Welcome to PMBA0608: Economics/Statistics Foundation

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Welcome toPMBA0608 Economics/Statistics
Foundation

• Fall 2006
• Session 7 September 30

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How was the exam?

• Any questions of exam questions?

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Next Class

• October 18
• I will be in Eastern campus
• Local students can meet with me at 700 pm (half
  hour before class)
• Study Chapters 3 and 4 of Stat book
• Study Chapter 5 of Econ book
• Send me your questions

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Assignment 3 (Due on or before October 14)

1. Application 3.17, Page 110 of Stat
2. Application 3.19, Page 110 of Stat
3. Application 3.27, Page 115 of Stat
4. Exercise 3.31, Page 123 of Stat
5. Application 3.33, Page 123 of Stat

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Stat Book Section 3.4Lets 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
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Lets 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.

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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

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Lets 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)
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Lets 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

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Unions versus Intersections

• P (AB) intersection of A and B Probability of
  A and B happening simultaneously
• P (AB) P (A) P (B\A)

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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

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Unions versus Intersections

• P (AUB) union Probability of A or B or both
• P (AUB) P (A) P (B) P (AB)

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And. Lets 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

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What is P(AUB) now?

• A and B are mutually exclusive
• P (AUB) P (A) P (B)

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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?

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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)

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Lets draw a map.
Positive, P 0.8
Cancer
P 0.01
Negative, P 0.2
Positive , P0.096
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 0.0950.103

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Now lets 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?

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Are you confused?Lets 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

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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.

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What is P (cancer\positive)?

• P (cancer\positive) Number of women with
  cancers/ total number of women with positive
  tests
• P (cancer\positive) 80/ (80950) 0.078 or
  7.8

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Bayes 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

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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
• Lets call that r
• Is r discrete or continuous?
• It is continuous

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Lets think of a discrete random variable

• Lets 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
• 1P (R) 0
• SP (R) 1

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Question What is your expected return
R P (R)
0 0.25
100 0.5
150 0.25
E(R) µ SR P (R) E(R) (0 0.25) (100
0.5) (150 0.25) E(R) 87.5 Note dont call
this average Average is for certain
outcomes Expected value is for uncertain outcomes
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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

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Variance and standard deviation of distribution
of returns

• Variance
• s2 S (R µ)2 P (R)
• What is it in our problem?
• Standard deviation square root of variance