Representing CS-480: Uncertainty
Chapter 13
Outline
Uncertainty Probability Syntax and Semantics Inference Independence and Bayes' Rule
Uncertainty
Define action At = leave for airport t minutes before flight to Hong Kong Q: Will At get me there on time? Problems:
1. 2. 3. 4. partial observability (road state, other drivers' plans, etc.) noisy sensors (traffic reports) uncertainty in action outcomes (flat tire, etc.) immense complexity of modeling and predicting traffic
Purely logical approach either
1. 2. risks falsehood: A95 will get me there on time, or leads to conclusions that are too weak for decision making:
A95 will get me there on time if there's no accident on the bridge and it doesn't rain and my tires remain intact etc etc. (A1440 might reasonably be said to get me there on time but I'd have to stay overnight in the airport )
Methods for handling uncertainty
Default or nonmonotonic logic:
Assume my car does not have a flat tire Assume A95 works unless contradicted by evidence
But What assumptions are reasonable? How to handle contradiction? Inference rules with confidence factors:
A95 |0.3 get there on time Sprinkler | 0.99 WetGrass WetGrass | 0.7 Rain
But how to combine these? e.g., Sprinkler causes Rain? Or can I infer WetGrass from Rain? Probability
Model agent's degree of belief with formal calculus for inference E.g., Given the available evidence,
A95 will get me there on time with probability 0.04
Two Views of Probability
Classical (or Frequentist) view:
Probability of 0.4 (say) means: In an infinite series of repetitions of the identical event, 40% of cases will give a certain result But: What of unique events? What constitute identical events?
Bayesian (or Subjectivist) view:
Probability is subjective degree of belief in a proposition given state of knowledge Define reasonable axioms to give a full calculus Requires assuming prior probability distributions (making arbitrary assumptions explicit)
Subjectivist Probability
Probabilistic assertions summarize effects of
laziness: failure to enumerate exceptions, qualifications, etc. ignorance: lack of relevant facts, initial conditions, etc.
Subjective probability: Probabilities relate propositions to agent's own state of knowledge e.g., P(A95 | no reported accidents) = 0.06 These are not assertions about the world Probabilities of propositions change with new evidence: e.g., P(A95 | no reported accidents, 5 a.m.) = 0.15
Making decisions under uncertainty
Suppose I believe the following:
P(A95 gets me there on time | ) P(A120 gets me there on time | ) P(A150 gets me there on time | ) P(A1440 gets me there on time | ) = 0.04 = 0.70 = 0.95 = 0.9999
Which action to choose? Depends on my preferences for missing flight vs. time spent waiting, etc.
Utility theory is used to represent and infer preferences Decision theory = probability theory + utility theory
Boolean Random Variables
We call A a boolean random variable if A denotes some event about which there is uncertainty as to its occurrence A = The president in 2013 will be female A = You will have a headache tomorrow A = You have the flu A = Five students were born on Tuesday
Probability: Intuitive
P(A) denotes fraction of possible worlds (given what I know) in which A is true
Event Space of all possible worlds Total area = 1 Worlds in which A is true We could discuss the philosophy of this idea for hours. but we wont!
P(A) = area of red oval Worlds in which A is false
Probability: Axioms
0 P(A) 1 P(true) = 1 P(false) = 0 P(A B) = P(A) + P(B) - P(A B)
Probability: Axioms
0 P(A) 1 P(true) = 1 P(false) = 0 P(A B) = P(A) + P(B) - P(A B)
Red oval cant get smaller than 0 Area of 0 means that A is true in no possible worlds
Worlds in which A is false
Probability: Axioms
0 P(A) 1 P(true) = 1 P(false) = 0 P(A B) = P(A) + P(B) - P(A B)
Red oval cant get larger than 1 Area of 1 means that A is true in all possible worlds
Worlds in which A is true
Worlds in which A is false
Probability: Axioms
0 P(A) 1 P(true) = 1 P(false) = 0 P(A or B) = P(A) + P(B) - P(A & B)
Worlds in which A is true Size of union is sum of sizes minus size of intersection
A&B are true
Worlds in which B is true
Thou Shalt Not Mess With The Axioms of Probability!
There have been many other attempted methodologies for uncertainty:
Fuzzy logic Dempster-Shafer theory Three-valued logic Nonmonotonic logics Etc., etc.,
But only probability has the property:
If you gamble according to the axioms of probability, you cant be unfairly exploited by someone using a different system. (di Finetti 1931)
Some Provable Facts
Axioms: 0 P(A) 1 P(true) = 1 P(false) = 0 P(A B) = P(A) + P(B) - P(A B)
We can show that: P(~A) = P(not A) = 1 - P(A) And furthermore: P(A) = P(A & B) + P(A & ~B)
Multivalued Random Variables
Suppose A can take on more than 2 values
e.g., Weather is one of {sunny,rainy,cloudy,snow}
Call A a random variable with arity k if it can take on one of k different values in some set {v1, v2, , vk} Thus: P(A=vi & A=vj) = 0 if i j P(A=v1 A=v2 A=vk) = 1
Easy Facts About Multivalued RVs
Axioms: 0 P(A) 1; P(true) = 1; P(false) = 0 P(A B) = P(A) + P(B) - P(A B) Recall: P(A=vi & A=vj) = 0 if i j ; P(A=v1 A=v2 A=vk) = 1
We can show that:
i
P(A = v1A = v2 L A = vi) = P(A = vj)
j=1
And therefore:
k
P(A = v1L A = vk) = P(A = vj)
j=1
More Facts About Multivalued RVs
Axioms: 0 P(A) 1; P(true) = 1; P(false) = 0 P(A B) = P(A) + P(B) - P(A B) Recall: P(A=vi & A=vj) = 0 if i j ; P(A=v1 A=v2 A=vk) = 1
We can show that:
i
P(B [A = v1L A = vi]) = P(B A = vj)
j=1
And therefore:
k
P(B) = P(B A = vj)
j=1
Conditional Probability
P(A|B) = probability of A given B = fraction of possible worlds with B true that also have A true
Worlds in which A is true P(Headache) = 0.1 P(Flu) = 0.02 P(Headache|Flu) = 0.5 Headaches are rare, Flu is much rarer, but if you have the Flu, you have a 50-50 chance of having a headache.
A&B are true
Worlds in which B is true
Conditional probability
Conditional or posterior probabilities
e.g., P(cavity | toothache) = 0.8 i.e., given that toothache is all I know
Notation for conditional distributions:
P(Cavity | Toothache) = 2-element vector of 2-element vectors
If we know more, e.g., cavity is also given, then we have
P(cavity | toothache,cavity) = 1
New evidence may be irrelevant, e.g.,
P(cavity | toothache, sunny) = P(cavity | toothache) = 0.8
Such simplification, based on domain knowledge, is crucial
Conditional Probability
Formal definition:
P(A B) P(A | B) = P(B)
Thus, the Chain Rule:
P(A B) = P(A | B)P(B)
P(A1A2 L An) = P(A1 | A2L An)P(A2 | A3L An)L P(An)
Syntax Summary
Basic element: random variable Similar to propositional logic: possible worlds defined by assignment of values to random variables. Boolean random variables
e.g., Cavity (do I have a cavity?)
Multivalued random variables
e.g., Weather is one of <sunny,rainy,cloudy,snow>
Domain values exhaustive and mutually exclusive Elementary proposition constructed by assignment of a value to a Complex propositions formed from elementary propositions and standard logical connectives e.g., Weather = sunny Cavity = false
Atomic Events
Atomic event: A complete specification of the state of the world about which the agent is uncertain
E.g., if the world consists of only two Boolean variables Cavity and Toothache, then there are 4 distinct atomic events:
Cavity = false Toothache = false Cavity = false Toothache = true Cavity = true Toothache = false Cavity = true Toothache = true
Atomic events are mutually exclusive and exhaustive
Prior probability
Prior or unconditional probabilities of propositions
e.g., P(Cavity = true) = 0.1 and P(Weather = sunny) = 0.72 correspond belief to prior to arrival of any (new) evidence
Probability distribution gives values for all possible assignments:
P(Weather) = <0.72,0.1,0.08,0.1> (normalized, i.e., sums to 1)
Joint probability distribution for a set of random variables gives the probability of every atomic event on those random variables
P(Weather,Cavity) = a 4 2 matrix of values:
Weather = Cavity = true Cavity = false
sunny 0.144 0.576
rainy 0.02 0.08
cloudy 0.016 0.064
snow 0.02 0.08
Every question about a domain can be answered by the joint distribution
Inference
Generally: Given some information about the probability distribution, determine the probability of some proposition = Cavity = Cavity & Toothache = ~Study & (GoodGrade or GoodJob)
Inference by enumeration
Start with the joint probability distribution:
For any proposition , sum the atomic events where it is true:
P() = : P()
Inference by enumeration
Start with the joint probability distribution:
For any proposition , sum the atomic events where it is true: P() = : P() P(toothache) = 0.108 + 0.012 + 0.016 + 0.064 = 0.2
Inference by enumeration
Start with the joint probability distribution:
For any proposition , sum the atomic events where it is true: P() = : P() P(toothache or cavity) = 0.108 + 0.012 + 0.016 + 0.064 +0.072 + 0.008= 0.28
Inference by enumeration
Start with the joint probability distribution:
Can also compute conditional probabilities:
P(cavity | toothache) = P(cavity toothache) P(toothache) = 0.016+0.064 0.108 + 0.012 + 0.016 + 0.064 = 0.4
Normalization
Denominator can be viewed as a normalization constant
= [P(Cavity,toothache,catch) + P(Cavity,toothache,catch)] = [<0.108,0.016> + <0.012,0.064>] = <0.12,0.08> = <0.6,0.4>
P(Cavity | toothache) = P(Cavity,toothache)
General idea: compute distribution on query variable (Cavity) by fixing evidence variables (Toothache) and summing over hidden variables (Catch)
Inference by enumeration
Typically, we want P(Y | E = e): posterior joint distribution of the query variables Y given specific values e for the evidence variables E Let the hidden variables be H = X \ (Y E) Then we can just sum over the hidden variables and normalize:
P(Y | E = e) = P(Y,E = e) =
[hP(Y,E= e, H = h)]
Terms are atomic events, because Y E H = X
Obvious problems: 1. Worst-case time complexity O(dn) where d is the largest arity 2. Space complexity O(dn) to store the joint distribution 3. How to find the numbers for O(dn) entries in the joint distribution?
4.
Independence
Two boolean random variables A and B are said to be independent if and only if P(A|B) = P(A) Equivalently P(B|A) = P(B) That is, the probability we give A (or B) is not affected by finding out that B (or A)
Independence Facts
If A and B are independent boolean RVs: P(A & B) = P(A | B) P(B) = P(A) P(B) P(~A|B) = 1-P(A|B) = 1-P(A) = P(~A) P(A|~B) = P(A & ~B) / P(~B) = P(~B | A)P(A) / P(~B) = P(~B)P(A) / P(~B) = P(A)
Multivalued Independence
For multivalued RVs A and B, A is independent of B iff
u,v : P(A = u | B = v) = P(A = u)
From which we can show, for example:
u,v : P(A = u B = v) = P(A = u)P(B = v) " u,v : P(B = v | A = u) = P(B = v)
So, suppose our domain knowledge allows us to make certain independence assumptions on our random variables:
Independence
P(Toothache, Catch, Cavity, Weather) = P(Toothache, Catch, Cavity) P(Weather) 16 entries reduced to 10 (really 8); For n independent biased coins, O(2n) O(n)
Absolute independence powerful but rare
Dentistry is a large field with hundreds of variables, none of which are really independent of each other. What to do?
Conditional independence
P(Toothache, Cavity, Catch) has 23 1 = 7 independent entries If I have a cavity, the probability that the probe catches in it doesn't depend on whether I have a toothache: (1) P(catch | toothache, cavity) = P(catch | cavity) The same independence holds if I haven't got a cavity: (2) P(catch | toothache,cavity) = P(catch | cavity) Catch is conditionally independent of Toothache given Cavity: P(Catch | Toothache,Cavity) = P(Catch | Cavity) Equivalent statements: P(Toothache | Catch, Cavity) = P(Toothache | Cavity) P(Toothache, Catch | Cavity) = P(Toothache | Cavity) P(Catch | Cavity)
Conditional independence
Get the full joint distribution using chain rule: P(Toothache, Catch, Cavity)
= P(Toothache | Catch, Cavity) P(Catch, Cavity) = P(Toothache | Catch, Cavity) P(Catch | Cavity) P(Cavity) = P(Toothache | Cavity) P(Catch | Cavity) P(Cavity)
Based on only 2 + 2 + 1 = 5 independent parameters In most cases, the use of conditional independence reduces the siz...