Transcript No Slide Title

Lesson 4:
Discrete Probability
Distributions
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-1
Outline
Random variables and probability distribution
Features of univariate probability distribution
Binomial Probability Distribution
Hypergeometric Probability Distribution
Poisson Probability Distribution
Features of bivariate probability distribution
Conditional distribution
Conditional expectation
Covariance and Correlation Coefficient
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-2
Random Variables and probability
distribution
 A random variable is a numerical value determined by the
outcome of an experiment. A random variable is often
denoted by a capital letter, e.g., X or Y.
 A probability distribution is the listing of all possible
outcomes of an experiment and the corresponding
probability.
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Lesson4-3
Types of Probability Distributions
 A discrete probability distribution can assume only
certain outcomes (need not be finite) – for random
variables that take discrete values.
 The number of students in a class.
 The number of children in a family.
 A continuous probability distribution can assume an
infinite number of values within a given range – for
random variables that take continuous values.
 The time it takes an executive to drive to work.
 The amount of money spent on your last haircut.
 The distance students travel to class.
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Lesson4-4
Types of Probability Distributions
Probability distribution may be classified according to the
number of random variables it describes.
Number of random variables
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Joint distribution
1
Univariate probability distribution
2
Bivariate probability distribution
3
Trivariate probability distribution
…
…
n
Multivariate probability distribution
ECON1003: Analysis of Economic Data
Lesson4-5
Discrete Random Variables
 Can only take on a countable number of values
Examples:
 Roll a die once
Let X be the number of times “4” comes up
(then X could be 0, or 1 time)
 Roll a die twice
Let X be the number of times “4” comes up
(then X could be 0, 1, or 2 times)
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Discrete Random Variables
 Can only take on a countable number of values
Examples:
 Toss a coin once.
Let X be the number of heads
(then X = 0 or 1)
 Toss a coin 5 times.
Let X be the number of heads
(then X = 0, 1, 2, 3, 4, or 5)
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Lesson4-7
Example 1: Tossing coin(s)
Discrete Probability Distribution
Experiment: Toss 1 Coin.
Let X = # heads.
Show P(x) , i.e., P(X = x) , for all values of x:
2 possible outcomes
T
H
Probability Distribution
X-value
Probability
0
1/2 =0.5
1
1/2 =0.5
0.5
0.4
0.3
0.2
0.1
0
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0
ECON1003: Analysis of Economic Data
1
Lesson4-8
Example 1: Tossing coin(s)
Discrete Probability Distribution
Experiment: Toss 2 Coins.
Let X = # heads.
Show P(x) , i.e., P(X = x) , for all values of x:
4 possible outcomes
T
T
T
Probability Distribution
X-value
Probability
0
1/4 =0.25
1
1/2 =0.5
2
1/4 =0.25
H
0.5
H
T
0.4
0.3
H
0.2
H
0.1
0
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0 Data 1
ECON1003: Analysis of Economic
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Lesson4-9
Example 1: Tossing coin(s)
Discrete Probability Distribution
 Consider a random experiment in which a coin is tossed three times.
Let x be the number of heads. Let H represent the outcome of a
head and T the outcome of a tail.
 The possible outcomes for such an experiment will be:
TTT, TTH, THT, THH,
HTT, HTH, HHT, HHH.
 Thus the possible values of x (number of heads) are
x=0:
x=1:
x=2:
x=3:
Ka-fu Wong © 2007
TTT
TTH, THT, HTT
THH, HTH, HHT
HHH
P(x=0) =1/8
P(x=1) =3/8
P(x=2) =3/8
P(x=3) =1/8
ECON1003: Analysis of Economic Data
Lesson4-10
Features of a Univariate Discrete
Distribution
 Let x1,…,xN be the list of all possible outcomes (N of them).
 The main features of a discrete probability distribution are:
 The probability of a particular outcome, P(xi), is
between 0 and 1.00.
 The sum of the probabilities of the various outcomes is
1.00. That is,
P(x1) + … + P(xN) = 1
 The outcomes are mutually exclusive. That is,
P(x1 and x2) = 0 and
Outcome
Prob.
P(x1 or x2) = P(x1)+ P(x2)
x1
P(x1)
Generally, for all i not equal to k.
x2
P(x2)
P(xi and xk) = 0.
…
…
P(xi or xk) = P(xi)+ P(xk)
xN
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P(xN)
Lesson4-11
Features of a Univariate Discrete
Distribution
Can the following be a probability distribution of a
random variable?
x
Prob.
x
Prob.
1
0.2
2
0.3
1
0.6
3
0.1
2
0.3
1
0.4
3
0.1
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event
Prob.
1 or 2
0.6
2 or 3
0.3
3 or 1
0.1
ECON1003: Analysis of Economic Data
Lesson4-12
Cumulative Probability Function
 The cumulative probability function, denoted F(c), shows
the probability that X is less than or equal to c
F(c) = P(X≤c)
 In other words,
F(c) = ∑x≤c P(X=x)
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Moments of a random variable
The n-th moment is defined as the expectation of
the n-th power of a random variable: E(Xn)
E(X)
First moment
E(X2)
Second moment
The n-th centralized moment is defined as:
E[X-E(X)]n
E(X-m)2
Second centralized moment
E(X-m)3
Third centralized moment
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ECON1003: Analysis of Economic Data
Lesson4-14
The Expectation (Mean) of Discrete
Probability Distribution
 The expectation or mean of a discrete random variable X, is
computed by the formula:
E(x)  Σ[xP(x)]
 x1P(x1 )  x 2P(x 2 )  ...  x nP(x n )
 Example:
Toss 2 coins, x = # of heads, what is the expected value of
x?
 P(0) = 0.25, P(1)=0.5, P(2)=0.25.
 E(X) = P(0)*0 + P(1)*1 + P(2)*2 = 1
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Transformation of Random variables
 A transformation of random variable(s) results in a new
random variable.
 For example, if X is a random variable, the following are
also random variables:
 Z=2X
 Z=3+2X
 Z=X2
 Z=log(X)
Example:
X = amount of insurance policy underwritten by an agent for the month
Z = income earned for the month
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ECON1003: Analysis of Economic Data
Lesson4-16
Transformation of Random variables
 Example: Z=3+2X
 X = amount of insurance policy underwritten by an agent for
the month
 Z = income earned for the month
x
P(x)
z
P(z)
1
P(1)
5
P(5)
2
P(2)
7
P(7)
3
P(3)
9
P(9)
P(x) = P(3+2x)
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ECON1003: Analysis of Economic Data
Lesson4-17
Expectation of a linear transformed
random variable
 If a and b are constants and X is a random variable, then
E(a) = a
E(bX) = bE(X)
E(a+bX) = a+bE(X)
E(a  bx)  Σ[(a  bx)P(a  bx)]
 Σ[(a  bx)P(x)]
 (a  bx1 )P(x1 )  (a  bx 2 )P(x 2 )  ...  (a  bx n )P(x n )
 aP(x1 )  bx1P(x1 )  aP(x 2 )  bx 2P(x 2 )  ...  aP(x n )  bx nP(x n )
 a[P(x1 )  P(x 2 )  ...  P(x n )]  b[x1P(x1 )  x 2P(x 2 )  ...  x nP(x n ) ]
 a  bE(x)
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ECON1003: Analysis of Economic Data
Lesson4-18
Expectation of a general functions of random
variables
 If P(x) is the probability function of a discrete random variable X,
and g(X) is some function of X , then the expected value of
function g is
 E[g(x)] = g( E(x) ) = g( ∑x xP(x) )
✓
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E[g(x)] = E[g(x)] = ∑x g(x)P(x)
ECON1003: Analysis of Economic Data
Lesson4-19
Expectation of a general functions of random
variables
Experiment: Toss 2 Coins.
Let X = # heads.
X-value
Probability
0
1/4 =0.25
1
1/2 =0.5
2
1/4 =0.25
g(x) = x2

E[g(x)] = g( E(x) ) = g( ∑x xP(x) )
= (0*0.25 + 1*0.5 + 2*0.25)2=12=1
✓
E[g(x)] = E[g(x)] = ∑x g(x)P(x)
= 02*0.25 + 12*0.5 + 22*0.25=1.5
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ECON1003: Analysis of Economic Data
Lesson4-20
Variance of a discrete random variable
 For univariate discrete probability distribution
Var ( X )  E[(X  μ)2 ]
 Σ[(x  μ)2 P(x)]
 (x 1  μ)2 P(x1 )  (x 2  μ)2 P(x 2 )  ...  (x n  μ)2 P(x n )
 Example:
Toss 2 coins, x = # of heads, what is the variance of x?
 P(0) = 0.25, P(1)=0.5, P(2)=0.25.
 E(X) = P(0)*0 + P(1)*1 + P(2)*2 = 1
 V(X) = P(0)*(0-1)2 + P(1)*(1-1)2 + P(2)*(2-1)2 = 0.5
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Lesson4-21
Variance of a linear transformed random
variable
 If a and b are constants and X is a random variable, then
Var(a) = 0
Var(bX) = b2Var(X)
Var(a+bX) = b2Var(X)
Var (a  bX)  E[ a  bX  (a  bμ ) ]2
 E[ bX  bμ ]2
 E[ b(X  μ) ]2
 E[ b 2 (X  μ)2 ]
 b 2 E[ (X  μ)2 ]
 b 2 Var(X)
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ECON1003: Analysis of Economic Data
Lesson4-22
The Binomial Distribution
Probability
Distributions
Discrete
Probability
Distributions
Binomial
Hypergeometric
Poisson
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Lesson4-23
Bernoulli Distribution
Consider only two outcomes: “success” or “failure”
Let P denote the probability of success
Let 1 – P be the probability of failure
Define random variable X:
x = 1 if success, x = 0 if failure
 Then the Bernoulli probability function is




P(0)  (1 P) and P(1)  P
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Bernoulli Distribution
Mean and Variance
P(0)  (1 P) and P(1)  P
 The mean is µ = P
μ  E(X)   xP(x)  (0)(1 P)  (1)P  P
X
 The variance is σ2 = P(1 – P)
σ 2  E[(X  μ)2 ]   (x  μ)2 P(x)
X
 (0  P) 2 (1 P)  (1 P) 2 P  P(1  P)
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Lesson4-25
Sequences of x Successes in n Trials
 The number of sequences with x successes in n independent
trials is:
C(n,x) = n!/([x!(n-x)!]
where n! = n*(n – 1)*(n – 2)* . . . *1 and 0! = 1
 These sequences are mutually exclusive, since no two can occur
at the same time
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Binomial Probability Distribution
 The binomial distribution has the following characteristics:
 An outcome of an experiment is classified into one of two
mutually exclusive categories, such as a success or failure.
 The data collected are the results of counts in a series of trials.
 The probability of success stays the same for each trial.
 The trials are independent.
 For example, tossing an unfair coin three times.
 H is labeled success and T is labeled failure.
 The data collected are number of H in the three tosses.
 The probability of H stays the same for each toss.
 The results of the tosses are independent.
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ECON1003: Analysis of Economic Data
Lesson4-27
Binomial Probability Distribution
 To construct a binomial distribution, let
 n be the number of trials
 x be the number of observed successes
 p be the probability of success on each trial
 The formula for the binomial probability distribution is:
P(x) = C(n,x) p x(1- p)n-x
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Binomial Probability Distribution
 The formula for the binomial probability distribution is:
P(x) = C(n,x) p x(1- p)n-x
TTT, TTH, THT, THH,
HTT, HTH, HHT, HHH.
C(n,x) = n!/([x!(n-x)!]
 X=number of heads
 The coin is fair, i.e., P(head) = 1/2.
 P(x=0) = C(3,0) 0.5 0(1- 0.5)3-0 =3!/(0!3!) (1) (1/8)=1/8
 P(x=1) = C(3,1) 0.5 1(1- 0.5)3-1 =3!/(1!2!) (1) (1/8)= 3/8
 P(x=2) = C(3,2) 0.5 2(1- 0.5)3-2 =3!/(2!1!) (1) (1/8)= 3/8
 P(x=3) = C(3,3) 0.5 3(1- 0.5)3-3 =3!/(3!0!) (1) (1/8)= 1/8
When the coin is not fair, simple counting rule will not work.
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ECON1003: Analysis of Economic Data
Lesson4-29
The density functions of binomial distributions
with n=20 and different success rates p
P(x) = C(n,x) px(1- p)n-x
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Lesson4-30
Example: Binomial
P(x) = C(n,x) px(1- p)n-x
x = number of students who will receive scholarships
among those who have a semester GPA>3.5
n = 4 , p = 0.1 , q = 1 – p = 1 - 0.1 = 0.9
Find the probability that 2 of the 4 students among those
who have a semester GPA>3.5
will receive scholarships.
4!
p(2)  P(x=2)=
(0.1) 2 (0.9) 4-2 =6(0.1) 2 (0.9) 2 =0.0486
2!(4-2)!
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ECON1003: Analysis of Economic Data
Lesson4-31
Example: Binomial
P(x) = C(n,x) px(1- p)n-x
After Asian Financial Crisis, Thailand’s Department of Labor
reports that 20% of the workforce is unemployed. From a
sample of 14 workers, calculate the following probabilities:
 Exactly three are unemployed.
 At least three are unemployed.
 At least one are unemployed.
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Lesson4-32
Example: Binomial
P(x) = C(n,x) px(1- p)n-x
 The probability of exactly 3:
P(3) C(14,3)(.2)3 (1 .2)11
 (364)(.008
)(.0859) .2501
 The probability of at least 3 is:
P(x  3)  C(14,3)(.20)3 (.80)11  ...  C(14,14)(.20)14 (.80)0
 .250 .172 ...  .000  .551
 The probability of at least one being unemployed:
P(x  1)  1  P(0)
 1  C(14,0)(.20)0 (1 .20)14
 1  .044  .956
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ECON1003: Analysis of Economic Data
Lesson4-33
Mean & Variance of the Binomial
P(x) = C(n,x) px(1- p)n-x
Distribution
 The mean is found by:
n
m   xP( x)  np
x 0
 The variance is found by:
n
 2   ( x  m ) 2 P( x)  np(1  p)
x 0
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ECON1003: Analysis of Economic Data
Lesson4-34
Example: Binomial
P(x) = C(n,x) px(1- p)n-x
 Since p =.2 and n=14.
 Hence, the mean is:
m= n p = 14(.2) = 2.8.
 The variance is:
2 = n p (1- p ) = (14)(.2)(.8) =2.24.
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ECON1003: Analysis of Economic Data
Lesson4-35
Using Binomial Tables
P(x) = C(n,x) px(1- p)n-x
N
x
…
p=.20
p=.25
p=.30
p=.35
p=.40
p=.45
p=.50
10
0
1
2
3
4
5
6
7
8
9
10
…
…
…
…
…
…
…
…
…
…
…
0.1074
0.2684
0.3020
0.2013
0.0881
0.0264
0.0055
0.0008
0.0001
0.0000
0.0000
0.0563
0.1877
0.2816
0.2503
0.1460
0.0584
0.0162
0.0031
0.0004
0.0000
0.0000
0.0282
0.1211
0.2335
0.2668
0.2001
0.1029
0.0368
0.0090
0.0014
0.0001
0.0000
0.0135
0.0725
0.1757
0.2522
0.2377
0.1536
0.0689
0.0212
0.0043
0.0005
0.0000
0.0060
0.0403
0.1209
0.2150
0.2508
0.2007
0.1115
0.0425
0.0106
0.0016
0.0001
0.0025
0.0207
0.0763
0.1665
0.2384
0.2340
0.1596
0.0746
0.0229
0.0042
0.0003
0.0010
0.0098
0.0439
0.1172
0.2051
0.2461
0.2051
0.1172
0.0439
0.0098
0.0010
Examples:
n = 10, x = 3, P = 0.35:
P(x = 3|n =10, p = 0.35) = .2522
n = 10, x = 8, P = 0.45:
P(x = 8|n =10, p = 0.45) = .0229
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ECON1003: Analysis of Economic Data
Lesson4-36
Finite Population
 A finite population is a population consisting of a fixed
number of known individuals, objects, or measurements.
Examples include:
 The number of students in this class.
 The number of cars in the parking lot.
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ECON1003: Analysis of Economic Data
Lesson4-37
Hypergeometric Distribution
 The hypergeometric distribution has the following
characteristics:
 There are only 2 possible outcomes.
 The probability of a success is not the same on each trial.
 It results from a count of the number of successes in a
fixed number of trials.
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ECON1003: Analysis of Economic Data
Lesson4-38
Example: Hypergeometric Distribution
In a bag containing 7 red chips and 5 blue chips you
select 2 chips one after the other without replacement.
6/11
7/12
5/12
R2
R1
5/11
B2
7/11
R2
B1
4/11
B2
The probability of a success (red chip) is not the same on each trial.
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ECON1003: Analysis of Economic Data
Lesson4-39
Hypergeometric Distribution
 The formula for finding a probability using the
hypergeometric distribution is:
C ( S , x)C ( N  S , n  x)
P( x) 
C ( N , n)
where N is the size of the population, S is the number
of successes in the population, x is the number of
successes in a sample of n observations.
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ECON1003: Analysis of Economic Data
Lesson4-40
Hypergeometric Distribution
 Use the hypergeometric distribution to find the probability
of a specified number of successes or failures if:
 the sample is selected from a finite population without
replacement (recall that a criteria for the binomial
distribution is that the probability of success remains
the same from trial to trial)
 the size of the sample n is greater than 5% of the size
of the population N .
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ECON1003: Analysis of Economic Data
Lesson4-41
The density functions of hypergeometric
distributions with N=100, n=20 and different
success rates p (=S/N).
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ECON1003: Analysis of Economic Data
Lesson4-42
Example: Hypergeometric Distribution
 The University Examination Board has a list of 10 reported
cheating cases. Suppose only 4 of the reported cheating
cases are real cheating and the University Examination
Board will only be able to investigate five of the cheating
cases. What is the probability that three of five reported
cheating cases randomly selected to be investigated are
real cheating cases?
 Number of reported cases, N=10
 Number of real cheating cases out of 10, S=4
 Number of cases investigated, n=5
C (4,3)C (10  4,5  3)
P(3) 
C (10,5)
C (4,3)C (6,2) 4(15)


 .238
C (10,5)
252
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ECON1003: Analysis of Economic Data
Lesson4-43
Example: Hypergeometric Distribution
■ 3 different computers are checked from 10 in the department. 4
of the 10 computers have illegal software loaded. What is the
probability that 2 of the 3 selected computers have illegal
software loaded?
■ N = 10, S=4, n=3, x=2
P(x  2 ) 
C(S,x)C(N-S, n-x) C( 4,2 )C( 6,1 ) ( 6 )( 6 )


 0.3
C(N,n)
C( 10,3 )
120
The probability that 2 of the 3 selected computers have illegal
software loaded is 0.30, or 30%.
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ECON1003: Analysis of Economic Data
Lesson4-44
Poisson Probability Distribution
The formula for the binomial probability distribution is:
P(x) = C(n,x) p x(1- p)n-x
 The binomial distribution becomes more skewed to the
right (positive) as the probability of success become
smaller.
 The limiting form of the binomial distribution where the
probability of success p is small and n is large is called
the Poisson probability distribution.
Let l=np, i.e., p=l/n.
limn  C(n,x) (l/n)x(1- l/n)n-x = [lxe-l]/x!
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ECON1003: Analysis of Economic Data
Lesson4-45
The Poisson Distribution
 Apply the Poisson Distribution when:
 You wish to count the number of times an event occurs
in a given continuous interval
 The probability that an event occurs in one subinterval
is very small and is the same for all subintervals
 The number of events that occur in one subinterval is
independent of the number of events that occur in the
other subintervals
 There can be no more than one occurrence in each
subinterval
 The average number of events per unit is l (lambda)
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ECON1003: Analysis of Economic Data
Lesson4-46
Poisson Probability Distribution
The Poisson distribution can be described
mathematically using the formula:
P( x) 
lx e  l
x!
where
 l is the mean number of successes in a particular
interval of time,
 e is the constant 2.71828, and
 x is the number of successes.
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-47
Poisson Distribution Characteristics
 Mean
μ  E(x)  λ
 Variance and Standard Deviation
σ2  E[( X  m )2 ]  λ
σ λ
where l = expected number of successes per unit
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-48
Poisson Probability Distribution
 The mean number of successes l can be determined in
binomial situations by n p, where n is the number of trials
and p the probability of a success.
 The variance of the Poisson distribution is also equal to n p.
 X, the number of success generally has no specific upper
limit.
 Probability distribution always skewed to the right.
 Becomes symmetrical when l gets large.
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-49
Example: Poisson Probability Distribution
 The Sylvania Urgent Care facility specializes in caring for
minor injuries, colds, and flu. For the evening hours of 610 PM the mean number of arrivals is 4.0 per hour. What is
the probability of 2 arrivals in an hour?
P( x) 
Ka-fu Wong © 2007
lx e  l
x!
4 2 e 4

 .1465
2!
ECON1003: Analysis of Economic Data
Lesson4-50
Example: Poisson Probabilities
x =
P( x) 
lx e  l
x!
number of Cleveland air traffic control errors
during one week
m = 0.4 (expected number of errors per week)
Find the probability that 3 errors will occur in a week.
e-0.4 (0.4)3
p(3)  P(x = 3) =
= .0072
3!
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-51
Several Poisson Distributions
As l gets large, the probability distribution becomes symmetric.
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-52
Using Poisson Tables
l
X
0
1
2
3
4
5
6
7
0.10
0.20
0.30
0.40
0.50
0.60
0.70
0.80
0.90
0.9048
0.0905
0.0045
0.0002
0.0000
0.0000
0.0000
0.0000
0.8187
0.1637
0.0164
0.0011
0.0001
0.0000
0.0000
0.0000
0.7408
0.2222
0.0333
0.0033
0.0003
0.0000
0.0000
0.0000
0.6703
0.2681
0.0536
0.0072
0.0007
0.0001
0.0000
0.0000
0.6065
0.3033
0.0758
0.0126
0.0016
0.0002
0.0000
0.0000
0.5488
0.3293
0.0988
0.0198
0.0030
0.0004
0.0000
0.0000
0.4966
0.3476
0.1217
0.0284
0.0050
0.0007
0.0001
0.0000
0.4493
0.3595
0.1438
0.0383
0.0077
0.0012
0.0002
0.0000
0.4066
0.3659
0.1647
0.0494
0.0111
0.0020
0.0003
0.0000
Example: Find P(X = 2) if l = .50
e  l lX e 0.50 (0.50)2
P( X  2) 

 .0758
X!
2!
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-53
What distributions to use?
 Poisson considers the number of times an event occurs
over an INTERVAL of TIME or SPACE. Note that we are not
considering a sample of given number of observations.
 Thus, if we are considering a sample of 10 observations
and we are asked to compute the probability of having
6 successes, we should not use Poisson. Instead, we
should consider Binomial or Hypergeometric.
 Hypergeometric consider the number of successes in a
sample when the probability of success varies across trials
due to “without replacement” sampling strategy. To
compute the Hypergeometric probability, one will need to
know N and S separately.
 Suppose we know that the probability of success is 0.3.
We are considering a sample of 10 observations and we
are asked to compute the probability of having 6
successes. We cannot use Hypergeometric because we
do not have N and S separately. Instead, we have to
use Binomial.
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-54
What distributions to use?
Example
In a shipment of 15 hard disks, 5 are defective. If 4 of the
disks are inspected, what is the probability that exactly 1 is
defective?
 First, we recognize that it is not Poisson because "4 of the disks
are inspected" (i.e., sample size =4).
 Second, it is sampling without replacement because if we were
to inspect four disks for defects, we will not want to sample with
replacement.
 Third, both N (15 hard disks) and S (5 are defective) are
given. Hence we will use Hypergeometric.
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-55
What distributions to use?
Example
From an inventory of 48 cars being shipped to local
automobile dealers, 12 have had defective radios installed.
What is the probability that one particular dealership
receiving 8 cars obtains all with defective radios?
 First, we recognize that it is not Poisson because 8 cars are
“inspected" (i.e., sample size =8).
 Second, it is sampling without replacement because if we were
to inspect all 8 cars for defects, we will not want to sample with
replacement.
 Third, both N (48 cars) and S (12 have defective radio) are
given. Hence we will use Hypergeometric.
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-56
What distributions to use?
Example
The number of claims for missing baggage for a well-known
airline in a small city averages nine per day. What is the
probability that, on a given day, there will be fewer than
three claims made?
 First, we recognize that it is likely Poisson because “on a given
day”.
 Second, we are asked to compute the probability of the number
of claims larger than some number. There is no limit on the
number of claims that can arrive in a given day.
 Third, “average per day” is given. Hence we will use Poisson.
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-57
What distributions to use?
Example
When a customer places an order with Rudy’s on-Line Office
Supplies, a computerized accounting information system (AIS)
automatically checks to see if the customer has exceeded his or her
credit limit. Past records indicate that the probability of customers
exceeding their credit limit is 0.05. Suppose that, on a given day,
20 customers place orders. What is the probability that zero
customers will exceed their limits?
 First, we recognize that it is not Poisson because 20 customers
place orders (i.e., sample size =20).
 Second, the probability of drawing a particular type of customers
appears the same across trials because “the probability of
customers exceeding their credit limit is 0.05”.
 Hence we will use Binomial.
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-58
Features of a Bivariate Discrete
Distribution
 If X and Y are discrete random variables, we may define their
joint probability function as PXY(xi,yi)
 Let (x1,…,xR) and (y1,…,yS) be the list of all possible outcomes for
X and Y respectively.
 The main features of a bivariate discrete probability distribution
are:
 The probability of a particular outcome, PXY(xi,yi) is between 0
and 1.
 The sum of the probabilities of the various outcomes is 1.00.
That is,
PXY(x1,y1) + PXY(x2,y1) +…+ PXY(xR,y1) + + … + PXY(xR,yS) = 1
 The outcomes are mutually exclusive. That is,
if xi not equal to xk, or yi not equal to yk
PXY((xi,yi) and (xk,yk)) = 0 and
PXY((xi,yi) or (xk,yk)) = PXY(xi,yi) + PXY(xk,yk)
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-59
Example: Bivariate Discrete Distribution
X takes 3 possible values and Y takes 4 possible values.
y1
y2
y3
y4
x1
P(x1,y1)
P(x1,y2)
P(x1,y3)
P(x1,y4)
x2
P(x2,y1)
P(x2,y2)
P(x2,y3)
P(x2,y4)
x3
P(x3,y1)
P(x3,y2)
P(x3,y3)
P(x3,y4)
joint probability
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-60
EXAMPLE: Bivariate distribution
The joint distribution of the movement of Hang Seng
Index (HSI) and weather is shown in the following table.
Rainy
Not Rainy
Totals
HSI falls
0.15
0.4
0.55
HSI rises
0.2
0.25
0.45
0.35
0.65
1.0
Totals
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-61
Hirshleifer, David; Shumway, Tyler (2003): “Good Day
Sunshine: Stock Returns and the Weather ,” Journal of
Finance, 58(2): 1009-32.
 Psychological evidence and casual intuition predict that sunny weather is
associated with upbeat mood.
 This paper examines the relationship between morning sunshine in the city
of a country's leading stock exchange and daily market index returns
across 26 countries from 1982 to 1997.
 Sunshine is strongly significantly correlated with stock returns. After
controlling for sunshine, rain and snow are unrelated to returns.
 Substantial use of weather-based strategies was optimal for a trader with
very low transactions costs. However, because these strategies involve
frequent trades, fairly modest costs eliminate the gains. These findings are
difficult to reconcile with fully rational price setting.
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-62
Marginal Distributions
Average
out the
impact of y
on the
probability.
Y
X
Rainy
Not Rainy
Totals
HSI falls
0.15
0.4
0.55
HSI rises
0.2
0.25
0.45
0.35
0.65
1.0
Totals
 The marginal probability function of X.
PX(x) = yPXY(x, y) = PXY(x, y1) +PXY(x, y2) +…+ PXY(x, yn)
P(HSI falls)= P(HSI falls and rainy) + P(HSI falls and not rainy)
P(HSI rises)= P(HSI rises and rainy) + P(HSI rises and not rainy)
 The double sum
xyPXY(x, y)
= P(HSI falls and rainy) + P(HSI falls and not rainy)+
P(HSI rises and rainy) + P(HSI rises and not rainy)
= P(HSI falls)+P(HSI rises)= 1
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-63
Marginal Distributions
Y
X
Rainy
Not Rainy
Totals
HSI falls
0.15
0.4
0.55
HSI rises
0.2
0.25
0.45
0.35
0.65
1.0
Totals
 The marginal probability function of X.
yPXY(x, y) = PX(x).
 The marginal probability function of Y.
xPXY(x, y) = PY(y).
 The double sum
yxPXY(x, y) = 1
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-64
Conditional Distributions
Y
X
Rainy
Not Rainy
Totals
HSI falls
0.15
0.4
0.55
HSI rises
0.2
0.25
0.45
0.35
0.65
1.0
Totals
 The conditional probability function of X given Y:
PX|Y=y(x ) = P(X = x | Y = y) = PXY(x,y)/PY(y) if P(Y = y) > 0
PX|Y=y(x ) =0
if P(Y = y) = 0
Note that PX|Y=y(x ) when P(Y = y) = 0 is undefined
using the top formula.
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-65
Conditional Distributions
Y
X
Rainy
Not Rainy
Totals
HSI falls
0.15
0.4
0.55
HSI rises
0.2
0.25
0.45
0.35
0.65
1.0
Totals
 For each fixed y this is a probability function for X, i.e.
the conditional probability function is non-negative and
XPX|Y=y(x ) = PX|Y=y(x1 ) + PX|Y=y(x2 )
= PX,Y(x1, y)/ PY(y) + PX,Y(x2, y)/ PY(y)
=[PX,Y(x1, y) + PX,Y(x2, y)]/ PY(y)
=1.
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-66
Conditional Distributions
Y
X
Rainy
Not Rainy
Totals
HSI falls
0.15
0.4
0.55
HSI rises
0.2
0.25
0.45
0.35
0.65
1.0
Totals
 By the definition of conditional probability,
PX|Y=y(x ) = PX,Y(x, y)/ PY(y).
E.g., P(HSI rises| Rainy) = 0.2/0.35.
 When X and Y are independent,
PX|Y=y(x ) is equal to PX(x).
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-67
Example: Conditional Distributions
Y
X HSI falls
HSI rises
Totals
Rainy
Not Rainy
Totals
0.15
0.4
0.55
0.15/0.35
0.4/0.65
0.2
0.25
0.45
0.2/0.35
0.25/0.65
0.35
0.65
1.0
1.0
1.0
0.4/0.55
1.0
0.25/0.45
1.0
P(Y|HSI falls) 0.15/0.55
P(Y|HSI rises)
PX|Y=y(x ) = PX,Y(x, y)/ PY(y).
0.2/0.45
P(X|Rainy) P(X| Not Rainy)
PY|X=x(y ) = PX,Y(x, y)/ PX(x).
Are X and Y statistically independent?
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-68
Conditional Mean of Bivariate Discrete
Probability Distribution
 For bivariate probability distribution, the conditional
expectation or conditional mean E(X|Y) is computed by
the formula:
E ( X | Y  yi )  Σ X xPX |Y  yi ( x)
 x1PX |Y  yi ( x1 ) x2 PX |Y  yi ( x2 )...  xn PX |Y  yi ( xn )
 Unconditional expectation or mean of X, E(X)
E(X)  ΣY E ( X|Y  yi ) PY ( yi )
 E[ E ( X|Y )]
 E[ m X|Y ]
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-69
Law of iterated expectations
 For a bivariate random variables, X and Y:
 E(Y) = E[E(Y|X)]
 For a trivariate random variables, X, Y and Z
 E(Y|X,Z) = conditional mean given X and Z.
 E(Y|Z) = E[E(Y|X,Z)|Z]
 E(Y|X) = E[E(Y|X,Z)|X]
 E(Y) = E{E[E(Y|X,Z)|Z]}
(note: taking expectations in several iterations.)
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-70
Law of iterated expectations
 Given E(e|X) = 0, find E(eX).
E(eX)
= E[E(eX|X)]
=E[E(e|X)X]
=E[0*X]
=0
 Given E(Y|X,Z) = 0, E(XY) = 2, E(Z) =4, find E(XYZ).
E(XYZ)
= E[E(YXZ|X,Z)]
= E[E(Y|X,Z) X Z]
= E[ 0* X Z]
= 0.
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-71
The Variance of discrete random variables
 For univariate discrete probability distribution
Var ( X )  E[(X  μ)2 ]
 Σ[(x  μ)2 P(x)]
 (x1  μ)2 P(x1 )  (x 2  μ)2 P(x 2 )  ... (x n  μ)2 P(x n )
 For bivariate discrete probability distribution
Var(X)  E[(X  μX )2 ]
 Σ Y Σ X [(x - μX )2 PX, Y (x, y)]
 (x 1 - μX )2 PX, Y (x 1 , y1 )  ( x 2 - μX )2 PX, Y (x 2 , y1 )  ...  ( x n - μX )2 PX, Y (x n , y1 )
 ...
(x 1 - μX )2 PX, Y (x 1 , y n )  ( x 2 - μX )2 PX, Y (x 2 , y n )  ...  ( x n - μX )2 PX, Y (x n , y n )
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-72
The Covariance of a Bivariate Discrete
Probability Distribution
Covariance measures how two random variables co-vary.
Cov(X,Y)  E[(X  μX )(Y  μY )]
 Σ Y Σ X [(x - μX )(Y  μY )PX, Y (x, y)]
Cov(X,Y)  E[(X  μ X )(Y  μ Y )]
 E[XY  μX Y  μ Y X  μX μ Y ]
 E[XY]  μ XE[Y]  μ YE[X]  μ X μ Y
 E[XY]  μ X μ Y  μ Y μ X  μ X μ Y
 E[XY]  μ X μ Y
 E[XY]  E[X]E[Y]
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-73
Covariance of linear transformed random
variables
 If a and b are constants and X is a random variable, then
Cov(a,b) = 0
Cov(a,bX) = 0
Cov(a+bX,Y) = bCov(X,Y)
Cov(a  bX, Y )  E[ a  bX  (a  bμX ) ](Y  μ Y )
 E[ bX  bμX ](Y  μ Y )
 E[ b(X  μX ) ](Y  μ Y )
 b E(X  μ)(Y  μ Y )
 bCov(X,Y)
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-74
Variance of a sum of random variables
 If a and b are constants and X and Y are random variables,
then
Var(X+Y) = Var(X) + Var(Y) + 2Cov(X,Y)
Var(aX+bY) =a2Var(X) + b2Var(Y) + 2abCov(X,Y)
Var(X  Y)  E[ X  Y  (μX  μY ) ]2
 E[ (X  μX )  (Y  μY )]2
 E[ (X  μX )2  (Y  μY )2  2(X  μX )(Y  μY )]
 E[ (X  μX )2 ]  E[(Y  μY )2  2E[(X  μX )(Y  μY )]
 Var(X)  Var(Y)  2Cov(X,Y)
Var(aX  bY)  E[ aX  bY  (aμX  bμY ) ]2
 E[ (aX  aμX )  (bY  bμY )]2
 E[ a2 (X  μX )2  b2 (Y  μY )2  2(aX  aμX )(bY  bμY )]
 a2E[ (X  μX )2 ]  b2E[(Y  μY )2  2abE[(X  μX )(Y  μY )]
 a2 Var(X)  b2 Var(Y)  2abCov(X,Y)
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-75
Correlation coefficient
 The strength of the dependence between X and Y is
measured by the correlation coefficient:
Cov(X,Y)
  Corr(X,Y) 
Var(X)Var(Y)
 ρ=0
 ρ>0
 when X
 ρ = +1
 ρ<0
 when X
 ρ = -1
Ka-fu Wong © 2007
no linear relationship between X and Y
positive linear relationship between X and Y
is high (low) then Y is likely to be high (low)
perfect positive linear dependency
negative linear relationship between X and Y
is high (low) then Y is likely to be low (high)
perfect negative linear dependency
ECON1003: Analysis of Economic Data
Lesson4-76
Portfolio Analysis
 Let random variable X be the price for stock A
 Let random variable Y be the price for stock B
 Suppose a portfolio consists of
 a shares of stock A, and
 b shares of stock B.
 The market value, W, for the portfolio is given by the linear
function
W = aX + bY
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-77
Portfolio Analysis
 The mean value for W is
 E(W) = E(aX + bY) = aE(X)+bE(Y)
 The variance for W is
 Var(W)= a2 Var(X) + b2 Var(Y) + 2abCov(X,Y)
 Var(W)= a2 Var(X) + b2 Var(Y) + 2abCorr(X,Y)[Var(X)Var(Y)]1/2
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-78
Example: Investment Returns
Return per $1,000 for two types of investments
P(xiyi)
Economic condition
Investment
Passive Fund X Aggressive Fund Y
.2
Recession
- $ 25
- $200
.5
Stable Economy
+ 50
+ 60
.3
Expanding Economy
+ 100
+ 350
E(x) = μx = (-25)(.2) +(50)(.5) + (100)(.3) = 50
E(y) = μy = (-200)(.2) +(60)(.5) + (350)(.3) = 95
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-79
Computing the Standard Deviation for
Investment Returns
P(xiyi)
Economic condition
Investment
Passive Fund X Aggressive Fund Y
0.2
Recession
- $ 25
- $200
0.5
Stable Economy
+ 50
+ 60
0.3
Expanding Economy
+ 100
+ 350
σ X  (-25  50)2 (0.2)  (50  50)2 (0.5)  (100  50)2 (0.3)
 43.30
σ y  (-200  95)2 (0.2)  (60  95)2 (0.5)  (350  95)2 (0.3)
 193.71
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-80
Covariance for Investment Returns
P(xiyi)
Economic condition
Investment
Passive Fund X Aggressive Fund Y
.2
Recession
- $ 25
- $200
.5
Stable Economy
+ 50
+ 60
.3
Expanding Economy
+ 100
+ 350
Cov(X, Y)  (-25  50)(-200  95)(.2)  (50  50)(60  95)(.5)
 (100  50)(350  95)(.3)
 8250
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-81
Interpreting the Results for Investment
Returns
 The aggressive fund has a higher expected return, but much more
risk
μy = 95 > μx = 50
but
σy = 193.21 > σx = 43.30
 The Covariance of 8250 indicates that the two investments are
positively related and will vary in the same direction
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-82
Portfolio Example
Investment X:
Investment Y:
μx = 50
μy = 95
σx = 43.30
σy = 193.21
σxy = 8250
Suppose 40% of the portfolio (P) is in Investment X and 60% is in
Investment Y:
E(P)  .4 (50)  (.6) (95)  77
σP  (.4)2 (43.30)2  (.6)2 (193.21)2  2(.4)(.6)(8250)
 133.04
The portfolio return and portfolio variability are between the values
for investments X and Y considered individually
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-83
Lesson 4:
Discrete Probability Distributions
- END -
Ka-fu Wong © 2007
ECON1003: Analysis of Economic Data
Lesson4-84