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I was reading a lecture regarding representing a join distribution as a graphical model. The lecture is available here (on page 4 of the slides):

http://www.maths.lth.se/matematiklth/personal/sminchis/mlc/lecture-6.pdf

In the lecture, it says:

l.h.s. (left hand side) is symmetric with respect to a,b,c. But right hand side is not symmetric with respect to a,b,c , could someone give me some hints or reasoning as to why the author says so?

My reasoning that they are the same is because:

P(a,b,c) = P(c|a,b) * P(a|b)*P(b) = P(a|b,c)*P(b|c)*P(c) = P(b|a,c)*P(a|c)*P(c)

We can factor the joint distribution in any order or way no??? and still have the same joint distribution right?

My question is I don't understand why the author says the R.H.S is not symmetric with respect to a,b,c. They must be some concepts here I am not getting.

Would be great if someone can explain little more to me.

Thank you.

john_w
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  • Notice the arrow. Event $a$ given $c$ gives you nothing. It's logically impossible to say$p(a,c)=p(a|c)p(c)$, because you can't get information about $a$ when $c$ happens. – Rowan Nov 27 '15 at 06:33

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LHS being symmetric wrt $a,b,c$ just means that in swapping $a,b,c$ in any way, we still have the same expression because $P(a,b,c)\equiv P(a,c,b)\equiv P(b,a,c)\equiv \cdots$.

On the RHS, swapping $a,b,c$ gives us a different expression. E.g. swapping $a,b$ changes the RHS from $P(c\mid b,a)P(b\mid a)P(a)$ to $P(c\mid a,b)P(a\mid b)P(b)$. Note that these two expressions have the same value but are different expressions.

I think the main point of this is that you have a choice of the way you order the nodes on the RHS. Some orderings will be simpler and easier and more logical to work with than others.

Mick A
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