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1Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Chapter 4:
Random Variables and
Probability Distributions
Hildebrand, Ott and Gray
Basic Statistical Ideas for Managers
Second Edition
2Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Learning Objectives for Ch. 4
• Defining a random variable and probability distribution
• Determining the expected value of a random variable
• Determining the standard deviation of a random variable
• Defining the joint probability distribution of two random variables
• Determining marginal probability distributions
• Determining conditional probability distributions
• Determining statistical independence
• Determining the covariance and correlation
• Determining the expected value and standard deviation of sums of random variables
3Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Section 4.1
Random Variable: Basic Ideas
4Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
• Concept: A random variable assigns numerical values to the possible outcomes of a random experiment.
Example: Today's closing price of a security relative to yesterday.
Let Y=1, if price goes up; = 0, if price is unchanged; = -1, if price goes down.
Y
Up +1 y1
Same 0 y2
Down -1 y3
• Notation: Y,X,Z (Upper case letters denote a r.v.)
• Values are determined by chance.
• Notation: y,x,z (Lower case letters denote a general value of a r.v.)
4.1 Random Variable: Basic Ideas
General notation
5Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
• Types of random variables:
4.1 Random Variable: Basic Ideas
-1 0 +1
8:45 9:00
discrete and continuous.
• A discrete r.v. has a countable number of values.
Real no. line (Closing Price of a Security)
• A continuous r.v. has an uncountable number of values.
Real no. line (Time at which you arrive for class)
6Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.1 Random Variable: Basic Ideas
Example (Game of Chuck-A-Luck):
Three dice are rolled. You wager on any number from 1 to 6, say a ‘4’. If any one of the three dice comes up with a ‘4’, you win the amount of your wager. You also get back your original stake. It could happen that more than one dice comes up with a ‘4’. For example, if you had a $1 bet on a ‘4’, and each of the three dice came up with a ‘4’, you would win $3. If two of the three dice came up with a ‘4’, you would win $2.
Let Y be the net amount you win in one play of the game.
Possible values of Y: -1, 1, 2, 3.
7Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Section 4.2
Probability Distribution:
Discrete Random Variables
8Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.2 Probability Distribution:
Discrete Random Variables
• A probability distribution for a discrete r.v. Y is the collection of possible values of Y (denoted by y) and their probabilities, P[Y = y].
• Notation: P[Y = y] = PY(y)
• Possible representations of a probability distribution:
• Table
• Graph
• Mathematical Formula
9Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.2 Probability Distribution:
Discrete Random Variables
Example (Game of Chuck-A-Luck):
• Y is the net amount won in one play of the game.
• Possible values of Y: -1, 1, 2, 3
Find the probability distribution of Y
and also find P(Y ≤ 1).
10Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.2 Probability Distribution:
Discrete Random Variables
y
P(Y=y)
3210-1
0.6
0.5
0.4
0.3
0.2
0.1
0.0
Probability Distribution of Y
P(Y ≤ 1) = P(Y=-1) + P(Y=+1) = 200/216
Sum = 1
(1/6) (1/6) (1/6) = 1/2163
3 (1/6) (1/6) (5/6) = 15/2162
3 (1/6) (5/6) (5/6) = 75/2161
(5/6) (5/6) (5/6) = 125/216-1
Py(Y)y
11Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.2 Probability Distribution:
Discrete Random Variables
• Requirements for a probability distribution:
• PY(y) ≥ 0, for all y
• 1)(y all
=∑ yPY
12Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.2 Probability Distribution:
Discrete Random Variables
• The cumulative distribution function (cdf) of a random variable Y is denoted FY(y).
• Gives the probability that Y is less than or equal to y,
for all values of y.
• Need to find P(Y ≤ y).
Example (Game of Chuck-A-Luck):
Find the cumulative distribution function of Y
and the P(Y ≤ 1) by using the cdf.
13Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Recall
4.2 Probability Distribution:
Discrete Random Variables
1Sum
1/2163
15/2162
75/2161
125/216-1
Py(Y)y
yP(Y=y)
3210-1
0.6
0.5
0.4
0.3
0.2
0.1
0.0
Probability Distribution of Y
For y < - 1, FY(y) = 0
For -1 ≤ y < 1, FY(y) = 125/216
For 1 ≤ y < 2, FY(y) = 200/216
For 2 ≤ y < 3, FY(y) = 215/216
For y ≥ 3, FY(y) = 1
FY(y)
1215/216
200/216
125/216
-1 0 1 2 3 y
14Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.2 Probability Distribution:
Discrete Random Variables
• Find P(Y ≤ 1)
• P(Y ≤ 1) = FY(1) = 200/216
• Same as before:
P(Y ≤ 1) = PY(-1) + PY(1) = 200/216
15Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Section 4.3
Expected Value
and Standard Deviation
16Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.3 Expected Value and Standard Deviation
• The expected value (or mean) of a discrete random variable Y is
Interpretation: Long-run average
µY = E[Y] = ∑ y PY(y)
Example (Game of Chuck-A-Luck):
Find the expected value of Y and provide an interpretation.
17Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.3 Expected Value and Standard Deviation
y PY(y)
-1 125/216
1 75/216
2 15/216
3 1/216
y PY(y)
-(125/216)
75/216
30/216
3/216
Interpretation: In the long run, you expect to lose about 8 cents out of every dollar wagered.
Suppose a game is defined to be fair if its expected value is 0.
Is the game of Chuck-A-Luck fair?
E(Y) = -(17/216) = - $.079
No!
18Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.3 Expected Value and Standard Deviation
• The variance of a discrete random variable Y is
y)(P)-(yV(Y) Y
2
Y
2
µσ ∑==Y
2
yY
2
y all
2
-y)(Py µσ ∑=Y
2
Y Yσ σ=
• An alternative computational formula for the variance is
• The standard deviation σY is the positive square root of the variance.
19Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.3 Expected Value and Standard Deviation
• A distribution that is compact has a small standard deviation, while a distribution that is dispersed has a large standard deviation.
Example: Matching fair coins
Two players each toss a coin. The equally likely outcomes are:
(H,H), (H,T) (T,H), (T,T).
One player calls “odds” or “evens.”
If the player calls “odds” and the coins differ, the player wins.
If the player calls “evens” and the coins match, the player wins.
20Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Suppose each player wagers $1. Let X be the net amount
you win. Find σX.
4.3 Expected Value and Standard Deviation
PX(x)
.5
-1 0 1 x
E(X) = 0
V(X) = (-1-0)2 (.5) + (1-0)2 (.5) = 1 ($)2
σX = $1
21Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Suppose each player wagers $10. Let Y be the net amount
you win. Find σY.
4.3 Expected Value and Standard Deviation
PY(y)
.5
-10 0 10 y
E(Y) = 0
V(Y) = (-10-0)2 (.5) + (10-0)2 (.5) = 100 ($)2
σY = $10
22Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.3 Expected Value and Standard Deviation
• The standard deviation measures the dispersion of the probability distribution relative to the expected value.
23Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.3 Expected Value and Standard Deviation
Example (Game of Chuck-A-Luck):
Find the standard deviation of Y.
(y -µ)2 PY(y) (y - µ)2 PY(y)
2
22
22
22
22
($)239.1
216
1
216
665
216
1
216
173
216
15
216
449
216
15
216
172
216
75
216
233
216
75
216
171
216
125
216
199
216
125
216
171
−−
−−
−−
−
−−−
V(Y) $1.11σ = =
24Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.3 Expected Value and Standard Deviation
Example (Game of Chuck-A-Luck):
Find the probability that Y takes on a value within onestandard deviation of the mean.
P[ µ − σ ≤ Y ≤ µ + σ ]
= P[ -.08 − 1.11 ≤ Y ≤ -.08 + 1.11 ]
= P[ -1.19 ≤ Y ≤ 1.03 ] = P(Y=-1) + P(Y=1)
= 200/216
25Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.3 Expected Value and Standard Deviation
Example: A financial analyst is trying to choose between two investments. Typical returns on the first investment, denoted by X, are: 0% and 20%, with probabilities 0.5 for each. Typical returns on the second investment, denoted by Y, are: -10%, 10%, 20%, and 30%, with probabilities 0.15, 0.35, 0.35 and 0.15, respectively. These probabilities have been subjectively determined.
The probability distributions of X and Y follow.
0.500.50P(X=x)
20%0%x
0.150.350.350.15P(Y=y)
30%20%10%-10%y
26Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.3 Expected Value and Standard Deviation
a. What is the expected return for investment 1?
What is the expected return for investment 2?
Obviously, E(X) = 10% { The probability distribution of X is symmetric about 10. }
E(Y) = -10(0.15) + 10(0.35) + 20(0.35) + 30(0.15)
= 13.5%
Using the expected value criterion, which investment would you choose?
Answer: Choose Y.
27Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.3 Expected Value and Standard Deviation
b. Calculate the standard deviation of the percentage returns for investments 1and 2.
V(X) =
= (0 – 10)2(.5) + (20 –10)2(.5) = 100 ⇒ σX = 10%
V(Y) =
= (-10 – 13.5)2( .15 ) + (10 – 13.5)2( .35 ) + (20 – 13.5)2(.35 )
+ (30 – 13.5)2( .15)
= 142.75 ⇒ σY = 11.95%
Using the criterion of minimizing risk, which investment would you choose?
Answer:
22
1
( ) ( )i X X i
i
x P xµ=
−∑
42
1
( ) ( )i Y Y i
i
y P yµ=
−∑
Choose X.
28Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.3 Expected Value and Standard Deviation
c. Using both the mean and standard deviation, which investment would you choose?
Are you a risk seeker? Risk averse? Risk neutral?
29Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Section 4.4
Joint Probability Distributions
and Independence
30Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.4 Joint Probability Distributions
and Independence
• The joint probability distribution for two discrete random variables X and Y, denoted by PXY (x,y), is the possible values of (x, y) together with their joint probabilities.
• Notation: PXY (x,y) = P(X = x, Y = y)
31Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.4 Joint Probability Distributions
and Independence
Example: Portfolio analysis
$1 is available to invest and there are two securities under consideration. How should one allocate the $1 between the two securities? The return on the first security, denoted by X, has two possible values: 0% and 20%. The return on the second security, denoted by Y, has four possible values: -10%, 10%, 20% and 30%.
• The joint probability distribution follows.
32Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.4 Joint Probability Distributions
and Independence
• The joint probability distribution of X and Y is given in the table below.
• Let P[X = x, Y = y] denote the joint probability that the random variable X takes on the value x and, at the same time, the random variable Y takes on the value y.
• Thus, P[X = 0, Y = -10] = .10
.10.20.15.0520
.05.15.20.100
302010-10y
x
33Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.4 Joint Probability Distributions
and Independence
• General requirements for a discrete joint probability distribution:
• PXY(x,y) ≥ 0 Probability can’t be negative.
• All the probabilities must add to 1. 1),(y
=∑∑ yxPx
XY
34Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.4 Joint Probability Distributions
and Independence
• The marginal probability distribution of X, denoted
PX(x), is the probability distribution of X by itself.
• Need to find for each value of X.
• Similarly, for each value of Y.
y) x,(P)(P XYy all
X ∑=x
y) x,(Py)(P XYxall
Y ∑=
35Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.4 Joint Probability Distributions
and Independence
Example: Portfolio analysis
a. Find the marginal distribution of the return (X) for security 1.
The joint probability distribution is given below:
= .10 +.20 +.15 +.05 = .50
= .05 +.15 + .20 +.10 =.50
.10.20.15.0520
.05.15.20.100
302010-10
y
x
y]Y0,P[X0]P[Xy
==∑==
y]Y0,2P[X0]2P[Xy
==∑==
36Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.4 Joint Probability Distributions
and Independence
The marginal probability distribution of X is:
.520
.50
P[X=x]x
37Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.4 Joint Probability Distributions
and Independence
b. Find the marginal distribution of the return (Y) for security 2.
.10.20.15.0520
P[Y=y]
.05.15.20.100
302010-10
y
x
.15 .35 .35 .15
38Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.4 Joint Probability Distributions
and Independence
c. What are the expected returns for securities 1 and 2 ?
Previously we saw that E(X) = 10% and E(Y) = 13.5%.
d. What are the standard deviations of the returns for securities 1 and 2 ?
Previously we saw that σX = 10% and σY = 11.95%.
• Only need the marginal distributions to find the expected values and standard deviations of X and Y, respectively.
39Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.4 Joint Probability Distributions
and Independence
• The conditional probability distribution of X given that Y = y is denoted PX|Y (x | y).
• Need to find for each value of X.
• Similarly, the conditional probability distribution of Y, given that X=x, is
y)(P
y) x,(Py)|x(P
Y
XYY|X =
x)(P
y) x,(Px)|y(P
X
XYX|Y =
40Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.4 Joint Probability Distributions
and Independence
Example: Portfolio analysis
e. Find the conditional probability distribution of the return for security 2, if the return for security 1 is 20%.
The joint and marginal probability distributions are given below:
.35
.20
.15
20
.50.10.15.0520
.15.35.15
.50.05.20.100
3010-10
y
x
P[X=x]
P[Y=y]
41Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.4 Joint Probability Distributions
and Independence
• General expression:
• In table format:
20]P[X
20]X andy P[Y20]X|yP[Y
===
===
.10/.50=.20
.20/.50=.40
.15/.50=.30
.05/.50=.10
P[Y=y|X=20]
302010-10y
42Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.4 Joint Probability Distributions
and Independence
• How many conditional distributions of Y are there?
One for each value of X.
• How many conditional distributions of X are there?
One for each value of Y.
43Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.4 Joint Probability Distributions
and Independence
• Sequence of generating the distributions
Joint Probability Distribution
Marginal Distributions
Conditional Distributions
• Can we go the other way?
44Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.4 Joint Probability Distributions
and Independence
• Random variables X and Y are independent if and only if
PXY (x, y) = PX (x) PY(y)
for all (x, y).
• Equivalently, X and Y are independent if and only if
PY|X (y | x) = PY (y)
for all (x, y).
45Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.4 Joint Probability Distributions
and Independence
Example: Portfolio Analysis
f. Are the returns from securities 1 and 2 independent?
Is P[Y = -10 | X = 20] = P[Y = -10]?
Therefore, the returns from securities 1 and 2 are not
independent.
.10 .15
46Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Section 4.5
Covariance and Correlation
of Random Variables
47Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.5 Covariance and Correlation
of Random Variables
• Concept: Covariance measures the degree of
dependence between 2 random variables
• The covariance of X and Y is denoted by Cov(X, Y).
• Need to find
• An alternative computational formula is
y) x,(P)-y)(-x( Y) Cov(X, XYYXyx
µµ∑∑=
YXXYyx
-y) x,(Pyx Y) Cov(X, µµ∑∑=
48Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.5 Covariance and Correlation
of Random Variables
Example: Portfolio analysis
g. Find the covariance between the returns for securities 1 and 2.
Cov(X,Y) = ∑xy PXY(x, y) - µx µy = 160 – (10)(13.5) = 25 {Units ?}
where ∑xy PXY (x, y) = 0 + 0 + 0 + 0
+ (20)(-10)(.05) + (20)(10)(.15)
+ (20)(20)(.20) + (20)(30)(.10)
.10.20.15.0520
.05.15.20.100
302010-10
y
x
49Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.5 Covariance and Correlation
of Random Variables
• Since the covariance is positive, the returns on the two securities tend to move in same direction.
• There is a tendency for both returns to be below their means at the same time or above their means at the same time.
50Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.5 Covariance and Correlation
of Random Variables
• Standardizing the covariance
• The coefficient of correlation is denoted by ρ.
• Need to find
• Measures the strength of the linear relationship between X and Y.
• Property:
( , )
X Y
Cov X Yρ
σ σ=
1 1ρ− ≤ ≤ +
51Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.5 Covariance and Correlation
of Random Variables
Example: Portfolio analysis
h. Find the correlation between the returns for securities 1 and 2.
= 0.21
( , )
X Y
Cov X Yρ
σ σ=
25
(10)(11.95)=
52Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.5 Covariance and Correlation
of Random Variables
Note: If X and Y are independent,
then Cov(X, Y) = 0 and ρ = 0.
53Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Example: Consider the following joint distribution.
Find the covariance between X and Y.
Are X and Y independent?
4.5 Covariance and Correlation
of Random Variables
1/402
01/41
01/4-1
1/40-2
41
Y
X
• E(X) = 0, by symmetry of PX(x)
• Cov(X, Y) = ∑ ∑ xy PXY(x, y) - µX µY = 0
• ρ = 0
• Are X and Y independent?
No, because PXY(-2,1) ≠ PX(-2)PY(1)
0 ≠ (1/4)(1/2)
• Are X and Y related?
Yes, because Y = X2
54Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.5 Covariance and Correlation
of Random Variables
Extensions:
For any two constants b and c,
E(bX + cY) = bE(X) + cE(Y)
V(bX + cY) = b2 V(X) + c2 V(Y) + 2 bc Cov(X,Y)
Equivalently,
V(bX + cY) = b2 V(X) + c2 V(Y) + 2 bc σxσy ρ.
55Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.5 Covariance and Correlation
of Random Variables
Example: Portfolio analysis
i. Suppose $0.40 is invested in security 1 and $0.60 in security 2.
What is the expected value of the percentage return on the portfolio?
How does this compare to the expected value for each security?
E[.4 X + .6 Y] = .4E(X) + .6E(Y) = .4(10%) + .6(13.5%) = 12.1%
vs.
E(X) = 10% and E(Y) = 13.5%
56Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.5 Covariance and Correlation
of Random Variables
Example: Portfolio analysis
j. What is the standard deviation of the percentage return on the portfolio? How does this compare to the risks of the individual securities?
V[.4 X + .6 Y] = (.4)2V(X) + (.6)2V(Y) + 2(.4)(.6)(25)
= 79.39 (%)2
⇒ σPORTFOLIO = = 8.91%
vs.
σX = 10% and σY = 11.95%
• Moral: “Don’t put all your eggs in one basket.”
79.39
57Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.5 Covariance and Correlation
of Random Variables
Example: Portfolio analysis
k. Sketch the investment frontier and specify your location as an investor.
11.95%142.80252513.50%13.50%110%0
11.01%121.172513.15%13.50%0.910%0.1
10.17%103.39362512.80%13.50%0.810%0.2
9.46%89.473232512.45%13.50%0.710%0.3
8.91%79.40892512.10%13.50%0.610%0.4
8.56%73.200362511.75%13.50%0.510%0.5
8.42%70.84842511.40%13.50%0.410%0.6
8.51%72.352232511.05%13.50%0.310%0.7
8.82%77.71212510.70%13.50%0.210%0.8
9.32%86.928032510.35%13.50%0.110%0.9
10.00%1002510.00%13.50%010%1
SD(P)V(P)Cov(X,Y)E(P)E(Y)%Asset2E(X)%Asset1
58Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.5 Covariance and Correlation
of Random Variables
Example: Portfolio analysis
The investment frontier is shown below.
Where are you?
Graph of Expected Value vs. Standard Deviation
0%
2%
4%
6%
8%
10%
12%
14%
0% 2% 4% 6% 8% 10% 12%
Standard Deviation
Exp
ecte
d V
alu
e
Ignore?
59Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.5 Covariance and Correlation
of Random Variables
Extensions (cont'd):
• For X - Y, b = +1 and c = -1
E(X - Y) = E(X) – E(Y)
V(X - Y) = V(X) + V(Y) -2Cov(x,y)
• If X and Y are independent,
V(bX + cY) = b2V(X) + c2V(Y)
60Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Keywords: Chapter 4
• Random variable
• Probability distribution
• Cumulative distribution function
• Expected value
• Standard deviation
• Joint probability distribution
• Marginal probability distribution
• Conditional probability distribution
• Covariance
• Correlation
• Expected value for sums of random variables
• Variance for sums of random variables
61Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Summary of Chapter 4
One Random Variable
• General Case – Discrete
Probability Distribution
µ = E(Y) = ΣyPY(y)
σ2 = V(Y) = Σ(y - µ)2PY(y)
y PY(y)
• •• •• •
62Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Summary of Chapter 4
Two or More Random Variables
• General Case – Discrete
Joint Probability Distribution
Y • • •X
• Joint probabilities• are in the cells of • the table.
63Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Summary of Chapter 4
• Joint dist. → Marginal dist.
• Joint dist.
& Conditional dist.Marginal dist.
P(X=x|Y=y) = P(X=x,Y=y)/P(Y=y)
64Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Summary of Chapter 4
• Cov(X,Y) : Do X and Y tend to vary together?
If there is a tendency for both Y and X to be below (or above) their means at the same time, then Cov(X,Y) > 0.
• Corr(X,Y) = Cov(X,Y)/σXσY
65Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Summary of Chapter 4
• X and Y are independent if
P(X = x, Y = y) = P(X = x) P(Y = y)
• If X and Y are independent,
Cov(X,Y) = 0
66Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc.
Summary of Chapter 4
• E(aX + bY) = aE(X) + bE(Y)
• V(aX + bY) = a2V(X) + b2 V(Y) + 2 ab Cov(X,Y)
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