Exercise 10.7 - Removing leaves in BN20 networks

Answers

For question (a), note that given 𝐳 , all components from 𝐱 are mutually independent, hence:

p ( 𝐙 | X 1 , X 2 , X 4 ) = 1 p ( X 1 , X 2 , X 4 ) p ( 𝐙 ) p ( X 1 , X 2 , X 4 | 𝐙 ) ,

according to Figure 10.16.(b), while that computed w.r.t. Figure 10.16.(a) is:

p ( 𝐙 | X 1 , X 2 , X 4 ) = x 3 , x 5 p ( 𝐙 , X 3 = x 3 , X 5 = x 5 | X 1 , X 2 , X 4 ) = 1 p ( X 1 , X 2 , X 4 ) x 3 , x 5 p ( 𝐙 ) p ( X 1 , X 2 , X 4 | 𝐙 ) p ( X 3 = x 3 , X 5 = x 5 | 𝐙 ) = 1 p ( X 1 , X 2 , X 4 ) p ( 𝐙 ) p ( X 1 , X 2 , X 4 | 𝐙 ) x 3 , x 5 p ( X 3 = x 3 , X 5 = x 5 | 𝐙 ) = 1 p ( X 1 , X 2 , X 4 ) p ( 𝐙 ) p ( X 1 , X 2 , X 4 | 𝐙 ) .

This completes the proof.

For question (b), consider the following decomposition of the joint probability:

p ( 𝐗 on , 𝐗 off , 𝐙 ) = p ( 𝐙 ) X 𝐗 on p ( X | 𝐙 ) Y 𝐗 off p ( Y | 𝐙 ) = Z 𝐙 p ( Z ) X 𝐗 on p ( X | 𝐙 ) Y 𝐗 off Z 𝐙 𝜃 Z , Y ,

where 𝜃 Z , Y is defined from the correlation between disease Z and symptom Y . If Z and Y are independent then 𝜃 Z , Y = 1 .

Now:

p ( 𝐗 on , 𝐗 off , 𝐙 ) = Z 𝐙 p ( Z ) [ Y 𝐗 off 𝜃 Z , Y ] X 𝐗 on p ( X | 𝐙 ) .

By changing the prior over Z from p ( Z ) into p ( Z ) [ Y 𝐗 off 𝜃 Z , Y ] we complete the proof.

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2021-03-24 13:42
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