GPCO 453: Quantitative Methods I Sec 05: Probability, continued Shane Xinyang Xuan 1 ShaneXuan.com October 29, 2017 1 Department of Political Science, UC San Diego, 9500 Gilman Drive #0521. 1 / 15 ShaneXuan.com
Contact Information Shane Xinyang Xuan xxuan@ucsd.edu The teaching staff is a team! Professor Garg Tu 1300-1500 (RBC 1303) Shane Xuan M 1100-1200 (SSB 332) M 1530-1630 (SSB 332) Joanna Valle-luna Tu 1700-1800 (RBC 3131) Th 1300-1400 (RBC 3131) Daniel Rust F 1100-1230 (RBC 3213) 2 / 15 ShaneXuan.com
Roadmap In this section, we cover the basics for probability: ◮ Basic relationship of probability 3 / 15 ShaneXuan.com
Roadmap In this section, we cover the basics for probability: ◮ Basic relationship of probability ◮ Independence and mutual exclusivity 3 / 15 ShaneXuan.com
Roadmap In this section, we cover the basics for probability: ◮ Basic relationship of probability ◮ Independence and mutual exclusivity ◮ Bayes’ Theorem 3 / 15 ShaneXuan.com
Roadmap In this section, we cover the basics for probability: ◮ Basic relationship of probability ◮ Independence and mutual exclusivity ◮ Bayes’ Theorem ◮ Z -score 3 / 15 ShaneXuan.com
Roadmap In this section, we cover the basics for probability: ◮ Basic relationship of probability ◮ Independence and mutual exclusivity ◮ Bayes’ Theorem ◮ Z -score ◮ Probability distribution 3 / 15 ShaneXuan.com
Basic Relationships of Probability ◮ Union (denoted by ∪ ): The union of A and B is the event containing all sample points belonging to A or B or both. 4 / 15 ShaneXuan.com
Basic Relationships of Probability ◮ Union (denoted by ∪ ): The union of A and B is the event containing all sample points belonging to A or B or both. ◮ Intersection (denoted by ∩ ): The intersection of A and B is the event containing the sample points belonging to both A and B. 4 / 15 ShaneXuan.com
Independence and Mutual Exclusion ◮ Events A and B are mutually exclusive if, when one event occurs, the other cannot occur; Pr( A ∪ B ) = Pr( A ) + Pr( B ) 5 / 15 ShaneXuan.com
Independence and Mutual Exclusion ◮ Events A and B are mutually exclusive if, when one event occurs, the other cannot occur; Pr( A ∪ B ) = Pr( A ) + Pr( B ) ◮ Two events A and B are independent if Pr( A | B ) = Pr( A ) (1) Pr( B | A ) = Pr( B ) (2) 5 / 15 ShaneXuan.com
More on Conditional Probability Pr( A | B ) = Pr( A ∩ B ) conditional probability (3) Pr( B ) Pr( B | A ) = Pr( A ∩ B ) conditional probability (4) Pr( A ) 6 / 15 ShaneXuan.com
More on Conditional Probability Pr( A | B ) = Pr( A ∩ B ) conditional probability (3) Pr( B ) Pr( B | A ) = Pr( A ∩ B ) conditional probability (4) Pr( A ) ◮ Multiplication law Pr( A ∩ B ) = Pr( B )Pr( A | B ) general case (5) Pr( A ∩ B ) = Pr( B )Pr( A ) independent events (6) 6 / 15 ShaneXuan.com
Bayes’ Theorem (Two-Event Case) Pr( A 1 ) Pr( B | A 1 ) Pr( A 1 | B ) = (7) Pr( A 1 ) Pr( B | A 1 ) + Pr( A 2 ) Pr( B | A 2 ) Pr( A 2 ) Pr( B | A 2 ) Pr( A 2 | B ) = (8) Pr( A 1 ) Pr( B | A 1 ) + Pr( A 2 ) Pr( B | A 2 ) 7 / 15 ShaneXuan.com
Bayes’ Theorem (Two-Event Case) Pr( A 1 ) Pr( B | A 1 ) Pr( A 1 | B ) = (7) Pr( A 1 ) Pr( B | A 1 ) + Pr( A 2 ) Pr( B | A 2 ) Pr( A 2 ) Pr( B | A 2 ) Pr( A 2 | B ) = (8) Pr( A 1 ) Pr( B | A 1 ) + Pr( A 2 ) Pr( B | A 2 ) ◮ Important to note that Pr( A 1 ) + Pr( A 2 ) = 1 (9) Pr( A 1 ∩ B ) + Pr( A 2 ∩ B ) = Pr( B ) (10) 7 / 15 ShaneXuan.com
Example: Bayes’ Theorem ◮ 983,764 putts were made and 629,470 putts were missed. 8 / 15 ShaneXuan.com
Example: Bayes’ Theorem ◮ 983,764 putts were made and 629,470 putts were missed. Pr( putt ) = 983764 1613234 ≈ . 61 ; Pr( ¬ putt ) = 1 − . 61 = . 39 8 / 15 ShaneXuan.com
Example: Bayes’ Theorem ◮ 983,764 putts were made and 629,470 putts were missed. Pr( putt ) = 983764 1613234 ≈ . 61 ; Pr( ¬ putt ) = 1 − . 61 = . 39 ◮ Suppose that a PGA Tour player has a par putt. It is known that of putts made, 64 . 0% were for par whereas for putts missed, 20 . 3% were for par. What is the revised probability of making a putt given the PGA Tour player has a par putt? 8 / 15 ShaneXuan.com
Example: Bayes’ Theorem ◮ 983,764 putts were made and 629,470 putts were missed. Pr( putt ) = 983764 1613234 ≈ . 61 ; Pr( ¬ putt ) = 1 − . 61 = . 39 ◮ Suppose that a PGA Tour player has a par putt. It is known that of putts made, 64 . 0% were for par whereas for putts missed, 20 . 3% were for par. What is the revised probability of making a putt given the PGA Tour player has a par putt? Pr( putt ) Pr( par | putt ) Pr( putt | par ) = Pr( putt ) Pr( par | putt ) + Pr( ¬ putt ) Pr( par |¬ putt ) . 61 × . 64 = . 61 × . 64 + . 39 × . 203 ≈ . 831 (11) 8 / 15 ShaneXuan.com
Z -score ◮ Sometimes we want to standardize a random variable Z ≡ X − µ X , (12) σ X � σ 2 where σ X = X . 9 / 15 ShaneXuan.com
Z -score ◮ Sometimes we want to standardize a random variable Z ≡ X − µ X , (12) σ X � σ 2 where σ X = X . ◮ Motivation of standardizing a random variable: Var ( Z ) = 1 Var ( X ) (13) σ 2 X = 1 σ 2 (14) X σ 2 X = 1 (15) 9 / 15 ShaneXuan.com
Z -score ◮ Sometimes we want to standardize a random variable Z ≡ X − µ X , (12) σ X � σ 2 where σ X = X . ◮ Motivation of standardizing a random variable: Var ( Z ) = 1 Var ( X ) (13) σ 2 X = 1 σ 2 (14) X σ 2 X = 1 (15) ◮ Outliers are defined as observations having a Z -score below − 3 or more than 3 . 9 / 15 ShaneXuan.com
Uniform Distribution 10 / 15 ShaneXuan.com
Uniform Distribution ◮ The density function is 1 f ( x ) = b − a, a ≤ x ≤ b (16) 11 / 15 ShaneXuan.com
Uniform Distribution ◮ The density function is 1 f ( x ) = b − a, a ≤ x ≤ b (16) ◮ E [ x ] = a + b 2 11 / 15 ShaneXuan.com
Uniform Distribution ◮ The density function is 1 f ( x ) = b − a, a ≤ x ≤ b (16) ◮ E [ x ] = a + b 2 ◮ var ( x ) = ( b − a ) 2 12 11 / 15 ShaneXuan.com
Uniform Distribution ◮ The density function is 1 f ( x ) = b − a, a ≤ x ≤ b (16) ◮ E [ x ] = a + b 2 ◮ var ( x ) = ( b − a ) 2 12 ◮ The expected time for a plane to land is 30 minutes. What is the probability of landing between 25 to 30 minutes? 11 / 15 ShaneXuan.com
Uniform Distribution ◮ The density function is 1 f ( x ) = b − a, a ≤ x ≤ b (16) ◮ E [ x ] = a + b 2 ◮ var ( x ) = ( b − a ) 2 12 ◮ The expected time for a plane to land is 30 minutes. What is the probability of landing between 25 to 30 minutes? 30 − 0 = 1 1 ◮ Density of probability is 30 11 / 15 ShaneXuan.com
Uniform Distribution ◮ The density function is 1 f ( x ) = b − a, a ≤ x ≤ b (16) ◮ E [ x ] = a + b 2 ◮ var ( x ) = ( b − a ) 2 12 ◮ The expected time for a plane to land is 30 minutes. What is the probability of landing between 25 to 30 minutes? 30 − 0 = 1 1 ◮ Density of probability is 30 ◮ Interval of success is [25 , 30] 11 / 15 ShaneXuan.com
Uniform Distribution ◮ The density function is 1 f ( x ) = b − a, a ≤ x ≤ b (16) ◮ E [ x ] = a + b 2 ◮ var ( x ) = ( b − a ) 2 12 ◮ The expected time for a plane to land is 30 minutes. What is the probability of landing between 25 to 30 minutes? 30 − 0 = 1 1 ◮ Density of probability is 30 ◮ Interval of success is [25 , 30] � 1 � = 1 ◮ Area under the curve is (5) 6 30 ���� length � �� � height 11 / 15 ShaneXuan.com
Binomial Distribution ◮ The probability of having x successes in n trials is � n � p x (1 − p ) n − x f ( x ) = (17) x 12 / 15 ShaneXuan.com
Binomial Distribution ◮ The probability of having x successes in n trials is � n � p x (1 − p ) n − x f ( x ) = (17) x – x : #(success) 12 / 15 ShaneXuan.com
Binomial Distribution ◮ The probability of having x successes in n trials is � n � p x (1 − p ) n − x f ( x ) = (17) x – x : #(success) – p : probability of success 12 / 15 ShaneXuan.com
Binomial Distribution ◮ The probability of having x successes in n trials is � n � p x (1 − p ) n − x f ( x ) = (17) x – x : #(success) – p : probability of success – n : #(trials) 12 / 15 ShaneXuan.com
Binomial Distribution ◮ The probability of having x successes in n trials is � n � p x (1 − p ) n − x f ( x ) = (17) x – x : #(success) – p : probability of success – n : #(trials) ◮ E [ x ] = np 12 / 15 ShaneXuan.com
Binomial Distribution ◮ The probability of having x successes in n trials is � n � p x (1 − p ) n − x f ( x ) = (17) x – x : #(success) – p : probability of success – n : #(trials) ◮ E [ x ] = np ◮ var ( x ) = np (1 − p ) 12 / 15 ShaneXuan.com
Binomial Distribution: Example ◮ A university found that 20% of its students withdraw without completing the introductory statistics course. Assume that 20 students registered for the course. Compute the probability that more than 2 will withdraw. Pr( X > 2) = 1 − Pr( X = 0) − Pr( X = 1) − Pr( X = 2) � � � � � � 20 20 20 . 2 0 (1 − . 2) 20 − 0 − . 2 1 ( . 8) 20 − 1 − . 2 2 ( . 8) 20 − 2 = 1 − 0 1 2 13 / 15 ShaneXuan.com
Recommend
More recommend
