Welcome to PMBA0608: Economics/Statistics Foundation

1. Welcome toPMBA0608: Economics/Statistics Foundation Fall 2006 Session 7: September 30

2. How was the exam? • Any questions of exam questions?

3. Next Class • October 18 • I will be in Eastern campus • Local students can meet with me at 7:00 pm (half hour before class) • Study Chapters 3 and 4 of Stat book • Study Chapter 5 of Econ book • Send me your questions

4. Assignment 3(Due on or before October 14) • Application 3.17, Page 110 of Stat • Application 3.19, Page 110 of Stat • Application 3.27, Page 115 of Stat • Exercise 3.31, Page 123 of Stat • Application 3.33, Page 123 of Stat

5. Stat Book: Section 3.4Let’s spin once P (1 given that red has occurred) = ? P (1 given that red has occurred) = ½= 0.5 P(1\red)= P (1 and red) / P (red) P (1\red)= 0.25 / 0.5 = 0.5 This is called conditional probability

6. Let’s practice • In Europe, 88% of all households have a television. 51% of all households have a television and a VCR. What is the probability that a household has a VCR given that it has a television? • 173% • 58% • 42% • None of the above.

7. Where did 58% come from? • P (TV) = 0.88 • P (TV & VCR) = 0.51 • P (VCR\TV) = 0.51/0.88 • P (VCR\TV) =0.58 or 58%

8. Let’s spin twice P (1 on second spin given that red has occurred on the first spin) = ? P (1 on second spin given that red has occurred on the first spin)= ¼ = 0.25 1 on the second spin is independent from red on the first spin so P (1 on second spin\red has occurred on the first spin) = P (1)

9. Let’s play cards now • A card is chosen at random from a deck of 52 cards. It is then replaced and a second card is chosen. What is the probability of choosing a jack the second time given that we chose an eight the first time? • P(jack\8) =  ? • P (jack\8) = P (jack)= 4/52

10. Unions versus Intersections • P (AB) = intersection of A and B = Probability of A and B happening simultaneously • P (AB) = P (A) P (B\A)

11. OK ready to practice? • 100 individuals • 50 male and 20 of them smoke • P (male & smoker) = ? • P (male & smoker) = P (male)* P (smoker/male) • P (male & smoker) = ½ * 2/5 = 2/10 =0.2 or 20%

12. Unions versus Intersections • P (AUB) = union =Probability of A or B or both • P (AUB) = P (A) + P (B) – P (AB)

13. And…. Let’s practice • 100 individuals • 50 male and 20 of them smoke • 50 female and 10 of them smoke • P (male or smoker) =? • P (male or smoker) = P (male) + P (smoker) – P (male and smoker) • P (male or smoker) = 0.5 + 0.3 - 0.2 = 0.6 or 60%

14. What is P(AUB) now? • A and B are mutually exclusive • P (AUB) = P (A) + P (B)

15. Please work on this problem with a partner • 1% of women at age forty who participate in routine screening have breast cancer.  80% of women with breast cancer will get positive mammographies.  9.6% of women without breast cancer will also get positive mammographies.  A woman in this age group had a positive mammography in a routine screening.  What is the probability that she actually has breast cancer?

16. Here is the problem • P (cancer) = 0.01 • P (positive \cancer) = 0.8 • P (positive \no cancer) =0.096 • P (cancer\positive) = ? • If we use conditional probability • P (cancer\positive) = P (cancer and positive)/P (positive)

17. Let’s draw a map. Positive, P = 0.8 Cancer P = 0.01 Negative, P =0.2 Positive , P=0.096 No Cancer P =0.99 Negative, P = 0.904 • P (cancer and positive) = 0.01 * 0.8 = 0.008 • P (positive) = P (cancer and positive) + P (no cancer and positive) = 0.008 + 0.095=0.103

18. Now let’s plug this into the conditional probability formula • P (cancer\positive) = P (cancer and positive)/P (positive) • P (cancer\positive) = 0.008/0.103 • P (cancer\positive) =0.078 • This is Bayes’ rule • If your mamo result is positive, there is only 7.8% chance that you have breast cancer • Not bad ha?

19. Are you confused?Let’s put it differently • 10,000 women • Group 1:  100 women with breast cancer. (1%) • 80% (80) positive • 20% (20) negative • Group 2:  9,900 women without breast cancer. • 9.6% (almost 950)positive mamo • The rest (8950) negative mamo

20. So, we have 4 groups of women • Group A:  80 women with breast cancer, and a positive mammography. • Group B:  20 women with breast cancer, and a negative mammography. • Group C:  950 women without  breast cancer, and a positive mammography. • Group D:  8,950 women without breast cancer, and a negative mammography.

21. What is P (cancer\positive)? • P (cancer\positive)= Number of women with cancers/ total number of women with positive tests • P (cancer\positive)= 80/ (80+950) = 0.078 or 7.8%

22. Baye’s Rule P (cancer\ positive) = p (positive\cancer)*p (cancer)divided by p (positive\cancer)*p (cancer) + p (positive\no cancer)*p (no cancer) P (cancer\positive) = (0.8 * 0.01)/ (0.8 * 0.01) + (0.096* 0.99) P (cancer\positive)= 0.008/(0.008 +0.095) = 0.078

23. What is a random variable? • Value of it depends on the outcome of an experiment • Example • The rate of return on the portfolio of your stocks is a random variable • Let’s call that r • Is r discrete or continuous? • It is continuous

24. Let’s think of a discrete random variable • Let’s suppose that there are only 3 possible outcomes for the return on your stock portfolio: $0, $100, and $150 • Now your return R is a discrete variables • Now suppose that there is 50% chance that you make $100 and 25% chance that you make $150. • What are the chances that you make $0? • 25% • R is a discrete random variable • 1≥P (R) ≥0 • ΣP (R) = 1

25. Question: What is your expected return E(R) = μ = ΣR * P (R) E(R) = (0 * 0.25) + (100 * 0.5) + (150* 0.25) E(R) =$87.5 Note: don’t call this average Average is for certain outcomes Expected value is for uncertain outcomes

26. Will you always make $87.5? • No • If you repeat this investment an infinite number of times on average you will make $87.5 • But each time you may make less or more • So there is a distribution of returns

27. Variance and standard deviation of distribution of returns • Variance • σ2= Σ (R – μ)2 P (R) • What is it in our problem? • Standard deviation = square root of variance