Chap 4-1 Statistics Please Stand By….. Chap 4-2 Chapter 4: Probability and Distributions Randomness General Probability Probability Models Random Variables.

Chap 4-1

Statistics

Please Stand By….

Chap 4-2

Chapter 4: Probability and Distributions

Randomness General Probability Probability Models Random Variables Moments of Random Variables

Chap 4-3

Randomness

The language of probability

Thinking about randomness

The uses of probability

Chap 4-4

Randomness

Chap 4-5

Randomness

Chap 4-6

Randomness

Chap 4-7

RandomnessTechnical Analysis 1 | More Examples | Another (Tipping Pts)

Chap 4-8

Chapter Goals

After completing this chapter, you should be able to:

Explain three approaches to assessing probabilities

Apply common rules of probability Use Bayes’ Theorem for conditional probabilities Distinguish between discrete and continuous

probability distributions Compute the expected value and standard

deviation for a probability distributions

Chap 4-9

Important Terms

Probability – the chance that an uncertain event will occur (always between 0 and 1)

Experiment – a process of obtaining outcomes for uncertain events

Elementary Event – the most basic outcome possible from a simple experiment

Randomness – Does not mean haphazard Description of the kind of order that emerges only in

the long run

Chap 4-10

Important Terms (CONT’D)

Sample Space – the collection of all possible elementary outcomes

Probability Distribution Function Maps events to intervals on the real line Discrete probability mass Continuous probability density

Chap 4-11

Sample Space

The Sample Space is the collection of all possible outcomes (based on an probabilistic experiment)

e.g., All 6 faces of a die:

e.g., All 52 cards of a bridge deck:

Chap 4-12

Events

Elementary event – An outcome from a sample space with one characteristic Example: A red card from a deck of cards

Event – May involve two or more outcomes simultaneously Example: An ace that is also red from a deck

of cards

Chap 4-13

Elementary Events

A automobile consultant records fuel type and vehicle type for a sample of vehicles

2 Fuel types: Gasoline, Diesel3 Vehicle types: Truck, Car, SUV

6 possible elementary events:

e1 Gasoline, Truck

e2 Gasoline, Car

e3 Gasoline, SUV

e4 Diesel, Truck

e5 Diesel, Car

e6 Diesel, SUV

Gasoline

Diesel

CarTruck

Truck

Car

SUV

SUV

e1

e2

e3

e4

e5

e6

Chap 4-14

Independent Events

E1 = heads on one flip of fair coin

E2 = heads on second flip of same coin

Result of second flip does not depend on the result of the first flip.

Dependent Events

E1 = rain forecasted on the news

E2 = take umbrella to work

Probability of the second event is affected by the occurrence of the first event

Independent vs. Dependent Events

Chap 4-15

Probability Concepts

Mutually Exclusive Events If E1 occurs, then E2 cannot occur

E1 and E2 have no common elements

Black Cards

Red Cards

A card cannot be Black and Red at the same time.

E1

E2

Chap 4-16

Coming up with Probability

Empirically From the data! Based on observation, not theory

Probability describes what happens in very many trials.

We must actually observe many trials to pin down a probability

Based on belief (Bayesian Technique)

Chap 4-17

Assigning Probability

Classical Probability Assessment

Relative Frequency of Occurrence

Subjective Probability Assessment

P(Ei) =Number of ways Ei can occur

Total number of elementary events

Relative Freq. of Ei =Number of times Ei occurs

N

An opinion or judgment by a decision maker about the likelihood of an event

Chap 4-18

Calculating Probability

Counting Outcomes

Observing Outcomes in Trials

Number of ways Ei can occur

Total number of elementary events

Chap 4-19

Counting

Chap 4-20

Counting

a b c d e …. ___ ___ ___ ___

1. N take n ___ ___ ___

2. N take k ___ ___

3. Order not important – less than permutations

Chap 4-21

Counting

Chap 4-22

Rules of Probability

S is sample space Pr(S) = 1 Events measured in numbers result in a

Probability Distribution

Chap 4-23

Rules of Probability

Rules for Possible Values

and Sum

Individual Values Sum of All Values

0 ≤ P(ei) ≤ 1

For any event ei

1)P(ek

1ii

where:k = Number of elementary events in the sample space

ei = ith elementary event

Chap 4-24

Addition Rule for Elementary Events

The probability of an event Ei is equal to the sum of the probabilities of the elementary events forming Ei.

That is, if:

Ei = {e1, e2, e3}

then:

P(Ei) = P(e1) + P(e2) + P(e3)

Chap 4-25

Complement Rule

The complement of an event E is the collection of all possible elementary events not contained in event E. The complement of event E is represented by E.

Complement Rule:

P(E)1)EP( E

E

1)EP(P(E) Or,

Chap 4-26

Addition Rule for Two Events

P(E1 or E2) = P(E1) + P(E2) - P(E1 and E2)

E1 E2

P(E1 or E2) = P(E1) + P(E2) - P(E1 and E2)Don’t count common elements twice!

■ Addition Rule:

E1 E2+ =

Chap 4-27

Addition Rule Example

P(Red or Ace) = P(Red) +P(Ace) - P(Red and Ace)

= 26/52 + 4/52 - 2/52 = 28/52Don’t count the two red aces twice!

BlackColor

Type Red Total

Ace 2 2 4

Non-Ace 24 24 48

Total 26 26 52

Chap 4-28

Addition Rule for Mutually Exclusive Events

If E1 and E2 are mutually exclusive, then

P(E1 and E2) = 0

So

P(E1 or E2) = P(E1) + P(E2) - P(E1 and E2)

= P(E1) + P(E2)

= 0

E1 E2

if mutually

exclusiv

e

Chap 4-29

Conditional Probability

Conditional probability for any

two events E1 , E2:

)P(E

)EandP(E)E|P(E

2

2121

0)P(Ewhere 2

Chap 4-30

What is the probability that a car has a CD player, given that it has AC ?

i.e., we want to find P(CD | AC)

Conditional Probability Example

Of the cars on a used car lot, 70% have air conditioning (AC) and 40% have a CD player (CD). 20% of the cars have both.

Chap 4-31

Conditional Probability Example

No CDCD Total

AC

No AC

Total 1.0

Of the cars on a used car lot, 70% have air conditioning (AC) and 40% have a CD player (CD). 20% of the cars have both.

.2857.7

.2

P(AC)

AC)andP(CDAC)|P(CD

(continued)

Chap 4-32

Conditional Probability Example

No CDCD Total

AC .2 .5 .7

No AC .2 .1 .3

Total .4 .6 1.0

Given AC, we only consider the top row (70% of the cars). Of these, 20% have a CD player. 20% of 70% is about 28.57%.

.2857.7

.2

P(AC)

AC)andP(CDAC)|P(CD

(continued)

Chap 4-33

For Independent Events:

Conditional probability for independent events E1 , E2:

)P(E)E|P(E 121 0)P(Ewhere 2

)P(E)E|P(E 212 0)P(Ewhere 1

Chap 4-34

Multiplication Rules

Multiplication rule for two events E1 and E2:

)E|P(E)P(E)EandP(E 12121

)P(E)E|P(E 212 Note: If E1 and E2 are independent, thenand the multiplication rule simplifies to

)P(E)P(E)EandP(E 2121

Chap 4-35

Tree Diagram Example

Diesel P(E2) = 0.2

Gasoline P(E1) = 0.8

Truck: P(E3|E1

) = 0.2

Car: P(E4|E1) = 0.5

SUV: P(E5|E1) = 0.3

P(E1 and E3) = 0.8 x 0.2 = 0.16

P(E1 and E4) = 0.8 x 0.5 = 0.40

P(E1 and E5) = 0.8 x 0.3 = 0.24

P(E2 and E3) = 0.2 x 0.6 = 0.12

P(E2 and E4) = 0.2 x 0.1 = 0.02

P(E3 and E4) = 0.2 x 0.3 = 0.06

Truck: P(E3|E2) = 0.6

Car: P(E4|E2) = 0.1

SUV: P(E5|E2) = 0.3

Chap 4-36

Get Ready….

More Probability Examples

Random Variables

Probability Distributions

Chap 4-37

Introduction to Probability Distributions

Random Variable – “X” Is a function from the sample space to

another space, usually Real line Represents a possible numerical value from

a random event Each r.v. has a Distribution Function – FX(x),

fX(x) based on that in the sample space Assigns probability to the (numerical)

outcomes (discrete values or intervals)

Chap 4-38

Random Variables

Not Easy to Describe

Chap 4-39

Random Variables

Chap 4-40

Random Variables

Not Easy to Describe

Chap 4-41

Introduction to Probability Distributions

Random Variable Represents a possible numerical value from

a random event

Random

Variables

Discrete Random Variable

ContinuousRandom Variable

Chap 4-42

A list of all possible [ xi , P(xi) ] pairs

xi = Value of Random Variable (Outcome)

P(xi) = Probability Associated with Value

xi’s are mutually exclusive (no overlap)

xi’s are collectively exhaustive (nothing left out)

0 P(xi) 1 for each xi

P(xi) = 1

Discrete Probability Distribution

Chap 4-43

Discrete Random Variables

Can only assume a countable number of values

Examples:

Roll a die twiceLet x be the number of times 4 comes up (then x could be 0, 1, or 2 times)

Toss a coin 5 times. Let x be the number of heads

(then x = 0, 1, 2, 3, 4, or 5)

Chap 4-44

Experiment: Toss 2 Coins. Let x = # heads.

T

T

Discrete Probability Distribution

4 possible outcomes

T

T

H

H

H H

Probability Distribution

0 1 2 x

x Value Probability

0 1/4 = .25

1 2/4 = .50

2 1/4 = .25

.50

.25

Pro

bab

ility

Chap 4-45

The Distribution Function

Assigns Probability to Outcomes F, f > 0; right-continuous and P(X<a)=FX(a)

Discrete Random Variable

Continuous Random Variable

( ) 0 asF X X

( ) 0 as

( ) 1 as

F X X

F X X

( ) 1 asF X X

Chap 4-46

Discrete Random Variable Summary Measures - Moments

Expected Value of a discrete distribution (Weighted Average)

E(x) = xi P(xi)

Example: Toss 2 coins, x = # of heads, compute expected value of x:

E(x) = (0 x .25) + (1 x .50) + (2 x .25) = 1.0

x P(x)

0 .25

1 .50

2 .25

Chap 4-47

Standard Deviation of a discrete distribution

where:

E(x) = Expected value of the random variable x = Values of the random variableP(x) = Probability of the random variable having

the value of x

Discrete Random Variable Summary Measures

P(x)E(x)}{xσ 2x

(continued)

Chap 4-48

Example: Toss 2 coins, x = # heads, compute standard deviation (recall E(x) = 1)

Discrete Random Variable Summary Measures

P(x)E(x)}{xσ 2x

.707.50(.25)1)(2(.50)1)(1(.25)1)(0σ 222x

(continued)

Possible number of heads = 0, 1, or 2

Chap 4-49

Two Discrete Random Variables

Expected value of the sum of two discrete random variables:

E(x + y) = E(x) + E(y) = x P(x) + y P(y)

(The expected value of the sum of two random variables is the sum of the two expected values)

Chap 4-50

Sums of Random Variables

Usually we discuss sums of INDEPENDENT random variables, Xi i.i.d.

Only sometimes is Due to Linearity of the Expectation operator,

E(Xi) = E(Xi) and Var(Xi) = Var(Xi)

CLT: Let Sn=Xi then (Sn - E(Sn))~N(0, var)

Xf f

Chap 4-51

Covariance

Covariance between two discrete random variables:

σxy = [xi – E(x)][yj – E(y)]P(xiyj)

where:

xi = possible values of the x discrete random variable

yj = possible values of the y discrete random variable

P(xi ,yj) = joint probability of the values of xi and yj occurring

Chap 4-52

Covariance between two discrete random variables:

xy > 0 x and y tend to move in the same direction

xy < 0 x and y tend to move in opposite directions

xy = 0 x and y do not move closely together

Interpreting Covariance

Chap 4-53

Correlation Coefficient

The Correlation Coefficient shows the strength of the linear association between two variables

where:

ρ = correlation coefficient (“rho”)σxy = covariance between x and yσx = standard deviation of variable xσy = standard deviation of variable y

yx

yx

σσ

σρ

Chap 4-54

The Correlation Coefficient always falls between -1 and +1

= 0 x and y are not linearly related.

The farther is from zero, the stronger the linear relationship:

= +1 x and y have a perfect positive linear relationship

= -1 x and y have a perfect negative linear relationship

Interpreting the Correlation Coefficient

Chap 4-55

Useful Discrete Distributions

Discrete Uniform

Binary – Success/Fail (Bernoulli)

Binomial

Poisson

Empirical Piano Keys Other “stuff that happens” in life

Chap 4-56

Useful Continuous Distributions

Finite Support Uniform fX(x)=c Beta

Infinite Support Gaussian (Normal) N() Log-normal Gamma Exponential

Chap 4-57

Section Summary

Described approaches to assessing probabilities

Developed common rules of probability

Distinguished between discrete and continuous probability distributions

Examined discrete and continuous probability distributions and their moments (summary measures)

Chap 4-58

Probability Distributions

Continuous Probability

Distributions

Binomial

Etc.

Poisson

Probability Distributions

Discrete Probability

Distributions

Normal

Uniform

Etc.

Chap 4-59

Some Discrete Distributions

Chap 4-60

P(x) = probability of x successes in n trials, with probability of success p on each trial

x = number of ‘successes’ in sample, (x = 0, 1, 2, ..., n)

p = probability of “success” per trial

q = probability of “failure” = (1 – p)

n = number of trials (sample size)

P(x)n

x ! n xp qx n x!

( )!

Example: Flip a coin four times, let x = # heads:

n = 4

p = 0.5

q = (1 - .5) = .5

x = 0, 1, 2, 3, 4

Binomial Distribution Formula

Chap 4-61

n = 5 p = 0.1

n = 5 p = 0.5

Mean

0.2.4.6

0 1 2 3 4 5

X

P(X)

.2

.4

.6

0 1 2 3 4 5

X

P(X)

0

0.5(5)(.1)npμ

0.6708

.1)(5)(.1)(1npqσ

2.5(5)(.5)npμ

1.118

.5)(5)(.5)(1npqσ

Binomial Characteristics

Examples

Chap 4-62

Using Binomial Tables

n = 10

x p=.15 p=.20 p=.25 p=.30 p=.35 p=.40 p=.45 p=.50

0

1

2

3

4

5

6

7

8

9

10

0.1969

0.3474

0.2759

0.1298

0.0401

0.0085

0.0012

0.0001

0.0000

0.0000

0.0000

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

10

9

8

7

6

5

4

3

2

1

0

p=.85 p=.80 p=.75 p=.70 p=.65 p=.60 p=.55 p=.50 x

Examples: n = 10, p = .35, x = 3: P(x = 3|n =10, p = .35) = .2522

n = 10, p = .75, x = 2: P(x = 2|n =10, p = .75) = .0004

Chap 4-63

The Poisson Distribution

Characteristics of the Poisson Distribution: The outcomes of interest are rare relative to the

possible outcomes The average number of outcomes of interest per time

or space interval is The number of outcomes of interest are random, and

the occurrence of one outcome does not influence the chances of another outcome of interest

The probability of that an outcome of interest occurs in a given segment is the same for all segments

Chap 4-64

Poisson Distribution Formula

where:

t = size of the segment of interest

x = number of successes in segment of interest

= expected number of successes in a segment of unit size

e = base of the natural logarithm system (2.71828...)

!x

e)t()x(P

tx

Chap 4-65

Poisson Distribution Shape

The shape of the Poisson Distribution depends on the parameters and t:

0.00

0.05

0.10

0.15

0.20

0.25

0 1 2 3 4 5 6 7 8 9 10

x

P(x

)

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0 1 2 3 4 5 6 7

x

P(x

)

t = 0.50 t = 3.0

Chap 4-66

Poisson Distribution Characteristics

Mean

Variance and Standard Deviation

http://www.math.csusb.edu/faculty/stanton/m262/poisson_distribution/Poisson_old.html

λtμ

λtσ2 λtσ where = number of successes in a segment of unit size

t = the size of the segment of interest

Chap 4-67

Graph of Poisson Probabilities

0.00

0.10

0.20

0.30

0.40

0.50

0.60

0.70

0 1 2 3 4 5 6 7

x

P(x

)X

t =

0.50

0

1

2

3

4

5

6

7

0.6065

0.3033

0.0758

0.0126

0.0016

0.0002

0.0000

0.0000P(x = 2) = .0758

Graphically:

= .05 and t = 100

Chap 4-68

Some Continuous Distributions

Chap 4-69

The Continuous Uniform Distribution:

otherwise 0

bxaifab

1

where

f(x) = value of the density function at any x value

a = lower limit of the interval

b = upper limit of the interval

The Uniform Distribution(continued)

f(x) =

Chap 4-70

Uniform Distribution

Example: Uniform Probability Distribution Over the range 2 ≤ x ≤ 6:

2 6

.25

f(x) = = .25 for 2 ≤ x ≤ 66 - 21

x

f(x)

Chap 4-71

Normal (Gaussian) Distribtion

Chap 4-72

2121( )

2

x

Xf x e

By varying the parameters μ and σ, we obtain different normal distributions

μ ± 1σencloses about 68% of x’sμ ± 2σ covers about 95% of x’s; μ ± 3σ covers about 99.7% of x’s

The chance that a value that far or farther

away from the mean is highly unlikely, given

that particular mean and standard deviation

Chap 4-73

f(x)

xμ

Probability as Area Under the Curve

0.50.5

The total area under the curve is 1.0, and the curve is symmetric, so half is above the mean, half is below

1.0)xP(

0.5)xP(μ 0.5μ)xP(

Chap 4-74

The Standard Normal Distribution

Also known as the “z” distribution = (x-)/

Mean is by definition 0 Standard Deviation is by definition 1

z

f(z)

0

1

Values above the mean have positive z-values, values below the mean have negative z-values

Chap 4-75

Comparing x and z units

z100

3.00250 x

Note that the distribution is the same, only the scale has changed. We can express the problem in original units (x) or in standardized units (z)

μ = 100

σ = 50

Chap 4-76

Finding Normal Probabilities

Suppose x is normal with mean 8.0 and standard deviation 5.0.

Now Find P(x < 8.6)

(continued)

Z

0.12

.0478

0.00

.5000 P(x < 8.6)

= P(z < 0.12)

= P(z < 0) + P(0 < z < 0.12)

= .5 + .0478 = .5478

Chap 4-77

Using Standard Normal Tables

Chap 4-78

Section Summary

Reviewed discrete distributions binomial, poisson, etc.

Reviewed some continuous distributions normal, uniform, exponential

Found probabilities using formulas and tables

Recognized when to apply different distributions

Applied distributions to decision problems