Infertat

Chapter 3

Discrete Probability
Distributions
 Chapter Goals
After completing this chapter, you should be able
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 find
probabilities
 Describe when to apply the binomial distribution
 Use Poisson discrete probability distributions to
find probabilities
 Introduction to Probability
Distributions

 Random Variable
 Represents a possible numerical value from
an uncertain event
Random
Variables

Ch. 3 Discrete Continuous Ch. 4

Random Variable Random Variable
 Discrete Random Variables
 Can only assume a countable number of values
Examples:

 Roll a die twice

Let 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)
 Discrete Probability Distribution

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

4 possible outcomes
Probability Distribution
T T X Value Probability
0 1/4 = .25
T H 1 2/4 = .50
2 1/4 = .25
H T
Probability

.50

H H .25

0 1 2 X
 Discrete Random Variable
Summary Measures
 Expected Value (or mean) of a discrete
distribution (Weighted Average)
N
  E(X)   Xi P( Xi )
i1

X P(X)
 Example: Toss 2 coins, 0 .25
X = # of heads, 1 .50

compute expected value of X: 2 .25

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

= 1.0
 Discrete Random Variable
Summary Measures
(continued)
 Variance of a discrete random variable
N
σ   [Xi  E(X)] P(Xi )
2 2

i1

 Standard Deviation of a discrete random variable

N
σ σ  2
 [X  E(X)] P(X )
i 1
i
2
i

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
 Discrete Random Variable
Summary Measures
(continued)

 Example: Toss 2 coins, X = # heads,

compute standard deviation (recall E(X) = 1)

σ  [X  E(X)] P(X )
i
2
i

σ  (0  1)2 (.25)  (1  1)2 (.50)  (2  1)2 (.25)  .50  .707

Possible number of heads

= 0, 1, or 2
 The Covariance

 The covariance measures the strength of the

linear relationship between two variables
 The covariance:
N
σ XY   [ Xi  E( X)][( Yi  E( Y )] P( Xi Yi )
i1

where: X = discrete variable X

Xi = the ith outcome of X
Y = discrete variable Y
Yi = the ith outcome of Y
P(XiYi) = probability of occurrence of the condition affecting
the ith outcome of X and the ith outcome of Y
 Computing the Mean for
Investment Returns
Return per $1,000 for two types of investments

Investment
P(XiYi) Economic condition 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

 Computing the Standard Deviation
for Investment Returns
Investment
P(XiYi) Economic condition Passive Fund X Aggressive Fund Y
.2 Recession - $ 25 - $200
.5 Stable Economy + 50 + 60
.3 Expanding Economy + 100 + 350

σ X  (-25  50)2 (.2)  (50  50)2 (.5)  (100  50)2 (.3)

 43.30

σ Y  (-200  95)2 (.2)  (60  95)2 (.5)  (350  95)2 (.3)

 193.71
 Computing the Covariance
for Investment Returns
Investment
P(XiYi) Economic condition Passive Fund X Aggressive Fund Y
.2 Recession - $ 25 - $200
.5 Stable Economy + 50 + 60
.3 Expanding Economy + 100 + 350

σ X, Y  (-25  50)(-200  95)(.2)  (50  50)(60  95)(.5)

 (100  50)(350  95)(.3)
 8250
 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
 The Sum of
Two Random Variables
 Expected Value of the sum of two random variables:

E(X  Y)  E( X)  E( Y )

 Variance of the sum of two random variables:

Var(X  Y)  σ 2
X Y  σ  σ  2σ XY
2
X
2
Y

 Standard deviation of the sum of two random variables:

σ X  Y  σ 2X  Y
 Portfolio Expected Return
and Portfolio Risk

 Portfolio expected return (weighted average

return):
E(P)  w E( X)  (1  w ) E( Y )

 Portfolio risk (weighted variability)

σ P  w 2σ 2X  (1  w )2 σ 2Y  2w(1 - w)σ XY

Where w = portion of portfolio value in asset X

(1 - w) = portion of portfolio value in asset Y
 Portfolio Example
Investment X: μX = 50 σX = 43.30
Investment Y: μY = 95 σY = 193.21
σXY = 8250

Suppose 40% of the portfolio is in Investment X and

60% is in Investment Y:
E(P)  .4 (50 )  (.6) (95 )  77

σ P  (.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
 Probability Distributions
Probability
Distributions

Ch. 5 Discrete Continuous Ch. 6

Probability Probability
Distributions Distributions

Binomial Normal

Hypergeometric Uniform

Poisson Exponential
 The Binomial Distribution
Probability
Distributions

Discrete
Probability
Distributions

Binomial

Hypergeometric

Poisson
 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
 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
 Possible Binomial Distribution
Settings

 A manufacturing plant labels items as

either defective or acceptable
 A firm bidding for contracts will either get a
contract or not
 A marketing research firm receives survey
responses of “yes I will buy” or “no I will
not”
 New job applicants either accept the offer
or reject it
 Rule of Combinations

 The number of combinations of selecting X

objects out of n objects is

n n!
  
 X  X! (n  X)!
where:
n! =n(n - 1)(n - 2) . . . (2)(1)
X! = X(X - 1)(X - 2) . . . (2)(1)
0! = 1 (by definition)
 Binomial Distribution Formula

n! X nX
P(X)  p (1-p)
X ! (n  X)!

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

with probability of success p on each trial
Example: Flip a coin four
times, let x = # heads:
X = number of ‘successes’ in sample,
n=4
(X = 0, 1, 2, ..., n)
p = 0.5
n = sample size (number of trials
or observations) 1 - p = (1 - .5) = .5
p = probability of “success” X = 0, 1, 2, 3, 4
 Example:
Calculating a Binomial Probability
What is the probability of one success in five
observations if the probability of success is .1?
X = 1, n = 5, and p = .1

n!
P( X  1)  p X (1  p)n X
X! (n  X)!
5!
 (.1)1(1  .1)51
1! (5  1)!

 (5)(.1)(.9)4
 .32805
 Binomial Distribution
 The shape of the binomial distribution depends on the
values of p and n
Mean P(X) n = 5 p = 0.1
.6
 Here, n = 5 and p = .1 .4
.2
0 X
0 1 2 3 4 5

.6
P(X) n = 5 p = 0.5
 Here, n = 5 and p = .5
.4
.2
0 X
0 1 2 3 4 5
 Binomial Distribution
Characteristics

 Mean
μ  E(x)  np
 Variance and Standard Deviation

σ  np(1 - p)
2

σ  np(1 - p)
Where n = sample size
p = probability of success
(1 – p) = probability of failure
 Binomial Characteristics
Examples
μ  np  (5)(.1)  0.5
Mean P(X) n = 5 p = 0.1
.6
.4
σ  np(1 - p)  (5)(.1)(1  .1) .2
0 X
 0.6708
0 1 2 3 4 5

μ  np  (5)(.5)  2.5 P(X) n = 5 p = 0.5

.6
.4
σ  np(1 - p)  (5)(.5)(1  .5) .2
0 X
 1.118
0 1 2 3 4 5
 Using Binomial Tables
n = 10
x … p=.20 p=.25 p=.30 p=.35 p=.40 p=.45 p=.50
0 … 0.1074 0.0563 0.0282 0.0135 0.0060 0.0025 0.0010 10
1 … 0.2684 0.1877 0.1211 0.0725 0.0403 0.0207 0.0098 9
2 … 0.3020 0.2816 0.2335 0.1757 0.1209 0.0763 0.0439 8
3 … 0.2013 0.2503 0.2668 0.2522 0.2150 0.1665 0.1172 7
4 … 0.0881 0.1460 0.2001 0.2377 0.2508 0.2384 0.2051 6
5 … 0.0264 0.0584 0.1029 0.1536 0.2007 0.2340 0.2461 5
6 … 0.0055 0.0162 0.0368 0.0689 0.1115 0.1596 0.2051 4
7 … 0.0008 0.0031 0.0090 0.0212 0.0425 0.0746 0.1172 3
8 … 0.0001 0.0004 0.0014 0.0043 0.0106 0.0229 0.0439 2
9 … 0.0000 0.0000 0.0001 0.0005 0.0016 0.0042 0.0098 1
10 … 0.0000 0.0000 0.0000 0.0000 0.0001 0.0003 0.0010 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
 Using PHStat

 Select PHStat / Probability & Prob. Distributions / Binomial…

 Using PHStat
(continued)

 Enter desired values in dialog box

Here: n = 10
p = .35

Output for X = 0
to X = 10 will be
generated by PHStat

Optional check boxes

for additional output
PHStat Output

P(X = 3 | n = 10, p = .35) = .2522

P(X > 5 | n = 10, p = .35) = .0949

 The Poisson Distribution
Probability
Distributions

Discrete
Probability
Distributions

Binomial

Hypergeometric

Poisson
 The Poisson Distribution

 Apply the Poisson Distribution when:

 You wish to count the number of times an event
occurs in a given area of opportunity
 The probability that an event occurs in one area of
opportunity is the same for all areas of opportunity
 The number of events that occur in one area of
opportunity is independent of the number of events
that occur in the other areas of opportunity
 The probability that two or more events occur in an
area of opportunity approaches zero as the area of
opportunity becomes smaller
 The average number of events per unit is  (lambda)
 Poisson Distribution Formula

 x
e 
P( X) 
X!
where:
X = number of successes per unit
 = expected number of successes per unit
e = base of the natural logarithm system (2.71828...)
 Poisson Distribution
Characteristics

 Mean
μλ
 Variance and Standard Deviation

σ λ
2

σ λ
where  = expected number of successes per unit
 Using Poisson Tables


X 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90

0 0.9048 0.8187 0.7408 0.6703 0.6065 0.5488 0.4966 0.4493 0.4066

1 0.0905 0.1637 0.2222 0.2681 0.3033 0.3293 0.3476 0.3595 0.3659
2 0.0045 0.0164 0.0333 0.0536 0.0758 0.0988 0.1217 0.1438 0.1647
3 0.0002 0.0011 0.0033 0.0072 0.0126 0.0198 0.0284 0.0383 0.0494
4 0.0000 0.0001 0.0003 0.0007 0.0016 0.0030 0.0050 0.0077 0.0111
5 0.0000 0.0000 0.0000 0.0001 0.0002 0.0004 0.0007 0.0012 0.0020
6 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0001 0.0002 0.0003
7 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

Example: Find P(X = 2) if  = .50

e   X e 0.50 (0.50)2
P( X  2)    .0758
X! 2!
 Graph of Poisson Probabilities
0.70

Graphically: 0.60

 = .50 0.50

= P(x) 0.40

X 0.50
0.30
0 0.6065
0.20
1 0.3033
2 0.0758 0.10

3 0.0126 0.00
0 1 2 3 4 5 6 7
4 0.0016
5 0.0002 x
6 0.0000
P(X = 2) = .0758
7 0.0000
 Poisson Distribution Shape

 The shape of the Poisson Distribution

depends on the parameter  :

0.70
 = 0.50 0.25
 = 3.00
0.60
0.20
0.50

0.15
0.40

P(x)
P(x)

0.30 0.10

0.20
0.05
0.10

0.00 0.00
0 1 2 3 4 5 6 7 1 2 3 4 5 6 7 8 9 10 11 12

x x
 Poisson Distribution
in PHStat
 Select:
PHStat / Probability & Prob. Distributions / Poisson…
 Poisson Distribution
in PHStat
(continued)

 Complete dialog box entries and get output …

P(X = 2) = 0.0758
 Chapter Summary

 Addressed the probability of a discrete random

variable
 Defined covariance and discussed its
application in finance
 Discussed the Binomial distribution
 Discussed and Reviewed the Poisson
distribution