程序代写代做代考 chain Uncertainty

Uncertainty

CISC 6525

Uncertainty

Chapter 13

Outline

• Uncertainty

• Probability

• Syntax and Semantics

• Inference

• Independence and Bayes’ Rule

Uncertainty

Let action At = leave for airport t minutes before flight

Will At get me there on time?

Problems:
1. partial observability (road state, other drivers’ plans, etc.)

2. noisy sensors (traffic reports)

3. uncertainty in action outcomes (flat tire, etc.)

4. immense complexity of modeling and predicting traffic

Hence a purely logical approach either
1. risks falsehood: “A25 will get me there on time”, or

2. leads to conclusions that are too weak for decision making:

“A25 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 A25 works unless contradicted by evidence

• Issues: What assumptions are reasonable? How to handle
contradiction?

• Rules with fudge factors:
– A25 |→0.3 get there on time

– Sprinkler |→ 0.99 WetGrass

– WetGrass |→ 0.7 Rain

• Issues: Problems with combination, e.g., Sprinkler causes Rain??

• Probability
– Model agent’s degree of belief

– Given the available evidence,

– A25 will get me there on time with probability 0.04

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(A25 | no reported accidents) = 0.06

These are not assertions about the world

Probabilities of propositions change with new evidence:

e.g., P(A25 | no reported accidents, 5 a.m.) = 0.15

Making decisions under

uncertainty
Suppose I believe the following:

P(A25 gets me there on time | …) = 0.04

P(A90 gets me there on time | …) = 0.70

P(A120 gets me there on time | …) = 0.95

P(A1440 gets me there on time | …) = 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

Syntax

• 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?)

• Discrete random variables
e.g., Weather is one of

• Domain values must be exhaustive and mutually exclusive

• Elementary proposition constructed by assignment of a value to a random
variable: e.g., Weather = sunny, Cavity = false

• (abbreviated as cavity)

• Complex propositions formed from elementary propositions and standard
logical connectives e.g., Weather = sunny  Cavity = false

Syntax

• 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

Axioms of probability

• For any propositions A, B

– 0 ≤ P(A) ≤ 1

– P(true) = 1 and P(false) = 0

– P(A  B) = P(A) + P(B) – P(A  B)

Prior probability

• Prior or unconditional probabilities of propositions
e.g., P(Cavity = true) = 0.1 and P(Weather = sunny) = 0.72 correspond to belief

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 = sunny rainy cloudy snow

Cavity = true 0.144 0.02 0.016 0.02

Cavity = false 0.576 0.08 0.064 0.08

• Every question about a domain can be answered by the joint distribution

Probability for continuous

variables
Express distribution as a parameterized function of value:

P(X =x) = U[18; 26](x) = uniform density between 18 and 26

Here P is a density; integrates to 1.

P(X =20:5) = 0:125 really means

lim P(20:5 X 20:5 + dx)=dx = 0:125

dx0

Gaussian density

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, allowing simplification, e.g.,
P(cavity | toothache, sunny) = P(cavity | toothache) = 0.8

• This kind of inference, sanctioned by domain knowledge, is crucial

Conditional probability

• Definition of conditional probability:
P(a | b) = P(a  b) / P(b) if P(b) > 0

• Product rule gives an alternative formulation:
P(a  b) = P(a | b) P(b) = P(b | a) P(a)

• A general version holds for whole distributions, e.g.,
P(Weather,Cavity) = P(Weather | Cavity) P(Cavity)

• (View as a set of 4 × 2 equations, not matrix mult.)

• Chain rule is derived by successive application of product rule:
P(X1, …,Xn) = P(X1,…,Xn-1) P(Xn | X1,…,Xn-1)

= P(X1,…,Xn-2) P(Xn-1 | X1,…,Xn-2) P(Xn | X1,…,Xn-1)

= …

= π
n

i= 1 P(Xi | X1, … ,Xi-1)

Inference by enumeration

• Start with the joint probability distribution:

• For any proposition φ, sum the atomic events where it is
true: P(φ) = Σω:ω╞φ P(ω)

Inference by enumeration

• Start with the joint probability distribution:

• For any proposition φ, sum the atomic events where it is
true: P(φ) = Σω:ω╞φ P(ω)

• P(toothache) = 0.108 + 0.012 + 0.016 + 0.064 = 0.2

Inference by enumeration

• Start with the joint probability distribution:

• For any proposition φ, sum the atomic events where it is
true: P(φ) = Σω:ω╞φ P(ω)

• P(toothache) = 0.108 + 0.012 + 0.016 + 0.064 = 0.2

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 α

P(Cavity | toothache) = α, P(Cavity,toothache)
= α, [P(Cavity,toothache,catch) + P(Cavity,toothache, catch)]

= α, [<0.108,0.016> + <0.012,0.064>]

= α, <0.12,0.08> = <0.6,0.4>

General idea: compute distribution on query variable by fixing evidence
variables and summing over hidden variables

Inference by enumeration,

contd.
Typically, we are interested in

the 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 the required summation of joint entries is done by summing out the
hidden variables:

P(Y | E = e) = αP(Y,E = e) = αΣhP(Y,E= e, H = h)

• The terms in the summation are joint entries because Y, E and H together
exhaust the set of random variables

• 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?

Independence

• A and B are independent iff

P(A|B) = P(A) or P(B|A) = P(B) or P(A, B) = P(A) P(B)

P(Toothache, Catch, Cavity, Weather)

= P(Toothache, Catch, Cavity) P(Weather)

• 32 entries reduced to 12; 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 independent. 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

contd.
• Write out 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)

I.e., 2 + 2 + 1 = 5 independent numbers

• In most cases, the use of conditional independence
reduces the size of the representation of the joint
distribution from exponential in n to linear in n.

• Conditional independence is our most basic and robust
form of knowledge about uncertain environments.

Bayes’ Rule

• Product rule P(ab) = P(a | b) P(b) = P(b | a) P(a)

 Bayes’ rule: P(a | b) = P(b | a) P(a) / P(b)

• or in distribution form

P(Y|X) = P(X|Y) P(Y) / P(X) = αP(X|Y) P(Y)

• Useful for assessing diagnostic probability from causal
probability:
– P(Cause|Effect) = P(Effect|Cause) P(Cause) / P(Effect)

– E.g., let M be meningitis, S be stiff neck:
P(m|s) = P(s|m) P(m) / P(s) = 0.8 × 0.0001 / 0.1 = 0.0008

– Note: posterior probability of meningitis still very small!

Bayes’ Rule and conditional

independence
P(Cavity | toothache  catch)

= αP(toothache  catch | Cavity) P(Cavity)

= αP(toothache | Cavity) P(catch | Cavity) P(Cavity)

• This is an example of a naïve Bayes model:
P(Cause,Effect1, … ,Effectn) = P(Cause) πiP(Effecti|Cause)

• Total number of parameters is linear in n

Wumpus World

Pij =true iff [i, j] contains a pit

Bij =true iff [i, j] is breezy

Include only B1;1 B1;2 B2;1 in the probability model

Specifying the probability model

The full joint distribution is P(P1;1 , … , P4;4, B1;1, B1;2, B2;1)

Apply product rule: P(B1;1, B1;2, B2;1 | P1;1 ,…, P4;4)P(P1;1,…, P4;4)

(Do it this way to get P( Effect | Cause ).)

First term: 1 if pits are adjacent to breezes, 0 otherwise

Second term: pits are placed randomly, probability 0.2 per square:

P(P1;1,…, P4;4) =

for n pits.

Observations and query

We know the following facts:

b =  b1;1  b1;2  b2;1
known =  p1;1   p1;2   p2;1

Query is P(P1;3 | known, b)

Define Unknown = Pijs other than P1;3 and known

For inference by enumeration, we have

P(P1;3 | known, b) =   unknown P(P1;3, unknown, known, b)

Grows exponentially with number of squares!

Using conditional independence

Basic insight: observations are conditionally independent of other hidden

squares given neighbouring hidden squares

Define Unknown = Fringe  Other

P(b | P1;3, Known, Unknown) = P(b | P1;3, Known, Fringe)

Manipulate query into a form where we can use this!

Using conditional independence

contd.

Using conditional independence

contd.

P(P1;3 | known, b) = ’ 0.2(0.04 + 0.16 + 0.16), 0.8(0.04 + 0.16) 

  0.31, 0.69 

P(P2;2 | known, b)   0.86, 0.14 

Summary

• Probability is a rigorous formalism for uncertain
knowledge

• Joint probability distribution specifies probability
of every atomic event

• Queries can be answered by summing over
atomic events

• For nontrivial domains, we must find a way to
reduce the joint size

• Independence and conditional independence
provide the tools