1. Chap 4-1 Statistics Please Stand By….

2. Chap 4-2 Chapter 4: Probability and Distributions Randomness General Probability Probability Models Random Variables Moments of Random Variables

3. Chap 4-3 Randomness The language of probability Thinking about randomness The uses of probability

4. Chap 4-4 Randomness

5. Chap 4-5 Randomness

6. Chap 4-6 Randomness

7. Chap 4-7 Randomness Technical Analysis 1 | More Examples | Another (Tipping Pts) Technical Analysis 1 More ExamplesAnother (Tipping Pts)

8. Chap 4-8 Chapter Goals After completing this chapter, you should be able to: Explain three approaches to assessing probabilities Apply common rules of probability Use Bayes’ Theorem for conditional probabilities Distinguish between discrete and continuous probability distributions Compute the expected value and standard deviation for a probability distributions

9. Chap 4-9 Important Terms Probability – the chance that an uncertain event will occur (always between 0 and 1) Experiment – a process of obtaining outcomes for uncertain events Elementary Event – the most basic outcome possible from a simple experiment Randomness – Does not mean haphazard Description of the kind of order that emerges only in the long run

10. Chap 4-10 Important Terms (CONT’D) Sample Space – the collection of all possible elementary outcomes Probability Distribution Function Maps events to intervals on the real line Discrete probability mass Continuous probability density

11. Chap 4-11 Sample Space The Sample Space is the collection of all possible outcomes (based on an probabilistic experiment) e.g., All 6 faces of a die: e.g., All 52 cards of a bridge deck:

12. Chap 4-12 Events Elementary event – An outcome from a sample space with one characteristic Example: A red card from a deck of cards Event – May involve two or more outcomes simultaneously Example: An ace that is also red from a deck of cards

13. Chap 4-13 Elementary Events A automobile consultant records fuel type and vehicle type for a sample of vehicles 2 Fuel types: Gasoline, Diesel 3 Vehicle types: Truck, Car, SUV 6 possible elementary events: e 1 Gasoline, Truck e 2 Gasoline, Car e 3 Gasoline, SUV e 4 Diesel, Truck e 5 Diesel, Car e 6 Diesel, SUV Gasoline Diesel Car Truck Car SUV e1e2e3e4e5e6e1e2e3e4e5e6

14. Chap 4-14 Independent Events E 1 = heads on one flip of fair coin E 2 = heads on second flip of same coin Result of second flip does not depend on the result of the first flip. Dependent Events E 1 = rain forecasted on the news E 2 = take umbrella to work Probability of the second event is affected by the occurrence of the first event Independent vs. Dependent Events

15. Chap 4-15 Probability Concepts Mutually Exclusive Events If E 1 occurs, then E 2 cannot occur E 1 and E 2 have no common elements Black Cards Red Cards A card cannot be Black and Red at the same time. E1E1 E2E2

16. Chap 4-16 Coming up with Probability Empirically From the data! Based on observation, not theory Probability describes what happens in very many trials. We must actually observe many trials to pin down a probability Based on belief (Bayesian Technique)

17. Chap 4-17 Assigning Probability Classical Probability Assessment Relative Frequency of Occurrence Subjective Probability Assessment P(E i ) = Number of ways E i can occur Total number of elementary events Relative Freq. of E i = Number of times E i occurs N An opinion or judgment by a decision maker about the likelihood of an event

18. Chap 4-18 Calculating Probability Counting Outcomes Observing Outcomes in Trials Number of ways E i can occur Total number of elementary events

19. Chap 4-19 Counting

20. Chap 4-20 Counting a b c d e …. ___ ___ ___ ___ 1. N take n___ ___ ___ 2. N take k___ ___ 3. Order not important – less than permutations

21. Chap 4-21 Counting

22. Chap 4-22 Rules of Probability S is sample space Pr(S) = 1 Events measured in numbers result in a Probability Distribution

23. Chap 4-23 Rules of Probability Rules for Possible Values and Sum Individual ValuesSum of All Values 0 ≤ P(e i ) ≤ 1 For any event e i where: k = Number of elementary events in the sample space e i = i th elementary event

24. Chap 4-24 Addition Rule for Elementary Events The probability of an event E i is equal to the sum of the probabilities of the elementary events forming E i. That is, if: E i = {e 1, e 2, e 3 } then: P(E i ) = P(e 1 ) + P(e 2 ) + P(e 3 )

25. Chap 4-25 Complement Rule The complement of an event E is the collection of all possible elementary events not contained in event E. The complement of event E is represented by E. Complement Rule: E E Or,

26. Chap 4-26 Addition Rule for Two Events P(E 1 or E 2 ) = P(E 1 ) + P(E 2 ) - P(E 1 and E 2 ) E1E2 P(E 1 or E 2 ) = P(E 1 ) + P(E 2 ) - P(E 1 and E 2 ) Don’t count common elements twice! ■ Addition Rule: E1E2 +=

27. Chap 4-27 Addition Rule Example P(Red or Ace) = P(Red) +P(Ace) - P(Red and Ace) = 26/52 + 4/52 - 2/52 = 28/52 Don’t count the two red aces twice! Black Color Type Red Total Ace 224 Non-Ace 24 48 Total 26 52

28. Chap 4-28 Addition Rule for Mutually Exclusive Events If E1 and E2 are mutually exclusive, then P(E1 and E2) = 0 So P(E 1 or E 2 ) = P(E 1 ) + P(E 2 ) - P(E 1 and E 2 ) = P(E 1 ) + P(E 2 ) = 0 E1E2 if mutually exclusive

29. Chap 4-29 Conditional Probability Conditional probability for any two events E 1, E 2 :

30. Chap 4-30 What is the probability that a car has a CD player, given that it has AC ? i.e., we want to find P(CD | AC) Conditional Probability Example Of the cars on a used car lot, 70% have air conditioning (AC) and 40% have a CD player (CD). 20% of the cars have both.

31. Chap 4-31 Conditional Probability Example No CDCDTotal AC No AC Total 1.0 Of the cars on a used car lot, 70% have air conditioning (AC) and 40% have a CD player (CD). 20% of the cars have both. (continued)

32. Chap 4-32 Conditional Probability Example No CDCDTotal AC.2.5.7 No AC.2.1.3 Total.4.6 1.0 Given AC, we only consider the top row (70% of the cars). Of these, 20% have a CD player. 20% of 70% is about 28.57%. (continued)

33. Chap 4-33 For Independent Events: Conditional probability for independent events E 1, E 2 :

34. Chap 4-34 Multiplication Rules Multiplication rule for two events E 1 and E 2 : Note: If E 1 and E 2 are independent, then and the multiplication rule simplifies to

35. Chap 4-35 Tree Diagram Example Diesel P(E 2 ) = 0.2 Gasoline P(E 1 ) = 0.8 Truck: P(E 3 |E 1 ) = 0.2 Car: P(E 4 |E 1 ) = 0.5 SUV: P(E 5 |E 1 ) = 0.3 P(E 1 and E 3 ) = 0.8 x 0.2 = 0.16 P(E 1 and E 4 ) = 0.8 x 0.5 = 0.40 P(E 1 and E 5 ) = 0.8 x 0.3 = 0.24 P(E 2 and E 3 ) = 0.2 x 0.6 = 0.12 P(E 2 and E 4 ) = 0.2 x 0.1 = 0.02 P(E 3 and E 4 ) = 0.2 x 0.3 = 0.06 Truck: P(E 3 |E 2 ) = 0.6 Car: P(E 4 |E 2 ) = 0.1 SUV: P(E 5 |E 2 ) = 0.3

36. Chap 4-36 Get Ready…. More Probability Examples Random Variables Probability Distributions

37. Chap 4-37 Introduction to Probability Distributions Random Variable – “X” Is a function from the sample space to another space, usually Real line Represents a possible numerical value from a random event Each r.v. has a Distribution Function – F X (x), f X (x) based on that in the sample space Assigns probability to the (numerical) outcomes (discrete values or intervals)

38. Chap 4-38 Random Variables Not Easy to Describe

39. Chap 4-39 Random Variables

40. Chap 4-40 Random Variables Not Easy to Describe

41. Chap 4-41 Introduction to Probability Distributions Random Variable Represents a possible numerical value from a random event Random Variables Discrete Random Variable Continuous Random Variable

42. Chap 4-42 A list of all possible [ x i, P(x i ) ] pairs x i = Value of Random Variable (Outcome) P(x i ) = Probability Associated with Value x i ’s are mutually exclusive (no overlap) x i ’s are collectively exhaustive (nothing left out) 0  P(x i )  1 for each x i  P(x i ) = 1 Discrete Probability Distribution

43. Chap 4-43 Discrete Random Variables Can only assume a countable number of values Examples: Roll a die twice Let x be the number of times 4 comes up (then x could be 0, 1, or 2 times) Toss a coin 5 times. Let x be the number of heads (then x = 0, 1, 2, 3, 4, or 5)

44. Chap 4-44 Experiment: Toss 2 Coins. Let x = # heads. T T Discrete Probability Distribution 4 possible outcomes T T H H HH Probability Distribution 0 1 2 x x Value Probability 0 1/4 =.25 1 2/4 =.50 2 1/4 =.25.50.25 Probability

45. Chap 4-45 The Distribution Function Assigns Probability to Outcomes F, f > 0; right-continuous and P(X<a)=F X (a) Discrete Random Variable Continuous Random Variable

46. Chap 4-46 Discrete Random Variable Summary Measures - Moments Expected Value of a discrete distribution (Weighted Average) E(x) =  x i P(x i ) Example: Toss 2 coins, x = # of heads, compute expected value of x: E(x) = (0 x.25) + (1 x.50) + (2 x.25) = 1.0 x P(x) 0.25 1.50 2.25

47. Chap 4-47 Standard Deviation of a discrete distribution where: E(x) = Expected value of the random variable x = Values of the random variable P(x) = Probability of the random variable having the value of x Discrete Random Variable Summary Measures (continued)

48. Chap 4-48 Example: Toss 2 coins, x = # heads, compute standard deviation (recall E(x) = 1) Discrete Random Variable Summary Measures (continued) Possible number of heads = 0, 1, or 2

49. Chap 4-49 Two Discrete Random Variables Expected value of the sum of two discrete random variables: E(x + y) = E(x) + E(y) =  x P(x) +  y P(y) (The expected value of the sum of two random variables is the sum of the two expected values)

50. Chap 4-50 Sums of Random Variables Usually we discuss sums of INDEPENDENT random variables, X i i.i.d. Only sometimes is Due to Linearity of the Expectation operator, E(  X i ) =  E(X i ) and Var(  X i ) =  Var(X i ) CLT: Let S n =  X i then (S n - E(S n ))~N(0, var)

51. Chap 4-51 Covariance Covariance between two discrete random variables: σ xy =  [x i – E(x)][y j – E(y)]P(x i y j ) where: x i = possible values of the x discrete random variable y j = possible values of the y discrete random variable P(x i,y j ) = joint probability of the values of x i and y j occurring

52. Chap 4-52 Covariance between two discrete random variables:  xy > 0 x and y tend to move in the same direction  xy < 0 x and y tend to move in opposite directions  xy = 0 x and y do not move closely together Interpreting Covariance

53. Chap 4-53 Correlation Coefficient The Correlation Coefficient shows the strength of the linear association between two variables where: ρ = correlation coefficient (“rho”) σ xy = covariance between x and y σ x = standard deviation of variable x σ y = standard deviation of variable y

54. Chap 4-54 The Correlation Coefficient always falls between -1 and +1  = 0 x and y are not linearly related. The farther  is from zero, the stronger the linear relationship:  = +1 x and y have a perfect positive linear relationship  = -1 x and y have a perfect negative linear relationship Interpreting the Correlation Coefficient

55. Chap 4-55 Useful Discrete Distributions Discrete Uniform Binary – Success/Fail (Bernoulli) Binomial Poisson Empirical Piano Keys Other “stuff that happens” in life

56. Chap 4-56 Useful Continuous Distributions Finite Support Uniform f X (x)=c Beta Infinite Support Gaussian (Normal) N(  ) Log-normal Gamma Exponential

57. Chap 4-57 Section Summary Described approaches to assessing probabilities Developed common rules of probability Distinguished between discrete and continuous probability distributions Examined discrete and continuous probability distributions and their moments (summary measures)

58. Chap 4-58 Probability Distributions Continuous Probability Distributions Binomial Etc. Poisson Probability Distributions Discrete Probability Distributions Normal Uniform Etc.

59. Chap 4-59 Some Discrete Distributions

60. Chap 4-60 P(x) = probability of x successes in n trials, with probability of success p on each trial x = number of ‘successes’ in sample, (x = 0, 1, 2,..., n) p = probability of “success” per trial q = probability of “failure” = (1 – p) n = number of trials (sample size) P(x) n x ! nx pq x n x ! ()!    Example: Flip a coin four times, let x = # heads: n = 4 p = 0.5 q = (1 -.5) =.5 x = 0, 1, 2, 3, 4 Binomial Distribution Formula

61. Chap 4-61 n = 5 p = 0.1 n = 5 p = 0.5 Mean 0.2.4.6 012345 X P(X).2.4.6 012345 X P(X) 0 Binomial Characteristics Examples

62. Chap 4-62 Using Binomial Tables n = 10 xp=.15p=.20p=.25p=.30p=.35p=.40p=.45p=.50 0 1 2 3 4 5 6 7 8 9 10 0.1969 0.3474 0.2759 0.1298 0.0401 0.0085 0.0012 0.0001 0.0000 0.1074 0.2684 0.3020 0.2013 0.0881 0.0264 0.0055 0.0008 0.0001 0.0000 0.0563 0.1877 0.2816 0.2503 0.1460 0.0584 0.0162 0.0031 0.0004 0.0000 0.0282 0.1211 0.2335 0.2668 0.2001 0.1029 0.0368 0.0090 0.0014 0.0001 0.0000 0.0135 0.0725 0.1757 0.2522 0.2377 0.1536 0.0689 0.0212 0.0043 0.0005 0.0000 0.0060 0.0403 0.1209 0.2150 0.2508 0.2007 0.1115 0.0425 0.0106 0.0016 0.0001 0.0025 0.0207 0.0763 0.1665 0.2384 0.2340 0.1596 0.0746 0.0229 0.0042 0.0003 0.0010 0.0098 0.0439 0.1172 0.2051 0.2461 0.2051 0.1172 0.0439 0.0098 0.0010 10 9 8 7 6 5 4 3 2 1 0 p=.85p=.80p=.75p=.70p=.65p=.60p=.55p=.50x Examples: n = 10, p =.35, x = 3: P(x = 3|n =10, p =.35) =.2522 n = 10, p =.75, x = 2: P(x = 2|n =10, p =.75) =.0004

63. Chap 4-63 The Poisson Distribution Characteristics of the Poisson Distribution: The outcomes of interest are rare relative to the possible outcomes The average number of outcomes of interest per time or space interval is The number of outcomes of interest are random, and the occurrence of one outcome does not influence the chances of another outcome of interest The probability of that an outcome of interest occurs in a given segment is the same for all segments

64. Chap 4-64 Poisson Distribution Formula where: t = size of the segment of interest x = number of successes in segment of interest = expected number of successes in a segment of unit size e = base of the natural logarithm system (2.71828...)

65. Chap 4-65 Poisson Distribution Shape The shape of the Poisson Distribution depends on the parameters and t: t = 0.50 t = 3.0

66. Chap 4-66 Poisson Distribution Characteristics Mean Variance and Standard Deviation http://www.math.csusb.edu/faculty/stanton/m262/poisson_distribution /Poisson_old.html http://www.math.csusb.edu/faculty/stanton/m262/poisson_distribution /Poisson_old.html where = number of successes in a segment of unit size t = the size of the segment of interest

67. Chap 4-67 Graph of Poisson Probabilities X t = 0.50 0123456701234567 0.6065 0.3033 0.0758 0.0126 0.0016 0.0002 0.0000 P(x = 2) =.0758 Graphically: =.05 and t = 100

68. Chap 4-68 Some Continuous Distributions

69. Chap 4-69 The Continuous Uniform Distribution: where f(x) = value of the density function at any x value a = lower limit of the interval b = upper limit of the interval The Uniform Distribution (continued) f(x) =

70. Chap 4-70 Uniform Distribution Example: Uniform Probability Distribution Over the range 2 ≤ x ≤ 6: 26.25 f(x) = =.25 for 2 ≤ x ≤ 6 6 - 2 1 x f(x)

71. Chap 4-71 Normal (Gaussian) Distribtion

72. Chap 4-72 By varying the parameters μ and σ, we obtain different normal distributions μ ± 1σ  encloses about 68% of x’s  μ ± 2σ covers about 95% of x’s; μ ± 3σ covers about 99.7% of x’s The chance that a value that far or farther away from the mean is highly unlikely, given that particular mean and standard deviation

73. Chap 4-73 f(x) x μ Probability as Area Under the Curve 0.5 The total area under the curve is 1.0, and the curve is symmetric, so half is above the mean, half is below

74. Chap 4-74 The Standard Normal Distribution Also known as the “z” distribution = (x-  )/  Mean is by definition 0 Standard Deviation is by definition 1 z f(z) 0 1 Values above the mean have positive z-values, values below the mean have negative z-values

75. Chap 4-75 Comparing x and z units z 100 3.00 250x Note that the distribution is the same, only the scale has changed. We can express the problem in original units (x) or in standardized units (z) μ = 100 σ = 50

76. Chap 4-76 Finding Normal Probabilities Suppose x is normal with mean 8.0 and standard deviation 5.0. Now Find P(x < 8.6) (continued) Z 0.12.0478 0.00.5000 P(x < 8.6) = P(z < 0.12) = P(z < 0) + P(0 < z < 0.12) =.5 +.0478 =.5478

77. Chap 4-77 Using Standard Normal Tables

78. Chap 4-78 Section Summary Reviewed discrete distributions binomial, poisson, etc. Reviewed some continuous distributions normal, uniform, exponential Found probabilities using formulas and tables Recognized when to apply different distributions Applied distributions to decision problems