Download Session 7

Welcome to
PMBA0608:
Economics/Statistics
Foundation
Fall 2006
Session 7: September 30
How was the exam?
Any questions of exam
questions?
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
Assignment 3
(Due on or before October 14)
1.
2.
3.
4.
5.
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
Stat Book: Section 3.4
Let’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
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?
1.
2.
3.
4.
173%
58%
42%
None of the above.
Where did 58% come from?
P
P
P
P
(TV) = 0.88
(TV & VCR) = 0.51
(VCR\TV) = 0.51/0.88
(VCR\TV) =0.58 or 58%
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)
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
Unions versus Intersections
 P (AB) = intersection of A and B =
Probability of A and B happening
simultaneously
 P (AB) = P (A) P (B\A)
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%




Unions versus Intersections
 P (AUB) = union = Probability of
A or B or both
 P (AUB) = P (A) + P (B) – P (AB)
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%





What is P(AUB) now?
 A and B are mutually exclusive
 P (AUB) = P (A) + P (B)
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?
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)
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
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?
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
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.
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%
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
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
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
Question: What is your expected
return
R
$0
$100
P (R)
0.25
0.5
E(R) = μ = ΣR * P (R)
$150
0.25
E(R) =$87.5
E(R) = (0 * 0.25) + (100 * 0.5) +
(150* 0.25)
Note: don’t call this average
Average is for certain outcomes
Expected value is for uncertain
outcomes
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
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