1. Tutorial 2
Simple examples of Bayesian networks, Queries, And the stories behind them
Tal Shor

2. Fatigue as a product of smokingH
0.2 h1
Values
ℎ1 – Has a history of smoking
𝑏1 – Has Bronchitis
𝑙1 - Has lung cancer
𝑓1 – Is fatigue
𝑐1 – Positive X-ray
B=1 H
0.25 h1
0.05 h2 B L
L=1 H
0.003 h1 h2
F=1 L B
0.75 l1 b1
0.10 l2
0.5 b2
0.05
C=1 L
0.6 l1
0.02 l2 F C

3. BN Attributes
𝐼 𝐷 𝐵, ∅, 𝐶 - we can clearly see from the graph that bronchitis has no influence on X-ray, and it is only natural
𝐼 𝐷 (𝐻, 𝐵,𝐿 , 𝐹) – given that an individual has both bronchitis and lung cancer – his/her smoking history is no longer relevant when estimating the individual’s fatigue.
𝐼 𝐷 (𝐻, 𝐿, 𝐶) – if we know an individual does not have lung cancer, it does not matter how much he smoked, we won’t find anything in the X-ray scans.

4. BN Attributes (2)
Those independencies create a probabilistic function that is easier to compute and much more compact than 𝐷𝑜𝑚 𝐻 ×𝐷𝑜𝑚 𝐵 ×𝐷𝑜𝑚 𝐿 ×Dom 𝐹 ×𝐷𝑜𝑚(𝐶) variables.
Without the independencies, those 5 variables would require a table with entries.
With them - 𝑃 𝐻, 𝐵, 𝐿. 𝐹, 𝐶 =𝑃 𝐻 ×𝑃 𝐵 𝐻 ×𝑃 𝐿 𝐵 ×𝑃 𝐹 𝐵,𝐿 ×𝑃(𝐶|𝐿) and there are only 11 entries ( )
For example : 𝑃 𝐻=ℎ1, 𝐵=𝑏2 , 𝐿=𝑙1. 𝐹,=𝑓1, 𝐶=𝑐2 =0.2×0.75×0.003×0.5×0.4

5. Simple Example of Bayesian Network
(*) 𝑃(𝑥,𝑦,𝑧) = 𝑃(𝑥) 𝑃(𝑦) 𝑃(𝑧|𝑥,𝑦)
In this BN, X and Y are independent. Follows from definition of BN by summing over values of Z in Eq. (*) yielding 𝑃(𝑥, 𝑦) = 𝑃(𝑥) 𝑃(𝑦) . X Y Z

6. Specifying Conditional Probability TablesBB AA OB OA OO 𝑌 𝑋 1 …

7. Specifying Conditional Probability Tables
Pr 𝑋=𝐴𝐴 = Pr 𝑋=𝐵𝐵 = Pr 𝑋=𝑂𝑂 = 1 9
Pr 𝑋=𝑂𝐴 = Pr 𝑋= 𝑂𝐵 = Pr 𝑋= 𝐴𝐵 = (twice the chance)
Pr 𝑌 =Pr⁡(𝑋)
Can you see where this is going?

8. Conditional Dependencies
Pr⁡(X=OA, Y=OA | Z=OO)= Pr Z=𝑂𝑂 X=𝑂𝐴, Y=𝑂𝐴)⋅Pr⁡(X=𝑂𝐴, Y=𝑂𝐴) Pr Z=𝑂𝑂 = 1 4 ⋅ = 2 18
Pr X=𝑂𝐴 Z=OO)= Pr Z=𝑂𝑂 X=𝑂𝐴)⋅Pr⁡(X=𝑂𝐴) Pr Z=𝑂𝑂 = 𝑦 Pr Z=𝑂𝑂 X=𝑂𝐴, Y=y)⋅ Pr 𝑌=𝑦 = = 4 18
⇒Pr⁡(X=OO, Y=AA | Z=OO)≠ Pr X=𝑂𝑂 Z=OO)× Pr Y=𝐴𝐴 Z=OO) X Y Z

9. Probability function
The probability function of 𝑋,𝑌,𝑍,𝑉,𝑊 in this graph is
𝑃 𝑋,𝑌,𝑍,𝑉,𝑊 =𝑃 𝑋 ∗𝑃 𝑌 ∗𝑃 𝑍 ∗𝑃 𝑉 𝑋,𝑌 ∗𝑃(𝑊|𝑌,𝑍)
And we can see that the probability function of the subset 𝑋,𝑌,𝑉,𝑊 is the same in the corresponding sub graph
𝑃 𝑋,𝑌,𝑉,𝑊 = 𝑍 𝑃 𝑋,𝑌,𝑍,𝑉,𝑊 = 𝑍 𝑃 𝑋 ∗𝑃 𝑌 ∗𝑃 𝑍 ∗𝑃 𝑉 𝑋,𝑌 ∗𝑃(𝑊|𝑌,𝑍)
= 𝑃 𝑋 ∗𝑃 𝑌 ∗𝑃 𝑉 𝑋,𝑌 ∗ 𝑍 𝑃 𝑊 𝑌,𝑍 ∗𝑃 𝑍 =𝑃 𝑋 ∗𝑃 𝑌 ∗𝑃 𝑉 𝑋,𝑌 ∗𝑃(𝑊|𝑌) X Y Z V W

10. Dependencies 𝐼 𝐷 𝑋,∅,𝑌 , 𝐼 𝐺 𝑋,∅,𝑍 , 𝐼 𝐺 𝑍,∅,𝑌 ¬ 𝐼 𝐷 (𝑋,𝑉,𝑌)
𝐼 𝐷 𝑋,∅,𝑌 , 𝐼 𝐺 𝑋,∅,𝑍 , 𝐼 𝐺 𝑍,∅,𝑌
¬ 𝐼 𝐷 (𝑋,𝑉,𝑌)
𝐼 𝐷 (𝑉,𝑌,𝑊) (Naïve Bayes)
𝐼 𝐷 (𝑋,∅,𝑊)
¬ 𝐼 𝐷 (𝑋,𝑉,𝑊) (the only path is blocked) X Y Z V W

11. The story behind the BN
G would represent the Genotype. It is the DNA code of an attribute that is inherited.
P would represent the Phenotype. It is the attribute itself, caused by the genotype
Blood type genotype can be one of 9 options {OO, OA, OB, AO, AA, AB, BO, BA, BB}
The phenotype can be one of 4 options {O, A, B, AB}
𝐺∈ 𝑂𝐴, 𝐴𝑂, 𝐴𝐴 ⇒𝑃=𝐴
𝐺∈ 𝑂𝐵, 𝐵𝑂, 𝐵𝐵 ⇒𝑃=𝐵
𝐺∈ 𝐴𝐵, 𝐵𝐴 ⇒𝑃=𝐴𝐵
𝐺=𝑂𝑂⇒𝑃=𝑂

12. Royal Blood The picture depicts the royal family’s pedigree
Each of the parents passes one of his {A, B, O} to his/her child at random (50-50).
So each individual, in the end, has 2 of {A, B, O}

13. Royal Blood
𝐺 𝐸𝑙𝑖𝑧𝑎𝑏𝑒𝑡ℎ 𝐼𝐼
𝐺 𝑃ℎ𝑖𝑙𝑖𝑝
𝑃 𝑃ℎ𝑖𝑙𝑖𝑝
𝑃 𝐸𝑙𝑖𝑧𝑎𝑏𝑒𝑡ℎ 𝐼𝐼
The blood type Bayesian network where G denotes the genotype and P the phenotype.
Each parent-child nodes pair have a directed edge between them. It represents the child genotype dependency of his parents
The phenotype of an individual is only dependent on his/her genotype.
𝐺 𝐷𝑖𝑎𝑛𝑛𝑎
𝐺 𝐶ℎ𝑎𝑟𝑙𝑒𝑠
𝐺 𝐴𝑛𝑛𝑒
𝐺 𝑀𝑎𝑟𝑘
𝑃 𝐷𝑖𝑎𝑛𝑛𝑎
𝑃 𝐶ℎ𝑎𝑟𝑙𝑒𝑠
𝐺 𝑊𝑖𝑙𝑙𝑖𝑎𝑚

14. Royal Blood
𝐺 𝐸𝑙𝑖𝑧𝑎𝑏𝑒𝑡ℎ ≜𝐴
𝐺 𝐺𝑒𝑜𝑟𝑔𝑒 𝑉𝐼 ≜𝐵
From the simple example, and the fact that Elizabeth II has an O type blood, we can learn that the chance that both Elizabeth and George are AO is 1 18
If Elizabeth II’s genotype wasn’t known, but one of her descendant’s genotype was, A and B would still be dependent given that condition. Less so, but still.
𝐺 𝐸𝑙𝑖𝑧𝑎𝑏𝑒𝑡ℎ 𝐼𝐼 ≜𝐶
𝐺 𝐶ℎ𝑎𝑟𝑙𝑒𝑠 ≜𝐷

15. Royal Blood 𝐺 𝐽𝑎𝑛𝑒 𝑆𝑒𝑦𝑚𝑜𝑎𝑟 𝐺 𝐻𝑒𝑛𝑟𝑦 𝑉𝐼𝐼𝐼 𝐺 𝐴𝑛𝑛𝑒 𝐵𝑜𝑙𝑒𝑦𝑛 𝑃 𝐴,𝐵,𝐶,𝐷,𝐸 =
𝑃 𝐴 ∗𝑃 𝐵 ∗𝑃 𝐶 ∗𝑃 𝐷 𝐴,𝐵 ∗𝑃(𝐸|𝐵.𝐶)
If Elizabeth I genotype is known, Edwards VI’s and Jane’s genotype are not independent anymore.
It blocks the only active path between them.
For Example, If Elizabeth I genotype was AB, So if Jane’s genotype was BB then Henry’s most likely would be AA giving an A to Edward.
𝐺 𝐸𝑙𝑖𝑧𝑎𝑏𝑒𝑡ℎ 𝐼
𝐺 𝐸𝑑𝑤𝑎𝑟𝑑 𝑉𝐼

16. Royal Blood 𝐺 𝐴𝑢𝑔𝑢𝑠𝑡𝑎 ≜𝐴 𝐺 𝐸𝑟𝑛𝑒𝑠𝑡 𝐼 ≜𝐵 𝐺 𝐸𝑑𝑤𝑎𝑟𝑑 ≜𝐶 𝐺 𝐴𝑙𝑏𝑒𝑟𝑡 ≜𝐷
𝑃 𝐴,𝐵,𝐶,𝐷,𝐸,𝐹 =
𝑃 𝐴 𝑃 𝐵 𝐴 𝑃 𝐶 𝐴 𝑃 𝐷 𝐵 𝑃 𝐸 𝐶 𝑃 𝐹 𝐷,𝐸
¬ 𝐼 𝐷 (𝐹,𝐺, 𝐴) – exists an active path
𝐼 𝐷 (𝐹, 𝐷,𝐸 ,𝐴) – removing parents from paths
𝐼 𝐷 (𝐹, 𝐵,𝐶 ,𝐴) – simply removing all active paths so it’s also d-seperated
𝐺 𝐴𝑙𝑏𝑒𝑟𝑡 ≜𝐷
𝐺 𝑉𝑖𝑐𝑡𝑜𝑟𝑖𝑎 ≜𝐸
𝐺 𝐸𝑑𝑤𝑎𝑟𝑑 𝑉𝐼𝐼 ≜𝐹