Uncertainty Uncertain Knowledge Probability Review Bayes’ Theorem Summary

Uncertainty

• Uncertain Knowledge• Probability Review• Bayes’ Theorem• Summary

Uncertain Knowledge

In many situations we cannot assign a value oftrue or false to world statements.

ExampleSymptom(p,Toothache) Disease(p,Cavity)

To generalize:Symptom(p,Toothache) Disease(p,Cavity) V Disease(p,GumDisease) V …

Uncertain Knowledge

Solution: Deal with degrees of belief.

We will use probability theory. Probability states a degree of belief based on evidence:

P(x) = 0.80 – based on evidence, 80% of the times in which the experiment is run, x occurs. It summarizes our uncertainty of what causes x.

Degree of truth – Fuzzy logic.

Utility Theory

Combine probability and decision theory

To make a decision (action) an agent needs tohave preferences between plans.

An agent should choose the action with highest expected utility averaged over

all possible outcomes.

Uncertainty


Random Variable

Definition: A variable that can take on several values, each value having a probability of occurrence.

There are two types of random variables:Discrete. Take on a countable number of values.Continuous. Take on a range of values.

Random Variable

Discrete Variables For every discrete variable X there will be a probability function P(x) = P(X = x).

Random Variable

Continuous Variables: For every continuous random variable X we will associate a probability density function f(x). It is the area under the density functions between two points that corresponds to the probability of the variable lying between the two values.

Prob(x1 < X <= x2) = ∫x1 f(x) dx x2

The Sample Space

The space of all possible outcomes of a given process or situation is called the sample space S.

SS

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An Event

An event A is a subset of the sample space.

SS

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Atomic Event

An atomic event is a single point in S.

Properties: Atomic events are mutually exclusive The set of all atomic events is exhaustive A proposition is the disjunction of the atomic events it covers.

The Laws of Probability

The probability of the sample space S is 1, P(S) = 1The probability of any event A is such that 0 <= P(A) <= 1. Law of Addition

If A and B are mutually exclusive events, then the probability that either one of them will occur is the sum of the individual probabilities:

P(A or B) = P(A) + P(B)


If A and B are not mutually exclusive:

P(A or B) = P(A) + P(B) – P(A and B)

AA BB

Prior Probability

It is called the unconditional or prior probabilityof event A.

P(A) -- Reflects our original degree of belief of X.

Conditional Probabilities

Given that A and B are events in sample space S, and P(B) is different of 0, then the conditional probability of A given B is

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

If A and B are independent then

P(A|B) = P(A)


Law of Multiplication

What is the probability that both A and B occur together?

P(A and B) = P(A) P(B|A) where P(B|A) is the probability of B conditioned on A.


If A and B are statistically independent:

P(B|A) = P(B) and then

P(A and B) = P(A) P(B)

Independence on Two Variables

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

If A and B are conditionally independent:

P(A|B,C) = P(A|C) and

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

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Exercises

Find the probability that the sum of the numbersFind the probability that the sum of the numberson two unbiased dice will be even by considering theon two unbiased dice will be even by considering theprobabilities that the individual dice will show an evenprobabilities that the individual dice will show an evennumber.number.

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Exercises

XX11 – first throw – first throw

XX22 – second throw – second throw

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Exercises

XX11 – first throw – first throw

XX22 – second throw – second throw

Pfinal = P(XPfinal = P(X11=1 & X=1 & X22=1) + P(X=1) + P(X11=1 & X=1 & X22=3) + P(X=3) + P(X11=1 & X=1 & X22=5) +=5) +

P(XP(X11=2 & X=2 & X22=2) + P(X=2) + P(X11=2 & X=2 & X22=4) + P(X=4) + P(X11=2 & X=2 & X22=6) +=6) +

P(XP(X11=3 & X=3 & X22=1) + P(X=1) + P(X11=3 & X=3 & X22=3) + P(X=3) + P(X11=3 & X=3 & X22=5) +=5) +

… … P(XP(X11=6 & X=6 & X22=2) + P(X=2) + P(X11=6 & X=6 & X22=4) + P(X=4) + P(X11=6 & X=6 & X22=6).=6).

PPfinalfinal = 18/36 = 1/2 = 18/36 = 1/2

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Exercises

Find the probabilities of throwing a sum of a) 3, b) 4Find the probabilities of throwing a sum of a) 3, b) 4with three unbiased dice.with three unbiased dice.

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Exercises

Find the probabilities of throwing a sum of a) 3, b) 4Find the probabilities of throwing a sum of a) 3, b) 4with three unbiased dice.with three unbiased dice.

X = sum of XX = sum of X11 and X and X22 and X and X33

P(X=3)?P(X=3)?

P(XP(X11=1 & X=1 & X22=1 & X=1 & X33=1) = 1/216=1) = 1/216

P(X=4)?P(X=4)?P(XP(X11=1 & X=1 & X22=1 & X=1 & X33=2) + P(X=2) + P(X11=1 & X=1 & X22=2 & X=2 & X33=1) + …=1) + …

P(X=4) = 3/216P(X=4) = 3/216

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Exercises

Three men meet by chance. What are the probabilities Three men meet by chance. What are the probabilities that a) none of them, b) two of them, c) all of themthat a) none of them, b) two of them, c) all of themhave the same birthday?have the same birthday?

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Exercises

None of them have the same birthdayNone of them have the same birthday

XX11 – birthday 1 – birthday 1stst person person

XX22 – birthday 2 – birthday 2ndnd person person

XX33 – birthday 3 – birthday 3rdrd person person

a)a) P(XP(X22 is different than X is different than X11 & X & X33 is different than X is different than X11 and X and X22))

PPfinalfinal = (364/365)(363/365) = (364/365)(363/365)

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Exercises

Two of them have the same birthdayTwo of them have the same birthday

P(XP(X11 = X = X22 and X and X33 is different than X is different than X11 and X and X22) +) +

P(XP(X11=X=X33 and X and X22 differs) + differs) +

P(XP(X22=X=X33 and X and X11 differs). differs).

P(XP(X11=X=X22 and X and X33 differs) = (1/365)(364/365) differs) = (1/365)(364/365)

PPfinalfinal = 3(1/365)(364/365) = 3(1/365)(364/365)

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Exercises

All of them have the same birthdayAll of them have the same birthday

P(XP(X11 = X = X22 = X = X33))

PPfinalfinal = (1/365)(1/365) = (1/365)(1/365)

Multivariate o Joint Distributions

P(x,y) = P( X = x and Y = y).

P’(x) = Prob( X = x) = ∑y P(x,y) It is called the marginal distribution of X The same can be done on Y to define the marginal distribution of Y, P”(y).

If X and Y are independent then P(x,y) = P’(x) P”(y)

Expectations: The Mean

Let X be a discrete random variable that takes the following values: x1, x2, x3, …, xn.

Let P(x1), P(x2), P(x3),…,P(xn) be their respective probabilities. Then the expected value of X, E(X), is defined as

E(X) = x1P(x1) + x2P(x2) + x3P(x3) + … + xnP(xn) E(X) = Σi xi P(xi)

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Exercises

Suppose that X is a random variable taking the valuesSuppose that X is a random variable taking the values{-1, 0, and 1} with equal probabilities and that Y = X{-1, 0, and 1} with equal probabilities and that Y = X22 . . Find the joint distribution and the marginal distributions Find the joint distribution and the marginal distributions of X and Y and also the conditional distributions of Xof X and Y and also the conditional distributions of Xgiven a) Y = 0 and b) Y = 1.given a) Y = 0 and b) Y = 1.

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Exercises

0 1/3 0

1/3 0 1/3YY

XX

1/31/3

2/32/3

1/3 1/3 1/31/3 1/3 1/3

-1 0 1-1 0 1

00

11

If Y = 0 then X= 0 with probability 1If Y = 0 then X= 0 with probability 1If Y = 1 then X is equally likely to be +1 or -1If Y = 1 then X is equally likely to be +1 or -1

Uncertainty


Bayes’ Theorem

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

The theorem:

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

More General Bayes’ Theorem

P(Y|X,e) = P(X|Y,e) P(Y|e) / P(X|e)

Where e: background evidence.

Thomas Bayes

Born in London (1701). Studied logic and theology (Univ. of Edinburgh).Fellow of the Royal Society (year 1742).

Given white and black balls in an urn, what is the prob. of drawing one or the other?

Given one or more balls, what can be said about the number of balls in the urn?

Uncertainty


Summary

• Uncertainty comes from ignorance on the true state of the world.

• Probabilities indicate our degree of belief on certain event.

• Concepts: random variable, prior probabilities, conditional probabilities, joint distributions, conditional independence, Bayes’ theorem.

Application: Predicting Stock Market

Bayesian Networks BNs have been exploited to predictthe behavior of the stock market. BNs can be constructed from daily stock returns over a certain amount of time.

Stocks can be analyzed from well-known repositories:e.g., S&P 500 index.

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Uncertainty Uncertain Knowledge Probability Review Bayes’ Theorem Summary