1. Probability Theory Longin Jan Latecki Temple University Slides based on slides by Aaron Hertzmann, Michael P. Frank, and Christopher Bishop

2. What is reasoning? How do we infer properties of the world? How should computers do it?

3. Aristotelian logic If A is true, then B is true A is true Therefore, B is true A: My car was stolen B: My car isn’t where I left it

4. Real-world is uncertain Problems with pure logic: Don’t have perfect information Don’t really know the model Model is non-deterministic So let’s build a logic of uncertainty!

5. Beliefs Let B(A) = “belief A is true” B(¬A) = “belief A is false” e.g., A = “my car was stolen” B(A) = “belief my car was stolen”

6. Reasoning with beliefs Cox Axioms [Cox 1946] 1.Ordering exists – e.g., B(A) > B(B) > B(C) 2.Negation function exists – B(¬A) = f(B(A)) 3.Product function exists – B(A  Y) = g(B(A|Y),B(Y)) This is all we need!

7. The Cox Axioms uniquely define a complete system of reasoning: This is probability theory!

8. “Probability theory is nothing more than common sense reduced to calculation.” - Pierre-Simon Laplace, 1814 Principle #1:

9. Definitions P(A) = “probability A is true” = B(A) = “belief A is true” P(A) 2 [0…1] P(A) = 1 iff “A is true” P(A) = 0 iff “A is false” P(A|B) = “prob. of A if we knew B” P(A, B) = “prob. A and B”

10. Examples A: “my car was stolen” B: “I can’t find my car” P(A) =.1 P(A) =.5 P(B | A) =.99 P(A | B) =.3

11. Sum rule: P(A) + P(¬A) = 1 Basic rules Example: A: “it will rain today” p(A) =.9 p(¬A) =.1

12. Sum rule:  i P(A i ) = 1 Basic rules when exactly one of A i must be true

13. Product rule: P(A,B) = P(A|B) P(B) = P(B|A) P(A) Basic rules

14. Conditioning  i P(A i ) = 1  i P(A i |B) = 1 P(A,B) = P(A|B) P(B) P(A,B|C) = P(A|B,C) P(B|C) Sum Rule Product Rule

15. Summary Product rule Sum rule All derivable from Cox axioms; must obey rules of common sense Now we can derive new rules P(A,B) = P(A|B) P(B)  i P(A i ) = 1

16. Example A = you eat a good meal tonight B = you go to a highly-recommended restaurant ¬B = you go to an unknown restaurant Model: P(B) =.7, P(A|B) =.8, P(A|¬B) =.5 What is P(A)?

17. Example, continued Model: P(B) =.7, P(A|B) =.8, P(A|¬B) =.5 1 = P(B) + P(¬B) 1 = P(B|A) + P(¬B|A) P(A) = P(B|A)P(A) + P(¬B|A)P(A) = P(A,B) + P(A,¬B) = P(A|B)P(B) + P(A|¬B)P(¬B) =.8.7 +.5 (1-.7) =.71 Sum rule Conditioning Product rule

18. Basic rules Marginalizing P(A) =  i P(A, B i ) for mutually-exclusive B i e.g., p(A) = p(A,B) + p(A, ¬B)

19. Syllogism revisited A -> B A Therefore B P(B|A) = 1 P(A) = 1 P(B) = P(B,A) + P(B, ¬A) = P(B|A)P(A) + P(B|¬A)P(¬A) = 1

20. Knowing P(A,B,C) is equivalent to knowing P(A|B,C), P(B|C), P(C), P(A), P(A|C), P(C|A), etc.

21. Given a complete model, we can derive any other probability Principle #2:

22. Model: P(B) =.7, P(A|B) =.8, P(A|¬B) =.5 If we know A, what is P(B|A)? (“Inference”) Inference P(A,B) = P(A|B) P(B) = P(B|A) P(A) P(B|A) = P(A|B) P(B) P(A) =.8.7 /.71 ≈.79 Bayes’ Rule

23. Inference Bayes Rule P(M|D) = P(D|M) P(M) P(D) Posterior Likelihood Prior

24. Describe your model of the world, and then compute the probabilities of the unknowns given the observations Principle #3:

25. Use Bayes’ Rule to infer unknown model variables from observed data Principle #3a: P(M|D) = P(D|M) P(M) P(D) Likelihood Prior Posterior

26. Bayes’ Theorem posterior  likelihood × prior Rev. Thomas Bayes 1702-1761

30. Example: diagnosis Jo takes a test for a nasty disease. The result of the test is either “positive” (T) or “negative” (¬T). The test is 95% reliable. 1% of people with Jo’s age/background have the disease. If the test result is “positive,” does Jo have the disease? [MacKay 2003]

31. Example: diagnosis Model: P(D) =.01, P(T|D) =.95, P(¬T|¬D) =.95 P(D|T) = P(T|D) P(D) P(T) ≈.16 Using P(T) = P(T|D)P(D) + P(T|¬D)P(¬D) =.95.01 + (1-.95).99 =.059

32. Example: diagnosis What if we tried a different test? 99.9% reliable test -> P(D|T 2 ) ≈ 91% 70% reliable test -> P(D|T 3 ) = 2.3 % The posterior merges information (could use multiple tests, e.g., P(D|T 2, T 3 ) )

33. Independence Definition: A and B are independent iff P(A,B) = P(A) P(B)

34. Example Suppose a red die and a blue die are rolled. The sample space: 1 2 3 4 5 6 1 x x x x x x 2 x x x x x x 3 x x x x x x 4 x x x x x x 5 x x x x x x 6 x x x x x x Are the events sum is 7 and the blue die is 3 independent?

35. |S| = 36 1 2 3 4 5 6 1 x x x x x x 2 x x x x x x 3 x x x x x x 4 x x x x x x 5 x x x x x x 6 x x x x x x |sum is 7| = 6 |blue die is 3| = 6 | in intersection | = 1 p(sum is 7 and blue die is 3) =1/36 p(sum is 7) p(blue die is 3) =6/36*6/36=1/36 Thus, p((sum is 7) and (blue die is 3)) = p(sum is 7) p(blue die is 3) The events sum is 7 and the blue die is 3 are independent:

36. Conditional Probability Let E, F be any events such that Pr( F) >0. Then, the conditional probability of E given F, written Pr( E | F), is defined as Pr( E | F) :≡ Pr( E  F) /Pr( F). This is what our probability that E would turn out to occur should be, if we are given only the information that F occurs. If E and F are independent then Pr( E | F) = Pr( E). Pr( E | F) = Pr( E  F) /Pr( F) = Pr( E) ×Pr( F) /Pr( F) = Pr( E)

37. Visualizing Conditional Probability If we are given that event F occurs, then – Our attention gets restricted to the subspace F. Our posterior probability for E (after seeing F ) corresponds to the fraction of F where E occurs also. Thus, p ′( E )= p ( E ∩ F )/ p ( F ). Entire sample space S Event F Event E Event E∩F

38. Suppose I choose a single letter out of the 26-letter English alphabet, totally at random. – Use the Laplacian assumption on the sample space {a,b,..,z}. – What is the (prior) probability that the letter is a vowel? Pr[Vowel] = __ / __. Now, suppose I tell you that the letter chosen happened to be in the first 9 letters of the alphabet. – Now, what is the conditional (or posterior) probability that the letter is a vowel, given this information? Pr[Vowel | First9] = ___ / ___. Conditional Probability Example a bc d e f g h i j k l m n o p q r s t u v w x y z 1 st 9 letters vowels Sample Space S

39. Example What is the probability that, if we flip a coin three times, that we get an odd number of tails (=event E ), if we know that the event F, the first flip comes up tails occurs? (TTT), (TTH), (THH), (HTT), (HHT), (HHH), (THT), (HTH) Each outcome has probability 1/4, p( E | F ) = 1/4+1/4 = ½, where E=odd number of tails or p( E | F ) = p( E  F )/p( F ) = 2/4 = ½ For comparison p( E ) = 4/8 = ½ E and F are independent, since p( E | F) = Pr( E ).

40. Example: Two boxes with balls Two boxes: first: 2 blue and 7 red balls; second: 4 blue and 3 red balls Bob selects a ball by first choosing one of the two boxes, and then one ball from this box. If Bob has selected a red ball, what is the probability that he selected a ball from the first box. An event E : Bob has chosen a red ball. An event F : Bob has chosen a ball from the first box. We want to find p( F | E )

41. What’s behind door number three? The Monty Hall problem paradox – Consider a game show where a prize (a car) is behind one of three doors – The other two doors do not have prizes (goats instead) – After picking one of the doors, the host (Monty Hall) opens a different door to show you that the door he opened is not the prize – Do you change your decision? Your initial probability to win (i.e. pick the right door) is 1/3 What is your chance of winning if you change your choice after Monty opens a wrong door? After Monty opens a wrong door, if you change your choice, your chance of winning is 2/3 – Thus, your chance of winning doubles if you change – Huh?

43. Monty Hall Problem C i - The car is behind Door i, for i equal to 1, 2 or 3. H ij - The host opens Door j after the player has picked Door i, for i and j equal to 1, 2 or 3. Without loss of generality, assume, by re-numbering the doors if necessary, that the player picks Door 1, and that the host then opens Door 3, revealing a goat. In other words, the host makes proposition H 13 true. Then the posterior probability of winning by not switching doors is P(C 1 |H 13 ).

44. The probability of winning by switching is P(C 2 |H 13 ), since under our assumption switching means switching the selection to Door 2, since P(C 3 |H 13 ) = 0 (the host will never open the door with the car) The posterior probability of winning by not switching doors is P(C 1 |H 13 ) = 1/3.

45. Discrete random variables Probabilities over discrete variables C 2 { Heads, Tails } P(C=Heads) =.5 P(C=Heads) + P(C=Tails) = 1 Possible values (outcomes) are discrete E.g., natural number (0, 1, 2, 3 etc.)

46. Terminology A (stochastic) experiment is a procedure that yields one of a given set of possible outcomes The sample space S of the experiment is the set of possible outcomes. An event is a subset of sample space. A random variable is a function that assigns a real value to each outcome of an experiment Normally, a probability is related to an experiment or a trial. Let’s take flipping a coin for example, what are the possible outcomes? Heads or tails (front or back side) of the coin will be shown upwards. After a sufficient number of tossing, we can “statistically” conclude that the probability of head is 0.5. In rolling a dice, there are 6 outcomes. Suppose we want to calculate the prob. of the event of odd numbers of a dice. What is that probability?

47. Random Variables A “random variable” V is any variable whose value is unknown, or whose value depends on the precise situation. – E.g., the number of students in class today – Whether it will rain tonight (Boolean variable) The proposition V = v i may have an uncertain truth value, and may be assigned a probability.

48. Example A fair coin is flipped 3 times. Let S be the sample space of 8 possible outcomes, and let X be a random variable that assignees to an outcome the number of heads in this outcome. Random variable X is a function X : S → X ( S ), where X ( S )={0, 1, 2, 3} is the range of X, which is the number of heads, and S ={ (TTT), (TTH), (THH), (HTT), (HHT), (HHH), (THT), (HTH) } X(TTT) = 0 X(TTH) = X(HTT) = X(THT) = 1 X(HHT) = X(THH) = X(HTH) = 2 X(HHH) = 3 The probability distribution (pdf) of random variable X is given by P(X=3) = 1/8, P(X=2) = 3/8, P(X=1) = 3/8, P(X=0) = 1/8.

49. Experiments & Sample Spaces A (stochastic) experiment is any process by which a given random variable V gets assigned some particular value, and where this value is not necessarily known in advance. – We call it the “actual” value of the variable, as determined by that particular experiment. The sample space S of the experiment is just the domain of the random variable, S = dom[ V ]. The outcome of the experiment is the specific value v i of the random variable that is selected.

50. Events An event E is any set of possible outcomes in S … – That is, E  S = dom [ V ]. E.g., the event that “less than 50 people show up for our next class” is represented as the set {1, 2, …, 49} of values of the variable V = (# of people here next class). We say that event E occurs when the actual value of V is in E, which may be written V  E. – Note that V  E denotes the proposition (of uncertain truth) asserting that the actual outcome (value of V ) will be one of the outcomes in the set E.

51. Probability of an event E The probability of an event E is the sum of the probabilities of the outcomes in E. That is Note that, if there are n outcomes in the event E, that is, if E = {a 1,a 2,…,a n } then

52. Example What is the probability that, if we flip a coin three times, that we get an odd number of tails? (TTT), (TTH), (THH), (HTT), (HHT), (HHH), (THT), (HTH) Each outcome has probability 1/8, p(odd number of tails) = 1/8+1/8+1/8+1/8 = ½

53. S Tail HH TT TH HT Sample Space S = {HH, HT, TH, TT} Venn Diagram Outcome Experiment: Toss 2 Coins. Note Faces. Event

54. Discrete Probability Distribution ( also called probability mass function (pmf) ) 1.List of All possible [x, p(x)] pairs –x = Value of Random Variable (Outcome) –p(x) = Probability Associated with Value 2.Mutually Exclusive (No Overlap) 3.Collectively Exhaustive (Nothing Left Out) 4. 0  p(x)  1 5.  p(x) = 1

55. Visualizing Discrete Probability Distributions Listing Table Graph Equation # Tails f(x) Count p(x) 01.25 12.50 21.25 px n xnx pp xnx () ! !()! ()    1.00.25.50 012 x p(x)

56. Marginal Probability Conditional Probability Joint Probability N is the total number of trials and n ij is the number of instances where X=x i and Y=y j

57. Sum Rule Product Rule

58. The Rules of Probability Sum Rule Product Rule

59. Continuous variables Probability Distribution Function (PDF) a.k.a. “marginal probability” p(x) P(a · x · b) = s a b p(x) dx x Notation: P(x) is prob p(x) is PDF

60. Continuous variables Probability Distribution Function (PDF) Let x 2 R p(x) can be any function s.t. s - 1 1 p(x) dx = 1 p(x) ¸ 0 Define P(a · x · b) = s a b p(x) dx

61. Continuous Prob. Density Function 1.Mathematical Formula 2.Shows All Values, x, and Frequencies, f(x) –f(x) Is Not Probability 3.Properties (Area Under Curve) Value (Value, Frequency) f(x) ab x fxdx fx () () All x a x b    1 0,

62. Continuous Random Variable Probability Probability Is Area Under Curve! Pcxdfxdx c d ()()  f(x) X cd

63. Probability mass function In probability theory, a probability mass function (pmf) is a function that gives the probability that a discrete random variable is exactly equal to some value. A pmf differs from a probability density function (pdf) in that the values of a pdf, defined only for continuous random variables, are not probabilities as such. Instead, the integral of a pdf over a range of possible values (a, b] gives the probability of the random variable falling within that range. Example graphs of a pmfs. All the values of a pmf must be non-negative and sum up to 1. (right) The pmf of a fair die. (All the numbers on the die have an equal chance of appearing on top when the die is rolled.)

64. Suppose that X is a discrete random variable, taking values on some countable sample space S ⊆ R. Then the probability mass function f X (x) for X is given by Note that this explicitly defines f X (x) for all real numbers, including all values in R that X could never take; indeed, it assigns such values a probability of zero. Example. Suppose that X is the outcome of a single coin toss, assigning 0 to tails and 1 to heads. The probability that X = x is 0.5 on the state space {0, 1} (this is a Bernoulli random variable), and hence the probability mass function is

65. Uniform Distribution 1. Equally Likely Outcomes 2. Probability Density 3. Mean & Standard Deviation Mean Median x f(x)f(x) dc

66. Uniform Distribution Example You’re production manager of a soft drink bottling company. You believe that when a machine is set to dispense 12 oz., it really dispenses 11.5 to 12.5 oz. inclusive. Suppose the amount dispensed has a uniform distribution. What is the probability that less than 11.8 oz. is dispensed?

67. Uniform Distribution Solution P(11.5  x  11.8)= (Base)(Height) = (11.8 - 11.5)(1) = 0.30 11.512.5 f(x)f(x)f(x)f(x) x 11.8 1.0

68. Normal Distribution 1.Describes Many Random Processes or Continuous Phenomena 2.Can Be Used to Approximate Discrete Probability Distributions –Example: Binomial 3.Basis for Classical Statistical Inference 4.A.k.a. Gaussian distribution

69. Normal Distribution 1.‘Bell-Shaped’ & Symmetrical 2.Mean, Median, Mode Are Equal 4. Random Variable Has Infinite Range Mean * light-tailed distribution

70. Probability Density Function f(x)=Frequency of Random Variable x  =Population Standard Deviation  =3.14159; e = 2.71828 x=Value of Random Variable (-  < x <  )  =Population Mean

72. Effect of Varying Parameters (  &  )

73. Normal Distribution Probability Probability is area under curve!

74. Infinite Number of Tables Normal distributions differ by mean & standard deviation. Each distribution would require its own table. That’s an infinite number!

75. Standardize the Normal Distribution One table! Normal Distribution Standardized Normal Distribution

76. Intuitions on Standardizing Subtracting  from each value X just moves the curve around, so values are centered on 0 instead of on  Once the curve is centered, dividing each value by  >1 moves all values toward 0, pressing the curve

77. Standardizing Example Normal Distribution

78. Standardizing Example Normal Distribution Standardized Normal Distribution

79. Why use Gaussians? Convenient analytic properties Central Limit Theorem Works well Not for everything, but a good building block For more reasons, see [Bishop 1995, Jaynes 2003]

80. Rules for continuous PDFs Same intuitions and rules apply “Sum rule”: s - 1 1 p(x) dx = 1 Product rule: p(x,y) = p(x|y)p(x) Marginalizing: p(x) = s p(x,y)dy … Bayes’ Rule, conditioning, etc.

81. Multivariate distributions Uniform: x » U (dom) Gaussian: x » N ( ,  )