6. Theory of Probability

50
Theory of Probability Prof Tasneem Chherawala NIBM

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Probability

Transcript of 6. Theory of Probability

Page 1: 6. Theory of Probability

Theory of Probability

Prof Tasneem Chherawala

NIBM

Page 2: 6. Theory of Probability

Uncertain Events

• See a Zebra the classroom today?

• Win in a lottery?

• Toss a coin and get heads / roll a die (or two) and get 5/ pick a card from a deck and get an Ace?

• Make a 10% profit from investment in shares over 1 month?

• Face a default on money lent?

• Earn positive returns on a fixed deposit?

• Count the total number of hours today and get 24?

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Probability

• Probability quantifies how uncertain we are about a

future outcome / event

• A probability refers to the percentage chance that

something will happen, from 0 (it is impossible) to 1 (it is

certain to occur), and the scale going from less likely to

more likely

• An interpretation based on data - Probability can be

interpreted as the relative frequency of the outcomes

(values) of uncertain events (variable) after a great many

(infinitely many) repetitions / parallel independent trials of

an experiment

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Why Measure Uncertainty

• Is something at stake?

• To make tradeoffs among uncertain events

• Measure combined effect of several uncertain events

• To communicate about uncertainty

• To draw inferences about a population from a sample

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Probability Theory in Finance

• Financial decisions – a game of chance!

• What chance?– Probability of making or losing money in an investment!

• Why chance? – Uncertainty and variability of future events (price movements

and value of investments)

• Probability concepts help define financial risk by quantifying the prospects for unintended and negative outcomes (losses)

• Probability also quantifies expected values of future events, which gives us a fair estimate of current value of investment

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Probability Theory in Finance

• Probability theory applications:– Make logical and consistent investment decisions

– Manage expectations in an environment of risk

• Hypothesis: No system for making financial choices from those offered to us can both (1) be certain to avoid losses and (2) have a reasonable chance of making us rich

• Expected values in a probability model are the prices of alternative financial decisions

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Some Definitions

• Random Experiment: a process that leads to one of several possible outcomes

• Outcome: result of the experiment– Examples: Head in a coin toss, two heads when tossing two

coins, win in a lottery, 10% returns from the stock in one day, Borrower repays Re 1 from Rs. 100 borrowed at 12% rate of interest

• Sample Space: a set of all possible outcomes of the experiment – relates to population data– Examples:

• Toss one coin: S = {H, T}

• Toss two coins / toss a coin twice: S = {HH, HT, TH, TT}

• Lottery: S = {Win, Lose}

• One day stock returns: S = {-100% to +Infinity?}

• Repayment on Rs. 100 loan by borrower at 12% rate of int.: S = {0,1,2,….., 100, 101,…,112}

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Sample Space

• Finite sample space: finite number of outcomes in the space S.

• Countable infinite sample space: ex. natural numbers.

• Discrete sample space: if it has finite or countable infinite number of outcomes.

• Continuous sample space: If the outcomes constitute a continuum. Ex. All the points in a line.

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Events

• Event: subset of a sample space – a combination of possible outcomes of an uncertain process – Examples

• Head in one toss of a coin: E1 = {H}

• One head and one tail in two coin tosses: E2 = {HT, TH}

• Mean of the dice values greater than or equal to 5 in two dice rolls: E3 = {4.6, 5.5, 5.6, 6.4, 6.5, 6.6}

• One day stock returns greater than equal to 10%: E4 = {10% to +Infinity}

• Loss to bank if Borrower doesn’t repay principal: D = {1%,2%,…..,99%,100%}

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Events

• Mutually Exclusive Events: events which have no outcome in common and thus cannot happen together

• If a list of events is mutually exclusive, it means that only one of them can possibly take place– Examples

• Toss one coin: E1 = {H}, E2 = {T}

• Toss two coins: E1: first toss is Head = {HT, HH}, E2: first toss is Tail = {TH, TT}

• Invest in a stock for one day: E1: One day stock returns greater than equal to 10% = {10% to +Infinity}, E2: One day stock returns between 2% to 5% = {2% to 5%}

– How would E1 compare with E3: One day stock returns less than or equal to 10% = {-100% to 10%)

• Lend money: E1: Borrower repays entire principal = {100,101,102,…112}, E2: Borrower repays 50% of principal = {50}

– How would E1 compare with E3: Borrower repays principal with 5% interest = {105} ?

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Properties of Probabilities defined

on a Sample Space• Sample space must be exhaustive: List all possible

outcomes

• Outcomes in the sample space must be mutually exclusive

• The exhaustive and mutually exclusive characteristic of a sample space together imply that

– The probability (likelihood of occurrence) of any one outcome or event must lie between 0 and 1

• 0 < P(E) <1

– The sum of the probabilities of all the outcomes in the sample space must be 1

• P(S) = 1

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Properties of Probabilities defined

on a Sample Space• By extension of the above rationale

– The probability of any event would lie between 0 (impossible) and 1 (certain), or 0 < P(E) < 1.

– There is no such thing as a negative probability (less than impossible?) or a probability greater than 1 (more certain than certain?).

– The sum of all probabilities of all outcomes would equal 1, provided the outcomes are both mutually exclusive and exhaustive.

– If outcomes are not mutually exclusive, the probabilities would add up to a number greater than 1, and if they were not exhaustive, the sum of probabilities would be less than 1.

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Classical / A Priori Approach to

Probabilities• When the range of possible uncertain outcomes in a

sample space is known and equally likely

• Probability for each outcome or event can be determined by logic

• Examples– Tossing a fair coin – outcomes can only be heads or tails and

both are equally likely

– Rolling a fair die – outcomes can only be 1,2…,6 and all six are equally likely

– Drawing a card from a pack – outcomes can only be 52, and all are equally likely

• The probability of each outcome in the above examples has been determined by construction of the coin/die/pack of cards

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Defining A Priori Probabilities

• In a sample of N mutually exclusive, exhaustive and equally likely outcomes we assign a chance (or weight) of 1/N to each outcome

• We define the probability of an event for such a sample as follows:

• The probability of an event E occurring is defined as:

• P(E) = n(E)/n(S)

– n(E) is the number of outcomes favourable to E and

– n(S) is the total number of equally likely outcomes in the sample space S of the experiment

• By extension, the probability of event E not occurring

• P(not E) = 1-P(E) = 1- n(E)/n(S)

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Example

• What is the probability that a card drawn at random from a deck of cards will be an ace?

• Total no. of outcomes in the sample space?

• Are they equally likely?

• Event of interest (E)?

• No. of Outcomes favourable to the Event?

• P(E)?

• What is the probability that a card drawn at random from the deck will not be an ace?

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A Priori Probabilities

• The same principle can be applied to the problem of determining the probability of obtaining different totals from a pair of dice.

• Possible Outcomes?

• Are they equally likely?

• Event 1 = A = sum of the two dice will equal 5

• Event 2 = B = the absolute difference will equal 1

• P (Event 1) =

• P (Event 2) =

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A Priori Probabilities

• In certain cases where outcomes are not equally likely,

one can still deduce rationally the a priori probability of

an event

• For example

– If we forecast that a company is 70% likely to win a

bid on a contract (irrespective of how this probability

is derived), and we know this firm has just one

business competitor, then we can also make an a

priori forecast that there is a 30% probability that the

bid will go to the competitor

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Empirical Probabilities

• In finance, we cannot depend upon the exactness of a process to determine a priori probabilities

• The financial analyst would have to depend upon historical occurrence of the event(s) or repeat an experiment multiple times to determine the probability of the event empirically

• For example, the range of outcomes of returns on a financial asset are virtually infinite and that too, not all outcomes are a priori, equally likely

• Thus, the financial analyst would have to observe many movements in asset prices to determine the probability of future price changes of a given magnitude

• Of course, we know that past performance does not guarantee future results, so a purely empirical approach has its drawbacks

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Defining Empirical Probabilities• The probability of an event (or outcome) is the proportion

of times the event occurs in a long run of repeated

experiment.

• The empirical probability of a given outcome / event Z is

defined as

– P(Z) = no. of Z occurrences/no. of trials of the experiment

• This is the same as analysis of relative frequency of

observations in a sufficiently large sample

• Consider a financial analyst who is interested in knowing

what will be the probable one day returns on a particular

stock.

– He tracks the past 100 days of price movements and returns of

a particular stock.

– Each of the 100 days would constitute trial and each day’s

returns would constitute the outcome

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Subjective Probabilities

• Probability under this approach is simply defined as the strength of belief that an event will occur

• It is based upon experience and judgment

• Such probabilities are applied to many business problems where a priori probabilities are not possible, nor are there sufficient empirical observations upon which to base probability estimates

• For example, subjective probability is incorporated in the forecasting of company profits by investment analysts

• Of course, subjective probabilities are unique to the person making them and to the specific assumptions made

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Rules of Probability

• There are a number of formal probability rules applied to

probability estimates

• Which of these rules is applicable will depend upon

whether

– We are concerned with a single event, in which case

the outcomes relate only to that event

– We are concerned with combinations of several

events, for example the changes in Sensex and

Exchange Rates together

– The combined events are independent or mutually

exclusive

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Rules of Probability

• The rules are

– Complement rule: When we are concerned with

whether an event A will not occur

– Multiplication rule: when we are concerned with event

A and B occurring together. This requires us to know

whether A and B are independent of each other

– Addition rule: when we are concerned with event A or

B happening. This requires us to know whether A and

B are mutually exclusive

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Intersection of Events and Joint

Probability

• Joint probability: The probability of both event A and

event B occurring: P(A and B) / P(A ∩ B) / P(AB)

• Intersection: Event defined as both A and B occur

• Example Using a priori probabilities:

– P(A and B) = No. of Outcomes favourable to the Joint Event /

Total no. of outcomes in the sample space

– Event A is you draw a spade from a deck of cards

Event B is you draw a king

– Intersection is event you draw a king of spades

– Joint probability = P(A and B) = 1/52

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Joint Probability Table

Job_statusDecision

accept reject

governmen 0.116 0.072

military 0.003 0.003

misc 0.000 0.003

private_s 0.266 0.352

retired 0.014 0.005

self_empl 0.035 0.051

student 0.005 0.008

unemploye 0.013 0.056

• Interpretation of 0.266?

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Marginal Probability

• Marginal probability: Probability of a single event

• When outcomes are exhaustive and mutually exclusive, can be calculated by adding all the joint probabilities containing the single event

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Marginal Probability

Job_statusDecision

accept reject Total

governmen 0.116 0.072 0.187

military 0.003 0.003 0.005

misc 0.000 0.003 0.003

private_s 0.266 0.352 0.617

retired 0.014 0.005 0.019

self_empl 0.035 0.051 0.086

student 0.005 0.008 0.013

unemploye 0.013 0.056 0.069

Total 0.451 0.549 1.000

• Interpretation of 0.187, 0.451?

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Conditional Probability

• A conditional probability is the probability of an event given that another event has occurred

• Conditional probability assumes that one event has taken place or will take place, and then asks for the probability of the other (A, given B)

• P(B | A): probability of B given A

• P(A | B): probability of A given B

• For example, – Probability of drawing a king given that a spade is drawn: P(king

| spade) = 1/13

– Probability of drawing a spade given that a king is drawn: P(spade | king) = 1/4

– What is the probability that the total of two dice will be greater than 8 given that the first die is a 6?

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Conditional Probability in Terms of

Joint Probability

Job_statusDecision

accept reject Total

governmen 0.116 0.072 0.187

military 0.003 0.003 0.005

misc 0.000 0.003 0.003

private_s 0.266 0.352 0.617

retired 0.014 0.005 0.019

self_empl 0.035 0.051 0.086

student 0.005 0.008 0.013

unemploye 0.013 0.056 0.069

Total 0.451 0.549 1.000

•Given that an application

is accepted, what is the

probability that the

customer is a Pvt. Sector

employee?

•What kind of probability

asked for?

•Imagine 1000 Customer

Applications. You would

expect 451 to be accepted

and 266 of those to be

Pvt. Sector employees

•P(Pvt. S Empl | Accept) =

116/451 = 0.257

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Conditional Probability in Terms of

Joint Probability• Joint Probability can be rescaled to find Conditional

Probability

P(A|B) = P(A and B) / P(B)

• Can think about this as rescaling the Accept Column so it sums to one

• Interpretation– P(Pvt. S Empl | Accept) = 266/451 = 0.59– P(Non Pvt. S Empl | Accept) = 185/451 = 0.41

– Means Pvt. S Employees are more likely to be accepted than Non Pvt. Sector Employees?

– P(Accept | Pvt. S Empl) = 266/617 = 0.43

– P(Accept | Non Pvt. S Empl) = 185/383 = 0.483

– Means more of the accepted applicants are Non Pvt. Sector Employees than Pvt. Sector Employees?

– If the bank’s objective was to specifically target a group for marketing of the product, which probabilities would it rely upon to give it an accurate picture?

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Joint Probability in Terms of

Conditional Probability• Another way to calculate Joint Probability of Events A

and B isP(A and B) = P(A) x P(B|A) = P(B) x P(A|B)

• where P(B|A) is the conditional probability of B given A and P(A|B) is the conditional probability of A given B

• Example: – If we believe that a stock is 70% likely to return 15% in the next

year, as long as GDP growth is at least 8%, then we have made our prediction conditional on a second event (GDP growth). In other words, event A is the stock will rise 15% in the next year; event B is GDP growth is at least 8%; and our conditional probability is P(A | B) = 0.7

– Now if we know that there is a 20% unconditional probability that GDP will grow at 8% or above P(B) = 0.2

– The probability that GDP will grow at least at 8% and stock will return 15% is P(A and B) = 0.7 *0.2 = 0.14

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Joint Probability Vs Conditional

Probability• Joint probability is not the same as conditional

probability, though the two concepts are often confused.

• Joint probability sets no conditions on the occurrence of events but simply provides the chance that both events will happen together

• In a problem, to help distinguish between the two, look for qualifiers that one event is conditional on the other (conditional) or whether they will happen concurrently (joint).

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Joint Probability and Independence

of Events• Independence of Event A and B: When Probability of

Event A occurring does not depend upon occurrence of Event B and vice versa

• Two events are independent if and only if

• P(A | B) = P(A) [implies P(B | A) = P(B)]

• If A and B are independent, then the probability that events A and B both occur is:P(A and B) = P(A) x P(B | A) = P(A) x P(B)

• Examples:– What is the probability that a fair coin will come up with heads

twice in a row?

– Now consider a similar problem: Someone draws a card at random out of a deck, replaces it, and then draws another card at random. What is the probability that the first card is the ace of clubs and the second card is a club (any club)?

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Joint Probability and Independence

of Events• Is the customer application being accepted independent

of being a Pvt. sector employee?

• P(Accept | Pvt. Sector employee) = 0.43

• P(Accept) = 0.451

• The rule generalizes for more than two events provided they are all independent of one another, so the joint probability of three events P(ABC) = P(A) * (P(B) * P(C), again assuming independence

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Joint Probability and Mutually

Exclusive Events• The joint probability of two mutually exclusive events

occurring is 0

• This is because by definition of mutually exclusive, events A and B cannot occur together

• For example:– Roll a die once: Event A = 6, Event B <5

– Are A and B mutually exclusive?

– P(A and B) = ?

– Roll 2 dice: Event A = {1,4}, Event B = {4,1}

– Are A and B mutually exclusive?

– P(A and B) = ?

– Roll 2 dice: Event A = {1,4}, Event B = at least one die shows 1

– Are A and B mutually exclusive?

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Union of Events and Probability

• Union: Union of events A and B is event that either A or B or both occur: (A or B) or (AUB)

• P(A or B) is interpreted as the probability that at one of the two events A and B will occur

• If events A and B are mutually exclusive, P(A and B) = 0

• P(A or B) = P(A) + P(B)

• Example:– What is the probability of rolling a die and getting either a 1 or a

6?

– Since it is impossible to get both a 1 and a 6, these two events are mutually exclusive

– P(1 or 6) = P(1) + P(6) = 1/6 + 1/6 = 1/3

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Union of Events and Probability

• If events A and B are not mutually exclusive, P(A and B) ≠ 0

• P(A or B) = P(A) + P(B) – P(A and B)

• The logic behind this formula is that when P(A) and P(B) are added, the occasions on which A and B both occur are counted twice. To adjust for this, P(A and B) is subtracted.

• Example:– What is the probability that a card selected from a deck will be

either an ace or a spade?

– P(Ace) = 4/52

– P(Spade) = 13/52

– P(Ace and Space) = ?

– P(Ace or Space) = ?

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Union of Events and Probability

• Consider the probability of rolling a die twice and getting a 6 on at least one of the rolls. The events are defined in the following way:

• Event A: 6 on the first roll: P(A) = 1/6

• Event B: 6 on the second roll: P(B) = 1/6

• P(A and B) = 1/6 x 1/6 (why?)

• P(A or B) = 1/6 + 1/6 - 1/6 x 1/6 = 11/36

• The same answer can be computed using the following admittedly convoluted approach:

• Getting a 6 on either roll is the same thing as not getting a number from 1 to 5 on both rolls.

• This is equal to: 1 - P(1 to 5 on both rolls) = 1 – 5/6 x 5/6 = 1 – 25/36 = 11/36

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Union of Events and Probability

• Despite the convoluted nature of this method, it has the advantage of being easy to generalize to three or more events.

• For example, the probability of rolling a die three times and getting a six on at least one of the three rolls is?

1 - 5/6 x 5/6 x 5/6 = 0.421

• In general, the probability that at least one of k independent events will occur is:

1 - (1 - α)k

where each of the events has probability α of occurring

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Union of Events and Probability

• Example:

• Assume that the bank had lent to 5 different borrowers, each of whom was assigned a probability of default = 0.1

• The bank wished to hedge the risk of its portfolio such that it bought insurance such that the loss against at least one borrower defaulting was made good

• How would the insurance company estimate the probability of at least one borrower defaulting, assuming that the default event of each borrower was independent?

• P(At least one default) = 1 – (1-0.1)5 = 0.41

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Union of Events and Probability

Job_statu

s

Decision

accept reject Total

governme

n 0.116 0.072 0.187

military 0.003 0.003 0.005

misc 0.000 0.003 0.003

private_s 0.266 0.352 0.617

retired 0.014 0.005 0.019

self_empl 0.035 0.051 0.086

student 0.005 0.008 0.013

unemploy

e 0.013 0.056 0.069

Total 0.451 0.549 1.000

•What is the probability that the

customer application is accepted

or the customer is a government

employee?

•What kind of probability asked

for?

•P(Accept) = 0.451

•P(Govt. Employee) = 0.187

•P(Accept and Govt. Employee =

0.116

•P(Accept or Govt. Employee) =

0.451 + 0.187 – 0.116 = 0.522

Page 41: 6. Theory of Probability

Union of Events and Probabilities

• Example:

• A Fund manager has invested in the stocks of two companies 1 and 2 and is interested in knowing what is the probability that the equity price of either company will rise

• Are the two events mutually exclusive?

• P(Co. 1 Equity ↑) = 0.55

• P(Co. 2 Equity ↑) = 0.35

• P(Co. 1 Equity ↑ and Co. 2 Equity ↑) = 0.3

• P(Co. 1 Equity ↑ or Co. 2 Equity ↑) = 0.55+0.35-0.30 = 0.60

Page 42: 6. Theory of Probability

Union of Events and Probability

• What if we want to know the probability of at least one

of 3 events A, B and C happening?

( ) ( ) ( ) ( ) ( ) ( ) ( ) ( )P A B C P A P B P C P AB P AC P BC P ABC

A B

C

Page 43: 6. Theory of Probability

Probability Rules Summary

• Multiplication Rule: Joint probability of any two events A and B is:– P(A and B) = P(A | B) * P(B)

– Follows from definition of conditional probability

• Multiplication Rule Independent Events: If A and B are independent, joint probability is:– P(A and B) = P(A) * P(B)

• Addition Rule mutually exclusive events:– P(A or B) = P(A) + P(B)

– Mutually exclusive if both cannot occur

• Addition Rule: Probability that event A or event B or both occur is– P(A or B) = P(A) + P(B) – P(A and B)

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• Are mutually exclusive events independent?

• Does pr(A|B) = pr(A)?

• NO!

• If B has happened, then A can not happen

• pr(A|B) = 0

• So

• Mutually exclusive events are DEPENDENT.

Page 45: 6. Theory of Probability

Probability Tree and Total

Probability Rule

Job_statusDecision

accept reject

Private

Sector 0.29 0.35

Non Pvt

Sector 0.16 0.20

Page 46: 6. Theory of Probability

Probability Tree

Accept: 0.45

Reject: 0.55

Pvt. Sector | Accept: 0.59

Not Pvt. Sector | Accept: 0.41

Pvt. Sector | Reject: 0.64

Not Pvt. Sector | Reject: 0.36

0.29

0.16

0.35

0.20

Pvt. S & Accept

Not Pvt. S & Accept

Pvt. S & Reject

Not Pvt. S & Reject

Page 47: 6. Theory of Probability

Total Probability Rule

• The Total Probability Rule: The total probability rule explains an unconditional probability of an event A, in terms of that event's conditional probabilities in a series of mutually exclusive, exhaustive scenarios of event B

• P(A) = P(A | B) x P(B) + P(A | not B) x P(not B)

• With the total probability rule, event A has a conditional probability based on each scenario of event B (i.e. the likelihood of event A, given that scenario), with each conditional probability weighted by the probability of that scenario for event B occurring

• Example: – What is the probability of a customer employed in the Private

Sector:

– P(Pvt. S. Empl) = P(Pvt. S Empl | Accept) x P(Accept) + P(Pvt. S Empl | Reject) x P(Reject) = 0.59*0.45+0.64*0.55 = 0.61

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Total Probability Rule

• A model that predicts whether a borrower has defaulted or not is 95 percent effective in predicting that a borrower has defaulted when it actually has. However, the model also yields a “false positive” result for 1 percent of the non-defaulted borrowers. That is, there is 1 percent chance that a borrower who has not defaulted will be identified as a defaulted borrower by the model.

• Q: If 0.5 percent of the bank’s portfolio has actually defaulted, what is the probability that a borrower has defaulted given that the model predicts default?

Page 49: 6. Theory of Probability

Probability Tree

Default: 0.005

No Default: 0.995

Def Pred | Default: 0.95

No Def Pred | Default: 0.05

Def Pred | No Default: 0.01

No Def Pred | No Default: 0.99

0.00475

0.00025

0.00995

0.98505

Page 50: 6. Theory of Probability

Total Probability Rule And Baye’s

Theorem

• Let D be the event that a borrower has defaulted

• Let E be the event that the model predicts default

• We need to estimate P(D|E)

• We know: P(E|D)=0.95, P(E|notD)=0.01, P(D)=0.005,

P(notD)=1-P(D)=0.995

0147.0995.001.0005.095.0)()()()()(

00475.0005.095.0)()()(

3231.00147.0

00475.0

)(

)()(

notDPnotDEPDPDEPEP

DPDEPEDP

EP

EDPEDP