Statistical Evidence Evaluation, fall semester 2015
Solutions to Exercises 2
1. a)
Pr E  E1 H  H1   1
Pr H  H1 E  E1  

(taken directly from probability table)
Pr E  E1 H  H1  Pr H  H1 

Pr E  E1 H  H1  Pr H  H1   Pr E  E1 H  H 2  Pr H  H 2 
1  0.6
 0.968
1  0.6  0.05  0.4
b)
Pr E  E1 H  H1   Pr E  E1 F  F1 , H  H1  Pr F  F1 H  H1  
 Pr E  E1 F  F2 , H  H1  Pr F  F2 H  H1  
 Pr E  E1 F  F1  Pr F  F1 H  H1  
 Pr E  E1 F  F2  Pr F  F2 H  H1  
 1  0.5  0.05  0.5  0.525
Pr H  H1 E  E1  

Pr E  E1 H  H1  Pr H  H1 

Pr E  E1 H  H1  Pr H  H1   Pr E  E1 H  H 2  Pr H  H 2 
Pr E  E1 H  H1   0.525
(from a) )
Pr E  E1 H  H 2   Pr E  E1 F  F1 , H  H 2  Pr F  F1 H  H 2  
 Pr E  E1 F  F2 , H  H 2  Pr F  F2 H  H 2  
 Pr E  E1 F  F1  Pr F  F1 H  H 2  
 Pr E  E1 F  F2  Pr F  F2 H  H 2  
 1  0.1  0.05  0.9  0.145
0.525  0.6
 Pr H  H1 E  E1  
 0.844
0.525  0.6  0.145  0.4
2. a)
Gender
Age
Female
Male
18-30
31-50
51-
All
40 %
60 %
35 %
30 %
35 %
Injured
30 %
70%
45%
20%
35%
Non-injured
?
?
?
?
?
Pr Female Injured   0.30 
Pr Injured Female 
Pr Female Injured  Pr Injured  0.30  0.01

 0.0075
Pr Female
0.40
Pr Non  Injured Female Pr Female
Pr Female Non  Injured  

Pr  Non  Injured 
1  0.0075  0.40  0.401

0.99
 Pr Male Non  Injured   0.599
Analogously,
Pr 18 - 30 Non  Injured  
Pr Non  Injured 18 - 30  Pr 18 - 30 

Pr  Non  Injured 
 0.45  0.01 
1 
  0.35
0.35 

 0.349
0.99
Pr Non  Injured 31 - 50  Pr 31 - 50 
Pr 31 - 50 Non  Injured  

Pr  Non  Injured 
Pr 51 - Non  Injured  
 0.20  0.01 
1 
  0.30
0.30 

 0.301
0.99
 1  Pr 18 - 30 Non  Injured   Pr 31 - 50 Non  Injured   0.350
Hence,
Gender
Age
Female
Male
18-30
31-50
51-
All
40 %
60 %
35 %
30 %
35 %
Injured
30 %
70%
45%
20%
35%
Non-injured
40.1 %
59.9 %
34.9 %
30.1 %
35.0 %
b)
Injured?
Pr
Injured
0.010
Non_injured
0.990
Pr
Gender:
Injured?
Injured
Non-injured
Female
0.300
0.401
Male
0.700
0.599
Pr
Age:
Injured?
Injured
Non-injured
A18_30
0.450
0.349
A31_50
0.200
0.301
A51_
0.350
0.350
c)
Node Injured? instantiated
Instantiating node Gender
(when node Inured? is
instantiated does not affect
node Age
d)
Instantiating node Gender to state Female gives the requested distribution in
node Age
e)
Instantiating both node Gender and node Age to the states of interest gives the
conditional probability for state Injured in node Injured?  0.78 % = 0.0078.
3. a) To build a network from the information given we need to assume that person
not having the disease do not have cough (not a matter of course, though), and that
a person not having cough would not ask for a prescript cough tincture
(reasonable).
Disease?
Pr
Yes
0.05
No
0.95
Pr
Disease?
Cough?:
Pr
Yes
No
Yes
0.30
0
No
0.70
1
Cough?
Prescription?:
Yes
No
Yes
0.10
0
No
0.90
1
b)
Instantiate node Cough? to state Yes
Due to the conditional probability that cough only occurs when the disease is
present this will also instantiate node Disease?  It cannot be changed and
hence cannot affect node Prescription?
Instantiate node Cough? to state No
No effect on node
Prescription? when
instantiating node Disease?
4. a)
b)
Gender
Pr
Temperature
Pr
Man
0.5
above20
0.3
Woman
0.5
between10_20
0.5
below10
0.2
Pr
Temperature
Gender
Hoodie?:
Yes
No
above20
Man
Woman
between10_20
Man
Woman
below10
Man
Woman
Enter unknowns:
Pr
Temperature
Hoodie?:
above20
between10_20
below10
Gender
Man
Woman
Man
Woman
Man
Woman
Yes
x
y
z
w
u
v
No
1–x
1–y
1-z
1–w
1–u
1-v


y  Pr Hoodie Woman, Above 20 C 
z  Pr Hoodie Man, Between 10 C and 20 C 
w  Pr Hoodie Woman, Between 10 C and 20 C 
u  Pr Hoodie Man, Below 10 C 
v  Pr Hoodie Woman, Below 10 C 
x  Pr Hoodie Man, Above 20 C







Known probabilities:


 Pr Hoodie Man, Above 20 C  Pr Man Above 20 C  
 Pr Hoodie Woman, Above 20 C  Pr Woman Above 20 C  
0.05  Pr Hoodie Above 20 C 




Gender assumed independent of temperature




 Pr Hoodie Woman, Above 20 C  Pr Woman 
 Pr Hoodie Man, Above 20 C  Pr Man  

 x  0.5  y  0.5 

x  y  0.1

0.10  Pr Hoodie Between 10 C and 20 C 

Analogously with above

 z  0.5  w  0.5 

z  w  0.2
0.30  Pr Hoodie Below 10 C  u  0.5  v  0.5  u  v  0.6
Alternatively…
0.20  Pr Hoodie Man  

 
 Pr Hoodie Man, Between 10 C and 20 C 

 Pr Hoodie Man, Above 20 C  Pr Above 20 C Man 




 Pr Hoodie Man, Below 10 C  Pr Below 10 C Man 
 Pr Between 10 C and 20 C Man 


Assuming independence between

gender and temperature, and
 x  0.20  z  0.50  u  0.30
with my assigned probabilities
0.10  Pr Hoodie Woman  

 

 Pr Hoodie Woman, Between 10 C and 20 C Woman 
 Pr Hoodie Woman, Above 20 C  Pr Above 20 C Woman 




 Pr Hoodie Woman, Below 10 C  Pr Below 10 C Woman 
 Pr Between 10 C and 20 C Woman 

 y  0.20  w  0.50  v  0.30

 x  y  0.1

 z  w  0.2
u  v  0.6

0.20 x  0.50 z  0.30u  0.20

0.20 y  0.50 w  0.30v  0.10
This system of equations is obviously not of full rank, but the rank is even lesser: 4 .
This is due to that the equations were obtained using decompositions with the same
terms.
However, since Gender and Temperature is naturally assumed independent provided
no observation made, we must require:
x y  z w  u v  0.2 0.1  2
which together with the first system defines a unique solution.

2 y  y  0.1  y  0.1 3  0.033 ; x  0.2 3  0.067
2 w  w  0.2  w  0.2 3  0.067 ; z  0.4 3  0.133
2v  v  0.6  v  0.6 3  0.2; u  1.2 3  0.4
Pr
Temperature
Hoodie?:
above20
between10_20
below10
Gender
Man
Woman
Man
Woman
Man
Woman
Yes
0.067
0.033
0.133
0.067
0.4
0.2
No
0.933
0.967
0.867
0.933
0.6
0.8
Note! We have actually preserved the independence between Gender and
Temperature when assigning the conditional probabilities.
Assigning them differently induces dependencies
 In a converging connection, even knowledge of the conditional probabilities of
states in the child node may induce dependency between the parents. It is not
necessary to instantiate the child node.
c)
Instantiating Gender to Man does not
affect the probabilities in node
Temperature.
If no state is given in node Hoodie? the
nodes Gender and Temperature are
independent.
Instantiating Hoodie? to Yes changes the
probability distributions in the parent
nodes...
…but are they still (conditionally)
independent?
Instantiating Gender to Man affects the
probabilities in node Temperature – even
if it is just slightly.
 Gender and Temperature are
conditionally dependent given a state in
Hoodie?
5. a)



PrE H   PrE F , H   PrF H   Pr E F , H  PrF H 


PrE   PrE   1; PrF   PrF   1; PrH   PrH   1
 Three nodes, H, F and E, with two states each.
A Bayesian network satisfies the Markov property  A node depends
“backwards” only on its set of parents.
 E must depend on both H and F and F must depend on H.
H
F
E
b)


PrE   PrE H , G   PrH   PrG   Pr E H , G  PrH   PrG  

 Pr E H , G  PrH   PrG   Pr E H , G  PrH   PrG 

PrE   PrE   1; PrH   PrH   1; PrG   PrG   1




 Three nodes, H, G and E, with two states each.
Markov property  E must depend on both H and G but G and H do not
depend on any parent node or on each other.
H
G
E
6. a)
PrE1 , E2 H   PrE1 , E2 F , G, H  PrF , G H  
 PrE1 , E2 F , G, H  PrF , G H  
 PrE1 , E2 F , G, H  PrF , G H  
 PrE1 , E2 F , G, H  PrF , G H 
PrE1 , E2 F , G, H  

Markov
property
Divergent
connection
 PrE1 F , G, H  PrE2 F , G, H  
 PrE1 F  PrE2 G 
Analogousl y...
PrE1 , E2 F , G, H   PrE1 F  PrE2 G 
PrE1 , E2 F , G, H   PrE1 F  PrE2 G 
PrE1 , E2 F , G, H   PrE1 F  PrE2 G 
Pr F , G H   Pr G F , H  Pr F H   Pr G F , H  Pr F 
Pr F , G H   Pr G F , H  Pr F 
Pr F , G H   Pr G F , H  Pr F 
Pr F , G H   Pr G F , H  Pr F 
 Pr E1 , E2 H   Pr E1 F  Pr E2 G  Pr G F , H  Pr F  
 Pr E1 F  Pr E2 G  Pr G F , H  Pr F  
 Pr E1 F  Pr E2 G  Pr G F , H  Pr F  
 Pr E1 F  Pr E2 G  Pr G F , H  Pr F  


 Pr E F  Pr F  Pr E G  Pr G F , H   Pr E G  Pr G F , H 
 Pr E , E H  
 Pr E F  Pr F   Pr E G  Pr G F , H   Pr E G  Pr G F , H 
 Pr E F  Pr F  Pr E G  Pr G F , H   Pr E G  Pr G F , H 
 Pr E1 F  Pr F   Pr E2 G  Pr G F , H   Pr E2 G  Pr G F , H  
1
2
2
1
2
2
1
2
2
1
2
b)
Pr E1 , E2 H  


 Pr E1 F  Pr F   Pr E2 G  Pr G F , H   Pr E2 G  Pr G F , H  

 

1
0




 Pr E1 F  Pr F  Pr E2 G  Pr G F , H   Pr E2 G  Pr G F , H  

 

0
1


 Pr E1 F  Pr F   Pr E2 G   Pr E1 F  Pr F  Pr E2 G 
Pr E1 , E2 H  


 Pr E1 F  Pr F   Pr E2 G  Pr G F , H   Pr E2 G  Pr G F , H  

 

0
1




 Pr E1 F  Pr F  Pr E2 G  Pr G F , H   Pr E2 G  Pr G F , H  

 

1
0


 Pr E1 F  Pr F   Pr E2 G   Pr E1 F  Pr F  Pr E2 G 