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Chap 4-1 EF 507 QUANTITATIVE METHODS FOR ECONOMICS AND FINANCE FALL 2008 Chapter 4 Probability

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Chap 4-2 Important Terms Random Experiment – a process leading to an uncertain outcome Basic Outcome – a possible outcome of a random experiment Sample Space – the collection of all possible outcomes of a random experiment Event – any subset of basic outcomes from the sample space

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Chap 4-3 Important Terms Intersection of Events – If A and B are two events in a sample space S, then the intersection, A ∩ B, is the set of all outcomes in S that belong to both A and B (continued) AB ABAB S

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Chap 4-4 Important Terms A and B are Mutually Exclusive Events if they have no basic outcomes in common i.e., the set A ∩ B is empty (continued) A B S

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Chap 4-5 Important Terms Union of Events – If A and B are two events in a sample space S, then the union, A U B, is the set of all outcomes in S that belong to either A or B (continued) AB The entire shaded area represents A U B S

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Chap 4-6 Important Terms Events E 1, E 2, … E k are Collectively Exhaustive events if E 1 U E 2 U... U E k = S i.e., the events completely cover the sample space The Complement of an event A is the set of all basic outcomes in the sample space that do not belong to A. The complement is denoted (continued) A S

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Chap 4-7 Examples Let the Sample Space be the collection of all possible outcomes of rolling one die: S = [1, 2, 3, 4, 5, 6] Let A be the event “Number rolled is even” Let B be the event “Number rolled is at least 4” Then A = [2, 4, 6] and B = [4, 5, 6]

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Chap 4-8 (continued) Examples S = [1, 2, 3, 4, 5, 6] A = [2, 4, 6] B = [4, 5, 6] Complements: Intersections: Unions:

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Chap 4-9 Mutually exclusive: A and B are not mutually exclusive The outcomes 4 and 6 are common to both Collectively exhaustive: A and B are not collectively exhaustive A U B does not contain 1 or 3 (continued) Examples S = [1, 2, 3, 4, 5, 6] A = [2, 4, 6] B = [4, 5, 6]

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Chap 4-10 Probability Probability – the chance that an uncertain event will occur (always between 0 and 1) 0 ≤ P(A) ≤ 1 For any event A Certain Impossible 0.5 1 0

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Chap 4-11 Assessing Probability There are three approaches to assessing the probability of an uncertain event: 1. classical probability (sample) Assumes all outcomes in the sample space are equally likely to occur

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Chap 4-12 Counting the Possible Outcomes Use the Combinations formula to determine the number of combinations of n things taken k at a time where n! = n(n-1)(n-2)…(1) 0! = 1 by definition

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Chap 4-13 Assessing Probability Three approaches (continued) 2. relative frequency probability (population) the limit of the proportion of times that an event A occurs in a large number of trials, n 3. subjective probability an individual opinion or belief about the probability of occurrence

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Chap 4-14 Probability Postulates 1. If A is any event in the sample space S, then 2. Let A be an event in S, and let O i denote the basic outcomes. Then (the notation means that the summation is over all the basic outcomes in A) 3.P(S) = 1

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Chap 4-15 Probability Rules The Complement rule: The Addition rule: The probability of the union of two events is

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Chap 4-16 A Probability Table B A Probabilities and joint probabilities for two events A and B are summarized in this table:

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Chap 4-17 Addition Rule Example Consider a standard deck of 52 cards, with four suits: ♥ ♣ ♦ ♠ Let event A = card is an Ace Let event B = card is from a red suit

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Chap 4-18 P(Red U Ace) = P(Red) + P(Ace) - P(Red ∩ 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 (continued) Addition Rule Example

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Chap 4-19 A conditional probability is the probability of one event, given that another event has occurred: The conditional probability of A given that B has occurred The conditional probability of B given that A has occurred Conditional Probability

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Chap 4-20 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.

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Chap 4-21 Conditional Probability Example No CDCDTotal AC 0.20.50.7 No AC 0.20.1 0.3 Total 0.40.61.0 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)

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Chap 4-22 Conditional Probability Example No CDCDTotal AC 0.2 0.50.7 No AC 0.20.1 0.3 Total 0.40.6 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 28.57%. (continued)

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Chap 4-23 Multiplication Rule Multiplication rule for two events A and B: also

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Chap 4-24 Multiplication Rule Example P(Red ∩ Ace) = P(Red| Ace)P(Ace) Black Color Type Red Total Ace 224 Non-Ace 24 48 Total 26 52

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Chap 4-25 Statistical Independence Two events are statistically independent if and only if (works in both directions) : Events A and B are independent when the probability of one event is not affected by the other event If A and B are independent, then if P(B)>0 if P(A)>0

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Chap 4-26 Statistical Independence Example No CDCDTotal AC 0.20.50.7 No AC 0.20.1 0.3 Total 0.40.61.0 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. Are the events AC and CD statistically independent?

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Chap 4-27 Statistical Independence Example No CDCDTotal AC 0.20.50.7 No AC 0.20.1 0.3 Total 0.40.61.0 (continued) P(AC ∩ CD) = 0.2 P(AC) = 0.7 P(CD) = 0.4 P(AC)P(CD) = (0.7)(0.4) = 0.28 P(AC ∩ CD) = 0.2≠ P(AC)P(CD) = 0.28 So the two events are not statistically independent

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Chap 4-28 Bivariate Probabilities B1B1 B2B2...BkBk A1A1 P(A 1 B 1 )P(A 1 B 2 )... P(A 1 B k ) A2A2 P(A 2 B 1 )P(A 2 B 2 )... P(A 2 B k ).............................. AhAh P(A h B 1 )P(A h B 2 )... P(A h B k ) Outcomes for bivariate events:

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Chap 4-29 Joint and Marginal Probabilities The probability of a joint event, A ∩ B: Computing a marginal probability: Where B 1, B 2, …, B k are k mutually exclusive and collectively exhaustive events

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Chap 4-30 Marginal Probability Example P(Ace) Black Color Type Red Total Ace 224 Non-Ace 24 48 Total 26 52

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Chap 4-31 Using a Tree Diagram Has AC Does not have AC Has CD Does not have CD Has CD Does not have CD P(AC)= 0.7 P(AC)= 0.3 P(AC ∩ CD) = 0.2 P(AC ∩ CD) = 0.5 P(AC ∩ CD) =0.1 P(AC ∩ CD) = 0.2 All Cars Given AC or no AC:

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Chap 4-32 Odds The odds in favor of a particular event are given by the ratio of the probability of the event divided by the probability of its complement The odds in favor of A are

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Chap 4-33 Odds: Example Calculate the probability of winning (=A) if the odds of winning are 3 to 1: Now multiply both sides by 1 – P(A) and solve for P(A): 3 x (1- P(A)) = P(A) 3 – 3P(A) = P(A) 3 = 4P(A) P(A) = 0.75

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Chap 4-34 Overinvolvement Ratio The probability of event A 1 conditional on event B 1 divided by the probability of A 1 conditional on activity B 2 is defined as the overinvolvement ratio: An overinvolvement ratio greater than 1 implies that event A 1 increases the conditional odds ration in favor of B 1 :

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Chap 4-35 Bayes’ Theorem Refers to situations where the outcome of an experiment depends on what happens in various intermediate stages. The probability that event A was reached via the r-th branch of the tree diagram, given that it was reached via one of its k branches, is the ratio of the probability associated with the r-th branch to the sum of the probabilities associated with all k branches of the tree. P(B 1 ) P(B 2 ) P(B K ) P(A/B 1 ) P(A/B 2 ) P(A/B K ) P(B 1 ) P(A/B 1 ) P(B 2 ) P(A/B 2 ) P(B K ) P(A/B K ) B1B1 B2B2 BKBK A A A

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Chap 4-36 Bayes’ Theorem where: E i = i th event of k mutually exclusive and collectively exhaustive events A = new event that might impact P(E i )

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Chap 4-37 Bayes’ Theorem Example A drilling company has estimated a 40% chance of striking oil for their new well. A detailed test has been scheduled for more information. Historically, 60% of successful wells have had detailed tests, and 20% of unsuccessful wells have had detailed tests. Given that this well has been scheduled for a detailed test, what is the probability that the well will be successful?

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Chap 4-38 Let S = successful well U = unsuccessful well P(S) = 0.4, P(U) =0.6 (prior probabilities) Define the detailed test event as D Conditional probabilities: P(D|S) = 0.6 P(D|U) = 0.2 Goal is to find P(S|D) Bayes’ Theorem Example (continued)

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Chap 4-39 Bayes’ Theorem Example (continued) Apply Bayes’ Theorem: So the revised probability of success (from the original estimate of 0.4), given that this well has been scheduled for a detailed test, is 0.667

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Chap 4-40 Bayes’ Theorem P(S) successful well P(U) unsuccessful well P(D/S) P(D/U) P(D/U) P(U) S U P(S/D) P(D/S) P(S)

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