Chapter 10 1

RESAMPLING AND PORTFOLIO ANALYSIS

Introduction

• Computer simulation is widely used in OR.

• Applications of simulation to statistics are

widespread.

• Topic of this chapter: simulation technique called the

“bootstrap” or “resampling”

• Will study the effects of estimation error on portfolio

selection.

• “bootstrap” from the phrase “pulling oneself up by

one’s bootstraps”

– “bootstrap” = “resampling”

Chapter 10 2

• Statistics from a random sample are random

variables.

– The sample is only one of many possible samples.

– Each possible sample gives a possible value of X.

– We only see one value of X

∗ But it was selected at random from the many

possible values.

– Thus, X is a random variable.

Chapter 10 3

• Confidence intervals and hypothesis tests are based

on the randomness of statistics.

– Example: confidence coefficient tells us the

probability that an interval constructed from a

random sample will contain the parameter.

– Confidence intervals are sometimes derived using

probability theory.

– Often the necessary probability calculations are

intractable.

– In that case we can replace theoretical calculations

by Monte Carlo simulation.

Chapter 10 4

• How do we simulate sampling from an unknown

population?

• We cannot do this exactly.

• However, a sample is a good representative of the

population.

• We can simulate sampling from the population by

sampling from the sample.

– this is usually called resampling

Chapter 10 5

X-

X-

X-

X-

X-

µ X-

X-

X-

X-

X-

Population Sample

Resample

Resample

ResampleResample

Resample

.

.

.

.

Samples not taken

.

.

.

Chapter 10 6

• Each resample has the same sample size, n, as the

original sample.

• We are trying to simulate the original sampling,

• We want the resampling to be as similar as possible

to the original sampling.

Chapter 10 7

• The resamples are drawn with replacement.

• Only sampling with replacement give independent

observations.

• We want the resamples to be i.i.d. just like the

original sample.

• If the resamples were drawn without replacement

then every resample would be exactly the same as the

original sample.

– So the resamples would show no random variation.

– This wouldn’t be very satisfactory, of course.

Chapter 10 8

• The number of resamples taken should be large.

• Just how large depends on context and will be

discussed more fully later.

• Sometimes, tens of thousands of resamples are taken.

• We will let B denote the number of resamples.

Chapter 10 9

• There is some good news and some bad news about

the bootstrap.

– The good news is that computer simulation

replaces difficult mathematics.

Chapter 10 10

• The bad news is that resampling is a new and

unfamiliar concept.

– It that takes some time to become comfortable

with resampling.

– The problem is not that resampling is all that

conceptually complex.

– Rather, the problem is that students don’t have

much experience with even a single random sample

from a population.

– Resampling is even more complex that that

∗ two layers of sampling and multiple resamples.

• However, studying resampling gives us a better

understanding of sampling.

Chapter 10 11

Confidence intervals for the mean

• Before resampling of the efficient frontier, we will look

at a simpler problem.

• Suppose we wish to construct a confidence interval for

the population mean.

• Start with the so-called t-statistic:

t =µ−X

s√n

. (1)

• The denominator of t, s/√

n, is just the standard

error of the mean.

Chapter 10 12

• Sampling from a normal population

– the probability distribution of t is the

t-distribution with n− 1 degrees of freedom.

• Denote by tα/2 the α/2 upper t-value,

– this is the 1− α/2 quantile of this distribution.

• Thus t in (1) has probability α/2 of exceeding tα/2.

• Because of the symmetry of the t-distribution, the

probability is also α/2 than t is less that −tα/2.

Chapter 10 13

• Therefore, for normally distributed data, the

probability is 1− α that

−tα/2 ≤ t ≤ tα/2. (2)

• Substituting (1) into (2), after a bit of algebra we find

that the probability is 1− α that

X − tα/2s√n≤ µ ≤ X + tα/2

s√n

, (3)

– which shows that

X ± s√n

tα/2

is a 1− α confidence interval for µ, assuming

normally distributed data.

Chapter 10 14

• What if we are not sampling from a normal

distribution?

• In that case, the distribution of t (1) will not be the t

distribution, but rather some other distribution that

is not known to us.

Chapter 10 15

• There are two problems.

– First, we don’t know the population distribution.

– Second, a difficult probability calculation is

necessary to get the distribution of the t-statistic

from the population distribution.

∗ This calculation has only be done for normal

populations.

Chapter 10 16

• Considering these difficulties, can we get a confidence

interval?

– “yes, by resampling.”

• Take a large number, say B, resamples from the

original sample.

Chapter 10 17

• Let Xboot,b and sboot,b be the sample mean and

standard deviation of the bth resample.

• Define

tboot,b =X −Xboot,b

sboot,b√n

. (4)

Chapter 10 18

• Notice that tboot,b is defined in the same way as t

except for two changes.

– First, X and s in t are replaced by Xboot,b and

sboot,b in tboot,b.

– Second, µ in t is replaced by X in tboot,b.

– The last point is a bit subtle, and you should stop

to think about it.

∗ A resample is taken using the original sample as

the population.

∗ Thus, for the resample, the population mean is

X!

Chapter 10 19

•• Resamples are independent

– Therefore, tboot,1, tboot,2, . . . is a random sample

from the t-statistic distribution.

• After B values of tboot,b have been calculated:

– we find the 2.5% and 97.5% percentiles of this

collection of tboot,b values.

– Call these percentiles tL and tU .

Chapter 10 20

• More specifically, we find tL and tU as follows:

– The B values of tboot,b are sorted from smallest to

largest.

– Then we calculate Bα/2 and round to the nearest

integer.

∗ example: if α = .05 and B = 1000, the

KL = (1000)(.05)/2 = 25

– Suppose the result is KL. Then the KLth sorted

value of tboot,b is tL.

– Similarly, let KU be B(1− α/2) rounded to the

nearest integer and then tU is the KUth sorted

value of tboot,b is tU .

Chapter 10 21

• If the original population is skewed, then we do not

necessarily expect that tL = −tU .

• However, this fact causes us no problem since the

bootstrap allows us to estimate tL and tU without

assuming any relationship between them.

• Now we replace −tα/2 and tα/2 in the confidence

interval (3) by tL and tU , respectively.

• Finally, the bootstrap confidence interval for µ is(X + tL

s√n

, X + tUs√n

).

Chapter 10 22

• The bootstrap has solved both problems mentioned

above.

– We do not need to know the population

distribution since we can estimate it by the sample.

– Moreover, we don’t need to calculate the

distribution of the t-statistic using probability

theory.

∗ Instead we can simulate from this distribution.

Chapter 10 23

We will use the notation

SE =s√n

and

SEboot =sboot√

n.

Chapter 10 24

Example: We start with a very small sample of size six

to illustrate how the bootstrap works.

• The sample is 82, 93, 99, 103, 104, 110.

– X = 98.50

– SE is 4.03.

• The first bootstrap sample is 82, 82, 93, 93, 103, 110.

– In this bootstrap sample,

∗ 82 and 93 were sampled twice,

∗ 103 and 110 were sampled once,

∗ the other elements of the original sample were

not sampled.

Chapter 10 25

• For this bootstrap sample

– Xboot = 93.83,

– SEboot = 4.57,

– tboot = (98.5− 93.83)/4.57 = 1.02.

Chapter 10 26

• The second bootstrap sample is 82, 103, 110, 110,

110, 110.

– In this bootstrap sample,

∗ 82 and 103 were sampled twice,

∗ 110 was sampled four times,

∗ the other elements of the original sample were

not sampled.

Chapter 10 27

It may seem strange at first that 110 was resample four

times.

• The number of times 110 appears in a resample is

binomial with parameters p = 1/6 and n = 6.

• The probability 110 occurs exactly four times in the

sample is

6!

4! 2!

(1

6

)4 (5

6

)2

= 0.00804.

• The probability that one of the six elements of the

original sample will occur exactly four times in a

resample is (6)(.00804) = .0482.

Chapter 10 28

• For this bootstrap sample

– Xboot = 104.17,

– SEboot = 4.58,

– tboot = (98.5− 104.17)/4.58 = −1.24.

Chapter 10 29

• The third bootstrap sample is 82, 82, 93, 99, 104, 110.

– For this bootstrap sample

∗ Xboot = 95.00,

∗ SEboot = 4.70,

∗ tboot = (98.5− 95.00)/4.570 = 1.02.

Chapter 10 30

• If this example were to continue:

– more bootstrap samples would be drawn

– all bootstrap t values would be saved in order

compute quantiles of the bootstrap t values.

• Since the sample size is so small, this example is not

very realistic and we will not continue it.

Chapter 10 31

Example: Suppose that we have a random sample with

a more realistic size of 40 from some population and

X = 107 and s = 12.6.

• Let’s find the “normal theory” 95% confidence

interval for the population mean µ.

• With 39 degrees of freedom, t.025 = 2.02.

• Therefore, the confidence interval for µ is

107± 2.0212.6√

40= (102.97, 111.03).

Chapter 10 32

• Suppose that we use resampling instead of normal

theory and that we use 1,000 resamples.

• This gives us 1,000 values of tboot,b. We rank them

from smallest to largest.

• The 25% percentile is the one with rank 25 =

(1000)(.025).

• Suppose the 25th smallest value of tboot,b is −1.98.

• The 97.5% percentile is the value of tboot,b with rank

975.

• Suppose that its value is 2.25.

Chapter 10 33

• Then

– tL = −1.98,

– tU = 2.25,

– the 95% confidence interval for µ is(107− 1.98

12.6√40

, 107 + 2.2512.6√

40

)= (103.06, 111.48).

Chapter 10 34

Example: Log-returns for MSCI-Argentina

• MSCI-Argentina is the Morgan Stanley Capital Index

for Argentina

– roughly comparable to the S&P 500 for the US.

• The log-returns for this index from January 1988 to

January 2002, inclusive, are used in this example.

• A normal plot is found later.

• There is evidence of non-normality, in particular, that

the log-returns are heavy-tailed, especially the left

tail.

Chapter 10 35

• The bootstrap was implemented with B = 10, 000.

• The t-values were

– tL = −1.93

– tU = 1.98.

• To assess the Monte Carlo variability, the bootstrap

was repeated two more times with results:

– tL = −1.98 and tU = 1.96

– tL = −1.94 and tU = 1.94.

• We see that B = 10, 000 gives reasonable accuracy

but that the third significant digit is still uncertain.

Chapter 10 36

• Also, using normal theory, tL = −t.025 = −1.974 and

tU = t.025 = 1.974, which are similar to the bootstrap

values that do not assume normality.

• Therefore, the use of the bootstrap in this example

confirms that the normal theory confidence interval is

satisfactory.

• In other example, particularly with strongly skewed

data and small sample sizes, the normal theory

confidence interval will be less satisfactory.

Chapter 10 37

Here is some MATLAB code that does resampling:

for b=1:2000 ;

select = ceil((n*rand(n,1)) ;

resample = sample(select,:) ;

% Put code in here to calculate

% the statistics that are needed in your application.

end ;

Chapter 10 38

• “select” which equals “ceil((n*rand(n,1))” will be n

random integers between 1 and n

• “resample” will be a resample from “sample” using

the indices in “select”

• For example, if n equals 6, the sample is

s1, s2, . . . , s6, and select equals [3 2 6 5 2 5], then

the resample is s3, s2, s6, s5, s2, s5.

• 2000 resamples are taken, that is, B = 2000.

Chapter 10 39

Here is some MATLAB code that does a 95% bootstrap-t

confidence interval by resampling:

xbar = mean(sample) ;

t = zeros(2000,1) ;

B = 2000 ;

for b=1:B ;

select = ceil((n*rand(n,1)) ;

resample = sample(select,:) ;

t(b) = (xbar - mean(resample)) / (std(resample)/sqrt(n)) ;

end ;

t=sort(t) ;

t_L = t( round(.05*B/2) ) ;

t_U = t( round((1-.05/2)*B) ) ;

Chapter 10 40

Resampling the efficient frontier

• One application of optimal portfolio selection is to

allocation of capital to different market segments.

• Michaud (1998) discusses a global asset allocation

problem where capital must be allocated to

– U.S. stocks and government/corporate bonds,

– Euros,

– the Canadian, French, German, Japanese, and

U.K. equity markets.

Chapter 10 41

Here we look at a similar example where we allocatecapital to the equity markets of ten different countries.Monthly log-returns for these markets were calculatedfrom:

• 1 = MSCI Hong Kong

• 2 = MSCI Singapore

• 3 = MSCI Brazil

• 4 = MSCI Argentina

• 5 = MSCI UK

• 6 = MSCI Germany

• 7 = MSCI Canada

• 8 = MSCI France

• 9 = MSCI Japan

• 10 = S&P 500

Chapter 10 42

• The data are in the file “countries.txt” on the course

web site and came from Datastream.

• “MSCI” means “Morgan-Stanley Capital Index.”

• The data are from January 1988 to January 2002,

inclusive, so there are 169 months (14 years and one

month) of data.

Chapter 10 43

0 50 100 150!1

!0.5

0

0.5

1Hong Kong

retu

rn

0 50 100 150!1

!0.5

0

0.5

1Singapore

retu

rn

0 50 100 150!1

!0.5

0

0.5

1Brazil

retu

rn

0 50 100 150!1

!0.5

0

0.5

1Argentina

retu

rn

0 50 100 150!1

!0.5

0

0.5

1UK

retu

rn

0 50 100 150!1

!0.5

0

0.5

1Germany

retu

rn

0 50 100 150!1

!0.5

0

0.5

1Canada

retu

rn

0 50 100 150!1

!0.5

0

0.5

1France

retu

rn

0 50 100 150!1

!0.5

0

0.5

1Japan

retu

rn

0 50 100 150!1

!0.5

0

0.5

1US

retu

rn

Chapter 10 44

0 5 10 15 20!0.2

!0.1

0

0.1

0.2Hong Kong

corr

0 5 10 15 20!0.2

!0.1

0

0.1

0.2Singapore

corr

0 5 10 15 20!0.2

!0.1

0

0.1

0.2Brazil

corr

0 5 10 15 20!0.2

!0.1

0

0.1

0.2Argentina

corr

0 5 10 15 20!0.2

!0.1

0

0.1

0.2UK

corr

0 5 10 15 20!0.2

!0.1

0

0.1

0.2Germany

corr

0 5 10 15 20!0.2

!0.1

0

0.1

0.2Canada

corr

0 5 10 15 20!0.2

!0.1

0

0.1

0.2France

corr

0 5 10 15 20!0.2

!0.1

0

0.1

0.2Japan

corr

0 5 10 15 20!0.2

!0.1

0

0.1

0.2US

corr

Chapter 10 45

!0.2 0 0.20.0030.01 0.02 0.05 0.10 0.25 0.50 0.75 0.90 0.95 0.98 0.99 0.997

Prob

abilit

y

Hong Kong

!0.2 !0.1 0 0.1 0.20.0030.01 0.02 0.05 0.10 0.25 0.50 0.75 0.90 0.95 0.98 0.99 0.997

Prob

abilit

y

Singapore

!0.8 !0.6 !0.4 !0.2 0 0.2 0.40.0030.01 0.02 0.05 0.10 0.25 0.50 0.75 0.90 0.95 0.98 0.99 0.997

Prob

abilit

y

Brazil

!1 !0.5 0 0.50.0030.01 0.02 0.05 0.10 0.25 0.50 0.75 0.90 0.95 0.98 0.99 0.997

Prob

abilit

y

Argentina

!0.1 !0.05 0 0.05 0.1 0.150.0030.01 0.02 0.05 0.10 0.25 0.50 0.75 0.90 0.95 0.98 0.99 0.997

Prob

abilit

y

UK

!0.2 !0.1 0 0.10.0030.01 0.02 0.05 0.10 0.25 0.50 0.75 0.90 0.95 0.98 0.99 0.997

Prob

abilit

y

Germany

!0.2 !0.1 0 0.10.0030.01 0.02 0.05 0.10 0.25 0.50 0.75 0.90 0.95 0.98 0.99 0.997

Prob

abilit

y

Canada

!0.1 0 0.1 0.20.0030.01 0.02 0.05 0.10 0.25 0.50 0.75 0.90 0.95 0.98 0.99 0.997

Prob

abilit

y

France

!0.2 !0.1 0 0.10.0030.01 0.02 0.05 0.10 0.25 0.50 0.75 0.90 0.95 0.98 0.99 0.997

Prob

abilit

y

Japan

!0.1 !0.05 0 0.05 0.10.0030.01 0.02 0.05 0.10 0.25 0.50 0.75 0.90 0.95 0.98 0.99 0.997

Prob

abilit

y

US

Chapter 10 46

• If we are planning to invest in a number of

international capital markets, then we might like to

know the efficient frontier,

– of course, we can never know it exactly.

• At best, we can estimate the efficient frontier using

estimated expected returns and the estimated

covariance matrix of returns.

• How close is the estimated efficient frontier to the

unknown true efficient frontier?

Chapter 10 47

• Each resample consists of 168 returns drawn with

replacement from the 168 returns of the original

sample.

• From the resampling perspective, the original sample

is treated as the population and the efficient frontier

calculated from the original sample is viewed as the

true efficient frontier.

• We can recalculate the efficient frontier using each of

the resamples and compare these re-estimated

efficient frontiers to the “true efficient frontier.”

Chapter 10 48

0 0.05 0.10

0.005

0.01

0.015

0.02

0.025

0.03

standard deviation of return (!P)ex

pect

ed re

turn

(µP)

achievedoptimal

0 0.05 0.10

0.005

0.01

0.015

0.02

0.025

0.03

standard deviation of return (!P)

expe

cted

retu

rn (µ

P)

achievedoptimal

0 0.05 0.10

0.005

0.01

0.015

0.02

0.025

0.03

standard deviation of return (!P)

expe

cted

retu

rn (µ

P)

achievedoptimal

0 0.05 0.10

0.005

0.01

0.015

0.02

0.025

0.03

standard deviation of return (!P)

expe

cted

retu

rn (µ

P)

achievedoptimal

0 0.05 0.10

0.005

0.01

0.015

0.02

0.025

0.03

standard deviation of return (!P)

expe

cted

retu

rn (µ

P)

achievedoptimal

0 0.05 0.10

0.005

0.01

0.015

0.02

0.025

0.03

standard deviation of return (!P)

expe

cted

retu

rn (µ

P)

achievedoptimal

Chapter 10 49

• To be more precise, let µ and Ω be the mean vector

and covariance matrix estimated from the original

sample.

• For a given target for the expected portfolio return,

µP , let ωµPbe the efficient portfolio weight vector

with g and h estimated from the original sample.

• Let ωµP ,b be the efficient portfolio weight vector with

g and h estimated from the bth resample.

Chapter 10 50

• Then the solid blue curves in the figure are

ωTµP

µ = µP plotted against√ωT

µpΩωµP

for a grid of µP values. The dashed red curves areωT

µP ,bµ plotted against√ωT

µP ,bΩωµP ,b.

• Unfortunately, the red resampled efficient frontier

curve lies below the blue true efficient frontier curve.

Chapter 10 51

• Because of estimation error, our estimated efficient

frontiers are suboptimal.

• Also, ωT

µP ,bµ does not, in general, equal µP .

• We do not achieve the expected return µP that we

have targeted because of estimation error when

estimating µ.

Chapter 10 52

0.03 0.035 0.04 0.045 0.05 0.055 0.06

0.01

0.015

0.02

0.025

0.03

standard deviation of return (!P)ex

pect

ed re

turn

(µP) achieved

optimaleff. frontier

!0.01 !0.005 0 0.005 0.01 0.015 0.020

50

100

150

!P ! !P,opt

frequ

ency

!6 !4 !2 0 2 4x 10!3

0

20

40

60

80

µP ! .012

frequ

ency

Chapter 10 53

• We concentrate on estimating only a single point on

the efficient frontier, the point where the expected

portfolio return is 0.012.

• This point is shown in the upper subplot as a large

black asterisk and is the point

(

√ωT

.012 Ω ω.012, .012).

• Each small blue asterisk in the upper subplot is the

estimate of this point from a resample and is

(

√ωT

.012,b Ωω.012,b, ωT

.012,bµ).

Chapter 10 54

• The middle subplot is a histogram of the values of√ωT

.012,b Ωω.012,b −

√ωT

.012Ωω.012.

• The lower subplot is a histogram of the values of

ωT

.012,bµ− .012.

Chapter 10 55

• Question: “what is the main problem here,

– mis-estimation of the expected returns,

– mis-estimation of the covariance matrix of the

returns,

– or both?”

One of the fun things we can do with resampling is to

play the game of “what if?”

• In particular, we can ask, “what would happen if we

knew the true expected returns and only had to

estimate the covariance matrix?”

• We can also ask the opposite question, “what would

happen if we knew the true covariance matrix and

only had to estimate the expected returns?”

Chapter 10 56

• By playing these “what if” games, we can address our

question about the relative effects of mis-estimation

of µ and mis-estimation of the covariance matrix.

• In next figure, when we estimate the efficient frontier

for each resample, we use the mean returns from the

original sample, which from the resampling

perspective are the population values.

• Only the covariance matrix is estimated from the

resample.

• The lower subplot is a histogram of the values of

ωT

.012,bµ− .012.

Chapter 10 57

0.03 0.035 0.04 0.045 0.05 0.055

0.01

0.015

0.02

0.025

0.03

standard deviation of return (!P)

expe

cted

retu

rn (µ

P)

achievedoptimaleff. frontier

0 0.5 1 1.5 2 2.5 3 3.5x 10!3

0

20

40

60

80

!P ! !P,opt

frequ

ency

Chapter 10 58

• In following figure, when we estimate the efficient

frontier for each resample, we use the covariance

matrix from the original sample.

• Only the expected returns are estimated from the

resample.

Chapter 10 59

0.03 0.035 0.04 0.045 0.05 0.055

0.01

0.015

0.02

0.025

0.03

standard deviation of return (!P)ex

pect

ed re

turn

(µP) achieved

optimaleff. frontier

!0.01 !0.005 0 0.005 0.01 0.015 0.020

50

100

150

!P ! !P,opt

frequ

ency

!6 !4 !2 0 2 4x 10!3

0

20

40

60

80

µP ! .012

frequ

ency

Chapter 10 60

• “What would happen if we had more data?”

Chapter 10 61

0.03 0.035 0.04 0.045 0.05 0.055

0.01

0.015

0.02

0.025

0.03

standard deviation of return (!P)ex

pect

ed re

turn

(µP) achieved

optimaleff. frontier

!0.01 !0.005 0 0.005 0.01 0.015 0.020

20

40

60

80

100

!P ! !P,opt

frequ

ency

!6 !4 !2 0 2 4x 10!3

0

20

40

60

80

µP ! .012

frequ

ency