Correlation Coefficient and Risk

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If You Liked This, You’re Sure to Love That By CLIVE THOMPSONPublished: November 21, 2008

THE “NAPOLEON DYNAMITE” problem is driving Len Bertoni crazy. Bertoni is a 51-year-old “semiretired” computer scientist who lives an hour outside Pittsburgh. In the spring of 2007, his sister-in-law e-mailed him an intriguing bit of news: Netflix, the Web-based DVD-rental company, was holding a contest to try to improve Cinematch, its “recommendation engine.” The prize: $1 million.

Chapter 4

Introduction to Probability

Experiment

A process that generates well-defined outcomes. On any single repetition of the experiment, one and only one of the possible experimental outcomes (or sample points) can occur.

Probability

A numerical measure of the likelihood that an event will occur.

Assigning Probabilities

1. The probability assigned to each experimental outcome must be between 0 and 1, inclusively.

0 < P(Ei) < 1 for all i2. The sum of the probabilities for all experimental

outcomes must equal 1. Given n possible outcomes:

P(E1) + P(E2) + . . . + P(En) = 1

Assigning Probabilities

Classical method – Assumes all the experimental outcomes are equally likely.

Relative frequency method – Assigns probabilities based on the frequency with which some event occurs in the data.

Subjective method – Assignment of a probability based on the degree of belief an event will occur.

Example

Which approach is being used to assign the probability in the following cases?• Political commentators believe there is a 60% chance Congress will pass health care legislation.• A baseball player has gotten 90 hits in 300 at-bats, so the probability he will get a hit at his next at-bat is 30%.• There is a 25% chance of flipping a coin twice and getting two heads.

Event

A collection of outcomes (or sample points).

Probability of an Event

The sum of the probabilities of the outcomes (sample points) in the event.

Example

Given a die with six sides, what is the probability of rolling the die and getting an even number?

xi P(xi)

1 1/62 1/63 1/64 1/65 1/66 1/6

P(E) = P(2) + P(4) + P(6) = 1/6 + 1/6 + 1/6 = 3/6 = 1/2

Complement of an Event

The complement of event A is the event consisting of all sample points that are not in A.

Probability of the Complement of an Event

P(AC) = 1 - P(A)

Venn Diagram

Event AComplement of A

Combinations of Events• Union of two events• Intersection of two events

Combinations of Events• Union of two events• Intersection of two events

Union of Two Events

The union of A and B is the event containing all outcomes (or sample points) belonging to A or B or both. The union is denoted by A U B.

Intersection of Two EventsGiven two events A and B, the intersection of A and B is the event containing the outcomes (sample points) belonging to both A and B. The intersection is denoted by A ∩ B.

Addition Law

The addition law is used to calculate the probability of the union of two events.

P(A U B) = P(A) + P(B) – P(A ∩ B)

Mutually Exclusive Events

Two events are said to be mutually exclusive if the events have no outcomes (sample points) in common.

Addition Law for Mutually Exclusive Events:P(A U B) = P(A) + P(B)

A B

Joint Probability

The probability of the intersection of two events.

Marginal Probability

The probability of an event occurring.

Cross TabulationNapoleon Dynamite

Like Dislike TotalLike 70 60 130Dislike 30 40 70Total 100 100 200

Transformers

Joint Probability TableNapoleon Dynamite

Like Dislike TotalLike .35 .30 .65Dislike .15 .20 .35Total .50 .50 1.00

Transformers

Practice

Assume event A is “likes Napoleon Dynamite” and event B is “likes Transformers”. Find the following probabilities of the events listed below:1. P(A)2. P(AC)3. P(A U B)4. P(AC U B)5. P(A U AC)6. P(A ∩ B)7. P(A ∩ AC)8. P(A ∩ BC)

Napoleon Dynamite

TransformersLike Dislike Total

Like .35 .30 .65Dislike .15 .20 .35Total .50 .50 1.00

PracticeA = “likes Napoleon Dynamite” B = “likes Transformers” 1. P(A) = .52. P(AC) = .53. P(A U B) = .5 + .65 - .35 = .84. P(AC U B) = .5 + .65 - .3 = .855. P(A U AC) = .5 + .5 = 16. P(A ∩ B) = .357. P(A ∩ AC) = 08. P(A ∩ BC) = .15

Napoleon Dynamite

TransformersLike Dislike Total

Like .35 .30 .65Dislike .15 .20 .35Total .50 .50 1.00

Conditional Probability

The probability of some event taking place given that some other event has already occurred.

P(A|B)

Probability they like Napoleon Dynamite (event A) given they like Transformers (event B) is:

P(A|B) = 70/(70+60) = 70/130 = 0.52

Conditional Probability, cont.

More generally:

)()()|(

BPBAPBAP

Practice #2

Assume event A is “likes Napoleon Dynamite” and event B is “likes Transformers”. Find the following probabilities listed below:1. P(B|A)2. P(AC|B)3. P(A|BC)4. P(AC|A)

Napoleon Dynamite

Transformers

Like Dislike TotalLike .35 .30 .65Dislike .15 .20 .35Total .50 .50 1.00

Practice #2

A = “likes Napoleon Dynamite” B = “likes Transformers” 1. P(B|A) = .35/.52. P(AC|B) = .3/.653. P(A|BC) = .15/.354. P(AC|A) = 0

Napoleon Dynamite

Transformers

Like Dislike TotalLike .35 .30 .65Dislike .15 .20 .35Total .50 .50 1.00

Independent Events

Two events are independent if the probability of A is the same regardless of whether or not B has occurred.

)()|(

)()|(

BPABPor

APBAP

Independent Events, cont.

Is liking Napoleon Dynamite independent of liking Transformers?P(A|B) = .52P(A) = .5P(A|B) ≠ P(A), so the events are not independent

Independent Events, cont.

Assume event A is “likes Napoleon Dynamite” and event B is “Birthday on odd-numbered day”. Are the events independent?

Napoleon Dynamite

Odd-numbered birthday

Like Dislike TotalYes .25 .25 .50No .25 .25 .50Total .50 .50 1.00

Independent Events, cont.

P(A|B) = .25/.50 = .50P(A) = .50Since P(A|B) = P(A) the events are independent

Multiplication Law

P(A ∩ B) = P(B)P(A|B)orP(A ∩ B) = P(A)P(B|A)

Tree Diagram

A graphic representation of a multistep experiment.

Each fork represents an experiment, each branch represents an outcome from the experiment.

Tree Diagram

Red, 2/4Red, 1/3

Black, 2/3

Black, 2/4Red, 2/3

Black, 1/3

Assume we take the four aces out of a deck of cards and we draw twice without replacement:

PracticeIf A is the probability of getting a red ace on the first draw, and B is the probability of getting a red ace on the second draw, using the multiplication rule find:1. P(A ∩ B)2. P(A ∩ BC) 3. P(AC ∩ B) 4. P(AC ∩ BC)

Practice, cont.1. P(A ∩ B) = (2/4)/(1/3) = 2/122. P(A ∩ BC) = (2/4)(2/3) = 4/123. P(AC ∩ B) = (2/4)(2/3) = 4/124. P(AC ∩ BC) = (2/4)/(1/3) = 2/12

Multiplication Rule, cont.

If events A and B are independent then P(A|B) = P(A)P(B). In this special case the multiplication rule reduces from:

P(A ∩ B) = P(B)P(A|B)to:

P(A ∩ B) = P(B)P(A)

Tree Diagram

Red, 2/4Red, 2/4

Black, 2/4

Black, 2/4Red, 2/4

Black, 2/4

Assume we take the four aces out of a deck of cards and we draw twice with replacement:

Are A and B statistically independent in this case?

Sampling and Statistical Independence

If we sample without replacement the outcomes will not be statistically independent.

However, if we are drawing from a large population the change in probability will be so small we can treat the draws as being statistically independent.

Graded Homework

P. 163, #29P. 170, #35, 37