Week 2

Question 1

What can cause uncertainty ?

Answer 1

Partial observability

Non determinism

Or both

Question 2

Belief state

Answer 2

Representation of the set of all possible world states that the agent may be in

Question 3

What is the qualification problem?

Answer 3

Accounting for every possible scenario

Eg, meteor hits

Question 4

3 problems of using logic in medical diagnosis

Answer 4

Laziness: too many possible antecedents and consequents

Theoretical ignorance: medical science has no complete theory for the domain

Practical ignorance: even with a complete theory not all of necessary tests will have been completed on everyone

Question 5

Limitation of agent knowledge

Answer 5

Degree of belief is best product

Question 6

Utility theory

Answer 6

Methodology for representing and reasoning with preferences of possible outcomes

Question 7

Decision theory =

Answer 7

Probability theory + utility theory

Question 8

An agent is rational iff

Answer 8

It chooses the action that yields the highest expected utility, averaged over all the possible outcomes of the action

Question 9

Maximum expected utility

Answer 9

The highest expected utility averaged over all the possible outcomes of the action

Question 10

Probabilities in the absence of any other information called

Answer 10

Unconditional or prior probabilities

Question 11

Product rule

Answer 11

P(a ^ b) = p(a|b)p(b)

Question 12

Inclusion exclusion principle

Answer 12

P(a v b) = p(a) + p(b) - p(a ^ b)

Question 13

Kolmogorov’s axioms

Answer 13

1) non negativity: P(A) >= 0

2) Unitarity: P(S) = 1 for sample space S

3) Additivity: P(A u B) = P(A) + P(B) for any mutually exclusive A,B

Question 14

Marginalisation

Answer 14

Adding probabilities of all scenarios of an event occurring eg

Adding all of cavity row to get marginal probability = 0.2

Question 15

Notation for marginalisation

Answer 15

Question 16

Conditioning

Answer 16

Using conditional probabilities instead of joint probabilities for marginalisation

Question 17

Normalización

Answer 17

Using conditional probabilities 1/p(a) can be considered a constant, α

Question 18

Independence

Answer 18

If P(A|B,C,D) = P(A)

Then A is marginally/absolutely independent of B,C,D

Question 19

Bayes’ rule

Answer 19

P(A|B) = P(B|A)P(A)/P(B)

Question 20

Causal and Diagnostic

Answer 20

Causal P(effect | cause)

Diagnostic P(cause | effect)

Question 21

X and Y are conditionally independent , given a 3rd variable Z, notationally?

Answer 21

P(X,Y|Z) = P(X|Z)P(Y|Z)

Question 22

Computing normalizing constant

Answer 22