Chapter 1
The Constant Expected Return
Model
Date: September 6, 2013
The first model of asset returns we consider is the very simple constant
expected return (CER) model. This model assumes that an asset’s return
over time is independent and identically normally distributed with a constant (time invariant) mean and variance. The model allows for the returns
on diﬀerent assets to be contemporaneously correlated but that the correlations are constant over time. The CER model is widely used in finance.
For example, it is used in mean-variance portfolio analysis, the Capital Asset Pricing model, and the Black-Scholes option pricing model. Although
this model is very simple, it provides important intuition about the statistical behavior of asset returns and prices and serves as a benchmark against
which more complicated models can be compared and evaluated. It allows us
to discuss and develop several important econometric topics such as Monte
Carlo simulation, estimation, bootstrapping, hypothesis testing, forecasting
and model evaluation.

1.1

CER Model Assumptions

Let  = ln( −1 ) denote the continuously compounded return on asset
 at time  We make the following assumptions regarding the probability
distribution of  for  = 1      assets over the time horizon  = 1     
Assumption 1
1

2CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL(i) Covariance stationarity and ergodicity: {1       } = { }=1 is acovariance stationary and ergodic stochastic process with [ ] =  
var( ) =  2 , cov(   ) =   and cor(   ) =  (ii) Normality:  ∼ (   2 ) for all  and (iii) No serial correlation: cov(   ) = cor(   ) = 0 for  6=  and  = 1     
Assumption 1 states that in every time period asset returns are jointly(multivariate) normally distributed, that the means and the variances of
all asset returns, and all of the pairwise contemporaneous covariances andcorrelations between assets are constant over time. In addition, all of theasset returns are serially uncorrelatedcor(   ) = cov(   ) = 0 for all  and  6= and the returns on all possible pairs of assets  and  are serially uncorrelatedcor(   ) = cov(   ) = 0 for all  6=  and  6= 
Clearly, these are very strong assumptions. However, they allow us to develop
a straightforward probabilistic model for asset returns as well as statistical
tools for estimating the parameters of the model, testing hypotheses about
the parameter values and assumptions.

1.1.1

Regression Model Representation

A convenient mathematical representation or model of asset returns can begiven based on Assumption 1. This is the CER regression model. For assets
 = 1      and time periods  = 1      the CER regression model is{ }=1=  +  ∼ GWN(0  2 )⎧⎨ =
cov(   ) =⎩ 0  6= 

(1.1)

The notation  ∼ GWN(0  2 ) stipulates that the stochastic process { }=1
is a Gaussian white noise process with [ ] = 0 and var( ) =  2  In

1.1 CER MODEL ASSUMPTIONS

3

addition, the random error term  is independent of  for all assets  6= and all time periods  6= .
Using the basic properties of expectation, variance and covariance, we
can derive the following properties of returns in the CER model:[ ] = [ +  ] =  + [ ] =  
var( ) = var( +  ) = var( ) =  2 
cov(   ) = cov( +    +  ) = cov(   ) =   cov(   ) = cov( +    +  ) = cov(   ) = 0  6= Given that covariances and variances of returns are constant over time implies that the correlations between returns over time are also constant: 
cov(   )==  cor(   ) = p var( )var( )0cov(   )== 0  6=   6= cor(   ) = pvar( )var( )Finally, since { }=1 ∼ GWN(0  2 ) it follows that { }=1 ∼  (   2 )
Hence, the CER regression model (1.1) for  is equivalent to the model
implied by Assumption 1.

1.1.2

Interpretation of the CER Regression Model

The CER model has a very simple form and is identical to the measurement
error model in the statistics literature.1 In words, the model states that each
asset return is equal to a constant  (the expected return) plus a normally
distributed random variable  with mean zero and constant variance. The
random variable  can be interpreted as representing the unexpected news
concerning the value of the asset that arrives between time  − 1 and time 
To see this, (1.1) implies that =  −  =  − [ ]
1

In the measurement error model,  represents the  measurement of some physical quantity  and  represents the random measurement error associated with themeasurement device. The value   represents the typical size of a measurement error.

4CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL
so that  is defined as the deviation of the random return from its expectedvalue. If the news between times  − 1 and  is good, then the realized
value of  is positive and the observed return is above its expected value  
If the news is bad, then  is negative and the observed return is less than
expected. The assumption [ ] = 0 means that news, on average, is neutral;neither good nor bad. The assumption that var( ) =  2 can be interpreted
as saying that volatility, or typical magnitude, of news arrival is constantover time. The random news variable aﬀecting asset ,   is allowed to becontemporaneously correlated with the random news variable aﬀecting asset   to capture the idea that news about one asset may spill over and aﬀect
another asset. For example, if asset  is Microsoft stock and asset  is Apple
Computer stock, then one interpretation of news in this context is generalnews about the computer industry and technology. Good news should leadto positive values of both  and   Hence these variables will be positively
correlated due to a positive reaction to a common news component. Finally,
the news on asset  at time  is unrelated to the news on asset  at time  forall times  6=  For example, this means that the news for Apple in January
is not related to the news for Microsoft in February.
Sometimes it is convenient to re-express the CER model (1.1) as{ }=1=  +  =  +   · ∼ GWN(0 1)

(1.2)

In this form, the random news shock is the  standard normal random variable  scaled by the “news” volatility    This form is particularly convenientfor value-at-risk calculations.

1.1.3

Time Aggregation and the CER Model

The CER model for continuously compounded returns has the following niceaggregation property with respect to the interpretation of  as news. Suppose that  represents months so that  is the continuously compoundedmonthly return on asset . Now, instead of the monthly return, suppose weare interested in the annual continuously compounded return  =  (12).
Since multi-period continuously compounded returns are additive,  (12) isthe sum of 12 monthly continuously compounded returns:
 =  (12) =

11
X
=0

− =  + −1 + · · · + −11 5

1.1 CER MODEL ASSUMPTIONSUsing the CER regression model (1.1) for the monthly return   we may
express the annual return  (12) as

11
11
X
X
 (12) =( +  ) = 12 ·  + =  +  (12)
=0

=0

where  = 12 ·  is the annual expected return on asset  and  (12) =P
11
=0 − is the annual random news component. The annual expected
return,   is simply 12 times the monthly expected return,  . The annual
random news component,  (12) , is the accumulation of news over the year.
¡ ¢2
As a result, the variance of the annual news component,   is 12 timethe variance of the monthly news component:var( (12)) = var

Ã 11
X

−

=0

=
=

11
X
=0
11
X

!

var(− ) since  is uncorrelated over time 2 since var( ) is constant over time

=0

¡ ¢2

= 12 ·  2 =  It follows that the standard deviation of the annual news is equal totimes the standard deviation of monthly news:

SD( (12)) =

√
12

√
√
12SD( ) = 12  This result is known as the square root of time rule. Similarly, due to theadditivity of covariances, the covariance between  (12) and  (12) is 12

6CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL
times the monthly covariance:
!
Ã 11
11
X
X
− −
cov( (12)  (12)) = cov
=0

=
=

11
X
=0
11
X

=0

cov(−  − ) since  and  are uncorrelated over time  since cov(   ) is constant over time

=0

= 12 ·   =   The above results imply that the correlation between the annual errors  (12)and  (12) is the same as the correlation between the monthly errors  and :cov( (12)  (12))
cor( (12)  (12)) = pvar( (12)) · var( (12))
12 ·  = q
12 2 · 12 2 ==  = cor(   )
 

1.1.4

The Random Walk Model of Asset Prices

The CER model of asset returns (1.1) gives rise to the so-called random
walk (RW) model for the logarithm of asset prices. To see this, recall thatthe continuously³´ compounded return,   is defined from asset prices via = ln −1 = ln( ) − ln(−1 ) Letting  = ln( ) and using therepresentation of  in the CER model (1.1), we can express the log-price as: = −1 +  +  (1.3)

The representation in (1.3) is known as the RW model for log-prices.2 It isa representation of the CER model in terms of log-prices.
2

The model (1.3) is technically a random walk with drift   A pure random walk has
zero drift ( = 0).

7

1.1 CER MODEL ASSUMPTIONSIn the RW model (1.3),  represents the expected change in the logprice (continuously compounded return) between months  − 1 and  and 
represents the unexpected change in the log-price That is,[∆ ] = [ ] =  
 = ∆ − [∆ ]
where ∆ =  − −1  Further, in the RW model, the unexpected changes
in log-price,   are uncorrelated over time (cov(   ) = 0 for  6= ) sothat future changes in log-price cannot be predicted from past changes in
the log-price.3
The RW model gives the following interpretation for the evolution of log
prices. Let 0 denote the initial log price of asset . The RW model saysthat the log-price at time  = 1 is1 = 0 +  + 1 
where 1 is the value of random news that arrives between times 0 and 1
At time  = 0 the expected log-price at time  = 1 is[1 ] = 0 +  + [1 ] = 0 +  
which is the initial price plus the expected return between times 0 and 1.
Similarly, by recursive substitution the log-price at time  = 2 is2 = 1 +  + 2= 0 +  +  + 1 + 2
2
X= 0 + 2 ·  +
 
=1

which is equal to the initial log-price, 0  plus the two period expected
P2 return,2 ·  , plus the accumulated random news over the two periods, =1   By
repeated recursive substitution, the log price at time  =  is = 0 +  ·  +
3


X

 

=1

The notion that future changes in asset prices cannot be predicted from past changes
in asset prices is often referred to as the weak form of the eﬃcient markets hypothesis.

8CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL
At time  = 0 the expected log-price at time  =  is[ ] = 0 +  ·  
which is the initial price plus the expected growth in prices over  periods.
The actual price,   deviates from the expected price by the accumulated
random news:X − [ ] =
 
=1

At time  = 0 the variance of the log-price at time  is
Ã Xvar( ) = var



=1

!

=  ·  2Hence, the RW model implies that the stochastic process of log-prices { } is
non-stationary because the variance of  increases with  Finally, because ∼  (0  2 ) it follows that (conditional on 0 )  ∼ (0 +     2 )
The term random walk was originally used to describe the unpredictable
movements of a drunken sailor staggering down the street. The sailor starts
at an initial position, 0  outside the bar. The sailor generally moves inthe direction described by  but randomly deviates from this direction after
each step  by an amountPequal to   After  steps the sailor ends up at
position  = 0 +· + =1   The sailor is expected to be at location 
but where he actuallyPends up depends on the accumulation of the randomchanges in direction =1   Because var( ) =  2  the uncertainty about
where the sailor will be increases with each step.
The RW model for log-price implies the following model for price: =  = 0  ·+

=1 

= 0   

=1 

P
where  = 0 +  + =1   The term   represents the expected exponential
growth rate in prices between times 0 and time  and the term


=1

represents the unexpected exponential growth in prices. Here, conditional on 0   is log-normally distributed because  = ln  is normally
distributed.

1.2 MONTE CARLO SIMULATION OF THE CER MODEL

1.1.5

9

The CER Model in Matrix Notation

Define the  × 1 vectors  = (1       )0 , μ = (1       )0  ε =(1       )0 and the  ×  symmetric covariance matrix⎞⎛2
  12 · · ·  1
⎟
⎜ 1
⎟
⎜
2
⎜  12  2 · · ·  2 ⎟⎟var(ε ) = Σ = ⎜⎜ ..
.. . .
.. ⎟ 
.
⎜ .
.
. ⎟
⎠
⎝
2
 1  2 · · ·  Then the CER model matrix notation isr = μ + ε ε ∼  (0 Σ)

(1.4)

which implies that  ∼  (μ Σ)

1.2

Monte Carlo Simulation of the CER Model

A simple technique that can be used to understand the probabilistic behaviorof a model involves using computer simulation methods to create pseudo datafrom the model. The process of creating such pseudo data is called Monte
Carlo simulation.4 To illustrate the use of Monte Carlo simulation, consider
creating pseudo return data from the CER model (1.1) for a single asset.
The steps to create a Monte Carlo simulation from the CER model are:
1. Fix values for the CER model parameters  and .
2. Determine the number of simulated values,  to create.
3. Use a computer random number generator to simulate   valuesof  from a (0  2 ) distribution. Denote these simulated values as̃1      ̃ 
4. Create the simulated return data ̃ =  + ̃ for  = 1     
4Monte Carlo refers to the famous city in Monaco where gambling is legal.

20Frequency

-0.1

0

-0.4

-0.3

10

-0.2

returns

0.0

0.1

30

0.2

40

0.3

10CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL

1993

1996
Ind e x

1999

-0 .4

-0 .2

0 .0

0 .2

re turns

Figure 1.1: Monthly continuously compounded returns on Microsoft. Source:
finance.yahoo.com.

To motivate plausible values for  and  in the simulation, Figure 1.1
shows the actual monthly continuously compounded return data on Microsoft
stock over the period July, 1992 through October, 2000. The parameter = [ ] in the CER model represents this central value, and  represents
the typical size of a deviation about . In Figure 1.1, the returns seem tofluctuate up and down about a central value near 0.03, or 3%, and the typicalsize of a return deviation about 0.03 is roughly 0.10, or 10%. In fact, thesample average return is 2.8% and the sample standard deviation is 10.7%,
respectively.
To mimic the monthly return data on Microsoft in the Monte Carlosimulation, the values  = 003 and  = 010 are used as the model’strue parameters and  = 100 is the number of simulated values (samplesize) Let {̃1      ̃100 } denote the 100 simulated values of the news variable1.2 MONTE CARLO SIMULATION OF THE CER MODEL

11 ∼ GWN(0 (010)2 )The simulated returns are then computed using5
̃ = 003 + ̃   = 1     100(1.5)

Example 1 Simulating observations from the CER model using R
To create and plot the simulated returns from (1.5) use> mu = 0.03
> sd.e = 0.10
> nobs = 100> set.seed(111)> sim.e = rnorm(nobs, mean=0, sd=sd.e)> sim.ret = mu + sim.e> par(mfrow=c(1,2))> ts.plot(sim.ret, main="",+xlab="months",ylab="return", lwd=2, col="blue")> abline(h=mu)> hist(sim.ret, main="", xlab="returns", col="slateblue1")> par(mfrow=c(1,1))
The simulated returns {̃ }100
=1 are shown in Figure 1.2. The simulated returndata fluctuate randomly about  = 003 and the typical size of the fluctuation is approximately equal to  = 010 The simulated return data looksomewhat like the actual monthly return data for Microsoft. The main difference is that the return volatility for Microsoft appears to have increasedin the latter part of the sample whereas the simulated data has constantvolatility over the entire sample. ¥Monte Carlo simulation of a model can be used as a first pass “reality
check” of the model. If simulated data from the model do not look like thedata that the model is supposed to describe, then serious doubt is cast on themodel. However, if simulated data look reasonably close to the actual datathen the first step reality check is passed. Ideally, one should consider many
simulated samples from the model because it is possible for a given simulated
sample to look strange simply because of an unusual set of random numbers.
Example 2 Simulating log-prices from the RW model
5

Alternatively, the returns can be simulated directly by simulating observations froma normal distribution with mean 005 and standard deviation 01020Frequency

0.0
-0.3

0

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10

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0.1

30

0.2

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40

12CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL

0

20

40

60

80

100

-0 .3

-0 .1

m o nths

0 .1

0 .3

re turns

Figure 1.2: Simulated returns from the CER model  = 003 +    ∼GWN(0 (010)2 )
The RW model for log-price based on the CER model (1.5) is = 0 + 003 ·  +


X
=1

   ∼ GWN(0 (010)2 )

A Monte Carlo simulation of this RW model with 0 = 1 can be created in
R using> mu = 0.03
> sd.e = 0.10
> nobs = 100> set.seed(111)> sim.e = rnorm(nobs, mean=0, sd=sd.e)> sim.p = 1 + mu*seq(nobs) + cumsum(sim.e)> sim.P = exp(sim.p)
Figure 1.3 shows the simulated values. The top panel shows the simulated
log price, ̃  the expected price [ ] = 1 + 003 ·  and the accumulated

0 1 2 3 4

13

p(t)
E[p(t)]
p(t)-E[p(t)]-2log price

1.2 MONTE CARLO SIMULATION OF THE CER MODEL

0

20

40

60

80

100

60

80

100

20
0

price

40

Time

0

20

40
Time

Figure 1.3: Simulated values from the RW model  = 0 + 003 ·  + ∼ GWN(0 (010)2 )

P

=1  P
random news ̃ − [ ] = =1 ̃ . The bottom panel shows the simulated
price levels ̃ = ̃  Figure 1.4 shows the actual log prices and price levels for
Microsoft stock. Notice the similarity between the simulated random walkdata and the actual data. ¥1.2.1

Simulating Returns on More than One Asset

Creating a Monte Carlo simulation of more than one return from the CER
model requires simulating observations from a multivariate normal distribution. This follows from the matrix representation of the CER model givenin (1.4). The steps required to create a multivariate Monte Carlo simulationare:
1. Fix values for  × 1 mean vector μ and the  ×  covariance matrixΣ.

14CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL

6

10

20

40

80

Log monthly prices on MSFT

1993

1994

1995

1996

1997

1998

1999

2000

2001

1998

1999

2000

2001

20

40

60

80 100Monthly Prices on MSFT

1993

1994

1995

1996

1997

Figure 1.4: Monthly prices and log-prices on Microsoft stock. Source: finance.yahoo.com.
2. Determine the number of simulated values,  to create.
3. Use a computer random number generator to simulate   values ofthe  × 1 random vector ε from the multivariate normal distribution
(0 Σ). Denote these simulated vectors as ε̃1      ε̃ 
4. Create the  × 1 simulated return vector ̃ = μ + ε̃ for  = 1     
To motivate the parameters for a multivariate simulation of the CER
model, consider the monthly continuously compounded returns for Microsoft,
Starbucks and the S & P 500 index over the ten-year period July, 1992
through October, 2000 illustrated in Figures 1.5 and 1.6. All returns seemto fluctuate around a mean value near zero. The volatilities of Microsoft
and Starbucks are similar with typical magnitudes around 0.10, or 10%. The
volatility of the S&P 500 index is considerably smaller at about 0.05, or
5%. The pairwise scatterplots show that all returns appear to be positively
related. The pairs (MSFT, SP500) and (SBUX, SP500) appear to be the mostcorrelated, and the shape of the scatter indicates correlation values around0.5. The pair (MSFT, SBUX) shows a weak positive correlation around 0.2.

15

sbux
sp500

0.0
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-0.05 0.00 0.05 0.10
-0.4

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msft

0.2

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0.0

0.2

1.2 MONTE CARLO SIMULATION OF THE CER MODEL

1993

1994

1995

1996

1997

1998

1999

2000

Index

Figure 1.5: Monthly continuously compounded returns on Microsoft, Starbucks and the S&P 500 index. Source: finance.yahoo.com.
Let  = (     500 )0 . Then, a plausible value for μ is μ = (0 0 0)0 
and a plausible value for Σ is⎛⎞2(010)(010)(010)(02) (010)(005)(05)
⎜
⎟
⎜
⎟
2
Σ = ⎜ (010)(010)(02)(010)(010)(005)(05) ⎟⎝⎠2(010)(005)(05) (010)(005)(05)(005)⎛⎞
00100 00020 00025⎜⎟⎜⎟= ⎜ 00020 00100 00025 ⎟ ⎝⎠
00025 00025 00025
Example 3 Monte Carlo simulation of CER model for three assets
Simulating values from the multivariate CER model (1.4) requires simulating
multivariate normal random variables. In R, this can be done using the16CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL

-0.2

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0.2

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0.0

0.2

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-0.2

sbux

-0.15

sp500

-0.05 0.00 0.05 0.10

-0.4

-0.2

msft

-0.4

-0.2

0.0

0.2

-0.15

-0.05 0.00 0.05 0.10

Figure 1.6: Scatterplot matrix of the monthly continuously compounded
returns on Microsoft, Starbucks and the S&P 500 index.
function rmvnorm() from the package mvtnorm. To create a Monte Carlosimulation from the CER model calibrated to the month continuously returns
on Microsoft, Starbucks and the S&P 500 index use> library(mvtnorm)> mu = rep(0,3)> sig.msft = 0.10
> sig.sbux = 0.10
> sig.sp500 = 0.05
> rho.msft.sbux = 0.2> rho.msft.sp500 = 0.5> rho.sbux.sp500 = 0.5> sig.msft.sbux = rho.msft.sbux*sig.msft*sig.sbux> sig.msft.sp500 = rho.msft.sp500*sig.msft*sig.sp500> sig.sbux.sp500 = rho.sbux.sp500*sig.sbux*sig.sp500> Sigma = matrix(c(sig.msft^2, sig.msft.sbux, sig.msft.sp500,+
sig.msft.sbux, sig.sbux^2, sig.sbux.sp500,

0.1
0.0
0.05
0.00

msft

-0.05

sp500

0.10

-0.2 -0.1

sbux

0.2

0.3 -0.2

-0.1

0.0

0.1

0.2

1.3 ESTIMATING THE PARAMETERS OF THE CER MODEL17

0

20

40

60

80

100

Index

Figure 1.7: Monte Carlo simulation of monthly continuously compounded
returns on Microsoft, Starbuck and the S&P 500 index from the multivariate
CER model.
+
sig.msft.sp500, sig.sbux.sp500, sig.sp500^2),+
nrow=3, ncol=3, byrow=TRUE)> nobs = 100> set.seed(123)> returns.sim = rmvnorm(nobs, mean=mu, sigma=Sigma)
The simulated returns are shown in Figure 1.7. They look fairly similar to
the actual returns given in Figure 1.5. ¥1.3

Estimating the Parameters of the CER
Model

The CER model of asset returns gives us a simple framework for interpretingthe time series behavior of asset returns and prices. At the beginning of every
month ,  is a random variable representing the continuously compounded18CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL

0.0

0.1

0.2

0.3

0.1

0.2

-0.2 -0.1

0.1

0.2

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-0.2

-0.1

0.0

msft

0.05

0.10

-0.2 -0.1

0.0

sbux

-0.05

0.00

sp500

-0.2

-0.1

0.0

0.1

0.2

-0.05

0.00

0.05

0.10

Figure 1.8: Scatterplot matrix of the simulated returns on Microsoft, Starbucks and the S&P 500 index.
return to be realized at the end of the month. The CER model states that ∼  (   2 ). Our best guess for the return at the end of the monthis [ ] =  , our measure 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 isconstant over time so that the normal distribution characterizing monthly
returns is the same every month.
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 monthly continuously compounded returns on  different assets over the horizon  = 1      It is assumed that the observed
returns are realizations of the stochastic process {1 }=1 , where  is de-1.3 ESTIMATING THE PARAMETERS OF THE CER MODEL19
scribed by the CER model (1.1). Under these assumptions, we can use theobserved returns to estimate the unknown parameters of the CER model.
However, before we describe the estimation of the CER model, it is necessary
to review some fundamental concepts in the statistical theory of estimation.

1.3.1

Estimators and EstimatesLet  denote some characteristic of the CER model (1.1) we are interested
in estimating. For example, if we are interested in the expected return, then =  ; if we are interested in the variance of returns, then  =  2 ; if we areinterested in the correlation between two returns then  =   The goal is toestimate  based on a sample of size  of the observed data.
Definition 4 Let {1       } denote a collection of  random returns fromthe CER model (1.1), and let  denote some characteristic of the model. An
estimator of  denoted ̂ is a rule or algorithm for estimating  as a functionof the random returns {1       } Here, ̂ is a random variable.
Definition 5 Let {1       } denote an observed sample of size  from the
CER model (1.1), and let  denote some characteristic of the model. An
estimate of  denoted ̂ is simply the value of the estimator for  based on
the observed sample. Here, ̂ is a number.
Example 6 The sample average as an estimator and an estimate
Let {1       } be described by the CER model (1.1), and supposeP
we are1interested in estimating  =  = [ ]. The sample average ̂ =  =1 is an algorithm for computing an estimate of the expected return  Before
the sample is observed, ̂ is 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 which produces
the estimate. For example, suppose  = 5 and the realized values of the
returns are 1 = 01 2 = 005 3 = 0025 4 = −01 5 = −005 Then theestimate of  using the sample average is1
̂ = (01 + 005 + 0025 + −01 + −005) = 0005
5¥

20CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL
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

1.3.2

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 todetermine 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 techniquesor the Central 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 Carlo approximation of  (̂) is the empirical distribution ̂ overthe diﬀerent simulated samples. For a given sample size  Monte Carlosimulation gives a very accurate approximation to  (̂) if the number ofsimulated samples is very 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 the estimate ̂ 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
the random variables {1       } will produce diﬀerent values of ̂ Some
values of ̂ 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.

1.3 ESTIMATING THE PARAMETERS OF THE CER MODEL21
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 7 The estimation error is the diﬀerence between the estimatorand the parameter being estimated:
error(̂ ) = ̂ − Definition 8 The bias of an estimator ̂ of  is the expected estimation
error:
bias(̂ ) = [error(̂ )] = [̂] − Definition 9 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 the estimator produces the correct answer “on average”, where “on average” meansover many hypothetical realizations of the random variables {1       }. It
is important to keep in mind that an unbiased estimator for  may not bevery 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 1.9. ̂1 is an unbiased
estimator of  with a large variance, and ̂2 is a biased estimator of  with asmall variance. Consider first, the pdf of ̂1 . The center of the distribution isat the true value  = 0 [̂1 ] = 0 but the distribution is very widely spreadout 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 thevalue of ̂1 can be quite a bit above or below  Hence, unbiasedness by itself
does not guarantee a good estimator of  Now 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 the value 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 

22CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL

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 1.9: Distributions of competiting estimators for  = 0 ̂1 is unbiased
but has high variance, and ̂2 is biased but has low variance.
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
error is small. The variability of the estimation error is captured by the meansquared error.
Definition 10 The mean squared error of an estimator ̂ of  is given bymse(̂ ) = [(̂ − )2 ] = [error(̂ )2 ]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´22
mse(̂ ) = [(̂ − [̂]) ] + [̂] −  = var(̂) + bias(̂ )2

1.3 ESTIMATING THE PARAMETERS OF THE CER MODEL23
The derivation of this result is straightforward and is given in the appendix.
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.
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 errorqrmse(̂ ) = mse(̂ )
and the standard errorse(̂) =1.3.3

qvar(̂)

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.
Consistency
Let ̂ be an estimator of  based on the random returns {1       }

24CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL
Definition 11 ̂ 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.
Statistical theorems known as Laws of Large Numbers are used to determine if an estimator is consistent or not. In general, we have the following
result: an estimator ̂ is consistent for  if
• bias(̂ ) = 0 as  → ∞³ ´• se ̂ = 0 as  → ∞That is, if  (̂) collapses to  as  → ∞ then ̂ is consistent for 6
Asymptotic Normality
Let ̂ be an estimator of  based on the random returns {1       }
Definition 12 An estimator ̂ is asymptotically normally distributed if
̂ ∼ ( se(̂)2 )(1.6)

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.7
6

Equivalently, ̂ is consistent if mse(̂ ) → 0 as  → ∞There are actually many versions of the CLT with diﬀerent assumptions. 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  andvariance  2 
7

1.3 ESTIMATING THE PARAMETERS OF THE CER MODEL25

1.3.4

Estimators for the Parameters of the CER ModelTo estimate the CER model parameters    2    and  we can use theplug-in principle from statistics:
Plug-in-Principle: Estimate model parameters using corresponding sample
statistics.
For the CER model, the plug-in principle estimates for the CER model
parameters    2    and  are the following sample descriptive statistics:
1X
 = ̄ 
̂ =
 =1

(1.7)

̂ 2

(1.8)



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


1 X̂  =
( − ̂ )( − ̂ )
 − 1 =1̂ =̂ ̂  ̂ (1.9)(1.10)(1.11)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.
Example 13 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
 = 100 monthly continuously compounded returns for Microsoft, Starbucks
and the S & P 500 index over the period July 1992 through October 2000 in
the matrix object returns.mat. These data are illustrated in Figures 1.5 and1.6. The estimates of  using (1.7) can be computed using the R functions
apply() and mean()> muhat.vals = apply(returns.mat,2,mean)> muhat.vals
sbux
msft
sp500
0.02777 0.02756 0.01253

26CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL
giving ̂ = 002777 ̂  = 002756 and ̂500 = 001253 The expected
return estimates for Microsoft and Starbucks are very similar at about 2.8%
per month, whereas the S&P 500 expected return estimate is smaller at only
1.25% per month. The estimates of the parameters  2     using (1.8) and(1.9) can be computed using apply(), var() and sd()
> sigma2hat.vals = apply(returns.mat,2,var)> sigma2hat.vals
sbux
msft
sp500
0.018459 0.011411 0.001432> sigmahat.vals = apply(returns.mat,2,sd)> sigmahat.vals
sbux
msft
sp500
0.13586 0.10682 0.03785
giving̂ 2 = 00185 ̂  = 01359
̂ 2  = 00114 ̂   = 01068̂ 2500 = 00014 ̂ 500 = 00379
Starbucks has the most variable monthly returns, and the S&P 500 index
has the smallest. The scatterplots of the returns are illustrated in Figure 1.6.
All returns appear to be positively related. The pairs (MSFT, SP500) and(SBUX, SP500) appear to be the most correlated. The estimates of   and using (1.10) and (1.11) can be computed using the functions var() andcor()> cov.mat
sbux
msft
sp500
sbux 0.018459 0.004031 0.002158
msft 0.004031 0.011411 0.002244
sp500 0.002158 0.002244 0.001432> cor.mat = cor(returns.z)> cor.matsbux
msft sp500
sbux 1.0000 0.2777 0.4198
msft 0.2777 1.0000 0.5551
sp500 0.4198 0.5551 1.0000

1.4 STATISTICAL PROPERTIES OF THE CER MODEL ESTIMATES27
Notice when var() and cor() are given a matrix of returns they return theestimated variance matrix and the estimated correlation matrix, respectively.
To extract the unique pairwise values use> covhat.vals = cov.mat[lower.tri(cov.mat)]> rhohat.vals = cor.mat[lower.tri(cor.mat)]> names(covhat.vals) <- names(rhohat.vals) <+ c("sbux,msft","sbux,sp500","msft,sp500")> covhat.vals
sbux,msft sbux,sp500 msft,sp500
0.004031
0.002158
0.002244> rhohat.vals
sbux,msft sbux,sp500 msft,sp500
0.27770.41980.5551The pairwise covariances and correlations are̂   = 00040 ̂  500 = 00022 ̂ 500 = 00022
̂  = 02777 ̂ 500 = 05551 ̂500 = 04198
These estimates confirm the visual results from the scatterplot matrix in
Figure 1.6.

1.4

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 stochastic process{1 }=1 where  is assumed to be generated by the CER model (1.1) for = 1     . We first consider the statistical properties of ̂ because thederivations are the most straightforward. We then summarize the propertiesof the remaining estimators.

1.4.1

Statistical Properties of ̂Consider the estimator ̂ given by (1.7) The following sub-sections give
derivations of the statistical properties of ̂ .

28CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL
Bias
P
To determine bias(̂ ) we must first compute [̂ ] = [ −1 =1  ] Using results about the expectation of a linear combination of random variables,
it follows that

#

1X

[̂ ] = 
 =1
"
#

1X
= 
( +  ) (since  =  +  ) =1 
"

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

1X
 (since [ ] = 0  = 1      )= =1 =1
 ·  =  Hence, the mean of (̂ ) is equal to   We have just proved that ̂ anunbiased estimator for  in the CER model.

Precision

Because Pbias(̂ ) = 0 the precision of ̂ is determined by var(̂ ) =

−1
var(
=1  ) Using the results about the variance of a linear combi-

1.4 STATISTICAL PROPERTIES OF THE CER MODEL ESTIMATES29
nation 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
var( ) (since  is independent over time) 2 =1


=

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

 2
1
2



=

2


Hence,
var(̂ ) =

 2



(1.12)

Notice var(̂ ) = var( ) and so is much smaller than var( ) Also, as
the sample size gets larger and larger, var(̂ ) gets smaller and smaller.
Using (1.12), the standard deviation of ̂ isSD(̂ ) =p
var(̂ ) = √ (1.13)The standard deviation of ̂ is often called the standard error of ̂ and isdenoted se(̂ ) :se(̂ ) = SD(̂ ) = √ (1.14)
The value of se(̂ ) is in the same units as ̂ and measures the precision of ̂
as an estimate. If se(̂ ) is small relative to ̂ then ̂ is a relatively preciseof  because (̂ ) will be tightly concentrated around  ; if se(̂ ) is large
relative to  then ̂ is a relatively imprecise estimate of  because  (̂ )will be spread out about   Figure 1.10 illustrates these relationships.

0.8

30CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL

0.4
0.0

0.2

pdf

0.6

pdf 1
pdf 2

-3

-2

-1

0

1

2

3

estimate value

Figure 1.10: Pdfs for ̂ with small and large values of SE(̂ ) True value of = 0
Unfortunately, se(̂ ) is not a practically useful measure of the precisionof ̂ because it depends on the unknown value of    To get a practicallyuseful measure of precision for ̂ we compute the estimated standard errorpbc ) = √se(̂b  ) = var(̂(1.15)


which is just (1.14) with the unknown value of   replaced by the estimateb given by (1.9) .

Example 14 se(̂b  ) values for Microsoft, Starbucks and the S&P 500 index

For the Microsoft, Starbucks and S&P 500 estimates of ̂ the values of se(̂b )using (1.15) are computed using the R commands> nobs = nrow(returns.mat)> se.muhat = sigmahat.vals/sqrt(nobs)> se.muhat
sbux
msft
sp500
0.013586 0.010682 0.003785

1.4 STATISTICAL PROPERTIES OF THE CER MODEL ESTIMATES31
giving01359se(̂b  ) = √= 001359
100
01068= 001068se(̂b   ) = √100
00379= 000379
se(̂b 500 ) = √100

Clearly, the mean return  is estimated more precisely for the S&P 500 index
than it is for Microsoft and Starbucks. This occurs because ̂ 500 is muchsmaller than ̂   and ̂   It is useful to compare the magnitude of se(̂b )to the value of ̂ to evaluate if ̂ is a precise estimate:̂ 002756== 2580se(̂b   )001068002777̂
== 2043se(̂b  )001359
̂500
001253== 3315se(̂b 500 )000378

Here we see that ̂  and ̂ are over two times their estimated standard
error values, and ̂500 is more than three times its estimated standard error
value. ¥The Sampling Distribution and Consistency
In the CER model,  ∼  (   2 ) and since ̂ is an average of these
normal random variables, it is also normally distributed. The mean of ̂ is 
2and its variance is   Therefore, we can express the probability distributionof ̂ asµ¶ 2̂ ∼   

(1.16)
The distribution for ̂ is centered at the true value   and the spread about
the average 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 than

32CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL

1.5
0.0

0.5

1.0

pdf1

2.0

2.5

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

-3

-2

-1

0

1

2

3

x.valsFigure 1.11: (0 1 ) sampling distributions for ̂ for  = 1 10 and 50
for 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 (1.16) 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
The distribution of ̂ , with  = 0 and  2 = 1 for various sample sizes is
illustrated in figure 1.11. Notice how fast the distribution collapses at  = 0
as  increases.

1.4.2

Confidence intervals for 

The precision of ̂ 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 likelihoodthat the interval covers  
The construction of an exact confidence interval for  is based on the1.4 STATISTICAL PROPERTIES OF THE CER MODEL ESTIMATES33
following statistical result (see the appendix for details).
Result: Let { }=1 denote a stochastic process generated from the CER
model (1.1). Define the -ratio as =̂ − 

se(̂
b )

(1.17)

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 (1.17) and the 1 − 2 quantile (critical value)  −1 (1 − 2) to give¶µ̂ − ≤  −1 (1 − 2) = 1 − Pr − −1 (1 − 2) ≤
se(̂
b )
which can be rearranged asb  ) ≤  ≤ ̂ +  −1 (1 − 2) · se(̂b  )) = 1 − Pr (̂ −  −1 (1 − 2) · se(̂Hence, the intervalb  ) ̂ +  −1 (1 − 2) · se(̂b  )][̂ −  −1 (1 − 2) · se(̂b )= ̂ ±  −1 (1 − 2) · se(̂(1.18)

̂ ± 2 · se(̂b  )

(1.19)

covers the true unknown value of  with probability 1 − For example, suppose we want to compute a 95% confidence interval for  In this case  = 005 and 1 −  = 095 Suppose further that  − 1 = 60
(five years of monthly return data) so that  −1 (1 − 2) = 60 (0975) = 2Then the 95% confidence for  is given by

34CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL
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 is to think about the game of horse shoes. The horse shoe is theconfidence 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 is0.95) then 95 of the 100 tosses should ring the post (95 of the constructedconfidence intervals should contain the true value  ).
Example 15 95% confidence intervals for  for Microsoft, Starbucks andthe S & P 500 index.
Consider computing 95% confidence intervals for  using (1.18) 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 = 99 The 97.5%
quantile, 99 (0975) can be computed using the R function qt()
> qt(0.975, df=99)[1] 1.984
Notice that this quantile is very close to 2 Then the exact 95% confidence
intervals are given by  : 002756 ± 1984 · 001068 = [00064 00488]
 : 002777 ± 1984 · 001359 = [00008 00547]
 500 : 001253 ± 1984 · 000379 = [00050 00200]
With probability 0.95, the above intervals will contain the true mean values
assuming the CER model is valid. The 95% confidence intervals for MSFT
and SBUX are fairly wide. The widths are almost 5%, with lower limits1.4 STATISTICAL PROPERTIES OF THE CER MODEL ESTIMATES35
near 0% and upper limits near 5%. This means that with probability 0.95,
the true monthly expected return is somewhere between 0% and 5%. The
economic implications of a 0% expected return and a 5% expected return are
vastly diﬀerent. In contrast, the 95% confidence interval for SP500 is about
half the width of the MSFT or SBUX confidence interval. The lower limit isnear 0.5% and the upper limit is near 2%. This clearly shows that the mean
return for SP500 is estimated much more precisely than the mean return for
MSFT or SBUX.

1.4.3

Interpreting [̂ ] se(̂ ) and Confidence Intervals
Using Monte Carlo Simulation

The exact meaning of unbiasedness, [̂ ] =   the interpretation of se(̂ )as a measure of precision, and the interpretation of the coverage probabilityof a confidence interval can be a bit hard to grasp at first. Strictly speaking,[̂ ] =  means that over an infinite number of repeated samples of { }=1
the average of the ̂ values computed over the infinite samples is equal to
the true value   The value of se(̂ ) represents the standard deviation ofthese ̂ values. And the 95% confidence intervals for  will actually contain in 95% of the samples We can think of these hypothetical samples as
diﬀerent Monte Carlo simulations of the CER model. In this way we canapproximate the computations involved in evaluating [̂ ] se(̂ ) and thecoverage probability of a confidence interval using a large, but finite, numberof Monte Carlo simulations.
To illustrate, consider the CER model = 005 +    = 1     100
 ∼  (0 (010)2 )(1.20)

Using Monte Carlo simulation, we can simulate  = 1000 samples ofsize  = 100 from (1.20) giving the sample realizations {̃ }100
=1 for  =1     1000 The first 10 of these simulated samples are illustrated in Figure
1.12. Notice that there is considerable variation in the simulated samples,but that all of the simulated samples fluctuate about the true mean valueof  = 005 and have a typical deviation from the mean of about 010 For
each of the 1000 simulated samples the estimate ̂ is formed giving 1000
mean estimates {̂1      ̂1000 } A histogram of these 1000 mean values is
illustrated in Figure 1.13. The histogram of the estimated means, ̂  can

0.3
Series 7
Series 8
Series 10

-0.1

0.1

0.3 -0.1

0.1

Series 9

0.3 -0.1 0.0 0.1 0.2

-0.1

0.1

-0.1

0.1

Series 6

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

Series 5

Series 4-0.1

Series 3

0.3-0.3

Series 2

0.3 -0.2

Series 1

36CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL

0

20

40

60

80

100

Index

0

20

40

60

80

100

Index

Figure 1.12: Ten simulated samples of size  = 100 from the CER model = 005 +    ∼  (0 (010)2 )
be thought of as an estimate of the underlying pdf, (̂) of the estimator ̂
which we know from (1.16) is a normal pdf centered at [̂] =  = 005 withse(̂ ) = √010= 001. This normal curve is superimposed on the histogram
100
in Figure 1.13. Notice that the center of the histogram is very close to thetrue mean value  = 005 That is, on average over the 1000 Monte Carlosamples 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 iscorrect. In fact, the average value of {̂1      ̂1000 } from the 1000 simulated
samples is
1 X 
̂ =
̂ = 004969
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 equal1.4 STATISTICAL PROPERTIES OF THE CER MODEL ESTIMATES37
to  = 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01041
̂ ̂ =
999 =1
Notice that this value is very close to se(̂ ) = √010= 001. If the number of100
simulated sample goes to infinity then
v
u

u 1 X
t
lim
(̂ − ̂)2 = se(̂ ) = 010→∞ − 1 =1

The coverage probability of the 95% confidence interval for  can also be
illustrated using Monte Carlo simulation. For each simulation  the interval
̂ ± 100 (0975) × se(̂b  ) is computed. The coverage probability is approximated by the fraction of intervals that contain (cover) the true  = 005
For the 1000 simulated samples, this fraction turns out to be 0.931. As thenumber of simulations goes to infinity, the Monte Carlo coverage probability
will be equal to 0.95.
Example 16 Monte Carlo simulation using R
The R code to perform the Monte Carlo simulation presented in this sectionis> mu = 0.05
> sd = 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)

20
0

10

Density

30

40

38CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL

0.02

0.03

0.04

0.05

0.06

0.07

0.08

muhat

Figure 1.13: Distribution of  computed from 1000 Monte Carlo simulations
from the CER model (1.20).
> qt.975 = qt(0.975, nobs-1)> for (sim in 1:n.sim) {+sim.ret = rnorm(n.obs,mean=mu,sd=sd)+
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+ }> mean(sim.means)[1] 0.04969> sd(sim.means)[1] 0.01041> in.interval = mu >= mu.lower & mu <= mu.upper> sum(in.interval)/n.sim[1] 0.931¥

1.4 STATISTICAL PROPERTIES OF THE CER MODEL ESTIMATES39

1.4.4

Evaluating Consistency of ̂ using Monte Carlo
Simulation

Monte Carlo simulations can also be used to illustrate the asymptotic property that ̂ is a consistent estimator of  
To be completed.

1.4.5

Statistical properties of the estimators of  2    and  

To determine the statistical properties of ̂ 2  ̂   ̂  and ̂ we need to treat
them as a functions of the random variables { }=1  We could straightforwardly derive the statistical properties of ̂  Unfortunately, the corresponding derivations for ̂ 2  ̂   ̂  and ̂ are much messier and in many cases we
do not have simple exact results. Hence, we only state the results and givereferences for the exact derivations.
Bias
Assuming that returns are generated by the CER model (1.1), the sample
variances and covariances are unbiased estimators,
[̂ 2 ] =  2 [̂  ] =   
but the sample standard deviations and correlations are biased estimators,
[̂ ] 6=   [̂ ] 6=  
The biases in ̂  and ̂  are very small and decreasing in  such thatbias(̂     ) = bias(̂   ) = 0 as  → ∞. The proofs of these results
are beyond the scope of this book and may be found, for example, in Goldberger (19xx). However, the results can be easily verified using Monte Carlo
methods.
Precision
The derivations of the variances of ̂ 2  ̂   ̂  and ̂ are complicated, andthe exact results are extremely messy and hard to work with. However, there

40CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL
are simple approximate formulas for the variances of ̂ 2  ̂  and ̂ based on
the CLT that are valid if the sample size,  is reasonably large.8 These large
sample approximate formulas are given by

√ 2
2
2
√  =p  

 2

se(̂ ) ≈ √ 2
(1 − 2 )√se( ) ≈se(̂ 2 ) ≈(1.21)(1.22)(1.23)

where “≈” denotes approximately equal. The approximations are such thatthe approximation error goes to zero as the sample size  gets very large.
As with the formula for the standard error of the sample mean, the formulas
for se(̂ 2 ) and se(̂  ) depend on  2  Larger values of  2 imply less precise
estimates of ̂ 2 and ̂   The formula for se( ) however, 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 the data (scatterplot will almost follow a straight line). Additionally,
the formulas for the standard errors above are inversely related to the squareroot of the sample size. Interestingly, se(̂  ) goes to zero the fastest andse(̂ 2 ) goes to zero the slowest. Hence, for a fixed sample size, these formulas
suggest that  is generally estimated more precisely than  2 and   and is estimated generally more precisely than  2 
The above formulas (1.21) - (1.23) 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 give8

The 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 ̂  

1.4 STATISTICAL PROPERTIES OF THE CER MODEL ESTIMATES41
rise to the estimated standard errors:
̂ 2
se(̂
b 2 ) ≈ p  
 2
̂ 
se(̂
b ) ≈ √ 2
(1 − ̂2 )√se(b  ) ≈

(1.24)(1.25)(1.26)b  ) and se(̂b  ) for Microsoft, Starbucks
Example 17 Computing se(̂b 2 ) se(̂and the S & P 500.
For the Microsoft, Starbucks and S&P 500 return data, the values of se(̂b 2 )
se(̂b  ) are computed using> se.sigma2hat = sigma2hat.vals/sqrt(nobs/2)> se.sigmahat = sigmahat.vals/sqrt(2*nobs)> se.sigma2hat
sbux
msft
sp500
0.0026105 0.0016138 0.0002026> se.sigmahat
sbux
msft
sp500
0.009607 0.007554 0.002676
which gives(01359)201359se(̂b 2 ) = p= 000261 se(̂b  ) = √= 000961
2 · 1001002(01068)201068= 000161 se(̂b   ) = √= 000755
se(̂b 2  ) = p
2 · 1001002(00379)2
00379= 000020 se(̂b 500 ) = √= 000268
se(̂b 2500 ) = p
2 · 100
1002

Notice that ̂ 500 is estimated much more precisely than ̂ and ̂   b )
Also notice that   is estimated more precisely than   The values of se(̂relative to ̂  are much smaller than the values of se(̂b  ) to ̂ 
The values of se(̂b  ) are computed using

42CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL> se.rhohat
sbux,msft sbux,sp500 msft,sp500
0.092290.08238
0.06919
which gives
1 − (02777)2√= 009229
100
1 − (04198)2√se(̂b 500 ) == 008238
100
1 − (05551)2√se(̂b  500 ) == 006919
100
se(̂
b   ) =These standard errors are moderate in size (relative to ̂ ) Notice that 500 has the smallest estimated standard error because ̂2 500 is
closest to one.
Sampling distributions
The exact distributions of ̂ 2  ̂  and ̂ are diﬃcult to derive.9 However,approximate normal distributions of the form (1.6) based on the CLT arereadily available:¶µ
4
2 4 

 

µ
¶
 2
̂  ∼    2!
Ã(1 − ̂2 )2̂ ∼   ̂2 ∼ 

These appoximate normal distributions can be used to compute approximate
confidence intervals for 2    and  
For example, the exact sampling distribution of ( − 1)̂ 2  2 is chi-square with  − 1
degrees of freedom.
91.4 STATISTICAL PROPERTIES OF THE CER MODEL ESTIMATES43
Approximate Confidence Intervals for  2    and 
Approximate 95% confidence intervals for 2    and  are given by̂2b 2 ) = ̂ 2 ± 2 · p  ̂ 2 ± 2 · se(̂
 2
̂ 
̂ ± 2 · se(̂b  ) = ̂  ± 2 · √2(1 − ̂2 )̂ ± 2 · se(̂b  ) = ̂ ± 2 · √

Example 18 95% confidence intervals for 2    and  for Microsoft, Starbucks and the S&P 500.
For the Microsoft, Starbucks and S&P 500 return data the approximate 95%
confidence intervals for  2 are  : 001141 ± 2 · (000161) = [000818 001464] : 001846 ± 2 · (000261) = [001324 002368]
 500 : 000143 ± 2 · (000020) = [000103 000184]
The approximate 95% confidence intervals for  are  : 01068 ± 2 · (0007554) = [009172 01219] : 01359 ± 2 · (0009607) = [011665 01551]
 500 : 00379 ± 2 · (0002676) = [003249 00432]
The approximate 95% confidence intervals for  are    : 02777 ± 2 · (009229) = [009314 04623]
   500 : 05551 ± 2 · (006919) = [041674 06935]
   500 : 04198 ± 2 · (008238) = [025499 05845]
Evaluating the Statistical Properties of ̂2 and ̂  by Monte Carlosimulation
We can evaluate the statistical properties of ̂ 2 and ̂  by Monte Carlo simulation in the same way that we evaluated the statistical properties of ̂ .

0

0

50

50

100

100

150

150

200

200

44CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL

0.004

0.006

0.008

0.010

0.012

0.014

0.016

Estimate of variance

0.08

0.10

0.12

Estimate of std. deviation

Figure 1.14: Histograms of ̂ 2 and ̂ computed from  = 1000 Monte Carlosamples from CER model.
Consider first the variability estimates ̂2 and ̂   We use the simulation
model (1.20) and  = 1000 simulated samples of size  = 50 to compute the¡ ¢1000¡ ¢1} and {̂ 1      ̂ 1000 } The histograms of theseestimates { ̂ 2      ̂ 2
values are displayed in figure 1.14. The histogram for the ̂ 2 values is bellshaped and slightly right skewed but is centered very close to 0010 =  2  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
1 X 2
̂ = 0009952
1000 =11000

1 X
̂ = 009928
1000 =1
1000

The sample standard deviation values of the Monte Carlo estimates of 2and  give approximations to SD(̂ 2 ) and SD(̂)

1.5 FURTHER READING

1.4.6

Evaluating the Statistical Properties of ̂ and̂ by Monte Carlo simulation

To be completed

1.4.7

Estimating Value-at-Risk in the CER Model

To be completed

1.5

45

Further Reading

To be completed

1.6

Appendix

1.6.1

Proofs of Some Technical Results³´2
Result: mse(̂ ) = [(̂ − [̂])2 ] + [̂] −  = var(̂) + bias(̂ )2
Proof. Recall, mse(̂ ) = [(̂ − )2 ] Write̂ −  = ̂ − [̂] + [̂] − 
Then
³
´2
³
´³
´ ³
´2
(̂ − )2 = ̂ − [̂] + 2 ̂ − [̂] [̂] −  + [̂] −  Taking expectations of both sides gives
∙³
h³
´i2
´2 ¸
mse(̂ ) =  ̂ − [̂]+  [̂] − = var(̂) + bias(̂ )2 

46CHAPTER 1 THE CONSTANT EXPECTED RETURN MODEL

1.6.2

The Chi-Square distribution with  degrees of
freedomLet 1  2       be independent standard normal random variables. Thatis, ∼  (0 1)  = 1     
Define a new random variable  such that = 12 + 22 + · · · 2 =


X

2 

=1

Then  is a chi-square random variable with  degrees of freedom. Such arandom variable is often denoted 2 and we use the notation  ∼ 2  The
pdf of  is illustrated in Figure xxx for various values of  Notice that 
is only allowed to take non-negative values. The pdf is highly right skewed
for small values of  and becomes symmetric as  gets large. The mean ofthe distribution is[] = [12 ] + [22 ] + · · · [2 ] = since [2 ] = var( ) = 1

1.6.3

Student’s t distribution with  degrees of freedom

Let  be a standard normal random variable,  ∼ (0 1) and let  bea chi-square random variable with  degrees of freedom,  ∼ 2  Assume
that  and  are independent. Define a new random variable  such that

= p



Then  is a Student’s t random variable with  degrees of freedom and we
use the notation  ∼  to indicate that  is distributed Student’s-t. Figurexxx shows the pdf of  for various values of the degrees of freedom  Noticethat the pdf is symmetric about zero and has a bell shape like the normal.
The tail thickness of the pdf is determined by the degrees of freedom. For
small values of  , the tails are quite spread out and are thicker than thetails of the normal. As  gets large the tails shrink and become close to the47

1.6 APPENDIX

normal. In fact, as  → ∞ the pdf of the Student’s t converges to the pdf
of the normal.
The Student’s-t distribution is used heavily in statistical inference andcritical values from the distribution are often needed. Let  () denote thecritical value such that

Pr(   ()) = For example, if  = 10 and  = 0025 then 10 (0025) = 2228; if  = 100
then 60 (0025) = 200 Since the Student’s-t distribution is symmetric about
zero, we have thatPr(− () ≤  ≤  ()) = 1 − 2For example, if  = 60 and  = 2 then 60 (0025) = 2 andPr(−60 (0025) ≤  ≤ 60 (0025)) = Pr(−2 ≤  ≤ 2) = 1 − 2(0025) = 095

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

49