Random Variables and Probability Distributions...4.2 Probability Distribution: Discrete Random Variables • Requirements for a probability distribution: • PY(y) ≥0, for all y

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


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


Section 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

• 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


• Types of random variables:


-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)



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.


Section 4.2

Probability Distribution:

Discrete Random Variables


4.2 Probability Distribution:


• 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





• 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).




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




• Requirements for a probability distribution:

• PY(y) ≥ 0, for all y

• 1)(y all

=∑ yPY




• 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).


Find the cumulative distribution function of Y

and the P(Y ≤ 1) by using the cdf.


Recall



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




• Find P(Y ≤ 1)

• P(Y ≤ 1) = FY(1) = 200/216

• Same as before:

P(Y ≤ 1) = PY(-1) + PY(1) = 200/216


Section 4.3

Expected Value

and Standard Deviation


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)


Find the expected value of Y and provide an interpretation.



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!



• 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.



• 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.


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


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



• The standard deviation measures the dispersion of the probability distribution relative to the expected value.




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 = =




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



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



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.



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.



c. Using both the mean and standard deviation, which investment would you choose?

Are you a risk seeker? Risk averse? Risk neutral?


Section 4.4

Joint Probability Distributions

and Independence


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)



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.



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



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



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 ∑=



and Independence


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

==∑==



and Independence

The marginal probability distribution of X is:

.520

.50

P[X=x]x



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



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.



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 =



and Independence


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]



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



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.



and Independence

• Sequence of generating the distributions

Joint Probability Distribution

Marginal Distributions

Conditional Distributions

• Can we go the other way?



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).



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


Section 4.5

Covariance and Correlation

of Random Variables


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∑∑=



of Random Variables


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



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.



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− ≤ ≤ +



of Random Variables


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)=



of Random Variables

Note: If X and Y are independent,

then Cov(X, Y) = 0 and ρ = 0.


Example: Consider the following joint distribution.

Find the covariance between X and Y.

Are X and Y independent?


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



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.



of Random Variables


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%



of Random Variables


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



of Random Variables


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



of Random Variables


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?



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)


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


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)

• •• •• •



Two or More Random Variables

• General Case – Discrete

Joint Probability Distribution

Y • • •X

• Joint probabilities• are in the cells of • the table.



• Joint dist. → Marginal dist.

• Joint dist.

& Conditional dist.Marginal dist.

P(X=x|Y=y) = P(X=x,Y=y)/P(Y=y)



• 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



• 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



• E(aX + bY) = aE(X) + bE(Y)

• V(aX + bY) = a2V(X) + b2 V(Y) + 2 ab Cov(X,Y)

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Random Variables and Probability Distributions...4.2 Probability Distribution: Discrete Random Variables • Requirements for a probability distribution: • PY(y) ≥0, for all y