Topic7_Jan13.pptw

7/27/2019 Topic7_Jan13.pptw

1/41

Discrete Probability

Distributions[Week 13]

TOPIC 7

1


2/41

Chap 7-2

Objectives

After completing this chapter, you should beable to:

Interpret the mean and standard deviation for a

discrete probability distribution Explain covariance and its application in finance

Use the binomial probability distribution to findprobabilities

Describe when to apply the binomial distribution Use the hypergeometric and Poisson discrete

probability distributions to find probabilities


3/41

Chap 7-3

Introduction to ProbabilityDistributions

Random Variable

Represents a possible numerical value froman uncertain event

Random

Variables

Discrete

Random Variable

Continuous

Random Variable


4/41

Chap 7-4

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)


5/41

Chap 7-5

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

.25Probability


6/41

Chap 7-6

Discrete Random VariableSummary Measures

Expected Value (or mean) of a discretedistribution (Weighted Average)

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

N

1i

ii )X(PXE(X)


7/41Chap 7-7

Variance of a discrete random variable

Standard Deviation of a discrete random variable

where:E(X) = Expected value of the discrete random variable X

Xi = the ith outcome of X

P(Xi

) = Probability of the ith occurrence of X


N

1i

i

2

i

2 )P(XE(X)][X

(continued)

N

1ii

2

i

2

)P(XE(X)][X


8/41Chap 7-8

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


)P(XE(X)][X i2

i .707.50(.25)1)(2(.50)1)(1(.25)1)(0 222

(continued)

Possible number of heads= 0, 1, or 2


9/41Chap 7-9

The Covariance

The covariance measures the strength of thelinear relationship between two variables

The covariance:

)YX(P)]Y(EY)][(X(EX[N

1i

iiiiXY

where: X = discrete variable XXi = the i

th outcome of XY = discrete variable YYi = the i

th outcome of YP(XiYi) = probability of occurrence of the condition affecting

the ith outcome of X and the ith outcome of Y


10/41Chap 7-10

Computing the Mean forInvestment Returns

Return per $1,000 for two types of investments

P(XiYi) Economic condition Passive Fund X Aggressive Fund Y.2 Recession - $ 25 - $200

.5 Stable Economy + 50 + 60

.3 Expanding Economy + 100 + 350

Investment

E(X) = X = (-25)(.2) +(50)(.5) + (100)(.3) = 50

E(Y) = Y = (-200)(.2) +(60)(.5) + (350)(.3) = 95


11/41Chap 7-11

Computing the Standard Deviationfor Investment Returns

P(XiYi) Economic condition Passive Fund X Aggressive Fund Y

.2 Recession - $ 25 - $200



Investment

43.30

(.3)50)(100(.5)50)(50(.2)50)(-25 222X

71.193

)3(.)95350()5(.)9560()2(.)95200-( 222Y


12/41Chap 7-12

Computing the Covariancefor Investment Returns

P(XiYi) Economic condition Passive Fund X Aggressive Fund Y

.2 Recession - $ 25 - $200



Investment

8250

95)(.3)50)(350(100

95)(.5)50)(60(5095)(.2)200-50)((-25 YX,


13/41Chap 7-13

Interpreting the Results forInvestment Returns

The aggressive fund has a higher expectedreturn, but much more risk

Y = 95 > X = 50

butY = 193.21 > X = 43.30

The Covariance of 8250 indicates that the twoinvestments are positively related and will varyin the same direction


14/41Chap 7-14

The Sum ofTwo Random Variables

Expected Value of the sum of two random variables:

Variance of the sum of two random variables:

Standard deviation of the sum of two random variables:

XY2Y

2X

2YX 2Y)Var(X

)Y(E)X(EY)E(X

2YXYX


15/41Chap 7-15

Portfolio Expected Returnand Portfolio Risk

Portfolio expected return (weighted averagereturn):

Portfolio risk (weighted variability)

Where w = portion of portfolio value in asset X

(1 - w) = portion of portfolio value in asset Y

)Y(E)w1()X(EwE(P)

XY2Y22X2P w)-2w(1)w1(w


16/41Chap 7-16

Portfolio Example

Investment X: X = 50 X = 43.30

Investment Y: Y = 95 Y = 193.21

XY = 8250

Suppose 40% of the portfolio is in Investment X and60% is in Investment Y:

The portfolio return and portfolio variability are between the values

for investments X and Y considered individually

77)95()6(.)50(4.E(P)

04.133

8250)2(.4)(.6)((193.21))6(.(43.30)(.4) 2222P


17/41Chap 7-17

Probability Distributions

Continuous

ProbabilityDistributions

Binomial

Poisson


Discrete


Normal

Uniform

Exponential


18/41Chap 7-18

Binomial Probability Distribution

A fixed number of observations, n e.g., 15 tosses of a coin; ten light bulbs taken from a warehouse

Two mutually exclusive and collectively exhaustive

categories e.g., head or tail in each toss of a coin; defective or not defective

light bulb

Generally called success and failure

Probability of success is p, probability of failure is 1 p

Constant probability for each observation e.g., Probability of getting a tail is the same each time we toss

the coin


19/41Chap 7-19

Binomial Probability Distribution(continued)

Observations are independent The outcome of one observation does not affect the

outcome of the other

Two sampling methods Infinite population without replacement

Finite population with replacement


20/41Chap 7-20

Possible Binomial DistributionSettings

A manufacturing plant labels items aseither defective or acceptable

A firm bidding for contracts will either get acontract or not

A marketing research firm receives surveyresponses of yes I will buy or no I will

not

New job applicants either accept the offeror reject it


21/41Chap 7-21

Rule of Combinations

The number ofcombinations of selecting Xobjects out of n objects is

)!Xn(!X

!n

X

n

where: n! =n(n - 1)(n - 2) . . . (2)(1)

X! = X(X - 1)(X - 2) . . . (2)(1)

0! = 1 (by definition)


22/41Chap 7-22

P(X) = probability ofX successes in n trials,

with probability of success pon each trial

X = number of successes in sample,

(X = 0, 1, 2, ..., n)n = sample size (number of trials

or observations)

p = probability of success

P(X)n

X ! n Xp (1-p)

X n X!

( )!

Example: Flip a coin fourtimes, let x = # heads:

n = 4

p = 0.5

1 - p = (1 - .5) = .5

X = 0, 1, 2, 3, 4

Binomial Distribution Formula


23/41

Chap 7-23

Example:Calculating a Binomial Probability

What is the probability of one success in fiveobservations if the probability of success is .1?

X = 1, n = 5, and p = .1

32805.

)9)(.1)(.5(

)1.1()1(.)!15(!1

!5

)p1(p)!Xn(!X

!n)1X(P

4

151

XnX


24/41

Chap 7-24

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

Binomial Distribution

The shape of the binomial distribution depends on thevalues of p and n

Here, n = 5 and p = .1

Here, n = 5 and p = .5


25/41

Chap 7-25

Binomial DistributionCharacteristics

Mean

Variance and Standard Deviation

npE(x)

p)-np(12

p)-np(1 Where n = sample size


(1 p) = probability of failure


26/41

Chap 7-26

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)np

0.6708

.1)(5)(.1)(1p)-np(1

2.5(5)(.5)np

1.118

.5)(5)(.5)(1p)-np(1

Binomial Characteristics

Examples


27/41

Chap 7-27

Using Binomial Tables

n = 10

x 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

0.1074

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=.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


28/41

Chap 7-28

The Poisson Distribution

Binomial

Poisson


Discrete



29/41

Chap 7-29

The Poisson Distribution

Apply the Poisson Distribution when:

You wish to count the number of times an eventoccurs in a given area of opportunity

The probability that an event occurs in one area ofopportunity is the same for all areas of opportunity

The number of events that occur in one area ofopportunity is independent of the number of eventsthat occur in the other areas of opportunity

The probability that two or more events occur in anarea of opportunity approaches zero as the area ofopportunity becomes smaller

The average number of events per unit is (lambda)


30/41

Chap 7-30

Poisson Distribution Formula

where:

X = number of successes per unit

= expected number of successes per unite = base of the natural logarithm system (2.71828...)

!X

e)X(P

x


31/41

Chap 7-31

Poisson DistributionCharacteristics

Mean

Variance and Standard Deviation

2

where = expected number of successes per unit


32/41

Chap 7-32

Using Poisson Tables

X

0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90

0

1

2

34

5

6

7

0.9048

0.0905

0.0045

0.00020.0000

0.0000

0.0000

0.0000

0.8187

0.1637

0.0164

0.00110.0001

0.0000

0.0000

0.0000

0.7408

0.2222

0.0333

0.00330.0003

0.0000

0.0000

0.0000

0.6703

0.2681

0.0536

0.00720.0007

0.0001

0.0000

0.0000

0.6065

0.3033

0.0758

0.01260.0016

0.0002

0.0000

0.0000

0.5488

0.3293

0.0988

0.01980.0030

0.0004

0.0000

0.0000

0.4966

0.3476

0.1217

0.02840.0050

0.0007

0.0001

0.0000

0.4493

0.3595

0.1438

0.03830.0077

0.0012

0.0002

0.0000

0.4066

0.3659

0.1647

0.04940.0111

0.0020

0.0003

0.0000

Example: Find P(X = 2) if = .50

.07582!

(0.50)e

!X

e)2X(P

20.50X


33/41

Chap 7-33

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

=0.50

0

1

2

34

5

6

7

0.6065

0.3033

0.0758

0.01260.0016

0.0002

0.0000

0.0000P(X = 2) = .0758

Graphically:

= .50


34/41

Chap 7-34

Poisson Distribution Shape

The shape of the Poisson Distributiondepends on the parameter :

0.00

0.05

0.10

0.15

0.20

0.25

1 2 3 4 5 6 7 8 9 10 11 12

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)

= 0.50 = 3.00


35/41

Chap 7-35

Using Poisson Distribution toApproximate Binomial Distribution

The Poisson Distribution can be used toapproximate the Binomial Distribution when:

n is large, and

p is very small

The larger the n and the smaller the p, the better is theapproximation.


36/41

Chap 7-36

where:

P(X) = probability of X successes given the parameters n and p

n = sample sizep = true probability of success

e = mathematical constant approximated by 2.71828

X = number of successes in the sample (X = 0, 1, 2, , n)

!

)()(

X

npeXP

xnp


U i P i Di ib i


37/41

Chap 7-37

Mean

Standard Deviation

npXE )(

np

where = expected number of successes per unit

n = sample size


(1 p) = probability of failure


U i P i Di t ib ti t


38/41

Chap 7-38


U i P i Di t ib ti t


39/41

Chap 7-39



40/41

Chap 7-40


41/41

Summary

Addressed the probability of a discrete random

variable

Defined covariance and discussed itsapplication in finance

Discussed the Binomial distribution

Discussed the Hypergeometric distribution Reviewed the Poisson distribution

Topic7_Jan13.pptw

Documents