ams-q01-sol-04-m.nb
1
AMS 691.02 - Portfolio Theory and
Capital Markets I
Solutions 4 - Factor Models
Robert J. Frey
Research Professor
Stony Brook University, Applied Mathematics and Statistics
frey@ams.sunysb.edu
Exercises for Class 3. The Chapters refer to Luenberger’s Investment Science.
February 16, 2005
Mathematica Tutorial
A matrix in Mathematica is represented by a list of equal length lists. The function "MatrixForm" is useful for displaying
both vectors and matrices in a user friendly form.
v = 81, 2, 3<; q = 881, 2, 0<, 84, 0, 6<, 80, 8, 9<<;
MatrixForm@vD
MatrixForm@qD
1y
i
j
z
j
z
j
z
j
j2z
z
j
z
k3{
i1
j
j
j
j
4
j
j
k0
2
0
8
0
6
9
y
z
z
z
z
z
z
{
In Mathematica vectors are treated as one-dimnsional objects not as row or column vectors per se. Thus, the bilinear
form vT Qv can be entered as
ams-q01-sol-04-m.nb
2
v.q.v
178
The "Solve" function attempts to solve an equation or system of equations symbolically. It returns its results as a list of a
list of production rules. Each top-level list represents a solution.
? Solve
Solve@eqns, varsD attempts to solve an equation
or set of equations for the variables vars. Solve@
eqns, vars, elimsD attempts to solve the equations
for vars, eliminating the variables elims. More…
For this system of equations there are four possible solutions. Each solution consists of two production rules.
Solve@8y1 == a x12 + b, y1 ã c x22 + d<, 8x1, x2<D
è!!!!!!!!!!!!!!!!
è!!!!!!!!!!!!!!!!
è!!!!!!!!!!!!!!!!
è!!!!!!!!!!!!!!!!
-b + y1
-d + y1
-b + y1
-d + y1
99x1 Ø - ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅ
,
x2
Ø
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅ
=,
9x1
Ø
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅ
,
x2
Ø
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
è!!!
è!!!
è!!!
è!!! ÅÅÅÅÅÅÅ =,
a
c
a
c
è!!!!!!!!!!!!!!!!
è!!!!!!!!!!!!!!!!
è!!!!!!!!!!!!!!!!
è!!!!!!!!!!!!!!!!
-b + y1
-d + y1
-b + y1
-d + y1
9x1 Ø ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
è!!! ÅÅÅÅÅÅÅ , x2 Ø - ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
è!!! ÅÅÅÅÅÅÅ =, 9x1 Ø ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
è!!! ÅÅÅÅÅÅÅ , x2 Ø ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
è!!! ÅÅÅÅÅÅÅ ==
a
c
a
c
"Outer" is usually used to perform an outer product of two vectors, although it generalizes to other operations and to
more complex list structures.
? Outer
Outer@f, list1, list2, ... D gives the generalized outer product of
the listi, forming all possible combinations of the lowestlevel elements in each of them, and feeding them as arguments
to f. Outer@f, list1, list2, ... , nD treats as separate
elements only sublists at level n in the listi. Outer@f,
list1, list2, ... , n1, n2, ... D treats as separate elements
only sublists at level ni in the corresponding listi. More…
ams-q01-sol-04-m.nb
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MatrixForm@Outer@Times, 8x1, x2, x3<, 8y1, y2<DD
x1 y1
i
j
j
j
j
j x2 y1
j
k x3 y1
x1 y2
x2 y2
x3 y2
y
z
z
z
z
z
z
{
A numerical matrix may be inverted with "Inverse".
? Inverse
Inverse@mD gives the inverse of a square matrix m. More…
qi = Inverse@qD; MatrixForm@qiD
MatrixForm@q.qiD
2
ÅÅÅ
i
5
j
j
j
j
3
j
ÅÅÅÅÅ
j
j
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k - ÅÅÅÅÅ
15
1
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j0
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k0
0
1
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10 y
z
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20
z
z
z
1
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{
15
y
z
z
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There are two common statistical functions used : "Mean" and "Variance".
? Mean
? Variance
Mean@listD gives the statistical mean of the elements in list. More…
Variance@listD gives the
statistical variance of the elements in list. More…
Chapter 7, Problem 2
Here's the data we are given in the problem.
ams-q01-sol-04-m.nb
4
nRiskFree = 0.1;
nMarketReturn = 0.18;
mnCovariance = 880.04, 0.01<, 80.01, 0.02<<;
vnMarketPortfolio = 80.5, 0.5<;
The market variance is a function of the covariance and the market portfolio, both of which are known.
nVarMarket = vnMarketPortfolio.mnCovariance.vnMarketPortfolio
0.02
The gradient of the Langrangian implies that Q x ∝ r – r f . Once we know the returns to within proportionality, we can
then solve for them using the CAPM and the fact that the market portfolio has a b of 1.
8pA, pB< = mnCovariance.80.5, 0.5<;
sol = First@
Solve@
8
l pA ã bA HnMarketReturn - nRiskFreeL,
l pB ã bB HnMarketReturn - nRiskFreeL,
1 ã 8bA, bB<.vnMarketPortfolio
<,
8l, bA, bB<D
D;
Once the b's are known, the CAPM can be used to compute the asset returns.
rA = nRiskFree + HbA ê. solL HnMarketReturn - nRiskFreeL
rB = nRiskFree + HbB ê. solL HnMarketReturn - nRiskFreeL
0.2
0.16
As a final check, we verify that the asset returns when taken in proportion to the market portfolio yield the market return.
nMarketReturn ã 8rA, rB<.vnMarketPortfolio
True
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Chapter 7, Problem 6
Here are the data provided in the table and text of the problem in form appropriate for computation.
vnAssetPrice = 81.50, 2.<;
vnAssetShares = 8100, 150<;
vnAssetReturn = 80.15, 0.12<;
vnAssetSdev = 80.15, 0.09<;
mnCorrelation = 881, 1 ê 3<, 81 ê 3, 1<<;
mnCovariance = Outer@Times, #, #D &@vnAssetSdevD mnCorrelation;
First, the market portfolio is computed as the proportion of each asset's market capitalizaion.
#
vnMarketPortfolio = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ &@vnAssetPrice vnAssetSharesD
Plus üü #
80.333333, 0.666667<
The expected rate of return and standard deviation are computed from the market portfolio applied to the expect asset
returns and asset covariances.
nMarketReturn = vnAssetReturn.vnMarketPortfolio
è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!!!!!!!!!!!!!!!!!
nMarketSdev = vnMarketPortfolio.mnCovariance.vnMarketPortfolio
0.13
0.09
Letting "rf" represent the risk free rate, we can solve for bA and bB in terms of the risk free rate.
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sol = First@
Solve@
8
vnAssetReturnP1T - rf ã bA HnMarketReturn - rfL,
vnAssetReturnP2T - rf ã bB HnMarketReturn - rfL
<,
8bA, bB<
D
D
1. H-0.15 + 1. rfL
1. H-0.12 + 1. rfL
9bA Ø - ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ , bB Ø - ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ
ÅÅÅÅÅÅÅÅÅÅÅÅÅÅ =
0.13 - 1. rf
0.13 - 1. rf
However, if we now weight these b's by the market portfolio, then we find the result is 1 independent of the value of "rf".
Simplify@H8bA, bB< ê. solL.vnMarketPortfolioD
1.
Any risk free rate works in satisfying the requirement that the market has a b of 1. Thus, the risk free rate is
indeterminate.
Chapter 7, Problem 8
The key returns are:
nMarketReturn = 0.33;
nRiskFree = 0.09;
Note that
p – êêp
1
1
E[I ÅÅÅÅÅÅÅÅ
ÅÅÅÅ M(rm – rêm )] = E[H ÅÅÅÅ1c L E[Hp – êê
pL(rm – rêm )] = ÅÅÅÅ12 ( ÅÅÅÅ
ÅÅ + ÅÅÅÅ
ÅÅ ) 20 s2m
c
20
16
and by the definition of b
p– p
1
1
1
ê
2
b = E[I ÅÅÅÅÅÅÅÅ
c ÅÅÅÅ M(rm – rm )] / sm = ÅÅÅÅ
2 ( ÅÅÅÅ
20ÅÅ + ÅÅÅÅ
16ÅÅ ) 20
êê
1
1
1 y
j
nBeta = ÅÅÅÅ i
j ÅÅÅÅÅÅÅ + ÅÅÅÅÅÅÅ z
z 20.
2 k 20
16 {
1.125
We can now determine the CAPM return.
ams-q01-sol-04-m.nb
nCapmReturn = nRiskFree + nBeta HnMarketReturn - nRiskFreeL
0.36
The actual rate of return projected is
1 i 24
24 y
nExpectedReturn = ÅÅÅÅ j
j ÅÅÅÅÅÅÅ + ÅÅÅÅÅÅÅ z
z - 1.
2 k 20
16 {
0.35
Thus, the project falls slightly short of the return required by the CAPM.
Chapter 8, Problem 2
The data for the problem are as follows.
vnReturn = 80.15, 0.20<;
nRiskFree = 0.10;
mnFactorLoadings = 881, 2, 1<, 81, 3, 4<<;
If we assume that the CAPM applies to the factors, then the excess factor returns are
vnFactorExcessReturns =
Inverse@mnFactorLoadingsPAll, 82, 3<TD.HvnReturn - nRiskFreeL
80.02, 0.01<
We can now construct l = {l0 , l1 , l2 } by simply prepending the risk free rate to the factor returns.
vnLambda = Prepend@vnFactorExcessReturns, nRiskFreeD
80.1, 0.02, 0.01<
When we apply l to the original APT factor loadings we recover the original expected returns.
7
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vnReturn ã mnFactorLoadings.vnLambda
True
Chapter 8, Problem 4
This the well-known statistical results for the sample mean and sample variance. Any text on mathematical statistics will
provide a proof. Intuitively, however, we are using the sample, rather than true, mean in the computation of the sample
variance and have lost a degree of freedom, i.e., we have n – 1 and not n independent data points.
Chapter 8, Problem 6
Here is the raw data for the problem: two-years of monthly returns, expressed as percentages.
vnMonthlyReturn =
81.0, 0.5, 4.2, -2.7, -2.0, 3.5, -3.1, 4.1, 1.7, 0.1, -2.4, 3.2,
4.2, 4.5, -2.5, 2.1, -1.7, 3.7, 3.2, -2.4, 2.7, 2.9, -1.9, 1.1<;
nSampleSize = Length@vnMonthlyReturnD;
The mean and standard deviation of return are
nMonthlyMean = Mean@vnMonthlyReturnD
è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
nMonthlySdev = Variance@vnMonthlyReturnD
1.
2.68166
Using the approximation suggested in the text, the annualized mean and standard deviation are
nAnnualizedMean = 12 nMonthlyMean
è!!!!!!!
nAnnualizedSdev = 12 nMonthlySdev
12.
9.28955
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The standard deviations of the monthly estimates are
è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
nMonthlyMeanSdev = nMonthlySdev ë nSampleSize
è!!!!
è!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
nMonthlyVarSdev = I 2 nMonthlySdev 2 M ë nSampleSize - 1
0.547392
2.1206
9