Solutions 4
Exercise 2.5:
See the Hugin network.
Exercise 2.22:
Part i:
When inserting and propagating the three pieces of evidence, Hugin
responds with the error message: "Propagation of inconsistent evidence
has been attempted, or underflow has occurred".
Part ii:
After modifying the potentials (see the resulting Hugin network) we get the following probability
distributions:
P(Cold?|e) = (0.03(y),0.97(n))
P(Angina?|Cold?=n,e) = (0,0.97,0.03)
P(Angina?|Cold?=y,e) = (0,1,0)
By multiplying the latter two probability distributions with
P(Cold?|e) we get P(Angina?,Cold?|e):
| no | mild | severe |
|---|
| yes | 0 | 0.03 | 0 |
|---|
| no | 0 | 0.94 | 0.03 |
|---|
Part iii:
By using max-propagation in Hugin (or by max-marginalizing out the
variables from e.g. the table above) we get: P(Cold?|e)=(0.03,0.94)
and P(Angina?|e)=(0,0.94,0.03). This shows that
(Cold?=no,Angina?=mild) is the most probable posterior configuration.
Part iv:
Conf(e)=log2((0.978*0.002*0.965)/7.513*10-7)=11.3
which indicates a conflict.
Part v:
If P(Angina?=mild,e)=x(s)=a*s+b and P(e)=y(s)=c*s+d, then
P(Angina=mild|e)=x(s)/y(s).
With s=0.01 we have P(e)=7.513*10-7 we get c*0.01+d=7.513*10-7. By
also entering Angina?=mild we get a*0.01+b=7.298*10-7.
Next we change s to 0.02 which gives us: c*0.02+d=1.48*10-6 and
a*0.02+b=1.46*10-6. We can now solve the two pairs of linear
equations, and we get: a=7.3*10-5, b=2.11*10-8, c= 7.287*10-5 and
d=2.3*10-8. Hence, P(Angina?=mild|e)=(7.3*s+0.00211)/(7.28*s+0.0023).