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[CS5340] Bayesian Network (Directed Graphical Model, DGM)

Conditional Independence#

NOTE

Random variables are often NOT fully independence, we use conditional independence to assume an intermediate degree of dependency among the random variables.

  • Definition: xaxcxbx_a ⊥ x_c | x_b
    • p(xa,xcxb)=p(xaxb)p(xcxb)p(x_a, x_c | x_b) = p(x_a|x_b)p(x_c|x_b)
    • p(xaxb,xc)=p(xaxb)p(x_a|x_b,x_c) = p(x_a|x_b) (learning the values of xcx_c does not change the prediction of xax_a once we know the value of xbx_b)

Markov Assumption#

  • xa(xnonDescxparent)xparentx_a ⊥ (x_{nonDesc} - x_{parent}) | x_{parent}
  • xax_a is dependent only on its parents when given its parents

Parent-child relationship#

  • p(xixparent,(xnonDescxparent)=p(xixparent)p(x_i | x_{parent}, (x_{nonDesc} - x_{parent}) = p(x_i | x_{parent})
  • parent-child example
  • conditional independent
  • by using parent-child relationship to represent **joint probability ** p(x1,x2,...,xn)p(x_1, x_2, ..., x_n), parameter reduce from O(Kn)O(K^n) to O(Kmi+1)O(K^{m_i + 1}), where "mm" is the # of parent of node xix_i, and "+1+1" is xix_i itself

How can we find conditional independent directly from DGM?

Three Canonical 3-Node Graph#

Head-to-Tail#

head to tail

  • p(a,b,c)=p(a)p(ca)p(bc)p(a,b,c)=p(a)p(c|a)p(b|c)
  • No observation: p(a,b)=p(a)cp(ca)p(bc)p(a,b)=p(a) \sum_c p(c|a)p(b|c)
    =p(a)cp(a)p(ca)p(bc)p(a)=p(a)cp(a,b,c)p(a)=p(ba)p(a)p(b)= p(a) \sum_c \frac {p(a)p(c|a)p(b|c)}{p(a)} = p(a) \sum_c \frac {p(a,b,c)}{p(a)} = p(b|a) ≠ p(a)p(b)
  • Observe cc: p(a,bc)=p(a,b,c)p(c)=p(a)p(ca)p(bc)p(c)=p(c)p(ac)p(bc)p(c)=p(ac)p(bc)p(a,b|c) = \frac {p(a,b,c)}{p(c)} = \frac {p(a)p(c|a)p(b|c)}{p(c)} = \frac {p(c)p(a|c)p(b|c)}{p(c)} = p(a|c)p(b|c), therefore ABCA⊥B|C

Tail-to-Tail#

tail to tail

  • p(a,b,c)=p(ac)p(bc)p(c)p(a,b,c) = p(a|c)p(b|c)p(c)
  • No observation: p(a,b)=cp(ac)p(bc)p(c)p(a)p(b)p(a,b) = \sum_c p(a|c)p(b|c)p(c) ≠ p(a)p(b)
  • Observe cc: p(a,bc)=p(ac)p(bc)p(c)p(c)=p(ac)p(bc)p(a,b|c) = \frac {p(a|c)p(b|c)p(c)}{p(c)} = p(a|c)p(b|c), therefore ABCA⊥B|C

Head-to-Head (V-Structure)#

head to head

  • p(a,b,c)=p(a)p(b)p(ca,b)p(a,b,c) = p(a)p(b)p(c|a,b)
  • No observation: p(a,b)=p(a)p(b)cp(ca,b)=p(a)p(b)p(a,b) = p(a)p(b) \sum_c p(c|a,b) = p(a)p(b), therefore ABA⊥B
  • Observe cc: p(a,bc)=p(a)p(b)p(ca,b)p(c)p(a,b|c) = \frac {p(a)p(b)p(c|a,b)}{p(c)}
Head-to-TailTail-to-TailHead-to-Head
No observationp(a,b)p(a)p(b)p(a,b)≠p(a)p(b)p(a,b)p(a)p(b)p(a,b)≠p(a)p(b)p(a,b)=p(a)p(b)p(a,b)=p(a)p(b)
Observe middle nodep(a,bc)=p(ac)p(bc)p(a,b\|c) = p(a\|c)p(b\|c) (CI)p(a,bc)=p(ac)p(bc)p(a,b\|c)=p(a\|c)p(b\|c) (CI)p(a,bc)p(ac)p(bc)p(a,b\|c)≠p(a\|c)p(b\|c) (No CI)
NOTE

If all paths from AA to BB are blocked by observing CC, AA is d-seperated from BB by CCABCA⊥B|C

How do we define “blocked”?

Bayes Ball Algorithm#

  • reachability test
  • If ball from node AA cannot reach BB by any path, then ABCA⊥B|C, else A⊥̸BCA \not\perp B|C
  • Bayes ball algorithm

Markov Blanket#

  • Markov Blanket of node {x_i} comprised the set of parents, children & co-parents
  • markov blanket
[CS5340] Bayesian Network (Directed Graphical Model, DGM)
https://itsjeremyhsieh.github.io/posts/cs5340-2-bayesian-network/
Author
Jeremy H
Published at
2024-08-19