week 1 quantitative analysis of financial markets basic statistics a · 2017-10-16 · michael...

Introduction Central Tendency Dispersion Portfolio Variance and Hedging Takeaways

Week 1Quantitative Analysis of Financial Markets

Basic Statistics A

Christopher Ting

Christopher Ting

http://www.mysmu.edu/faculty/christophert/

k: [email protected]: 6828 0364

ÿ: LKCSB 5036

October 14, 2017

Christopher Ting QF 603 October 14, 2017 1/36

http://www.mysmu.edu/faculty/christophert/

mailto:[email protected]


Table of Contents

1 Introduction

2 Central Tendency

3 Dispersion

4 Portfolio Variance and Hedging

5 Takeaways



Introduction

' Statistics as a discipline refers to the methods we use to analyzedata. Statistical methods fall into one of two categories:descriptive statistics or inferential statistics.

' Descriptive statistics summarize the important characteristics oflarge data sets. The goal is to consolidate numerical data intouseful information.

' Inferential statistics pertain to the procedures used to makeforecasts, estimates, or judgments about a large set of data on thebasis of the statistical characteristics of a smaller set (a sample).

' In risk management, we often need to describe the relationshipbetween two random variables. Is there a relationship betweenthe returns of an equity and the returns of a market index?



Learning Outcomes of QA02

Chapter 3.Michael Miller, Mathematics and Statistics for Financial Risk Management, 2nd Edition(Hoboken, NJ: John Wiley & Sons, 2013).

' Interpret and apply the mean, standard deviation, and variance ofa random variable.

' Calculate the mean, standard deviation, and variance of a discreterandom variable.

' Calculate and interpret the covariance and correlation betweentwo random variables.

' Calculate the mean and variance of sums of variables.



Population vs. Sample

* A population is defined as the set of all possible members of astated group.

* Examples

1 Cross Section: All stocks listed on the Nasdaq

2 Time Series: Dow Jones Industrial Average Index

* It is frequently too costly or time consuming to obtainmeasurements for every member of a population, if it is evenpossible.

* A sample is a subset randomly drawn from the population.



Central Tendency: Mean

* Population mean of N entities

E(X):= µ =

1

N

N∑i=1

Xi, ∀ i = 1, 2, . . . , N.

* Sample mean is an estimate of the true mean µ:

Xn =1

n

n∑j=1

Xj , ∀ j = i1, i2, . . . , in.

* The law of large numbers for a random variable X states that, fora sample of independently realized values, x1, x2, . . . , xn,

limn→∞

1

n

n∑i=1

xi = µ,

* What is the point to calculate the sample mean?Christopher Ting QF 603 October 14, 2017 6/36


Estimator

* The (functional) form of the estimator X in Slide 6 is an estimator.

* You can also define the sample average alternative as

︷︸︸︷X :=

1

n+ 1

n∑i=1

xi.

* Which is better, X or︷︸︸︷X ?



Unbiasedness and Consistency

* UnbiasednessAn estimator θ̂n of a population statistic θ is said to be unbiasedwhen E

(θ̂n

)= θ.

* ConsistencyAn estimator θ̂n of a population statistic θ is said to be unbiasedwhen limn→∞ θ̂n → θ in probability. Namely, for any ε > 0.

limn→∞

P(∣∣θ̂n − θ∣∣ ≥ ε) = 0.

* Exercise: Show that the estimator X̂ for the mean µ is unbiased

but estimator︷︸︸︷X is biased.

* Exercise: Show that the estimators X̂ and︷︸︸︷X are consistent.

* Is θ̂7 unbiased? Consistent?



Independently and Identically Distributed (i.i.d.)

* The concept of independence is a strong condition. Two randomvariables are independent when they are not related in any way.

* “Identical” means that the two random variables are the samefrom the statistical standpoint. They follow the same probabilitydistribution, hence the same descriptive statistics.

* X is a random variable, but it has many copies. With the subscripti, Xi is a copy of X when it will realize a value for the i-th time.

* Question: Is the sample mean Xn a random variable?



Expected Value: Discrete and Continuous

* For a discrete random variable with n possible outcomes, supposethe probabilities are

P(X = xi) = pi, i = 1, 2, . . . , n.

The mean is

µ = E(X)=

n∑i=1

pixi.

* For a continuous random variable with probability density functionf(x) and cumulative distribution function F (x), the mean is givenby the integration:

µ = E(X)=

∫ ∞−∞

x f(x) dx =

∫ ∞−∞

x dF.



Linearity of Expectation

* The expectation operator E(·)

is linear. That is, for two randomvariables, X and Y , and a constant, c, the following two equationsare true:

E(cX)= cE

(X)

E(X + Y

)= E

(X)+ E

(Y)

* Having introduced two constants a and b, show that

E(aX + bY

)= aE

(X)+ bE

(Y).



Probability and Expected Value

* Tossing a fair coin and the random variable X.

X =

{1, if ω = ‘Head’;0, if ω = ‘Tail’.

* Suppose P(ω = ‘Head’

)=

1

2= P

(ω = ‘Tail’

).

* Quiz: What is the value of E(X)?

* Suppose the coin is not fair and P(ω = ‘Head’

)= 0.6.

1 What is the value of E(X)?

2 Suppose x1, x2, . . . , xn are the results of tossing the unfair coin n

times. What is the value of1

n

n∑i=1

xi if n is very large?



Probability and Expected Value (cont’d)

* Consider the indicator variable 1A, which is defined as

1A =

{1, if ω ∈ A;0, if ω ∈ Ac.

What is E(1A)?



Central Tendency: Median and Mode

* The median of a discrete random variable is the value such thatthe probability that a value is less than or equal to the median isequal to 50%.

P(X ≤ m

)= P

(x ≥ m

)=

1

2.

* The median is found by first ordering the data and then separatingthe ordered data into two halves.

* The mode of a sample is the value that has the highest frequencyof occurrences.

* Question: Calculate the mean, median, and mode of the followingdata set.

−20%,−10%,−5%,−5%, 0%, 10%, 10%, 10%, 19%



Sample Problem

* At the start of the year, a bond portfolio consists of two bonds,each worth $100. At the end of the year, if a bond defaults, it willbe worth $20. If it does not default, the bond will be worth $100.The probability that both bonds default is 20%. The probabilitythat neither bond defaults is 45%. What are the mean, median,and mode of the year-end portfolio value?

* Solution1 If both bonds default, then the portfolio value V will be $20 + $20 =

$40. The problem says that P(V = $40

)= 20%.

2 If neither bond defaults, then V will be $100 + $100 = $200. Theproblem says that P

(V = $200

)= 45%.

3 So the probability of one of the two bonds defaults isP(V = $120

)= 1− 0.2− 0.45 = 35%.

4 Hence, E(V)= 0.2× $40 + 0.35× $120 + 0.45× $200 = $140.

5 The mode is $200, as it occurs with the highest probability of 45%.6 The median is $120; half of the outcomes are less than or equal to

$120.Christopher Ting QF 603 October 14, 2017 15/36


Another Sample Problem

* Recall the probability density function f(x) =8

9x for x ∈

(0,

3

2

].

* To calculate the median, we need to find m, such that the integralof f(x) from the lower bound of f(x), zero, to m is equal to 0.50.∫ m

0f(x) dx = 0.5.

Solving for m, we find m =1

2

√3

2* To find the mean, we compute

µ =

∫ 3/2

0x f(x) dx

to find that µ = 1.



A Measure of Dispersion

# Variance is defined as the expected value of the differencebetween the variable and its mean squared:

σ2 := E((X − µ)2

)=: V

(X)

The symbol σ2 is often used to denote the variance of the randomvariable X with mean µ.

# The square root of variance, σ, is the standard deviation.# The mean µ of investment return is often referred to as the

expected return. The Standard deviation of investment return R isreferred to as volatility.

# Volatility is not risk.# Exercise: Show that

σ2 = E(X2)− µ2.

# Exercise: Compute the variance of X in Slide 12.Christopher Ting QF 603 October 14, 2017 17/36


A Property of Variance

# Prove that, with c being a constant,

V(cX)= c2V

(X).

# Proof:1 Let Y = cX.2 µY = E

(cX)= cE

(X)=: cµX

3 By definition, σ2Y = E

((Y − µY

)2).4 Therefore,

σ2Y = E

((cX − cµX

)2)= E

(c2(X − µX

)2)= c2 E

((X − µX

)2)= c2 V

(X).

5 Since σ2Y = V

(cX), the proof is complete.



Sample Variance

# The sample average or sample mean of a random variable X is

Xn =1

n

n∑i=1

Xi.

# The sample variance, as an estimator, is defined as

σ̂2n =1

n− 1

n∑i=1

(Xi −Xn

)2.

# Why divided by n− 1 and not n? Alternative estimator of samplevariance is

σ̃2n =1

n

n∑i=1

(Xi −Xn

)2.

# Which is the “correct” one?Christopher Ting QF 603 October 14, 2017 19/36


Sample Variance σ̂2n is Unbiased.

# First we show thatn∑

i=1

(Xi −Xn

)2=

n∑i=1

(X2

i − 2XiXn +X2n

)=

n∑i=1

X2i − 2X

n∑i=1

Xi + nX2n.

# Since nXn =n∑

i=1

Xi, we have

n∑i=1

(Xi −Xn

)2=

n∑i=1

X2i − nX

2.

# Note that E(X2i ) = σ2 + µ2.



Sample Variance σ̂2n is Unbiased. (cont’d)

# Next, we need to calculate E(X

2n

). Let Y := Xn.

# Note that for any random variable Y , E(Y 2)= V

(Y)+ µ2.

# The variance of the sample average is, by the assumption of theindependence of Xi,

V(Y)= V

(1

n

n∑i=1

Xi

)=

1

n2V

(n∑

i=1

Xi

)

=1

n2

n∑i=1

V(Xi

)=

1

n2× nσ2

=σ2

n.



Sample Variance σ̂2n is Unbiased. (cont’d)

# If follows that

E(Y 2)= E

(X

2n

)=σ2

n+ µ2.

# To show unbiasedness, we need to prove that E(σ̂2)= σ2. Noting

that E(X2i ) = σ2 + µ2, we find

E(σ̂2)=

1

n− 1

(n∑

i=1

E(X2

i

)− nE

(X

2n

))

=1

n− 1

(n∑

i=1

(σ2 + µ2

)− n

(σ2

n+ µ2

))

=1

n− 1

((n− 1)σ2

)= σ2.



Variance for a Continuous Random Variable

# Definitionσ2 =

∫ ∞−∞

(x− µ

)2f(x) dx

# Exercise: Suppose the probability density function is f(x) =8

9x

for x ∈(0,

3

2

]. Compute the variance.



Standardized Variables

# Suppose X is a random variable with constant mean µ andvariance σ2. Since volatility σ 6= 0 for a random variable, we candefine

Y :=X − µσ

.

# The variable Y has mean zero and variance 1.

# Quiz: If X is a stochastic process dXt = µdt+ σ dBt, what is thestochastic process for Y t?



Covariance

# Covariance is a generalized version of variance. It is defined as

C(X,Y ) ≡ σXY := E((X − µX

)(Y − µY

)).

# Variance is a special case: C(X,X) = σXX = V(X).

# Whereas variance is strictly positive, covariance can be positive,negative, and zero.

# If X and Y are independent, then it must be that C(X,Y ) = 0.# If C(X,Y ) = 0, it is not necessarily true that X and Y are

independent.# Exercise: Show that

1 σXY = E(XY

)− µXµY .

2 C(X,Y

)= C

(Y ,X

).

3 C(X + Y , Z

)= C

(X,Z

)+ C

(Y , Z

).



Estimators of Covariance

# Given the paired data, (xi, yi), i = 1, 3, . . . , n, the samplecovariance is defined as

σ̂XY =1

n− 1

n∑i=1

(xi − µ̂X

)(yi − µ̂Y

).

# C(Xi, Yj

)= 0, if i 6= j.

Proof: Suppose Y and X are related by a mapping f(·), i.e.,Yj = f(Xj). Note that the mapping involves the paired copiesbecause each Yj is independent and does not relate at all with Yi.Otherwise, if Yj = f(Xi, Xj), then Yj may depend on Yi indirectlythrough since Yi = f(Xh, Xi).

# Homework Assignment: Show that

1

n∑i=1

(Xi − µ̂X

)(Yi − µ̂Y

)=

n∑i=1

XiYi − nXnY n.

2 Use these results to show that the sample covariance is unbiased.Christopher Ting QF 603 October 14, 2017 26/36


Linear Combination of Two Random Variables

# Suppose X and Y are a pair random variables with meansµX = E(X) and µY = E(Y ), respectively. Also, suppose a and bare two constants. Prove that

V(aX + bY

)= a2V

(X)+ b2V

(Y)+ 2abC

(X,Y

).

# Proof1 V

(aX + bY

)= E

((aX + bY )2

)− (aµX + bµY )

2.2 Expanding the two quadratic term and collecting the expanded

terms accordingly, we obtain

a2 E(X2)− a2µ2

X + b2 E(Y 2)− b2µ2

Y + 2abE(XY

)− 2abµXµY ,

which is

a2(E(X2)− µ2

X

)+ b2

(E(Y 2)− µ2

Y

)+ 2ab

(E(XY

)− µXµY

).



Correlation: Normalized Covariance

# The normalization of covariance gives rise to correlation, which isdefined as ρXY :=

σXY

σXσY.

# Correlation has the nice property that it varies between -1 and +1.If two variables have a correlation of +1 (-1), then we say they areperfectly correlated (anti-correlated).

# If one random variable causes the other random variable, or thatboth variables share a common underlying driver, then they arehighly correlated.

# But in general, high correlation does not imply causation of onevariable on the other.

# If two variables are uncorrelated, it does not necessarily followthat they are unrelated.

# So what does correlation really tell us?



Sample Problem

# If X has an equal probability of being -1, 0, or +1, what is thecorrelation between X and Y if Y = X2?

# First, we calculate the respective means of both variables:

E(X)=

1

3(−1) + 1

3(0) +

1

3(1) = 0.

E(Y)=

1

3((−1)2) + 1

3(02) +

1

3(12) =

2

3.

# The covariance can be found as follows:

σXY =1

3

((−1− 0)((−1)2 − 2/3) + (0− 0)(02 − 2/3)

+ (1− 0)(12 − 2/3))= 0.

# So, even though X and Y are clearly related (Y = X2), theircorrelation is zero!



Portfolio Variance

e If we have two securities with random returns XA and XB, withmeans µA and µB and standard deviations σA and σB,respectively, we can calculate the variance of XA plus XB asfollows:

σ2A+B = σ2A + σ2B + 2ρABσAσB,

where ρAB is the correlation between XA and XB.

e If the securities are uncorrelated, then σ2A+B = σ2A + σ2B.

e In general, suppose Y =

n∑i=1

Xi. The portfolio’s variance is

σ2Y =

n∑i=1

m∑i=j

ρijσiσj .



Square Root Rule

e Suppose Xi is a copy of X such that σi = σ for all i, and that all ofthe Xi’s are uncorrelated, i.e., ρij = 0 for i 6= j. Then,

σY =√nσ.

e Consider the time series of weekly i.i.d. returns. The volatility isσ = 2.06%. What is the annualized volatility?AnswerAssume that one year has 52 weeks. Using the square root rule,we obtain 2.06%×

√52 = 14.85%.

e If i.i.d. fails to hold, square root rule may lead to a misleadingvalue.



Application: Static Hedging

e If the portfolio P is a linear combination of XA and XB, i.e.,P = aXA + bXB, then

σ2P = a2σ2A + b2σ2B + 2abρABσAσB.

e Correlation is central to the problem of hedging.

e Let a = 1, i.e., XA is our primary asset. What should the hedgeratio b be such that the portfolio variance σ2P is the smallestpossible?

e The first-order condition with respect to b is

dσ2Pdb

= 2bσ2B + 2ρABσAσB = 0.



Application: Static Hedging (cont’d)

e The optimal hedge ratio is

b? = −ρABσAσB

= −C(XA, XB

)V(XB

) .

e If b? is positive (negative), long (short) the asset B.

e Substituting b? back into our original equation, the smallestvolatility you can achieve for the hedged portfolio is

σ?P = σA

√1− ρ2AB.

e When ρAB equals zero (i.e., when the two securities areuncorrelated), the optimal hedge ratio is zero. You cannot hedgeone security with another security if they are uncorrelated.



Puzzling?

e Adding an uncorrelated security to a portfolio will always increaseits variance!

e For example, $100 of Security A plus $20 of uncorrelated SecurityB will have a higher dollar standard deviation.

e But if Security A and Security B are uncorrelated and have thesame standard deviation, then replacing some of Security A withSecurity B will decrease the dollar standard deviation of theportfolio.

e For example, $80 of Security A plus $20 of uncorrelated SecurityB will have a lower dollar standard deviation than $100 of SecurityA.



Demystifying the Puzzle

e Let RA and RB be the returns of Security A and Security B,respectively. Let σ2A

(= V(RA)

)and σ2B

(= V(RB)

)be the

variances of these returns. Moreover, suppose σA = σB = σ.

e The dollar value of Security A will become $100RA. If the portfoliois constructed by investing $100 in Security A, then the volatility of

the portfolio value in dollars is√

V(100RA

)= $100σ.

e But if the portfolio is made by having $80 invested in Security Aand $20 invested in uncorrelated Security B, then the volatility ofthe portfolio value in dollars is

√V(80RA

)+ V

(20RB

), which is√

6,400σ2A + 400σ2B =√

6,800σ < 100σ.



Important Lessons

& Mean-variance analysis is a cornerstone of investment, eventrading.

& All sample estimators such as sample average and samplevariance are random variables due to sampling randomness.

& Sample mean, sample variance, and sample covariance

& Unbiasedness of the three sample estimates

& Diversification is more subtle than you thought.


week 1 quantitative analysis of financial markets basic statistics a · 2017-10-16 · michael...

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