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Chapter 4:
Random Variables and
Probability Distributions
Hildebrand, Ott and Gray
Basic Statistical Ideas for Managers
Second Edition
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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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
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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Section 4.1
Random Variable: Basic Ideas
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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4.1 Random Variable: Basic Ideas
• 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
General notation
• 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.)
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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4.1 Random Variable: Basic Ideas
• Types of random variables: discrete and continuous.
•
A discrete r.v. has a countable number of values.
-1
•
0
Real no. line (Closing Price of a Security)
+1
A continuous r.v. has an uncountable number of
values.
8:45
Real no. line (Time at which you arrive for class)
9:00
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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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.
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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Section 4.2
Probability Distribution:
Discrete Random Variables
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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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
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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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).
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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4.2 Probability Distribution:
Discrete Random Variables
Probability Distribution of Y
Py(Y)
0.6
-1
(5/6) (5/6) (5/6)
1
3 (1/6) (5/6) (5/6) = 75/216
2
3 (1/6) (1/6) (5/6) = 15/216
3
(1/6) (1/6) (1/6)
0.5
= 125/216
= 1/216
0.4
P(Y=y)
y
0.3
0.2
Sum = 1
0.1
0.0
-1
0
1
y
2
3
P(Y ≤ 1) = P(Y=-1) + P(Y=+1) = 200/216
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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4.2 Probability Distribution:
Discrete Random Variables
• Requirements for a probability distribution:
•
PY(y) ≥ 0, for all y
•
∑ P ( y ) =1
Y
all y
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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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.
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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4.2 Probability Distribution:
Discrete Random Variables
Recall
Probability Distribution of Y
0.6
Py(Y)
-1
125/216
1
75/216
2
15/216
3
1/216
Sum
1
0.5
0.4
P(Y=y)
y
0.3
0.2
0.1
0.0
-1
0
1
y
2
3
FY(y)
For y < - 1, FY(y) = 0
For -1 ≤ y < 1, FY(y) = 125/216
For 1 ≤ y < 2, FY(y) = 200/216
1
215/216
200/216
125/216
For 2 ≤ y < 3, FY(y) = 215/216
For y ≥ 3, FY(y) = 1
-1
0
1
2
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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
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Section 4.3
Expected Value
and Standard Deviation
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4.3 Expected Value and Standard Deviation
• The expected value (or mean) of a discrete random
variable Y is
µY = E[Y] = ∑ y PY(y)
Interpretation: Long-run average
Example (Game of Chuck-A-Luck):
Find the expected value of Y and provide an interpretation.
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4.3 Expected Value and Standard Deviation
y PY(y)
y
PY(y)
-1
125/216
-(125/216)
1
75/216
75/216
2
15/216
30/216
3
1/216
3/216
E(Y) = -(17/216) = - $.079
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? No!
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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4.3 Expected Value and Standard Deviation
• The variance of a discrete random variable Y is
2
2
σ Y = V(Y) = ∑(y - µ Y ) PY ( y)
• An alternative computational formula for the variance is
2
2
2
σ Y = ∑ y P Y ( y) - µ y
all y
• The standard deviation σY is the positive square root of
the variance.
σ Y = σ Y2
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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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.
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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4.3 Expected Value and Standard Deviation
Suppose each player wagers $1. Let X be the net amount
you win. Find σX.
PX(x)
.5
-1
0
1
x
E(X) = 0
V(X) = (-1-0)2 (.5) + (1-0)2 (.5) = 1 ($)2
σX = $1
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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4.3 Expected Value and Standard Deviation
Suppose each player wagers $10. Let Y be the net amount
you win. Find σY.
PY(y)
.5
-10
0
10
y
E(Y) = 0
V(Y) = (-10-0)2 (.5) + (10-0)2 (.5) = 100 ($)2
σY = $10
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4.3 Expected Value and Standard Deviation
• The standard deviation measures the
dispersion of the probability distribution
relative to the expected value.
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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4.3 Expected Value and Standard Deviation
Example (Game of Chuck-A-Luck):
Find the standard deviation of Y.
(y -µ)2

 17  
 − 1 −  − 216  



  17 
1 −  − 216 

 
2
2
  17  
2 −  − 216  

 
  17 

3 −  −
  216 
2
2
PY(y)
(y - µ)2 PY(y)
125
216
 199   125 
−
 

 216   216 
75
216
 233   75 

 

 216   216 
2
2
2
15
216
 449   15 

 

 216   216 
1
216
 665   1 

 

 216   216 
2
1.239 ($)
2
σ = V(Y) = $1.11
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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4.3 Expected Value and Standard Deviation
Example (Game of Chuck-A-Luck):
Find the probability that Y takes on a value within one
standard 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
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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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.
x
0%
20%
P(X=x)
0.50
0.50
y
-10%
10%
20%
30%
P(Y=y)
0.15
0.35
0.35
0.15
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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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.
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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4.3 Expected Value and Standard Deviation
b. Calculate the standard deviation of the percentage returns for
investments 1and 2.
2
( xi − µ X )
V(X) = ∑
i =1
= (0 –
2
PX ( xi )
10)2(.5)
+ (20 –10)2(.5) = 100 ⇒ σX = 10%
4
( yi − µY ) PY ( yi )
V(Y) = ∑
i =1
= (-10 – 13.5)2( .15 ) + (10 – 13.5)2( .35 ) + (20 – 13.5)2(.35 )
+ (30 – 13.5)2( .15)
2
= 142.75 ⇒ σY = 11.95%
Using the criterion of minimizing risk, which investment would you
choose?
Answer: Choose X.
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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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?
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Section 4.4
Joint Probability Distributions
and Independence
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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)
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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.
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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4.4 Joint Probability Distributions
and Independence
• The joint probability distribution of X and Y is given in the
table below.
y
x
-10
10
20
30
0
.10
.20
.15
.05
20
.05
.15
.20
.10
• 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
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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4.4 Joint Probability Distributions
and Independence
• General requirements for a discrete joint probability
distribution:
•
•
PXY(x,y) ≥ 0
∑∑ P
XY
y
Probability can’t be negative.
( x, y ) = 1
All the probabilities must add to 1.
x
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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 PX ( x) = ∑ PXY ( x, y) for each value of X.
all y
• Similarly, PY ( y) = ∑ PXY ( x, y) for each value of Y.
all x
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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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:
y
x
-10
10
20
30
0
.10
.20
.15
.05
20
.05
.15
.20
.10
P[X = 0] = ∑ P[X = 0, Y = y]
= .10 +.20 +.15 +.05 = .50
y
P[X = 20] = ∑ P[X = 20, Y = y] = .05 +.15 + .20 +.10 =.50
y
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4.4 Joint Probability Distributions
and Independence
The marginal probability distribution of X is:
x
P[X=x]
0
.5
20
.5
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4.4 Joint Probability Distributions
and Independence
b. Find the marginal distribution of the return (Y) for security 2.
y
x
0
-10
10
20
30
.10
.20
.15
.05
20
.05
.15
.20
.10
P[Y=y]
.15
.35
.35
.15
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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.
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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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 PX| Y ( x | y) =
PXY ( x, y)
for each value of X.
PY ( y)
• Similarly, the conditional probability distribution of Y,
given that X=x, is
PY|X ( y | x) =
PXY ( x, y)
PX ( x)
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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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:
y
x
0
20
P[Y=y]
-10
10
20
30
P[X=x]
.10
.20
.15
.05
.50
.05
.15
.20
.10
.50
.15
.35
.35
.15
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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4.4 Joint Probability Distributions
and Independence
• General expression:
P[Y = y | X = 20] =
P[Y = y and X = 20]
P[X = 20]
• In table format:
y
-10
10
20
30
P[Y=y|X=20]
.05/.50
=.10
.15/.50
=.30
.20/.50
=.40
.10/.50
=.20
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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.
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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?
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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).
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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]?
.10
.15
Therefore, the returns from securities 1 and 2 are not
independent.
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Section 4.5
Covariance and Correlation
of Random Variables
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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
Cov(X, Y) = ∑ ∑ ( x - µ X )( y - µ Y ) PXY ( x, y)
x
y
• An alternative computational formula is
Cov(X, Y) = ∑ ∑ x y PXY ( x, y) - µ X µ Y
x
y
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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4.5 Covariance and Correlation
of Random Variables
Example: Portfolio analysis
g. Find the covariance between the returns for securities 1 and 2.
y
x
-10
10
20
30
0
.10
.20
.15
.05
20
.05
.15
.20
.10
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)
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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.
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4.5 Covariance and Correlation
of Random Variables
• Standardizing the covariance
• The coefficient of correlation is denoted by ρ.
• Need to find ρ =
Cov( X , Y )
σ Xσ Y
• Measures the strength of the linear relationship between
X and Y.
• Property: −1 ≤
ρ ≤ +1
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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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.
ρ=
Cov( X , Y )
σ XσY
25
(10)(11.95)
=
= 0.21
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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2005 Brooks/Cole, a division of Thomson Learning, Inc.
51
4.5 Covariance and Correlation
of Random Variables
Note: If X and Y are independent,
then Cov(X, Y) = 0 and ρ = 0.
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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2005 Brooks/Cole, a division of Thomson Learning, Inc.
4.5 Covariance and Correlation
of Random Variables
Example: Consider the following joint distribution.
Find the covariance between X and Y.
Are X and Y independent?
Y
X
• E(X) = 0, by symmetry of PX(x)
1
4
-2
0
1/4
-1
1/4
0
1
1/4
0
2
0
1/4
• 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
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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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.
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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2005 Brooks/Cole, a division of Thomson Learning, Inc.
54
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%
55
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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©
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 =
79.39 = 8.91%
vs.
σX = 10% and σY = 11.95%
•
Moral: “Don’t put all your eggs in one basket.”
56
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
Copyright
©
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.
%Asset1
E(X)
%Asset2
E(Y)
E(P)
Cov(X,Y)
V(P)
SD(P)
1
10%
0
13.50%
10.00%
25
100
10.00%
0.9
10%
0.1
13.50%
10.35%
25
86.92803
9.32%
0.8
10%
0.2
13.50%
10.70%
25
77.7121
8.82%
0.7
10%
0.3
13.50%
11.05%
25
72.35223
8.51%
0.6
10%
0.4
13.50%
11.40%
25
70.8484
8.42%
0.5
10%
0.5
13.50%
11.75%
25
73.20036
8.56%
0.4
10%
0.6
13.50%
12.10%
25
79.4089
8.91%
0.3
10%
0.7
13.50%
12.45%
25
89.47323
9.46%
0.2
10%
0.8
13.50%
12.80%
25
103.3936
10.17%
0.1
10%
0.9
13.50%
13.15%
25
121.17
11.01%
0
10%
1
13.50%
13.50%
25
142.8025
11.95%
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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2005 Brooks/Cole, a division of Thomson Learning, Inc.
57
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
14%
12%
Expected Value
10%
8%
Ignore?
6%
4%
2%
0%
0%
2%
4%
6%
8%
10%
12%
Standard Deviation
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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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)
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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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
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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60
Summary of Chapter 4
One Random Variable
• General Case – Discrete
Probability Distribution
y
PY(y)
•
•
•
•
•
•
µ = E(Y) = ΣyPY(y)
σ2 = V(Y) = Σ(y - µ)2PY(y)
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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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.
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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2005 Brooks/Cole, a division of Thomson Learning, Inc.
Summary of Chapter 4
• Joint dist. → Marginal dist.
• Joint dist.
&
Marginal dist.
Conditional dist.
P(X=x|Y=y) = P(X=x,Y=y)/P(Y=y)
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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2005 Brooks/Cole, a division of Thomson Learning, Inc.
63
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
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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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
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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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)
Hildebrand, Ott & Gray, Basic Statistical Ideas for Managers, 2nd edition, Chapter 4
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2005 Brooks/Cole, a division of Thomson Learning, Inc.
66