Quantitative Analysis I

8/13/2019 Quantitative Analysis I

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Quantitative Analysis- I


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Agenda

Probability

o Counting Principle (1 hr)

o Probability Concepts, Bayes Theorem (1 hr)

Moments (1 hr)

Probability distributions (2 hr)


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Counting Principle

Number of ways of selecting r objects out of n objects

o nCr

o n!/(r!)*(n-r)!

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

o N^r


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Example-Counting Principle

In how many ways 3 stocks can be chosen out of 10 stocks in a portfolio?(Combination)

o Choosing 3 out of 10 stocks is basically the number of combinations of 3 objects out of 10

o Therefore, the number of ways are

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?

o First stock can be sold in 10 ways

o Second can be sold in again 10 ways

o Third stock can again be sold in 10 ways

o Therefore total number of ways become =103 =1000


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Question

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. 80

B. 700

C. 1,716

D. 100,800


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Solution

Answer: B

o Solution: C(5, 3) x C(8, 4)


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Probability - Definitions

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

Examples of Definitions:

o Experiment: Roll a die two hundred times noting the outcomes

o Event of interest: A six faces upwards

o Trial: Roll the die once

o Number of trials: 200

o 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

o For example, if the event is "occurrence of an even number when a die is rolled

o 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

o 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.

o Notation for joint probability takes the form:

P(X Y) or P(X,Y)

Which reads the joint probability of X and Y.

o The following table shows the joint probability of different events. Lets say an economist is

predicting the market scenario and the price of IBM stock fro the next year.

o Next year market can be Good, Bad or Neutral

o IBM stock may go up or go down

o The probability of IBM stock being Up and Market being Good is 10%

o Similarly, the probability of IBM stock being down and Market being neutral is 40%


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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


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Conditional and Unconditional Probabilities

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?


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Some definitions and properties of Probability

Definitions

o Mutually Exclusive: If one event occurs, then other cannot occur

o Exhaustive: all exhaustive events taken together form the complete sample space (Sum of

probability =1)

o 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

o For example, if the probability of a company going bankrupt within one year period is 20%,

then the

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


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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%


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Solution

Answer:

o Both would avoid default only if None defaults

o This implies that first does not default AND second does not default

o = (1PD (first)) x (1PD (second))


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Some Properties of Probability

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.


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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 stock?

I. 0.5382

II. 0.7821III. 0.6435

IV. None of these


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Solution

Answer: III

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 II is correct.


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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.

o 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.

o We know that P(AB) = P(B/A).P(A)

o Also P(BA)= P(A/B).P(B)

o Now equating both P(AB) and P(BA) we get:

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


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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 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. 30

B. 40

C. 37.5

D. 42.5


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Solution

Answer: C

o 25% are given new treatment =>75% are given old treatment.

o P(Cure) = P(Cure/New) * P(New) + P(Cure/Old) * P(Old) = 0.375

o So out of 100. 37.5 will get cured


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Example

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. 0.375

B. 0.425

C. 0.4

D. 0.425


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Solution

Answer: C

o Apply Bayes Theorem


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Question

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 necessarilydefault if the subsidiary defaults.

Assume that the parent had a 1 year PD = 0.5% and the subsidiary has 1 year PD of 0.9%.

A. 0.45%

B. 0.5%

C. 0.545%

D. 0.55%


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Solution

Answer: B

o P (S| P) = 1 = P(P & S)/P(P)

o P(P & S) = P(P) = 0.5%


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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 functionthat 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 two times.

There are four possible outcomes: HH, HT, TH, and TT.

Let the variable X represent the number of Heads that result from this experiment

o It can take on the values 0, 1, or 2.

o 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 statisticalexperiment with its probability of occurrence



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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



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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 variableo otherwise, it is called a discrete variable

If a random variable is a discrete variable, its probability distribution is called a discrete probability

o For example, tossing of a coin & noting the number of heads (random variable) can take a

discrete value

o Binomial probability distribution, Poisson probability distribution

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

probability distribution

o The probability that a continuous random variable will assume a particular value is zero

o A continuous probability distribution cannot be expressed in tabular form.

o An equation or formula is used to describe a continuous probability distribution (called a

probability density function or density function or PDF)

o 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

o Normal probability distribution, Student's t distribution are examples of continuous probability

istributions



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Agenda

Probability

Moments

o Mean, Variance & Covariance, Skewness and Kurtosis (1hr)

Probability distributions



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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:

o Mean

o Variance

o Skewness

o 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 3 different distributions.



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Variance & Standard deviation

Variance is the squared dispersion around the mean.

The standard deviation is more convenient to use, as it has the same units as the original variable X

o SD(X) =

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Covariance and Correlation

Covariance is the measure of how two variable move with each other.

o The mathematical expression for covariance of two variables X and Y is given as:

o 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 variable.

o Correlation() is actually the normalized form of covariance.

o Correlation is given by the following expression:



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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



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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.


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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.



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Example

Assuming that the distribution of ABC stock returns is a population, what is the population standard

deviation?

A. 5.0

B. 6.8

C. 45.22

D. 80.2



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Solution

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

2 = [(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:


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Example

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. 13

B. 29

C. 35

D. 41


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Solution

Answer: C

o Var(X + Y) = Var(X) + Var(Y) +2*Cov(x,y)

o Var(X - Y) = Var(X) + Var(Y) -2*Cov(x,y)

o Var(cX) = c^2 * Var(X)

o Cov (ax,by) = abCov(x,y)

o So, Var(2X + 3Y) = 22 Var(X) +32 Var(Y) +2*2*3*Cov(x,y)

o Var(2X + 3Y) = 4*2 + 9*3 + 12*0.5 = 41


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Question

You are given the following information about the returns of stock P and stock Q:

o Variance of return of stock P=100.0

o Variance of return of stock Q=225.0

o 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%


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Answer

Answer: B


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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 meanand 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


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Answer

Answer: D

By definition, a platykurtic distribution has thinner tails than both the normal distribution andany 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 willnecessarily 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

Important Probability Distributions

Discrete Probability Distributions (1hr)

Continuous Distribution (1hr)


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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)


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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) = np (1p)


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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%


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Solution

Answer: C

one particular bond defaults and other nine do not with the probability 0.05* (0.95)^9 canhappen in 10 different ways.

= 10*0.05* (0.95)^9 = 32%


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Question

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%


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Answer

Answer: C

Let the probability of not defaulting in 1 quarter is (nd). Then the probability of not defaulting for afull year is (nd)4. This implies that the probability of defaulting within 1 year time is {1 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%


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Answer

Answer: 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%


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Poisson Distribution

Assumptions:

The probability of observing a single event over a small interval is of that interval.

The probability of an event within a certain interval

The probability of an event in one interval is independent 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 =np

Number of trials n

A Binomial distribution will become a Poisson distribution

E[X] = , Var(X) =


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Plots of Poisson Distribution


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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"?

When the mean of the Poisson distribution is very small.

When the variance of the Poisson distribution is very small.

When the number of observations is very large and the success rate is close to 1.

When the number of observations is very large and the success rate is close to 0.

As sample size gets large (typically > 30)

Sampling distribution becomes almost normal regardless of shape of population


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Answer

Answer: C

The Normal distribution can approximate the distribution of a Poisson random variable with alarge lambda parameter(). This will be the case when both the number observations (n) is very

large and the success rate (p) is close to 1 since = 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(). But since = n*p, where n is the number

observations and p is the success rate, will not be large if p is close to 0.


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Example

A. 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. 1

B. 0.1

C. 0.5

D. 0


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Solution

Answer: B

Use Poisson distribution

P(X = x) = [x *e-]/ !

Is there any other intuitive way as well???


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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


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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.


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Solution

Answer: C


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Standard Normal Distribution

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


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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.

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


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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.


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Answer

Answer: A

The answer requires understanding of distributions and standard deviation. The key is that everydistribution 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 represent

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)?

0.5683

0.8794

0.7831

0.9145


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Solution

Answer: D

In the diagram given below, the area representing the region P(-1.87 Z 1.23) or P(|Z| 1.6) isshown 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.


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Agenda

Appendix


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Appendix: Example FRM EXAM

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 x

B. Plot the probability of v exceeding x standard deviations against Log of x

C. Plot the Log of the probability of v exceeding x standard deviations against x

D. Plot the Log of the probability of v exceeding x standard deviations against the Log of x


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Solution

Answer: D

The mathematical relationship in the question can be rewritten (by taking the logs on bothsides): 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 calculatethe probability of picking a red ball from Andrewsbox.

After the exchange, Tom stole a ball from one of the boxes and found that its white. If you

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


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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 Andrewsbox 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(Jacksbox/If the balls is White)= Probability of white balls in Jacksbox/(Probability of white ball

in Jacksbox +Probability of white ball in Andrewsbox)

Similarly, P(Andrewsbox/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.


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Appendix: Question

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

be awarded?A. 120

B. 720

C. 1,440

D. 604,800


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Solution

Answer: B

10P3 = 720


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Bayes Theorem Problem

There is a city which hosts two taxi-cab companies, the Blue 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 inthe city are cabs have been involved in 70% of all traffic accidents in the city that involved cabs, and

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 doesnt know, but rather misidentifies it as being of the other color). The witness says

the taxi accident was 'blue.' On these facts, and strictly assuming the taxi is the approximate

probability that the taxi-cab involved in the accident belonged to the Blue Cab Co. and the Green CabCo. Blue cabs Green cabs. Moreover, historically speaking, Blue Green cabs taxi-cab involved in the

taxi-cab was not from some other city, what Cab Co.


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Analysis


Solution

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


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Understanding Bayes Theorem

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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FRM Exam 2006

Which of the following statements is the most accurate about the relationship between a normal

distribution and a Students t-distribution that have the same mean and standard deviation?A. They have the same skewness and the same kurtosis.

B. The Students t-distribution has larger skewness and larger kurtosis.

C. The kurtosis of a Students t -distribution converges to that of the normal distribution as the

number of degrees of freedom increases.

D. The normal distribution is a good approximation for the Students t-distribution when the

number of degrees of freedom is small.


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Answer

Answer: C

The skewness of both distributions is zero and the kurtosis of the Students t distribution converges tothat of the normal distribution as the number of degrees of freedom increases.

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Quantitative Analysis I