a possible approach as you showed in P(E|A,B,C,D)=P(E|A,B). However, it seems I still need to do some guessing by making random groups.... How to do it systematically? – user6023 Apr 28 '11 at 15:53
@user8991 Sorry but your second bulleted point is wrong: condition 3 does not imply P(E|A,B,C,D)=P(E|A,B). One would need E to be independent on C,D given A,B and one only knows that E is independent on C,D given A. – Did May 1 '11 at 7:21

These conditions are not enough to determine P(A,B,C,D,E) and in particular they do not imply the formula you suggest (and condition 4 is useless since it is implied by condition 3). If these were enough, conditions 1 and 2 alone would yield an expression for P(A,B,C,D). But these conditions yield formulas like P(A,B,C,D)=P(D|A,C)P(A|B)P(C|B)P(B) and nothing much simpler I am afraid.

In fact you could make more precise the kind of formula you are looking for.

Edit As already explained above, if the formula the OP suggests was true then it would be true when E is the sure event. Hence it is enough to show that in this restricted case the formula the OP suggests does not hold to prove that it does not hold in general. (This point is pure logic and has nothing to do with probability.)

The formula P(A,B,C,D)=P(D|A,C)P(A|B)P(C|B)P(B) that I indicated above in the restricted case coincide with the formula the OP proposes in the general case if and only if P(D|A,C)=P(D|A). Since this last relation is not a consequence of conditions 1 to 4, the formula the OP proposes cannot hold in the general case.

Here is an example. Assume U, V and W are independent symmetric signs and let A=[U=+1], B=[V=+1], C=[W=+1], D=[UW=+1] and E the sure event. Then 1 holds because A, B and C are independent, 2 holds because D is (A,C) measurable and B is independent from (A,C) and 3 and 4 hold because E is the sure event. Now, P(D|A,C)=1 and P(D|A)=P(W=+1)=1/2.

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I am sorry but.... you have ignored E in your answer.... – user6023 Apr 28 '11 at 15:50
Read again, I have not. – Did Apr 28 '11 at 15:51
I am terribly sorry.... From P(D|A,C)P(A|B)P(C|B)P(B)/P(A,C), I get P(D|AC)P(ABC)/P(AC)=P(D|AC)P(B|AC)=P(BD|AC).... and I really figure out where E is. Please give a little hint.... – user6023 Apr 28 '11 at 16:01
and why my answer is wrong.... Sorry for so many follow-up questions.... – user6023 Apr 28 '11 at 16:06
See edit.   – Did Apr 30 '11 at 21:51