SS1 Quantitative Analysis I

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Quantitative Analysis - I

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Agenda

• Probability– Counting Principle(1hr)– Probability Concepts, Bayes Theorem(1hr)

• Moments• Probability distributions

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Counting Principle

• Number of ways of selecting r objects out of n objects– nCr

– n!/(r!)*(n-r)!

• Number of ways of giving r objects to n people, such that repetition is allowed– Nr

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Example-Counting Principle

• In how many ways 3 stocks can be chosen out of 10 stocks in a portfolio?(Combination)– Choosing 3 out of 10 stocks is basically the number of combinations of 3 objects out of 10– Therefore, the number of ways are 10C3= =120

• In how many ways 3 stocks can be sold, if the sold stock is bought back in the portfolio before the next stock is sold?– First stock can be sold in 10 ways– Second can be sold in again 10 ways– Third stock can again be sold in 10 ways– Therefore total number of ways become =103 =1000

)!310!*(3!10


• You wish to choose a portfolio of 3 bonds and 4 stocks from a list of 5 bonds and 8 stocks. How many different 7 asset portfolio can you make from this list.

A. 80B. 700C. 1,716D. 100,800

Question

5


• B – Solution:

Solution

70070*101*2*3*45*6*7*8*

1*24*5

)!48(!4!8*

)!35(!3!5* 4

83

5

CC

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Probability - Definitions

• A probability experiment involves performing a number of trials to enable us to measure the chance of an event occurring in the future. A trial is a process by which an outcome is noted.

• Examples of Definitions:– Experiment: Roll a die two hundred times noting the outcomes– Event of interest: A six faces upwards– Trial: Roll the die once– Number of trials: 200– Outcomes: 1, 2, 3, 4, 5 or 6

• Probability of an event to occur is defined as number of cases favorable for the event, over the number of total outcomes possible in unbiased experiment– For example, if the event is "occurrence of an even number when a die is rolled“– The probability is given by 3/6=1/2, since 3 faces out of the 6 have even numbers

and each face has the same probability of appearing.

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Conditional & Joint Probability

• Joint Probability– A statistical measure where the likelihood of two events occurring together and at the same point in time

are calculated. Joint probability is the probability of event Y occurring at the same time event X occurs.– Notation for joint probability takes the form:

P(X ∩Y) or P(X,Y) Which reads the joint probability of X and Y.

– The following table shows the joint probability of different events. Let’s say an economist is predicting the market scenario and the price of IBM stock from the next year.

– Next year market can be Good, Bad or Neutral– IBM stock may go up or go down

Joint Probability Table:

– The probability of IBM stock being Up and Market being Good is 10%– Similarly, the probability of IBM stock being down and Market being neutral is 40%

MarketIBMUP 10% 30% 5% 45%

DOWN 0% 15% 40% 55%Total 10% 45% 45% 100%

Good Bad Neutral Total

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Conditional & Joint Probability

• Conditional Probability– Probability of an event or outcome based on the occurrence of a previous event or outcome.

Conditional probability is calculated by multiplying the probability of the preceding event by the updated probability of the succeeding event

– The probability of event A given that the event B has occurred is P(A/B), which is equal to the ratio of joint probability of A and B, and unconditional probability of B.

• The unconditional probability of market being Neutral is 45%. Then using the table below we can find 3 conditional probabilities.– P(Up/Neutral) = 0.05 / 0.45– P(Up/Good) = 0.1/0.1– P(Down/Bad) = 0.15/0.45

Joint Probability Table:

)()()/(

BPBAPBAP

MarketIBMUP 10% 30% 5% 45%

DOWN 0% 15% 40% 55%Total 10% 45% 45% 100%

Good Bad Neutral Total

Unconditional Probabilities of IBM stockbeing UP/Down

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Conditional and Unconditional Probabilities


MARKET

Good Bad Neutral

10% 0%

UP DOWNIBM

30% 15% 5% 40%

10% 45% 45%

UP DOWN UP DOWN

Unconditional Probability of Market

Calculate:Unconditional Probability of market to be good next year?Conditional Probability of IBM stock rising when the market is neutral?Conditional Probability of market being good when IBM stock is down?


• Definitions– Mutually Exclusive: If one event occurs, then other cannot occur– Exhaustive: all exhaustive events taken together form the complete sample space

(Sum of probability = 1)– Independent Events: One event occurring has no effect on the other event

• The probability of any event A:

• If the probability of happening of event A is P(A), then the probability of A not happening is (1-P(A)).– For example, if the probability of a company going bankrupt within one year period is 20%, then the

probability of company surviving within next one year period is 80%.

Some definitions and properties of Probability

]1,0[)( AP

)(1)( APAP

11


Question

• For a bond with “B” rating, assume 1 year probability of default for each issuer is 6%, and that default probability of each issuer are independent. What is the probability that both issuers avoid default during the 1st year.

A. 88%B. 88.4%C. 94%D. 96.4%


• B Both would avoid default only if None defaultsThis implies that first does not default AND second does not default= (1 – PD (first)) x (1 – PD (second))= (1 – 0.06) x (1 – 0.06) = 0.884 = 88.4%

Solution

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• The probability of happening of event A or event B can be given as the sum of the three portions defined by the figure below.

Some Properties of Probability

)(AP )(BP)( BAP

)(AP )(BP

0)( BAP

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Question

• Jensen, a portfolio manager is managing two portfolios. One for High Net Worth Individuals (HNI) and second for Low Net Worth Individuals (LNI).

• HNI portfolio contains 5 bonds and 7 stocks and LNI contains 6 bonds and 11 stocks.• One instrument from HNI is transferred to LNI portfolio.• Now Jensen selects an instrument from LNI, what is the probability that instrument selected is bond?

A. 0.5382B. 0.7821C. 0.6435D. None of these


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Solution

• B• Here required probability = [P(stock transferred from HNI) AND P(Stock selected from LNI)] OR

[P(bond transferred from HNI) AND P(Stock selected from LNI)]• So, the required probability = (7/12) × (12/18) + (5/12) × (11/18) = 139/216 = 0.6435• Hence option ‘B’ is correct.



Sum Rule & Bayes’ Theorem

• The unconditional probability of event B is equal to the sum of joint probabilities of event (A,B) and the probability of event (A’,B). Here A’ is the probability of not happening of A.– The joint probability of events A and B is the product of conditional probability of B, given A has occurred

and the unconditional probability of event A.

– We know that P(AB) = P(B/A) * P(A)– Also P(BA)= P(A/B) * P(B) – Now equating both P(AB) and P(BA) we get:

– P(B) can be further broken down using sum rule defined above:

)()(*)/()/(

BPAPABPBAP

)()/()()/()()()( ccc APABPAPABPBAPBAPBP

)()/()()/()()/()/( cc APABPAPABP

APABPBAP

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Example

• Use the given information to answer next two questions• Out of a group of 100 patients being treated for chronic back trouble, 25% are chosen at random to

receive a new, experimental treatment as opposed to the more usual muscle relaxant-based therapy which the remaining patients receive. Preliminary studies suggest that the probability of a cure with the standard treatment is 0.3, while the probability of a cure from the new treatment is 0.6.

• How many patients (on an average) out of the 100 patients selected at random would be cured?A. 30B. 40C. 37.5D. 42.5


Solution

• C– 25% are given new treatment =>75% are given old treatment.– P(Cure) = P(Cure/New) * P(New) + P(Cure/Old) * P(Old) = 0.375– So out of 100 patients 37.5 will get cured

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• Some time later, one of the patients returns to thank the staff for her complete recovery. What is the probability that she was given the new treatment?

A. 0.375B. 0.425C. 0.4D. 0.425

Example

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• C– Apply Bayes’ Theorem– P (New / Cure) = P (Cure / New) * P (New) / P (Cure) = 0.6 * 0.25 / 0.375 = 0.40

Solution

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• Calculate the probability of a subsidiary and parent company both defaulting over the next year. Assume that the subsidiary will default if the parent defaults, but the parent will not necessarily default if the subsidiary defaults.

• Assume that the parent had a 1 year PD = .5% and the subsidiary has 1 year PD of .9%.A. 0.45%B. 0.5%C. 0.545%D. 0.55%

Question


• B – P (S| P) = 1 = P(P & S)/P(P)– P(P & S) = P(P) = 0.5%

Solution


• There is a city which hosts two taxi-cab companies, the Blue Cab Co. and the Green Cab Co. Blue cabs are blue and Green cabs are green; they are otherwise identical. 70 percent of the cabs in the city are Blue cabs, and 30 percent of the cabs in the city are Green cabs. Moreover, historically speaking, Blue cabs have been involved in 70% of all traffic accidents in the city that involved cabs, and Green cabs have been involved in 30% of all traffic accidents in the city that involved cabs. One night, there is a traffic accident involving a taxi-cab in the city, to which there is one witness. Authorities perform extensive tests on the witness, and determine that his ability to recognize cabs by their color at night is approximately 80 percent accurate and 20 percent inaccurate (meaning that when he is wrong he does not say he doesn't know, but rather misidentifies it as being of the other color). The witness says the taxi-cab involved in the accident was 'blue.' On these facts, and strictly assuming the taxi-cab was not from some other city, what is the approximate probability that the taxi-cab involved in the accident belonged to the Blue Cab Co.

Bayes’ Theorem Problem


• Let P(R) be the probability of witness being accurate. Then P(R) = 0.8 which implies P(W) = 0.2 i.e. probability of witness being wrong.

• Let P(B/R) = Probability of accident by a blue car, conditional on the fact that the statement is a right statement.

• Then P(B/R) = 0.7 Also P(B/W)=0.3, Similarly P(G/R) = 0.3• Here we need to find P(R/B) i.e. If the witness has said that it was a blue car, then what's the probability

that it was actually blue.• Applying Bayes Theorem now:

P(R/B) = P(B/R) * P(R)/P(B)• Here we know all except P(B):

P(B) = P(B/R)*P(R) +P(B/W)*P(W)= 0.7*0.8 + 0.3*0.2= 0.56 + 0.06

Therefore: P(R/B) = 0.7*0.8/0.62 = 0.903

Solution


Understanding Bayes’ Theorem

0.7 0.3

Witness Identifies it as:

0.7*.8=0.56

0.3*0.2=0.060.7*0.2

=0.14

0.3*0.8=0.24

Witness is rightWitness is wrong Witness is wrong Witness is right

Total number of caseswhen witness identifies

accident by blue cab(0.56+0.06)=0.62

Cases when witnessrightly identifies it

as blue=0.56

Therefore the probability of the car being actually blue, when the witness identified it as blue equals: (0.56/0.62)=0.903

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Agenda

• Probability• Moments

– Mean, Variance & Covariance, Skewness and Kurtosis (1hr)• Probability distributions



Moments

• A random variable is characterized by its distribution function. Instead of having to report the whole function, it is convenient to summarize it by a few parameters, or moments.

• There are 4 moments:– Mean– Variance– Skewness– Kurtosis

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Mean

• The expected value(Mean) measures the central tendency, or the center of gravity of the population.• It is given by :

• A family has 4 members, father, mother and 2 kids Hemal and Rishi who are twins. The average age of the family members is 20 years. Age of mother and father is 30 and 32 respectively. Can you tell the age of Hemal?

• The graph shows the mean of normal distributions.

N

xXE

n

ii

1)(

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

-4 -2 0 2 4

Standard Normal Distribution

μ=0,σ=2

μ=1,σ=1

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Variance & Standard deviation

• Variance is the squared dispersion around the mean.

• The standard deviation, which is the square root of the Variance, is more convenient to use,as it has the same units as the original variable X

– SD(X) =

N

xVAR

n

ii

1

2)(

)(xVARN

xn

ii

1

2)(


Covariance and Correlation

• Covariance is the measure of how two variables move with each other.– The mathematical expression for covariance of two variables X and Y is given as:

– Here E(X) is the expected value of X and E(Y) is the expected value of Y

• Similarly correlation measures the linear relationship between two variables.– Correlation(ρ) is actually the normalized form of covariance.– Correlation is given by the following expression:

YEYXEXEYXCov ii )(),(

YX

YXCovYX

),(),(

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Some Properties of Variance

• Variance of a constant = 0

• Covariance between same variables is also their variance

• For independent or uncorrelated variables, – covariance or correlation = 0

)()( 2 XVarabaXVar

),(2)()()( 22 YXabCovYVarbXVarabYaXVar

n

ii

n

ii XVarXVar

11)()(

n

i

n

jji

n

ii XXCovXVar

1 11),()(


Skewness

• Skewness describes departures from symmetry

• Skewness can be negative or positive.

• Negative skewness indicates that the distribution has a long left tail, which indicates a high probability of observing large negative values.

• If this represents the distribution of profits and losses for a portfolio, this is a dangerous situation.

31

3)(

n

ii

k

xS

Normal Distribution

Positively Skewed Distribution

Negatively Skewed Distribution


Kurtosis

• Kurtosis describes the degree of “flatness” of a distribution, or width of its tails.

• Because of the fourth power, large observations in the tail will have a large weight and hence create large kurtosis. Such a distribution is called leptokurtic, or fat tailed.

• A kurtosis of 3 is considered average.

• High kurtosis indicates a higher probability of extreme movements.

41

4)(

n

iix

K

0

0.05

0.1

0.15

0.2

0.25

0.3

0.35

0.4

0.45

-4 -3 -2 -1 0 1 2 3 4

PlatykurticK<3Mesokurtic

K=3

LeptokurticK>3


Example

• Assuming that the distribution of ABC stock returns is a population, what is the population variance and standard deviation?

A. 5.0B. 6.8C. 45.22D. 80.2

ABC Annual stock prices

2001 2002 2003 2004 2005 2006

12% 5% -7% 11% 2% 11%


Solution

• B • The population variance is given by taking the mean of all squared deviations from the mean.

– σ2 = [(12-5.67)2 + (5-5.67)2 + (-7-5.67)2 + (11-5.67)2 + (2-5.67)2 + (11-5.67)2] / 6= 45.22 (%2)

The standard deviation is the square root of the variance:

%72.6%22.45

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• The random variables X and Y have variances of 2 and 3 respectively, and covariance of 0.5. The variance of 2X + 3Y is:

A. 13B. 29C. 35D. 41

Example

37


• D– Var(X + Y) = Var(X) + Var(Y) +2*Cov(x,y)– Var(X - Y) = Var(X) + Var(Y) -2*Cov(x,y)– Var(cX) = c^2 * Var(X)– Cov (ax,by) = abCov(x,y)– So, Var(2X + 3Y) = 22 Var(X) +32 Var(Y) +2*2*3*Cov(x,y)– Var(2X + 3Y) = 4*2 + 9*3 + 12*0.5 = 41

Solution

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Question

• You are given the following information about the returns of stock P and stock Q:– Variance of return of stock P=100.0– Variance of return of stock Q=225.0– Covariance between the return of stock P and the return of stock Q=53.2

• At the end of 1999, you are holding USD 4 million in stock P. You are considering a strategy of shifting USD 1 million into stock Q and keeping USD 3 million in stock P. What percentage of risk, as measured by standard deviation of return, can be reduced by this strategy?

A. 0.50%B. 5.00%C. 7.40%D. 9.70%


Solution

• B


FRM Exam 2009

• Which type of distribution produces the lowest probability for a variable to exceed a specified extreme value ‘X’ which is greater than the mean assuming the distributions all have the same mean and variance?A. A leptokurtic distribution with a kurtosis of 4.B. A leptokurtic distribution with a kurtosis of 8.C. A normal distribution.D. A platykurtic distribution


Solution

• D– By definition, a platykurtic distribution has thinner tails than both the normal distribution and any leptokurtic

distribution. Therefore, for an extreme value X, the lowest probability of exceeding it will be found in the distribution with the thinner tails.

– A. Incorrect. A leptokurtic distribution has fatter tails than the normal distribution. The kurtosis indicates the level of fatness in the tails, the higher the kurtosis, the fatter the tails. Therefore, the probability of exceeding a specified extreme value will be higher .

– B. Incorrect. Since answer A. has a lower kurtosis, a distribution with a kurtosis of 8 will necessarily produce a larger probability in the tails.

– C. Incorrect. By definition, a normal distribution has thinner tails than a leptokurtic distribution and larger tails than a platykurtic distribution.

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Agenda

• Probability• Moments• Probability distributions

– Discrete Probability Distributions (1hr)– Continuous Distribution (1hr)



Probability Distribution

• A Random Variable is a function, which assigns unique numerical values to all possible outcomes of a random experiment under fixed conditions. A random variable is not a variable but rather a function that maps events to numbers

– Probability distribution describes the values and probabilities that a random event can take place. The values must cover all of the possible outcomes of the event, while the total probabilities must sum to exactly 1, or 100%

• Example– Suppose you flip a coin twice. – There are four possible outcomes: HH, HT, TH, and TT. – Let the variable X represent the number of Heads that result from this experiment

– It can take on the values 0, 1, or 2. – X is a random variable (its value is determined by the outcome of a statistical experiment)

– A probability distribution is a table or an relation that links each outcome of a statistical experiment with its probability of occurrence

Number of heads (X) Probability P(X=x)

0 0.25

1 0.50

2 0.25

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Continuous & Discrete Probability Distributions

• If a variable can take on any value between two specified values, it is called a continuous variable– otherwise, it is called a discrete variable

• If a random variable is a discrete variable, its probability distribution is called a discrete probability– For example, tossing of a coin & noting the number of heads (random variable) can take a discrete

value– Binomial probability distribution, Poisson probability distribution

• If a random variable is a continuous variable, its probability distribution is called a continuous probability distribution

– The probability that a continuous random variable will assume a particular value is zero– A continuous probability distribution cannot be expressed in tabular form.– An equation or formula is used to describe a continuous probability distribution (called a probability

density function or density function or PDF)– The area bounded by the curve of the density function and the x-axis is equal to 1, when computed over

the domain of the variable– Normal probability distribution, Student's t distribution are examples of continuous probability

distributions


Probability distribution

• Cumulative Probability is a rule or equation which describes the sum of all the probabilities till that observation.

• Take the previous example of flipping of coin twice. The following table gives the probability of occurrence of heads and the cumulative probability as well.

• The point to note here is that the cumulative probability of the first event is equal to the probability of that event.

• The cumulative probability of the last event is always 1

Number of heads (X) Probability P(X=x) Cumulative Probability: P(X < x)

0 0.25 0.25

1 0.50 0.75

2 0.25 1


Bernoulli Distribution

• Assumptions– The outcome of an experiment can either be success (i.e., 1) and failure (i.e., 0).

• Pr(X=1) = p, Pr(X=0) = 1-p, or• The expected value E[X] of the event is equal to the probability of success(p)

E[X] = p • The variance of the event is the product of the probability of success and probability of failure:

Var(X) = p(1-p)


Binomial Distribution

• Assumptions:– There are n trials.– Each trial has two possible outcomes, “success” or “failure”.– The probability of success p is the same for each trial.– Each trial is independent.

• If we take n Bernoulli trials, and say out of those n trials we have total number of “x” successes, then the probability of such an event can be given as:

• The expected number of successes E[X] = n * p

• The variance of number of successes Var (X) = n * p * (1-p)

)()1(**)( xnxx

n ppCxXP


Example

• There are 10 bonds in a credit default swap basket; the probability of default for each of the bonds is 5%. The probability of any one of the bond defaulting is completely independent of what happens to the other bonds in the basket. What is the probability exactly one bond default?

A. 5%B. 50%C. 32%D. 3%


Solution

• C– one particular bond defaults and other nine do not with the probability 0.05* (0.95)^9 can happen in 10

different ways. – = 10 * 0.05^1* (0.95)^9 = 32%

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• Company ABC was incorporated on January 1, 2004. it has expected annual default rate of 10%. Assuming a constant quarterly default rate, what is the probability that company ABC will not have defaulted by April 1, 2004?

A. 2.4%B. 2.5%C. 97.4%D. 97.5%

Question


Solution

• C• Let the probability of not defaulting in 1 quarter is (nd). Then the probability of not defaulting for a full

year is (nd)4. This implies that the probability of defaulting within 1 year time is {1-(nd)4}, which is given as 10%.

1-(nd)4=0.1 which implies (nd)=0.91/4

= 97.4%

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Question

• A corporate bond will mature in 3 years. The marginal probability of default in year one is 0.03%. The marginal probability of default in year 2 is 0.04%. The marginal probability of default in year 3 is 0.06%. What is the cumulative probability that default will occur during the 3 year period?

A. 0.1247%B. 0.1276%C. 0.1299%D. 0.1355%


Solution

• C– The cumulative probability of default= 1-{Product of marginal probabilities of not defaulting}

– =1-{(1-0.0003)*(1-.0004)*(1-0.0006)}– =0.001299– Therefore the cumulative probability of default is 0.1299%


Poisson Distribution

• Assumptions:– The probability of observing a single event over a small interval is approximately proportional to the size

of that interval.– The probability of an event within a certain interval does not change over different intervals.– The probability of an event in one interval is independent of the probability of an event in any other

interval which is not overlapping.• Poisson distribution is a special case of Binomial distribution when the probability of success (p)

becomes very small and the number of events (n) becomes very large in such a way that the product of both gives a constant ().

– Fix the expectation =n * p– Number of trials n– A Binomial distribution will become a Poisson distribution

• E[X] = , Var(X) =

otherwise0

0!)( xe

xxXPx

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Plots of Poisson Distribution

0

0.050.1

0.150.2

0.250.3

0.350.4

0 2 4 6 8 10 12 14 16 18 20 22 240

0.020.040.060.08

0.10.120.140.160.18

0.2

0 2 4 6 8 10 12 14 16 18 20 22 24 26 28

0

0.02

0.04

0.06

0.08

0.1

0.12

0 2 4 6 8 10 12 14 16 18 20 22 240

0.010.020.030.040.050.060.070.080.09

0 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 54

λ=1 λ=5

λ=25λ=15


Question

• When can you use the Normal distribution to approximate the Poisson distribution, assuming you have "n" independent trials each with a probability of success of "p"? A. When the mean of the Poisson distribution is very small.B. When the variance of the Poisson distribution is very small. C. When the number of observations is very large and the success rate is close to 1.D. When the number of observations is very large and the success rate is close to 0.


Solution

• C– The Normal distribution can approximate the distribution of a Poisson random variable with a large lambda

parameter (λ). This will be the case when both the number observations (n) is very large and the success rate (p) is close to 1 since λ = n*p.

– INCORRECT: A, The mean of a Poisson distribution must be large to allow approximation with a Normal distribution.

– INCORRECT: B, The variance of a Poisson distribution must be large to allow approximation with a Normal distribution.

– INCORRECT: D, The Normal distribution can approximate the distribution of a Poisson random variable with a large lambda parameter (λ). But since λ = n*p, where n is the number observations and p is the success rate, λ will not be large if p is close to 0.

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Example

• The number of false fire alarms in a suburb of Houston averages 2.1 per day. What is the (approximate) probability that there would be 4 false alarms on 1 day?

A. 1B. 0.1C. 0.5D. 0


Solution

• B– Use Poisson distribution– P(X = x) = [x *e-]/ !– Is there any other intuitive way as well???


Uniform distribution

• The simplest continuous distribution function is the uniform distribution. This is defined over a range of values for x, a ≤ x ≤ b.

• The density function is,– f (x) =1 /(b− a) , a ≤ x ≤ b

• Its mean and variance are given by– E(X) =(a + b)/ 2– V(X) = (b− a)2 /12


Question

• Assume we use continuous uniform distribution U(0,10) to generate a series of random numbers. Which of the following statements is Correct?

A. The number 5 is likely to be observed much more often than any other number.B. Numbers between 4 and 6 are more likely to occur than the number between 6 and 10, because the first

interval is closer to the center of the distribution.C. Numbers between 1 and 3 are as likely as the number between 4 and 6. D. Numbers between 1 and 3 are less likely than the number between 4 and 6, due to skewness of the

distribution.


Solution

• C


Standard Normal Distribution

The standard normal distribution has mean = 0 and standard deviation sigma=1


Normal (Gaussian) Distribution

• The normal distribution is defined by first two moments, mean () and variance (2)• The probability density function P(x) of normally distributed variable is given by:

• The probability of the value lying between a and b is given by:

• The expected value of a normally distributed variable: E[X]= , • The variance of normally distributed variable: Var(X)= 2

• If two variables are individually normally distributed, then the linear combination of the both is also normally distributed.

– Lets take an example of two variable X1 and X2 which are normally distributed such that:– X1~N(1,1) and X2~N(2,2)– Then X= a.X1+ b.X2 is also normally distributed.

2

2

2 2)(exp

21)(

xxP

b

a

dxxPbXaP ).()(

The skewness of normal distribution is = 0 and the kurtosis is =3

65


Question

• Suppose the standard deviation of a normal population is known to be 10 and the mean is hypothesized to be 8. Suppose a sample size of 100 is considered. What is the range of sample means that allows the hypothesis to be accepted at a level of significance of 0.05?

A. Between -11.60 and 27.60B. Between 6.04 and 9.96C. Between 6.355 and 9.645D. Between -8.45 and 24.45


Solution

• B– To accept the hypothesis at a 0.05 significance level, the test statistic Z must fall between -1.96 and 1.96

– Z = (X - 8) / (10 / Sqrt(100)) and -1.96 <=z<=1.96

– which implies that the sample mean X must be between 6.04 and 9.96.

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Example: FRM EXAM 2002

• Consider a stock with an initial price of $100. Its price one year from now is given by S = 100×exp(r), where the rate of return r is normally distributed with a mean of 0.1 and a standard deviation of 0.2. With 95% confidence, after rounding, S will be between

A. $67.57 and $147.99B. $70.80 and $149.20C. $74.68 and $163.56D. $102.18 and $119.53


Solution

• C. – Note that this is a two-tailed confidence band, so that α = 1.96. – We find the extreme values from $100exp(µ ± ασ). – The lower limit is then V1 = $100 * exp(0.10 − 1.96 * 0.2) = $100 * exp(−0.292) = $74.68. The upper limit

is– V2 = $100 * exp(0.10 + 1.96 × 0.2) = $100 * exp(0.492) = $163.56.

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Question

• Let X be a uniformly distributed random variable between minus one and one so that the standard deviation of X is 0.577. What percentage of the distributions will be less than 1.96 standard deviations above the mean:

A. 100%B. 97.5%C. 95%D. Insufficient information provided.


Solution

• A– The answer requires understanding of distributions and standard deviation. The key is that every

distribution has a standard deviation. However the number of standard deviations associated with different probabilities are different for each distribution. In this case 1.96 standard deviation represents a move of 1.12 or less. As the total distribution is defined as falling between minus one and one the correct answer is ‘A’.

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Question

• For the standard normal distribution, calculate the value of P(-1.87 ≤ Z ≤ 1.23) or P(|Z| ≤ 1.6)?A. 0.5683B. 0.8794C. 0.7831D. 0.9145


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Solution

• D• In the diagram given below, the area representing the region P(-1.87 ≤ Z ≤ 1.23) or P(|Z| ≤ 1.6) is

shown below. The area will be from Z = -1.87 to Z = 1.6 and common area is from Z = 1.6 to Z = 1.23.• P(-1.87 ≤ Z ≤ 1.23) or P(|Z| ≤ 1.6) = P(Z ≥ 1.87) + P(Z ≥ 1.6) = 0.4693 + 0.4452 = 0.9145• Hence option ‘D’ is correct.



Agenda

• Appendix


Appendix: Lognormal Distribution

• A random variable X is said to have a lognormal distribution if its logarithm Y = ln(X) is normally distributed.

• The lognormal density function has the following expression:

2

2

2 2))(ln(exp

21P(x)

xx

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Appendix: The Chi-square Distribution

• The chi-square distribution is a family of distributions, depending on degrees of freedom:

• d.f. = n - 1

0 4 8 12 16 20 24 28

d.f. = 15

20 4 8 12 16 20 24 28

d.f. = 5

20 4 8 12 16 20 24 28

d.f. = 1

2


• Assume you have empirical data showing historical returns (v) for a given financial variable (e.g.: Forex rate), how could you perform a quick test of the validity of the power law Prob(v > x) = K * x-a where x is large, as a good model of the tail of the distribution?

A. Plot the probability of v exceeding x standard deviations against xB. Plot the probability of v exceeding x standard deviations against Log of xC. Plot the Log of the probability of v exceeding x standard deviations against xD. Plot the Log of the probability of v exceeding x standard deviations against the Log of x

Appendix: Example FRM EXAM

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Solution

• D– The mathematical relationship in the question can be rewritten (by taking the logs on both sides):

Log(Prob(v > x)) = Log(K) - aLog(x), i.e. the plot of the Log of the probability of v exceeding x standard deviations against the log of x should be a straight (decreasing) line if the relationship strictly holds. The intercept is an estimate of Log of K and the slope of the line yields the parameter a.

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Appendix: Sum Rule & Bayes’ Theorem

• Jack has 3 white balls and 2 red balls in his box while his friend Andrew has 4 white and 5 red balls.– Andrew took 1 white ball from jack and gave him 1 red ball in compensation. Now calculate the

probability of picking a red ball from Andrew’s box.– After the exchange, Tom stole a ball from one of the boxes and found that it’s white. If you have to tell

who lost his white ball, what should be your say?


Solution

• Initially Jack has 3 white balls and 2 red balls in his box while his friend Andrew has 4 white and 5 red balls.

• After the exchange Jack has 2 white and 3 red balls, and Andrew has 5 white and 4 red balls. Therefore the probability of picking a red ball from Andrew’s box is:

P(RAndrew)=4/(5+4)=4/9• Now Tom stole a white ball from one of the two boxes. To make a calculated guess about

who lost 1 white ball, we need to calculate the conditional probabilities.• P(Jack’s box/If the balls is White)= Probability of white balls in Jack’s box/(Probability of white

ball in Jack’s box +Probability of white ball in Andrew’s box)•

• Similarly, P(Andrew’s box/White)

• Point to note here is that the white ball can come from 2 boxes only, so the sum of conditional probabilities of the boxes, given the ball is white should sum to 1, which is (18/43+25/43) =1 in our case.

43/189/55/2

5/2

43/259/55/2

9/5

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• There are 10 sprinters in the Olympic finals. How many ways can the gold, silver, and bronze medals be awarded?

A. 120B. 720C. 1,440D. 604,800

Appendix: Question


• B – 10P3 = 720

Solution

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SS1 Quantitative Analysis I