Explain how $ p(a,b,c,d) = \frac{\phi(a,b,c)\phi(a,b,c)}{Z} $ leads to $ Zp(a,b,c) = \phi(a,b,c) \sum_d \phi(b,c,d) $

Question

this might be a quiet basic question:

Let $ \phi(\chi^i) $ be a potential.

Then we have

$ p(a,b,c,d) = \frac{\phi(a,b,c)\phi(b,c,d)}{Z} $

By summing we have:

$ Zp(a,b,c) = \phi(a,b,c) \sum_d \phi(b,c,d) $

and

$ Zp(b,c,d) = \phi(b,c,d) \sum_a \phi(a,b,c) $

I don't understand how these two equations can result out of the given. How and why does the "summation step" work?

Got this from Bayesian Reasoning and Machine Learning by C. Barber:

Answer 1

Once the typos in your post are corrected, these identities correspond to the marginalizations $$ p(a,b,c)=\sum_dp(a,b,c,d),\qquad p(b,c,d)=\sum_ap(a,b,c,d). $$