Chapter 1
Estimation of The CER Model
Date: February 17, 2015
The CER model of asset returns presented in the previous chapter gives us
a simple framework for interpreting the time series behavior of asset returns
and prices. At the beginning of time  − 1, R is an  × 1 random vector
representing the returns (simple or continuously compounded) on assets  =1      to be realized at time . The CER model states that R ∼ (μ Σ). Our best guess for the return at  on asset  is [ ] =  , ourmeasure of uncertainty about our best guess is captured by SD( ) =   and our measures of the direction and strength of linear association between
 and  are   = cov(   ) and  = cor(   ) respectively. The
CER model assumes that the economic environment is constant over time sothat the multivariate normal distribution characterizing returns is the samefor all time periods .
Our life would be very easy if we knew the exact values of    2    and  the parameters of the CER model. In actuality, however, we do not know
these values with certainty. Therefore, a key task in financial econometrics
is estimating these values from a history of observed return data.
Suppose we observe returns on  diﬀerent assets over the sample  =1      Denote this sample {r1      r } = {r }=1  where r = (r1      r )0is the  × 1 vector of returns on  assets observed at time  It is assumed
that the observed returns are realizations of the random variables {R }=1 ,where R = (1       )0 is a vector of  asset returns described by the
CER modelR = μ + ε  ε ∼  (0 Σ)
(1.1)
Under these assumptions, we can use the observed returns {r }=1 to estimate
1

2

CHAPTER 1 ESTIMATION OF THE CER MODEL

the unknown parameters in μ and Σ of the CER model. However, before we
describe the estimation of the CER model in detail, it is necessary to review
some fundamental concepts in the statistical theory of estimation.
In this chapter we use the R packages PerformanceAnalytics, tseries
and zoo. Make sure these packages are installed and loaded prior to runningthe examples in the chapter.

1.1

Estimators and EstimatesLet  be the return on a single asset described by the CER model and let 
denote some characteristic (parameter) of the CER model we are interested
in estimating. For simplicity, assume that  ∈ R is a single parameter. For
example, if we are interested in the expected return on the asset, then  = ;if we are interested in the variance of asset  returns, then  = 2 ; if we areinterested in the first lag autocorrelation then  = 1  The goal is to estimate
 based on a sample of size  of the observed data.
Definition 1 Let {1       } denote a collection of  random returns fromthe CER model, and let  denote some characteristic of the model. An estimator of  denoted ̂ is a rule or algorithm for estimating  as a functionof the random variables {1       } Here, ̂ is a random variable.
Definition 2 Let {1       } denote an observed sample of size  from the
CER model, and let  denote some characteristic of the model. An estimateof  denoted ̂ is simply the value of the estimator for  based on the observed
sample {1       }. Here, ̂ is a number.
Example 3 The sample average as an estimator and an estimate
Let  be the return on a single asset described by the CER model, andsuppose we are interested in estimating  =  = [P] from the sample of1
observed returns { }=1 . The sample average ̂ =  =1  is an algorithm
for computing an estimate of the expected
return  Before the sample isP
observed, we can think of ̂ = 1 =1  as a simple linear function of
the random variables { }=1 and so is itself a random variable. After the
sample is observed, the sample average can be evaluated using the observed
data { }=1 which produces the estimate of . For example, suppose  = 5

1.1 ESTIMATORS AND ESTIMATES

3and the realized values of the returns are 1 = 01 2 = 005 3 = 0025 4 =−01 5 = −005 Then the estimate of  using the sample average is1
̂ = (01 + 005 + 0025 + −01 + −005) = 0005
5¥
The example above illustrates the distinction between an estimator andan estimate of a parameter . However, typically in the statistics literature
we use the same symbol, ̂, to denote both an estimator and an estimate.
When ̂ is treated as a function of the random returns it denotes an estimatorand is a random variable. When ̂ is evaluated using the observed data it
denotes an estimate and is simply a number The context in which we discuss̂ will determine how to interpret it

1.1.1

Properties of Estimators

Consider ̂ as a random variable. In general, the pdf of ̂ (̂) depends
on the pdf’s of the random variables { }=1  The exact form of (̂) may
be very complicated. Sometimes we can use analytical calculations to determine the exact form of  (̂) In general, the exact form of (̂) is often too
diﬃcult to derive exactly. When  (̂) is too diﬃcult to compute we can often approximate (̂) using either Monte Carlo simulation techniques or theCentral Limit Theorem (CLT). In Monte Carlo simulation, we use the computer the simulate many diﬀerent realizations of the random returns { }=1
and on each simulated sample we evaluate the estimator ̂ The Monte Carloapproximation of (̂) is the empirical distribution ̂ over the diﬀerent simulated samples. For a given sample size  Monte Carlo simulation gives avery accurate approximation to (̂) if the number of simulated samples isvery large. The CLT approximation of (̂) is a normal distribution approximation that becomes more accurate as the sample size  gets very large.
An advantage of the CLT approximation is that it is often easy to compute.
The disadvantage is that the accuracy of the approximation depends on theestimator ̂ and sample size For analysis purposes, we often focus on certain characteristics of  (̂)
like its expected value (center), variance and standard deviation (spread
about expected value) The expected value of an estimator is related to theconcept of estimator bias, and the variance/standard deviation of an estimator is related to the concept of estimator precision. Diﬀerent realizations of

4

CHAPTER 1 ESTIMATION OF THE CER MODEL

the random variables { }=1 will produce diﬀerent values of ̂ Some valuesof ̂ will be bigger than  and some will be smaller. Intuitively, a good estimator of  is one that is on average correct (unbiased) and never gets too
far away from  (small variance). That is, a good estimator will have small
bias and high precision.
Bias
Bias concerns the location or center of (̂) in relation to  If (̂) is centeredaway from  then we say ̂ is a biased estimator of . If  (̂) is centered at
 then we say that ̂ is an unbiased estimator of . Formally, we have thefollowing definitions:Definition 4 The estimation error is the diﬀerence between the estimatorand the parameter being estimated:
error(̂ ) = ̂ − 

(1.2)

Definition 5 The bias of an estimator ̂ of  is the expected estimation
error:
bias(̂ ) = [error(̂ )] = [̂] − 
(1.3)
Definition 6 An estimator ̂ of  is unbiased if bias(̂ ) = 0; i.e., if [̂] = or [error(̂ )] = 0
Unbiasedness is a desirable property of an estimator. It means that theestimator produces the correct answer “on average”, where “on average”
means over many hypothetical realizations of the random variables { }=1 .
It is important to keep in mind that an unbiased estimator for  may not
be very close to  for a particular sample, and that a biased estimator may
be actually be quite close to . For example, consider two estimators of ̂1 and ̂2  The pdfs of ̂1 and ̂2 are illustrated in Figure 1.1. ̂1 is anunbiased estimator of  with a large variance, and ̂2 is a biased estimatorof  with a small variance. Consider first, the pdf of ̂1 . The center of the
distribution is at the true value  = 0 [̂1 ] = 0 but the distribution isvery widely spread out about  = 0 That is, var(̂1 ) is large. On average(over many hypothetical samples) the value of ̂1 will be close to  but
in any given sample the value of ̂1 can be quite a bit above or below Hence, unbiasedness by itself does not guarantee a good estimator of  Now

5

1.1 ESTIMATORS AND ESTIMATES

0.0

0.5

pdf

1.0

1.5

theta.hat 1
theta.hat 2

-3

-2

-1

0

1

2

3

estimate value

Figure 1.1: Distributions of competiting estimators for  = 0 ̂1 is unbiased
but has high variance, and ̂2 is biased but has low variance.
consider the pdf for ̂2  The center of the pdf is slightly higher than  = 0i.e., bias(̂2  )  0 but the spread of the distribution is small. Although thevalue of ̂2 is not equal to 0 on average we might prefer the estimator ̂2 over̂1 because it is generally closer to  = 0 on average than ̂1 
While unbiasedness is a desirable property of an estimator ̂ of , it byitself, is not enough determine if ̂ is a good estimator. Being correct on
average means that ̂ is seldom exactly correct for any given sample. In
some samples ̂ is less than  and some samples ̂ is greater than  In
addition we need to know how far ̂ typically is from . That is, we need toknow about the magnitude of the spread of the distribution of ̂ about its
average value. This will tell us the precision of ̂Precision
An estimate is, hopefully, our best guess of the true (but unknown) value of. Our guess most certainly will be wrong, but we hope it will not be too
far oﬀ. A precise estimate is one in which the variability of the estimation

6

CHAPTER 1 ESTIMATION OF THE CER MODEL

error is small. The variability of the estimation error is captured by the meansquared error.
Definition 7 The mean squared error of an estimator ̂ of  is given bymse(̂ ) = [(̂ − )2 ] = [error(̂ )2 ](1.4)

The mean squared error measures the expected squared deviation of ̂ from If this expected deviation is small, then we know that ̂ will almost always
be close to  Alternatively, if the mean squared is large then it is possible
to see samples for which ̂ is quite far from  A useful decomposition ofmse(̂ ) is³´2
mse(̂ ) = [(̂ − [̂])2 ] + [̂] −  = var(̂) + bias(̂ )2

The derivation of this result is straightforward. Write̂ −  = ̂ − [̂] + [̂] − 
Then

³
´2
³
´³
´ ³
´2
(̂ − )2 = ̂ − [̂] + 2 ̂ − [̂] [̂] −  + [̂] −  Taking expectations of both sides gives

∙³
h³
´i2
³
´³
´
´2 ¸
mse(̂ ) =  ̂ − [̂]+ 2 [̂] − [̂] [̂] −  +  [̂] − 
∙³
´2 ¸
h³
´i2
+  [̂] − =  ̂ − [̂]= var(̂) + bias(̂ )2 

The result states that for any estimator ̂ of  mse(̂ ) can be split intoa variance component, var(̂) and a bias component, bias(̂ )2  Clearly,mse(̂ ) will be small only if both components are small. If an estimator isunbiased then mse(̂ ) = var(̂) = [(̂ − )2 ] is just the squared deviationof ̂ about  Hence, an unbiased estimator ̂ of  is good, if it has a small
variance.

1.1 ESTIMATORS AND ESTIMATES

7

The mse(̂ ) and var(̂) are based on squared deviations and so are notin the same units of measurement as  Measures of precision that are in thesame units as  are the root mean square error

and the standard errorqrmse(̂ ) = mse(̂ )

(1.5)

q
se(̂) = var(̂)

(1.6)

If bias(̂ ) ≈ 0 then the precision of ̂ is typically measured by se(̂)
Good Estimators
With the concepts of bias and precision in hand, we can state what defines
a good estimator.
Definition 8 A good estimator ̂ of  has a small bias (1.3) and a small
standard error (1.6).

1.1.2

Asymptotic Properties of Estimators

Estimator bias and precision are finite sample properties. That is, they
are properties that hold for a fixed sample size  Very often we are alsointerested in properties of estimators when the sample size  gets very large.
For example, analytic calculations may show that the bias and mse of an
estimator ̂ depend on  in a decreasing way. That is, as  gets very large
the bias and mse approach zero. So for a very large sample, ̂ is eﬀectively
unbiased with high precision. In this case we say that ̂ is a consistentestimator of  In addition, for large samples the CLT says that (̂) can
often be well approximated by a normal distribution. In this case, we saythat ̂ is asymptotically normally distributed. The word “asymptotic” means“in an infinitely large sample” or “as the sample size  goes to infinity”. Of
course, in the real world we don’t have an infinitely large sample and so theasymptic results are only approximations. How good these approximation are
for a given sample size  depends on the context. Monte Carlo simulations
can often be used to evaluate asymptotic approximations in a given context.

8

CHAPTER 1 ESTIMATION OF THE CER MODEL

Consistency
Let ̂ be an estimator of  based on the random returns { }=1 
Definition 9 ̂ is consistent for  (converges in probability to ) if for any
0
lim Pr(|̂ − |  ) = 0
 →∞Intuitively, consistency says that as we get enough data then ̂ will eventuallyequal  In other words, if we have enough data then we know the truth.
Theorems in probability theory known as Laws of Large Numbers areused to determine if an estimator is consistent or not. In general, we havethe following result: an estimator ̂ is consistent for  if
• bias(̂ ) = 0 as  → ∞³ ´• se ̂ = 0 as  → ∞Equivalently, ̂ is consistent for  if mse(̂ ) → 0 as  → ∞ Intuitively, if (̂) collapses to  as  → ∞ then ̂ is consistent for 
Asymptotic Normality
Let ̂ be an estimator of  based on the random returns { }=1 
Definition 10 An estimator ̂ is asymptotically normally distributed if
̂ ∼ ( se(̂)2 )(1.7)

for large enough 
Asymptotic normality means that (̂) is well approximated by a normal
distribution with mean  and variance se(̂)2  The justification for asymptotic
normality comes from the famous Central Limit Theorem.

1.1 ESTIMATORS AND ESTIMATES

9Central Limit Theorem There are actually many versions of the CLTwith diﬀerent assumptions.1 In its simplist form, the CLT says that the sample average of a collection of iid random variables 1       with [ ] = and ( ) =  2 is asymptotically normal with mean  and variance  2 
In particular, the CDF of the standardized sample mean¶µ̄ −  √̄ − 
̄ − = √ = se(̄) converges to the CDF of a standard normal random variable  as  → ∞This result can be expressed as¶µ√̄ − ∼  ∼ (0 1)

for large enough  Equivalently,

µ
¶
¡
¢
2

̄ ∼  + √ ×  ∼  =   se(̄)2 

for large enough  This form shows that ̄ is asymptotically normal withmean  and variance 2 
Asymptotic Confidence Intervals
For an asymptotically normal estimator ̂ of  the precision of ̂ is measureby se(b ̂) but is best communicated by computing a (asymptotic) confidence
interval for the unknown value of  A confidence interval is an interval
estimate of  such that we can put an explicit probability statement about
the likelihood that the interval covers 
The construction of an asymptotic confidence interval for  uses the asymptotic normality result̂ − 
se(̂)=  ∼ (0 1)

(1.8)

Then, for  ∈ (0 1) we compute a (1 − ) · 100% confidence interval for using (1.8) and the 1 − 2 standard normal quantile (critical value) (1−2)
to give
Ã
!
̂
−



Pr −(1−2)= 1 − ≤ (1−2)≤se(̂)
1

White (1984) gives a comprehensive discussion of CLTs useful in econometrics.

10

CHAPTER 1 ESTIMATION OF THE CER MODEL

which can be rearranged as³´Pr ̂ − (1−2)· se(̂) ≤  ≤ ̂ + (1−2)· se(̂) = 1 − Hence, the interval[̂ − (1−2)· se(̂) ̂ + (1−2)· se(̂)] = ̂ ± (1−2)· se(̂)(1.9)

covers the true unknown value of  with probability 1 − In practice, typical values for  are 005 and 001 for which (0975)= 196and (0995) = 258 Then, approximate 95% and 99% asymptotic confidence
intervals for  have the form ̂ ± 2 · se(̂) and ̂ ± 25 · se(̂) respectively.

1.2

Estimators for the Parameters of the CER
Model

Let {r }=1 denote a sample of size  of observed returns on  assets from the
CER model (1.1). To estimate the unknown CER model parameters   2   and  from {r }=1 we can use the plug-in principle from statistics:

Plug-in-Principle: Estimate model parameters using corresponding sample
statistics.
For the CER model parameters, the plug-in principle estimates are thefollowing sample descriptive statistics discussed in Chapter xxx (Descriptive
Statistics):
1X
 
̂ =
 =1

(1.10)

̂ 2

(1.11)


1 X=( − ̂ )2 
 − 1 =1q̂  = ̂ 2 


1 X̂  =
( − ̂ )( − ̂ )
 − 1 =1̂ =̂ ̂  ̂ (1.12)(1.13)(1.14)

1.2 ESTIMATORS FOR THE PARAMETERS OF THE CER MODEL11
The plug-in principle is appropriate because the the CER model parameters   2    and  are characteristics of the underlying distribution of returns
that are naturally estimated using sample statistics.
The plug-in principle sample statistics (1.10) - (1.14) are given for a single asset and the statistics (1.13) - (1.14) are given for one pair of assets.
However, these statistics can be computed for a collection of  assets using
the matrix sample statistics
⎛
⎞
̂

⎜ 1 ⎟
1X
⎜ .. ⎟(1.15)μ̂ =r = ⎜ . ⎟ 
 =1
⎝
⎠
( ×1)
̂⎞⎛̂ 21 ̂ 12 · · · ̂ 1
⎟
⎜
⎟
⎜

2
X
⎜ ̂ 12 ̂ 2 · · · ̂ 2 ⎟
1
0
⎟
⎜
Σ̂ =
(r − μ̂)(r − μ̂) = ⎜ .
.. . .
.. ⎟  (1.16)(×) − 1 =1
.
. . ⎟
⎜ .
.
⎠
⎝
2
̂ 1 ̂ 2 · · · ̂ 

Here μ̂ is called the sample mean vector and Σ̂ is called the sample covariancematrix. The sample variances are the diagonal elements of Σ̂ and the sample
covariances are the oﬀ diagonal elements of Σ̂ To get the sample correlations,define the  ×  diagonal matrix⎛⎞
̂ 1 0 · · · 0⎜⎟⎜⎟⎜ 0 ̂ 2 · · · 0 ⎟⎜⎟D̂ = ⎜ . . .
.. ⎟ 
.
.
.
. . ⎟
⎜ . .
⎝
⎠
0 0 · · · ̂ 
Then the sample correlation matrix R̂ is computed as⎞⎛1 ̂12 · · · ̂1⎟⎜⎟⎜⎜ ̂12 1 · · · ̂2 ⎟−1−1⎟
R̂ = D̂ Σ̂D̂ = ⎜⎜ ..
.. . .
. ⎟
. .. ⎟⎜ .
.
⎠⎝̂1 ̂2 · · · 1(1.17)

12

0.0
0.0
-0.2
-0.05
-0.15

SP500

0.05

-0.4

SBUX

0.2

-0.4

-0.2

MSFT

0.2

CHAPTER 1 ESTIMATION OF THE CER MODEL

1999

2000

2001

2002

2003

2004

2005

2006

2007

2008

2009

2010

2011

2012

Index

Figure 1.2: Monthly cc returns on Microsoft stock, Starbucks stock, and the
S&P 500 index, over the period January 1998 through May 2012.
Here, the sample correlations are the oﬀ diagonal elements of R̂
Example 11 Estimating the CER model parameters for Microsoft, Starbucks and the S&P 500 index.
To illustrate typical estimates of the CER model parameters, we use data on
monthly continuously compounded returns for Microsoft, Starbucks and the
S & P 500 index over the period January 1998 through May 2012. The data is
the same as that used in Chapters xxx (Descriptive Statistics & CER Model)
and is retrieved from Yahoo! using the tseries function get.hist.quote()
as follows> msftPrices = get.hist.quote(instrument="msft", start="1998-01-01",+
end="2012-05-31", quote="AdjClose",+
provider="yahoo", origin="1970-01-01",

1.2 ESTIMATORS FOR THE PARAMETERS OF THE CER MODEL13+
compression="m", retclass="zoo")> sbuxPrices = get.hist.quote(instrument="sbux", start="1998-01-01",+
end="2012-05-31", quote="AdjClose",+
provider="yahoo", origin="1970-01-01",+
compression="m", retclass="zoo")> sp500Prices = get.hist.quote(instrument="^gspc", start="1998-01-01",+
end="2012-05-31", quote="AdjClose",+
provider="yahoo", origin="1970-01-01",+
compression="m", retclass="zoo")> colnames(msftPrices) = "MSFT"> colnames(sbuxPrices) = "SBUX"> colnames(sp500Prices) = "SP500"> index(msftPrices) = as.yearmon(index(msftPrices))> index(sbuxPrices) = as.yearmon(index(sbuxPrices))> index(sp500Prices) = as.yearmon(index(sp500Prices))> cerPrices = merge(msftPrices, sbuxPrices, sp500Prices)> msftRetS = Return.calculate(msftPrices, method="simple")> sbuxRetS = Return.calculate(sbuxPrices, method="simple")> sp500RetS = Return.calculate(sp500Prices, method="simple")> cerRetS = Return.calculate(cerPrices, method="simple")> msftRetS = msftRetS[-1]> sbuxRetS = sbuxRetS[-1]> sp500RetS = sp500RetS[-1]> cerRetS = cerRetS[-1]> msftRetC = log(1 + msftRetS)> sbuxRetC = log(1 + sbuxRetS)> sp500RetC = log(1 + sp500RetS)> cerRetC = merge(msftRetC, sbuxRetC, sp500RetC)> colnames(cerRetC) = c("MSFT", "SBUX", "SP500")
These data are illustrated in Figures 1.2 and 1.3.
The estimates of  ( =   500) using (1.10) or (1.15) can becomputed using the R functions apply() and mean()> muhat = apply(cerRetC,2,mean)> muhat
MSFT
SBUX
SP500
0.004127 0.014657 0.001687

14

CHAPTER 1 ESTIMATION OF THE CER MODEL

-0.2

0.0

0.2

0.0

0.2

-0.4

0.0

0.2

-0.4

-0.2

MSFT

0.05

-0.4

-0.2

SBUX

-0.15

-0.05

SP500

-0.4

-0.2

0.0

0.2

-0.15

-0.05

0.05

Figure 1.3: Scatterplot matrix of the monthly cc returns on Microsoft stock,
Starbucks stock, and the S&P 500 index.
Starbucks has the highest average monthly return at 1.5% and the S&P 500
index has the lowest at 0.2%.
The estimates of the parameters  2     using (1.11) and (1.12) can becomputed using apply(), var() and sd()
> sigma2hat = apply(cerRetC,2,var)> sigma2hat
MSFT
SBUX
SP500
0.010051 0.012465 0.002349> sigmahat = apply(cerRetC,2,sd)> sigmahat
MSFT
SBUX
SP500
0.10026 0.11164 0.04847

1.2 ESTIMATORS FOR THE PARAMETERS OF THE CER MODEL15
Starbucks has the most variable monthly returns at 11%, and the S&P 500
index has the smallest at 5%.
The scatterplots of the returns are illustrated in Figure 1.3. All returns
appear to be positively related. The covariance and correlation matrix estimates using (1.16) and (1.17) can be computed using the functions var()(or cov()) and cor()> covmat = var(cerRetC)> covmat
MSFT
SBUX
SP500
MSFT 0.010051 0.003819 0.003000
SBUX 0.003819 0.012465 0.002476
SP500 0.003000 0.002476 0.002349> cormat = cor(cerRetC)> cormatMSFT
SBUX SP500
MSFT 1.0000 0.3412 0.6173
SBUX 0.3412 1.0000 0.4575
SP500 0.6173 0.4575 1.0000
To extract the unique pairwise values of  and  from the matrix objects
covmat and cormat use> covhat = covmat[lower.tri(covmat)]> rhohat = cormat[lower.tri(cormat)]> names(covhat) <- names(rhohat) <+ c("msft,sbux","msft,sp500","sbux,sp500")> covhat
msft,sbux msft,sp500 sbux,sp500
0.003819
0.003000
0.002476> rhohat
msft,sbux msft,sp500 sbux,sp500
0.34120.61730.4575The pairs (MSFT, SP500) and (SBUX, SP500) are the most correlated.
These estimates confirm the visual results from the scatterplot matrix in
Figure 1.3. ¥

16

1.3

CHAPTER 1 ESTIMATION OF THE CER MODEL

Statistical Properties of the CER Model
Estimates

To determine the statistical properties of plug-in principle estimators ̂  ̂2  ̂   ̂ and ̂ in the CER model, we treat them as functions of the random variables{R }=1 where R is assumed to be generated by the CER model (1.1).

1.3.1

Bias

Assuming that returns are generated by the CER model (1.1), ̂  ̂ 2 and ̂ 
are unbiased estimators,

[̂ ] = [̂ 2 ] =  2 [̂  ] =   

but ̂  and ̂ are biased estimators,

[̂  ] 6=   [̂ ] 6=  
It can be shown that the biases in ̂  and ̂  are very small and decreasing in such that bias(̂     ) = bias(̂   ) = 0 as  → ∞. The proofs of theseresults are beyond the scope of this book and may be found, for example, inGoldberger (1991). As we shall see, these results about bias can be easily
verified using Monte Carlo methods.
It is instructive to illustrate how to derive the result [̂ ] =   Using
results about the expectation of a linear combination of random variables, it1.3 STATISTICAL PROPERTIES OF THE CER MODEL ESTIMATES17
follows that#
1X
[̂ ] = 

 =1
"
#

1X
= 
( +  ) (since  =  +  ) =1 
"



1X
1X
 +
[ ] (by the linearity of [·])= =1   =1

1X=
 (since [ ] = 0  = 1      ) =1 =1
 ·  =  The derivation of the results [̂2 ] =  2 and [̂  ] =   are similar but areconsiderably more involved and so are omitted.

1.3.2

Precision

Because the CER model estimators are either unbiased or the bias is verysmall, the precision of these estimators is measured by their standard errors.
The standard error for ̂  se(̂ ) can be calculated exactly and is given byse(̂ ) = √ (1.18)

18

CHAPTER 1 ESTIMATION OF THE CER MODEL

The derivation of this result is straightforward. Using the results about thevariance of a linear combination of uncorrelated random variables, we havevar(̂ ) = var

Ã

Ã

1X

 =1


!

!

1X
= var
( +  )(since  =  +  ) =1 !
Ã

1X
(since  is a constant)
= var

 =1

1 X
= 2
var( ) (since  is independent over time)
 =1


1 X 2

 2 =1 ==(since var( ) =  2   = 1      )

 2
1
2


=

2 


Then se(̂ ) = SD(̂ ) = √  We make the following remarks:
1. The value of se(̂ ) is in the same units as ̂ and measures the precisionof ̂ as an estimate. If se(̂ ) is small relative to ̂ then ̂ is a relatively
precise of  because  (̂ ) will be tightly concentrated around  ; ifse(̂ ) is large relative to  then ̂ is a relatively imprecise estimateof  because (̂ ) will be spread out about  
2. The magnitude of se(̂ ) depends positively on the volatility of returns,
 = SD( ) For a given sample size  , assets with higher return
volatility have larger values of se(̂ ) than assets with lower return
volatility. In other words, estimates of expected return for high volatility assets are less precise than estimates of expected returns for low
volatility assets.
3. For a given return volatility    se(̂ ) is smaller for larger sample sizes
 In other words, ̂ is more precisely √estimated for larger samples.
Moreover, se(̂ ) → 0 as  → ∞ at rate 1.3 STATISTICAL PROPERTIES OF THE CER MODEL ESTIMATES19
The derivations of the standard errors for ̂ 2  ̂   ̂  and ̂ are complicated, and the exact results are extremely messy and hard to work with.
However, there are simple approximate formulas for the standard errors of̂ 2  ̂  and ̂ based on the CLT that are valid if the sample size,  is reasonably large.2 These large sample approximate formulas are given by
√ 2
2
2
2
se(̂  ) ≈ √  = p  (1.19) 2
se(̂ ) ≈ √ (1.20)
2
(1 − 2 )√se( ) ≈(1.21)

where “≈” denotes approximately equal. The approximations are such thatthe approximation error goes to zero as the sample size  gets very large.
We make the following remarks:
1. As with the formula for the standard error of the sample mean, theformulas for se(̂ 2 ) and se(̂  ) depend on  2  Larger values of  2 imply
less precise estimates of ̂ 2 and ̂  
2. The formula for se( ) does not depend on  2 but rather depends on
2 and is smaller the closer 2 is to unity. Intuitively, this makes sense
because as 2 approaches one the linear dependence between  and becomes almost perfect and this will be easily recognizable in thedata (scatterplot will almost follow a straight line).
3. The formulas for the standard errors√ above are inversely related to thesquare root of the sample size,  , which means that larger sample
sizes imply smaller values of the standard errors.
4. Interestingly, se(̂  ) goes to zero the fastest and se(̂ 2 ) goes to zero theslowest. Hence, for a fixed sample size, these formulas suggest that   isgenerally estimated more precisely than  2 and   and  is estimatedgenerally more precisely than  2 
2The large sample approximate formula for the variance of ̂  is too messy to workwith so we omit it here. In practice, we can use the bootstrap to provide an estimatedstandard error for ̂ 

20

CHAPTER 1 ESTIMATION OF THE CER MODEL

The above formulas (1.19) - (1.21) are not practically useful, however,
because they depend on the unknown quantities 2    and   Practicallyuseful formulas replace 2    and  by the estimates ̂ 2  ̂  and ̂ and giverise to the estimated standard errors:b(1.22)
se(̂b ) = √

̂ 2
se(̂
b 2 ) ≈ p  (1.23)
 2
̂ 
se(̂
b ) ≈ √ (1.24)
2
(1 − ̂2 )√
(1.25)se(b  ) ≈It is good practice to report estimates together with their estimated standard errors. In this way the precision of the estimates is transparent to theuser. Typically, estimates are reported in a table with the estimates in one
column and the estimated standard errors in an adjacent column.
Example 12 se(̂b  ) values for Microsoft, Starbucks and the S&P 500 index
For Microsoft, Starbucks and S&P 500, the values of se(̂b  ) are easily
computed in R using> n.obs = nrow(cerRetC)> seMuhat = sigmahat/sqrt(n.obs)b  ) shown together are
The values of ̂ and se(̂> cbind(muhat, seMuhat)
muhat seMuhat
MSFT 0.00413 0.00764
SBUX 0.01466 0.00851
SP500 0.00169 0.00370

For Microsoft and Starbucks, the values of se(̂b  ) are similar because thevalues of ̂  are similar, and se(̂b  ) is smallest for the S&P 500 index. Thisoccurs because ̂ 500 is much smaller than ̂   and ̂   Hence, ̂ isestimated more precisely for the S&P 500 index (a highly diversified portfolio)
than it is for Microsoft and Starbucks stock (individual assets).
It is tempting to compare the magnitude of se(̂b  ) to the value of ̂ toevaluate if ̂ is a precise estimate

1.3 STATISTICAL PROPERTIES OF THE CER MODEL ESTIMATES21> seMuhat/muhat
MSFT SBUX SP500
1.852 0.581 2.191
Here we see that se(̂b   ) and se(̂b 500 ) are about twice as large as ̂ and ̂500  respectively, whereas se(̂b  ) is about half the size of ̂ .
This seems to indicate that ̂ is most precisely estimated. However,this comparison is misleading. To see why, consider the range of values
determined by ̂ ± 2 × se(̂b  )

> upper = muhat + 2*seMuhat> lower = muhat - 2*seMuhat> width = upper - lower> cbind(lower, upper, width)
lower
upper width
MSFT -0.01116 0.01942 0.0306
SBUX -0.00237 0.03168 0.0341
SP500 -0.00570 0.00908 0.0148

For normally distributed ̂  this range contains the true value of  withprobability around 0.95. For all assets, this range contains both positive
and negative values but the range is smallest for the S&P 500 index. ¥Example 13 Computing se(̂b 2 ) se(̂b  ) and se(̂b  ) for Microsoft, Starbucks
and the S&P 500.
For Microsoft, Starbucks and S&P 500, the values of se(̂b 2 ) se(̂b  ) and se(̂b  )
2(together with the estimates ̂   ̂  and ̂ ) are> seSigma2hat = sigma2hat/sqrt(n.obs/2)> seSigmahat = sigmahat/sqrt(2*n.obs)> cbind(sigma2hat, seSigma2hat, sigmahat, seSigmahat)sigma2hat seSigma2hat sigmahat seSigmahat
MSFT
0.01005
0.001084
0.10030.00541
SBUX
0.012460.001344
0.11160.00602
SP500
0.002350.000253
0.04850.00261

Notice that 2 and  for the S&P 500 index are estimated much more precisely than the values for Microsoft and Starbucks. Also notice that   isestimated more precisely than  for all assets: the values of se(̂b  ) relativeto ̂  are much smaller than the values of se(̂b  ) to ̂ 
The values of se(̂b  ) (together with ̂ ) are

22

CHAPTER 1 ESTIMATION OF THE CER MODEL> seRhohat = (1-rhohat^2)/sqrt(n.obs)> cbind(rhohat, seRhohat)rhohat seRhohat
msft,sbux
0.341
0.0674msft,sp500 0.617
0.0472
sbux,sp500 0.457
0.0603The values of se(̂b  ) are moderate in size (relative to ̂ ) Notice that̂500 has the smallest estimated standard error because ̂2500 is closest to one. ¥1.3.3

Sampling Distributions and Confidence Intervals

Sampling Distribution for ̂P
In the CER model,  ∼  (   2 ) and since ̂ = 1 =1  is anaverage of these normal random variables, it is also normally distributed.
2
The mean of ̂ is  and its variance is   Therefore, the exact probability
distribution³ of2 ´̂   (̂ ) for a fixed sample size  is the normal distribution̂ ∼     where(̂ ) =µ2 2


¶−12

½
exp −

¾
1
2
(̂ −  ) 2 2  

(1.26)The probability curve (̂ ) is centered at the true value   and the spread
about  depends on the magnitude of  2  the variability of   and thesample size,  . For a fixed sample size,  , the uncertainty in ̂ is larger
for larger values of  2  Notice that the variance of ̂ is inversely related tothe sample size  Given  2  var(̂ ) is smaller for larger sample sizes thanfor smaller sample sizes. This makes sense since we expect to have a more
precise estimator when we have more data. If the sample size is very large (as → ∞) then var(̂ ) will be approximately zero and the normal distributionof ̂ given by (1.26) will be essentially a spike at   In other words, if thesample size is very large then we essentially know the true value of   Hence,
we have established that ̂ is a consistent estimator of  as the sample size
goes to infinity
Example 14 Sampling distribution of ̂ with diﬀerent sample sizes.

1.3 STATISTICAL PROPERTIES OF THE CER MODEL ESTIMATES23

1.5
0.0

0.5

1.0

mu

2.0

2.5

pdf: T=1
pdf: T=10
pdf: T=50

-3

-2

-1

0

1

2

3

x.valsFigure 1.4: (0 1 ) sampling distributions for ̂ for  = 1 10 and 50
The distribution of ̂ , with  = 0 and  2 = 1 for various sample sizes is
illustrated in figure 1.4. Notice how fast the distribution collapses at  = 0
as  increases.

Confidence intervals for 
The precision of ̂ is measure by se(̂b  ) but is best communicated by computing a confidence interval for the unknown value of   A confidence interval
is an interval estimate of  such that we can put an explicit probability
statement about the likelihood that the interval covers  
The construction of an exact confidence interval for  is based on thefollowing statistical result (see the appendix for details).
Result: Let { }=1 be generated from the CER model (1.1). Define

24

CHAPTER 1 ESTIMATION OF THE CER MODEL

the -ratio as =̂ − 
̂ − =  √ 
se(̂
b )
̂ 

(1.27)

Then  ∼  −1 where  −1 denotes a Student’s  random variable with  − 1
degrees of freedom.
The Student’s  distribution with   0 degrees of freedom is a symmetric
distribution centered at zero, like the standard normal. The tail-thickness(kurtosis) of the distribution is determined by the degrees of freedom parameter  For values of  close to zero, the tails of the Student’s  distribution
are much fatter than the tails of the standard normal distribution. As  getslarge, the Student’s  distribution approaches the standard normal distribution.
For  ∈ (0 1) we compute a (1 − ) · 100% confidence interval for 
using (1.27) and the 1 − 2 quantile (critical value)  −1 (1 − 2) to giveµ¶̂ − Pr − −1 (1 − 2) ≤≤  −1 (1 − 2) = 1 − 
se(̂
b )
which can be rearranged as

Pr (̂ −  −1 (1 − 2) · se(̂b  ) ≤  ≤ ̂ +  −1 (1 − 2) · se(̂b  )) = 1 − Hence, the intervalb  ) ̂ +  −1 (1 − 2) · se(̂b  )][̂ −  −1 (1 − 2) · se(̂b )= ̂ ±  −1 (1 − 2) · se(̂(1.28)

covers the true unknown value of  with probability 1 − 
Example 15 Computing 95% confidence intervals for 

Suppose we want to compute a 95% confidence interval for   In this case
 = 005 and 1 −  = 095 Suppose further that  − 1 = 60 (e.g., five yearsof monthly return data) so that  −1 (1 − 2) = 60 (0975) = 2 This can beverified in R using the function qt()

1.3 STATISTICAL PROPERTIES OF THE CER MODEL ESTIMATES25
Then the 95% confidence for  is given byb  )
̂ ± 2 · se(̂(1.29)

The above formula for a 95% confidence interval is often used as a rule ofthumb for computing an approximate 95% confidence interval for moderatesample sizes. It is easy to remember and does not require the computationof the quantile  −1 (1 − 2) from the Student’s  distribution. It is also
an approximate 95% confidence interval that is based the asymptotic normality of ̂  Recall, for a normal distribution with mean  and variance  2
approximately 95% of the probability lies between  ± 2 ¥The coverage probability associated with the confidence interval for  isbased on the fact that the estimator ̂ is a random variable. Since confidence
interval is constructed as ̂ ± −1 (1−2)· se(̂b  ) it is also a random variable.
An intuitive way to think about the coverage probability associated with theconfidence interval is to think about the game of “horseshoes”.3 The horseshoe is the confidence interval and the parameter  is the post at which
the horse shoe is tossed. Think of playing game 100 times (i.e, simulate 100
samples of the CER model). If the thrower is 95% accurate (if the coverage
probability is 0.95) then 95 of the 100 tosses should ring the post (95 of theconstructed confidence intervals should contain the true value  ).
Example 16 95% confidence intervals for  for Microsoft, Starbucks andthe S & P 500 index.
Consider computing 95% confidence intervals for  using (1.28) based on
the estimated results for the Microsoft, Starbucks and S&P 500 data. The
degrees of freedom for the Student’s  distribution is  − 1 = 171 The 97.5%
quantile, 99 (0975) can be computed using the R function qt()
> t.975 = qt(0.975, df=(n.obs-1))> t.975[1] 1.97
Notice that this quantile is very close to 2 Then the exact 95% confidence
intervals are given by
3

Horse shoes is a game commonly played at county fairs.
http://en.wikipedia.org/wiki/Horseshoes for a complete description of the game.

See

26

CHAPTER 1 ESTIMATION OF THE CER MODEL> lower = muhat - t.975*seMuhat> upper = muhat + t.975*seMuhat> width = upper - lower> cbind(lower, upper, width)
lower
upper width
MSFT -0.01096 0.01922 0.0302
SBUX -0.00215 0.03146 0.0336
SP500 -0.00561 0.00898 0.0146
With probability 0.95, the above intervals will contain the true mean values
assuming the CER model is valid. The 95% confidence intervals for Microsoft
and Starbucks are fairly wide (about 3%) and contain both negative and positive values. The confidence interval for the S&P 500 index is tighter but also
contains negative and positive values. For Microsoft, the confidence intervalis [−11% 19%] This means that with probability 0.95, the true monthly
expected return is somewhere between -1.1% and 1.9%. The economic implications of a -1.1% expected monthly return and a 1.9% expected returnare vastly diﬀerent. In contrast, the 95% confidence interval for the SP500 is
about half the width of the intervals for Microsoft or Starbucks. The lower
limit is near -0.5% and the upper limit is near 1%. This result clearly showsthat the monthly mean return for the S&P 500 index is estimated much more
precisely than the monthly mean returns for Microsoft or Starbucks.
Sampling distributions for ̂ 2  ̂  and ̂The exact distributions of ̂ 2  ̂  and ̂ based on a fixed sample size  are
diﬃcult to derive.4 However, approximate normal distributions of the form
(1.7) based on the CLT are readily available:
¶
µ
4
¡ 2
¢
2
2 2
2 4 ̂  ∼     se(̂  ) =    
(1.30)

¶
µ
¡
¢
 2
2

(1.31)
̂  ∼     se(̂  ) =    
2
!
Ã
2 2
¢
¡
)
(1
−
̂


(1.32)
̂ ∼    se(̂ )2 =   For example, the exact sampling distribution of ( − 1)̂ 2  2 is chi-square with  − 1
degrees of freedom.
4

1.3 STATISTICAL PROPERTIES OF THE CER MODEL ESTIMATES27
These appoximate normal distributions can be used to compute approximate
confidence intervals for  2    and  
Approximate Confidence Intervals for  2    and 
Approximate 95% confidence intervals for 2    and  are given byb 2 )
̂ 2 ± 2 · se(̂̂ 22= ̂  ± 2 · p


 2
̂ 
b  ) = ̂  ± 2 · √ ̂  ± 2 · se(̂2(1 − ̂2 )b  ) = ̂ ± 2 · √̂ ± 2 · se(̂(1.33)(1.34)(1.35)

Example 17 Approximate 95% confidence intervals for  2    and  forMicrosoft, Starbucks and the S&P 500.
Using (1.33) - (1.34), the approximate 95% confidence intervals for  2 and  ( = Microsoft, Starbucks, S&P 500) are> lowerSigma2 = sigma2hat - 2*seSigma2hat> upperSigma2 = sigma2hat + 2*seSigma2hat> widthSigma2 = upperSigma2 - lowerSigma2> cbind(lowerSigma2, upperSigma2, widthSigma2)
lowerSigma2 upperSigma2 widthSigma2
MSFT
0.007880.012220.00434
SBUX
0.009780.015150.00538SP500
0.00184
0.002860.00101> upperSigma = sigmahat + 2*seSigmahat> widthSigma = upperSigma - lowerSigma> cbind(lowerSigma, upperSigma, widthSigma)
lowerSigma upperSigma widthSigma
MSFT
0.08940.1111
0.0216SBUX
0.09960.12370.0241SP500
0.0432
0.05370.0105The 95% confidence intervals for  and  2 are larger for Microsoft and Starbucks than for the S&P 500 index. For all assets, the intervals for  are

28

CHAPTER 1 ESTIMATION OF THE CER MODEL

fairly narrow (2% for Microsoft and Starbucks and 1% for S&P 500 index)
indicating that  is precisely estimated.
The approximate 95% confidence intervals for  are> cbind(lowerRho, upperRho, widthRho)
lowerRho upperRho widthRho
msft,sbux
0.206
0.476
0.269
msft,sp500
0.523
0.712
0.189
sbux,sp500
0.337
0.578
0.241
The 95% confidence intervals for  are not too wide and all contain justpositive values away from zero. The smallest interval is for  500 because̂ 500 is closest to 1

1.4

Using Monte Carlo Simulation to Understand the Statistical Properties of Estimators

Let  be the return on a single asset described by the CER model, let 
denote some characteristic (parameter) of the CER model we are interested
in estimating, and let ̂ denote an estimator for  based on a sample ofsize  The exact meaning of estimator bias, bias(̂ ) the interpretationof se(̂) as a measure of precision, the sampling distribution  (̂) and theinterpretation of the coverage probability of a confidence interval for  can
all be a bit hard to grasp at first. If bias(̂ ) = 0 so that [̂] =  then overan infinite number of repeated samples of { }=1 the average of the ̂ values
computed over the infinite samples is equal to the true value  The valueof se(̂) represents the standard deviation of these ̂ values. The sampling
distribution (̂) is the smoothed histogram of these ̂ values. And the 95%
confidence intervals for  will actually contain  in 95% of the samples We canthink of these hypothetical samples as diﬀerent Monte Carlo simulations of
the CER model. In this way we can approximate the computations involvedin evaluating [̂] se(̂)  (̂) and the coverage probability of a confidence
interval using a large, but finite, number of Monte Carlo simulations.

0.1

-0.1

0.1

0.3
100

-0.1

Series 680

0.3 -0.1 0.0 0.1 0.2

Series 7
60

Index

0.1

Series 840

0.3 -0.1

Series 920

0.1

Series 10
0

-0.1

0.2
0.0
0.3 -0.2
0.1
-0.1
0.1
-0.1
0.2

Series 4

0.0
0.2 -0.2
0.0

Series 5-0.2

Series 3

0.3-0.3

Series 2

Series 1

1.4 USING MONTE CARLO SIMULATION TO UNDERSTAND THE STATISTICAL PR

0

20

40

60

80

100

Index

Figure 1.5: Ten simulated samples of size  = 100 from the CER model = 005 +    ∼  (0 (010)2 )

1.4.1

Evaluating the Statistical Properties of ̂ Using
Monte Carlo Simulation

Consider the CER model = 005 +    = 1     100
 ∼ GWN(0 (010)2 )

(1.36)

Here, the true parameter values are  = 005 and  = 010 Using Monte
Carlo simulation, we can simulate  = 1000 diﬀerent samples of size  = 100
from (1.36) giving the sample realizations { }100
=1 for  = 1     1000 The
first 10 of these simulated samples are illustrated in Figure 1.5. Notice that
there is considerable variation in the appearance of the simulated samples,but that all of the simulated samples fluctuate about the true mean value of = 005 and have a typical deviation from the mean of about 010 For each

30

CHAPTER 1 ESTIMATION OF THE CER MODELof the 1000 simulated samples we can estimate ̂ giving 1000 mean estimates{̂1      ̂1000 } A histogram of these 1000 mean values is illustrated in Figure??. The histogram of the estimated means, ̂  can be thought of as an
estimate of the underlying pdf, (̂) of the estimator ̂ which we know from(1.26) is a normal pdf centered at [̂] =  = 005 with se(̂ ) = √010= 001.
100
This normal curve (solid orange line) is superimposed on the histogram in
Figure 1.6. Notice that the center of the histogram (white dashed verticalline) is very close to the true mean value  = 005 That is, on average overthe 1000 Monte Carlo samples the value of ̂ is about 0.05. In some samples,the estimate is too big and in some samples the estimate is too small but on
average the estimate is correct. In fact, the average value of {̂1      ̂1000 }from the 1000 simulated samples is
1 X ̂ = 00497
1000 =1
1000

̂ =

which is very close to the true value 005. If the number of simulated samples
is allowed to go to infinity then the sample average ̂ will be exactly equalto  = 005 :
1 X 
̂ = [̂] =  = 005
lim
→∞ 
=1

The typical size of the spread about the center of the histogram representsse(̂ ) and gives an indication of the precision of ̂  The value of se(̂ ) may
be approximated by computing the sample standard deviation of the 1000̂ values:
v
u
1000
u 1 X
t
(̂ − 004969)2 = 00104
̂ ̂ =
999 =1= 001. If the number of
Notice that this value is very close to se(̂ ) = √010100
simulated sample goes to infinity then
v
u

u 1 X
t
(̂ − ̂)2 = se(̂ ) = 010lim →∞ − 1 =1

The coverage probability of the 95% confidence interval for  can also be
illustrated using Monte Carlo simulation. For each simulation  the interval

20
0

10

Density

30

40

1.4 USING MONTE CARLO SIMULATION TO UNDERSTAND THE STATISTICAL PR

0.02

0.03

0.04

0.05

0.06

0.07

0.08

muhat

Figure 1.6: Distribution of ̂ computed from 1000 Monte Carlo simulations
from the CER model (1.36). White dashed line is the average of the  values,
and orange curve is the true (̂)
̂±100 (0975)×se(̂b  ) is computed. The coverage probability is approximatedby the fraction of intervals that contain (cover) the true  = 005 For the1000 simulated samples, this fraction turns out to be 0.931. As the numberof simulations goes to infinity, the Monte Carlo coverage probability will beequal to 0.95.
Example 18 Monte Carlo simulation to evaluate [̂] se(̂) and 95% confidence intervals for 
The R code to perform the Monte Carlo simulation presented in this sectionis> mu = 0.05

32

CHAPTER 1 ESTIMATION OF THE CER MODEL> sigma = 0.10
> n.obs = 100> n.sim = 1000
> set.seed(111)> sim.means = rep(0,n.sim)> mu.lower = rep(0,n.sim)> mu.upper = rep(0,n.sim)> qt.975 = qt(0.975, nobs-1)> for (sim in 1:n.sim) {+sim.ret = rnorm(n.obs,mean=mu,sd=sigma)+
sim.means[sim] = mean(sim.ret)+
se.muhat = sd(sim.ret)/sqrt(n.obs)+
mu.lower[sim] = sim.means[sim]-qt.975*se.muhat+
mu.upper[sim] = sim.means[sim]+qt.975*se.muhat+ }The 1000 × 1 vectors sim.means, mu.lower and mu.upper contain the valuesof ̂  ̂ −100 (0975)×se(̂b  ) and ̂ +100 (0975)×se(̂b  ) computed from eachof the simulated samples  = 1     1000 The mean and standard deviationof {̂1      ̂1000 } are> mean(sim.means)[1] 0.0497> sd(sim.means)[1] 0.0104
To evaluate the converage probability of the 95% confidence intervals, wecount the number of times each interval actually contains the true value of 
> in.interval = mu >= mu.lower & mu <= mu.upper> sum(in.interval)/n.sim[1] 0.931¥

1.4.2

Evaluating the Statistical Properties of ̂2 and ̂ 
Using Monte Carlo simulation

We can evaluate the statistical properties of ̂ 2 and ̂  by Monte Carlo simulation in the same way that we evaluated the statistical properties of ̂ . We use

1.4 USING MONTE CARLO SIMULATION TO UNDERSTAND THE STATISTICAL PR
the simulation model (1.36) and  = 1000 simulated samples of size  = 100¡ ¢1000¡ ¢1} and {̂ 1      ̂1000 } The histo compute the estimates { ̂ 2      ̂ 2
tograms of these values, with the asymptotic normal distributions overlayed,
are displayed in Figure 1.7. The histogram for the ̂ 2 values is bell-shaped
and slightly right skewed but is centered very close to  2 = 0010 The histogram for the ̂ values is more symmetric and is centered near  = 010
The average values of ̂ 2 and ̂ from the 1000 simulations are̂ 2 =

1000
1 X ¡ 2 ¢
= 000999
̂1000 =1

1 X 
̂ =
̂ = 00997
1000 =1
1000

The Monte Carlo estimate of the bias for ̂ 2 is 000999 − 001 = −00000,
and the estimate of bias for ̂ is 00997 − 0010 = −00003 This confirmsthat ̂2 is unbiased and that the bias in ̂ is extremely small. If the numberof simulated samples,  goes to infinity then ̂ 2 → [̂2 ] =  2 = 001 and̂ → [̂] =  + bias(̂ )
The sample standard deviation values of the Monte Carlo estimates of  2
and  give approximations to se(̂ 2 ) and se(̂)
v
u
1000
u 1 X
t
((̂ 2 ) − 000999)2 = 000135
̂ ̂2 =999 =1
v
u
1000
u 1 X
t
(̂  − 00997)2 = 000676
̂ ̂ =
999 =1
The approximate values for se(̂ 2 ) and se(̂) based on the CLT are(010)2se(̂2 ) = p= 000141
1002
010
= 000707
se(̂2 ) = √2 × 100
Notice that the Monte Carlo estimates of se(̂ 2 ) and se(̂) are a bit diﬀerent
from the CLT based estimates. The reason is that the CLT based estimates

34

CHAPTER 1 ESTIMATION OF THE CER MODEL
sigma hat

0

0

50

10

20

30

Density150
100

Density

200

40

250

50

300

60

sigma2 hat

0.006

0.010

0.014

sigma2 hat values

0.08

0.10

0.12

sigma hat values

Figure 1.7: Histograms of ̂ 2 and ̂ computed from  = 1000 Monte Carlosamples from CER model.

are approximations that hold when the sample size  is large. Because
 = 100 is not too large, the Monte Carlo estimates of se(̂ 2 ) and se(̂) arelikely more accurate (and will be more accurate if the number of simulations
is larger).
¡ ¢For each simulation  the approximate 95% confidence intervals ̂ 2 ±¡ ¢
2 × se(b ̂ 2 ) and ̂  ± 2 × se(̂b  ) are computed. The coverage probabilitiesof these intervals is approximated by the fractions of intervals that contain(cover) the true values  2 = 001 and  = 010 respectively For the 1000
simulated samples, these fractions turn out to be 0.951 and 0.963, respectively. As the number of simulations and the sample size goes to infinity, theMonte Carlo coverage probability will be equal to 0.95.

1.4 USING MONTE CARLO SIMULATION TO UNDERSTAND THE STATISTICAL PR

1.4.3

Evaluating the Statistical Properties of ̂ byMonte Carlo simulation

To evaluate the statistical properties of ̂ = cor(   ) we must simulate from the CER model in matrix form (1.1). For example, consider thebivariate CER model
⎛
⎝

1

⎞

⎛

⎠=⎝

005

⎞

⎛

⎠+⎝

1

⎞

⎠   = 1     100

(1.37)
2
003
2
⎞
⎛⎛ ⎞ ⎛⎞⎞⎛
2(075)(010)(005)
0(010)
⎠⎠ ⎝ 1 ⎠ ∼   ⎝⎝ ⎠  ⎝(075)(010)(005)(005)2
2
0(1.38)
where 1 = 005 2 = 003  1 = 010  2 = 005 and 12 = 075 We
use the simulation model (1.37)-(1.38) with  = 1000 simulated samples ofsize  = 100 to compute the estimates {̂112      ̂1000
12 } The histogram ofthese values, with the asymptotic normal distribution overlaid, is displayedin Figure 1.8. The histogram for the ̂12 values is bell-shaped with a mild left
skewness and centered close to 12 = 075 The sample mean and standard
deviation values of ̂12 across the 1000 simulations are, respectively,
1 X 
̂12 =
̂ = 0747
1000 =1 12
v
u
1000
u 1 X
t
̂ ̂12 =(̂12 − 0747)2 = 0045
999 =1
1000

There is very slight downward bias in ̂12 . The Monte Carlo standard
deviation, ̂√̂12  is very close to the approximate standard error se(̂12 ) =(1 − 0752 ) 100 = 0044 For each simulation  the approximate 95% confib 12 ) is computed. The coverage probability of thisdence interval ̂12 ± 2 × se(̂interval is approximated by the fractions of intervals that contain (cover) thetrue values 12 = 075 For the 1000 simulated samples, this fractions turns
out to be 0.952.

36

CHAPTER 1 ESTIMATION OF THE CER MODEL

0

2

4

Density6

8rhohat

0.55

0.60

0.65

0.70

0.75

0.80

0.85

rhohat values

Figure 1.8: Histograms of ̂12 computed from  = 1000 Monte Carlo samples
from the bivariate CER model (1.37) - (1.38).

1.4.4

Estimating Value-at-Risk in the CER Model

Consider the CER model for the simple return  . From the location-scale
representation
 =  +  ×  
the -quantile of  is =  +  ×  
where  is the -quantile of  ∼  (0 1) Then, for an initial investment
0 and  ∈ (0 1) the  · 100% value-at-risk (VaR ) is given by
VaR = 0 ×  = 0 ( +  ×  )

1.4 USING MONTE CARLO SIMULATION TO UNDERSTAND THE STATISTICAL PR
Given the CER model estimators ̂ and ̂ the plug-in principle estimator
for VaR isd  = 0 × ̂ = 0 (̂ + ̂ ×  )
VaR(1.39)d  is a linear function of ̂  which itself is a linear function of ̂
Here VaRd  can then be easily derived from theand ̂ The statistical properties of VaRstatistical properties of ̂ and ̂
If the CER model is applied to the cc return  then
³
³ 
´
´


+×
VaR = 0 ×  − 1 = 0 
−1 
and the estimate of VaR is³³´´d  = 0 × ̂ − 1 = 0 ̂+̂× − 1 VaR(1.40)d  is a nonlinear function of ̂ (and ̂ and ̂) In this case, theHere, VaRd  cannot be easily derived from the statisticalstatistical properties of VaRproperties of ̂ and ̂
Statistical Properties of ̂ for Simple Returns
Regarding bias we have[̂ ] = [] +  × [̂] ≈  +  ×  (1.41)

Hence, ̂ is approximately unbiased for  because ̂ is unbiased for  and̂ is approximately unbiased for 
To derive results regarding precision and the sampling distribution we
require the following result.
Result. In the CER model⎛ ⎞⎞⎞⎛⎛ ⎞ ⎛0̂se(̂)2⎝ ⎠ ∼  ⎝⎝ ⎠  ⎝⎠⎠(1.42)20se(̂)
̂
for large enough  Hence, ̂ and ̂ are (asymptotically) jointly normally
distributed and cov(̂ ̂) = 0 which implies that they are also independent

38

CHAPTER 1 ESTIMATION OF THE CER MODEL
Using (1.42) we haveThenvar(̂ ) = var(̂ + ̂ ×  )¡ ¢2= var(̂) +  var(̂) + 2 cov(̂ ̂)¡ ¢2= var(̂) +  var(̂) (since cov(̂ ̂) = 0)
¡  ¢2 2


2
=
+ 
 ∙
2
¸
1 ¡  ¢2
2
1+


=

2 
se(̂ ) =

∙
¸12
p
1 ¡  ¢2
var(̂ ) = √ 1 +
2 

(1.43)

Using the above results, the sampling distribution of ̂ can be approximatedby the normal distribution̂ ∼ (  se(̂ )2 )

(1.44)

for large enough Remarks
1. se(̂ ) increases with  and   and decreases with  In particular,se(̂ ) increases as  goes to zero (show plot)
2. The formula for se(̂ ) is not practically useful because it depends on
the unknown value  The practically useful estimated standard error
replaces the unknown value of  with the estimate ̂ and is given by∙¸12
1 ¡  ¢2̂
se(̂b ) = √ 1 +

(1.45)
2 [Insert figure of se(̂b  ) vs.  here]

Example 19 Estimating ̂ for Microsoft, Starbucks and the S&P 500 index.
The estimates of  and se(̂ ), for  = 005 and  = 001 from thesimple monthly returns for Microsoft, Starbucks and the S&P 500 index are1.4 USING MONTE CARLO SIMULATION TO UNDERSTAND THE STATISTICAL PR> n.obs = length(msftRetC)> muhatS = colMeans(cerRetS)> sigmahatS = apply(cerRetS, 2, sd)> qhat.05 = muhatS + sigmahatS*qnorm(0.05)> qhat.01 = muhatS + sigmahatS*qnorm(0.01)> seQhat.05 = (sigmahatS/sqrt(n.obs))*sqrt(1 + 0.5*qnorm(0.05)^2)> seQhat.01 = (sigmahatS/sqrt(n.obs))*sqrt(1 + 0.5*qnorm(0.01)^2)> cbind(qhat.05, seQhat.05, qhat.01, seQhat.01)
qhat.05 seQhat.05 qhat.01 seQhat.01
MSFT -0.1578
0.01187 -0.227
0.01490
SBUX -0.1587
0.01276 -0.233
0.01602
SP500 -0.0758
0.00559 -0.108
0.00702
For Microsoft and Starbucks, the values of se(̂b  ) are about 1.2%, 1.5%
and for  = 005 and  = 001 respectively. For the S&P 500 index, thecorresponding values of se(̂b  ) are only 0.5% and 0.7%, respectively. To seethat these standard errors values are actually fairly large consider the 95%
confidence intervals for 005
and 001> lowerQhat.05 = qhat.05 - 2*seqhat.05> upperQhat.05 = qhat.05 + 2*seqhat.05> widthQhat.05 = upperQhat.05 - lowerQhat.05> cbind(lowerQhat.05, upperQhat.05, widthQhat.05)
lowerQhat.05 upperQhat.05 widthQhat.05
MSFT-0.197-0.11810.0794SBUX-0.203-0.11440.0884SP500-0.095-0.0566
0.0384> lowerQhat.01 = qhat.01 - 2*seqhat.01> upperQhat.01 = qhat.01 + 2*seqhat.01> widthQhat.01 = upperQhat.01 - lowerQhat.01> cbind(lowerQhat.01, upperQhat.01, widthQhat.01)
lowerQhat.01 upperQhat.01 widthQhat.01
MSFT-0.277-0.1772
0.0996SBUX-0.289-0.17750.1110SP500-0.132-0.08420.0482For example, the 95% confidences for 005
and 001
for Microsoft are [−197% −118%]and [−277% −177%] respectively, which are quite large. Hence, the 5% and
1% simple monthly return quantiles are not estimated very precisely. ¥40

CHAPTER 1 ESTIMATION OF THE CER MODEL

Statistical Properties of ̂ for CC Returns
To be completed
The Delta Method To be completed
Statistical Properties of VaRFirst, consider estimated VaR for simple returns (1.39). Then, using (1.41)- (1.44)d  ] = 0 × [̂ ] ≈ 0 ×  = VaR [VaRd  ) = 0 × se(̂ )
se(VaRd  ∼  (VaR  se(VaRd  )2 )
VaRd  is approximately unbiased and normally distributed with
Hence, VaR∙¸121 ¡  ¢2dse(VaR ) = 0 × √ 1 +2 


The practically useful estimated standard error replaces the unknown valueof  with the estimate ̂ and is given by∙¸12
1 ¡  ¢2̂dse(VaR ) = 0 × √ 1 +2 

Example 20 Estimating VaR for Microsoft, Starbucks and the S&P 500
index.
Consider a $100 000 investment for one month in Microsoft, Starbucksd  ), for  = 005
and the S&P 500 index. The estimates of VaR and se(VaR
and  = 001 from the simple monthly returns for Microsoft, Starbucks andthe S&P 500 index are> w0=100000> VaR.05 = qhat.05*w0> seVaR.05 = w0*seQhat.05> VaR.01 = qhat.01*w0> seVaR.01 = w0*seQhat.01

1.5 FURTHER READING> cbind(VaR.05, seVaR.05, VaR.01, seVaR.01)
VaR.05 seVaR.05 VaR.01 seVaR.01
MSFT -15780
1187 -22696
1490
SBUX -15865
1276 -23303
1602
SP500 -7577
559 -10834
702

The 95% confidence intervals for VaR005 and VaR001 are> lowerVaR.05 = VaR.05 - 2*seVaR.05> upperVaR.05 = VaR.05 + 2*seVaR.05> widthVaR.05 = upperVaR.05 - lowerVaR.05> cbind(lowerVaR.05, upperVaR.05, widthVaR.05)
lowerVaR.05 upperVaR.05 widthVaR.05
MSFT-18154-134054748
SBUX-18418-13312
5106
SP500
-8695
-6459
2236
> lowerVaR.01 = VaR.01 - 2*seVaR.01> upperVaR.01 = VaR.01 + 2*seVaR.01> widthVaR.01 = upperVaR.01 - lowerVaR.01> cbind(lowerVaR.01, upperVaR.01, widthVaR.01)
lowerVaR.01 upperVaR.01 widthVaR.01
MSFT-25676-197175959
SBUX-26507-200996408
SP500-12237
-9431
2806

1.5

Further Reading

To be completed

41

Bibliography[1] Campbell, Lo and MacKinley (1998). The Econometrics of Financial
Markets, Princeton University Press, Princeton, NJ.

43