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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”
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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.
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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.
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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
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Chapter 10 5
X-
X-
X-
X-
X-
µ X-
X-
X-
X-
X-
Population Sample
Resample
Resample
ResampleResample
Resample
.
.
.
.
Samples not taken
.
.
.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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.
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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)
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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!
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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 .
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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 .
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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
).
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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.
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Chapter 10 23
We will use the notation
SE =s√n
and
SEboot =sboot√
n.
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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.
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Chapter 10 25
• For this bootstrap sample
– Xboot = 93.83,
– SEboot = 4.57,
– tboot = (98.5− 93.83)/4.57 = 1.02.
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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.
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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.
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Chapter 10 28
• For this bootstrap sample
– Xboot = 104.17,
– SEboot = 4.58,
– tboot = (98.5− 104.17)/4.58 = −1.24.
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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.
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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.
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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).
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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.
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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).
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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.
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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.
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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.
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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 ;
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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.
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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) ) ;
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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.
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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
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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.
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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
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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
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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
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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?
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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.”
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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
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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.
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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.
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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.
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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
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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).
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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.
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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?”
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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.
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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
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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.
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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
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Chapter 10 60
• “What would happen if we had more data?”
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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