A Solution Manual for Statistics and Data Analysis for Financial Engineering 2nd edition by David Ru

SOLUTION MANUAL

A Solution Manual for: Statistics and Data Analysis for Financial Engineering: 2nd Edition

by David Ruppert and David Matteson John L. Weatherwax∗ March 24, 2021

∗

wax@alum.mit.edu

1

Text copyright ©2019 John L. Weatherwax All Rights Reserved Please Do Not Redistribute Without Permission from the Author

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To my family.

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Introduction Here you’ll find some notes that I wrote up as I worked through this excellent book. For some of the problems I used R to perform any needed calculations. The code snippets for various exercises can be found at the following location: http://www.waxworksmath.com/Authors/N_Z/Ruppert/SNDAFFE_2nd_Edition/ruppert.html

I’ve worked hard to make these notes as good as I can, but I have no illusions that they are perfect. If you feel that that there is a better way to accomplish or explain an exercise or derivation presented in these notes; or that one or more of the explanations is unclear, incomplete, or misleading, please tell me. If you find an error of any kind – technical, grammatical, typographical, whatever – please tell me that, too. I’ll gladly add to the acknowledgments in later printings the name of the first person to bring each problem to my attention.

Acknowledgments Special thanks to (most recent comments are listed first): Chenguang Wang, Reina Shintaku, Ajay Julian, Som Sowmyan, Duc Truyen Dam, Martin Vedani, Patrick Ahearn, John Chung, and Xu Ren for helping improve these notes and solutions.

Contents 1 Chapter 2 (Returns)

6

2 Chapter 3 (Fixed Income Securities)

17

3 Chapter 4 (Exploratory Data Analysis)

31

4 Chapter 5 (Modeling Univariate Distributions)

45

5 Chapter 6 (Resampling)

64

6 Chapter 7 (Multivariate Statistical Models)

71

4

7 Chapter 8 (Copulas)

79

8 Chapter 9 (Regression: Basics)

93

9 Chapter 10 (Regression: Troubleshooting)

102

10 Chapter 11 (Regression: Advanced Topics)

109

11 Chapter 12 (Time Series Models: Basics)

118

12 Chapter 13 (Time Series Models: Further Topics)

137

13 Chapter 14 (GARCH Models)

154

14 Chapter 15 (Cointegration)

171

15 Chapter 16 (Portfolio Theory)

176

16 Chapter 17 (The Capital Asset Pricing Model)

185

17 Chapter 18 (Factor Models and Principal Components)

195

18 Chapter 19 (Risk Management)

207

19 Chapter 20 (Bayesian Data Analysis and MCMC)

220

20 Chapter 21 (Nonparametric Regression and Splines)

228

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0.1

0.15 0.10

0.0

0.05

−0.1

MRK return

0.00

FReturn

−0.05

−0.2

−0.10 −0.15

−0.2

−0.1

0.0

0.1

−0.3

GMReturn

−0.2

−0.1

0.0

0.1

0.2

MSFT return

Figure 1: Left: A scatter plot of the returns of Ford as a function of the returns of GM. Left: A scatter plot of the returns of Merck as a function of the returns of Microsoft.

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Chapter 2 (Returns)

R Lab See the R script Rlab.R where the problem for this chapter are worked.

Problem 1 (GM and Ford returns) In Figure 1 (left) we show the scatter plot of the returns of Ford as a function of the returns of GM. We notice that these returns do appear to be correlated (they are distributed somewhat symmetrically about a line) and the outliers of each stocks return do appear together.

Problem 2 In the accompanying R code we plot the two returns. The two sets of points lie almost exactly on the line y = x. They have a correlation (using the R function cor) given by 0.9995408.

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Problem 3 (MSFT and MRK returns) In Figure 1 (right) we show the scatter plot of the returns of Merck as a function of the returns of Microsoft. We notice that these returns are much less correlated (they have a correlation coefficient of 0.2710328) than the returns between GM and Ford (which have a correlation coefficient of 0.6139335).

Problem 4 I get with a certainty of 51% that the value of the stock will be below $950000 at the close of at least one of the 45 trading days.

Problem 5-8 I get that the hedge fund will make a profit with a probability of 0.38775. I get that the hedge fund will suffer a loss with a probability of 0.58844. I get that the hedge funds expected profit is given by 9922.63 but the expected return (in units of days) -0.01783836.

Problem 9 Based on the expression used to compute the logr variable the mean log-return for one year is 0.05 or 5% and the standard deviation of the log-returns for one year is 0.2 or 20%

Prob1em 10 Looking at the plots generated many of them seem to “trend”. The trending is “real” in the sense that the mean return each day is positive and each step should result in a larger price (in expectation). The variance of this process is very large (as is true with financial data) which adds noise to the time series.

Problem 11 That command simulates the prices of an geometric random walk where Pt = P0 exp(rt + rt−1 + rt−2 + · · · + r1 ) .

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Problem 12 In the R code for this problem we plot the log-returns of McDonald’s as a function of the return. The two are indistinguishable. The correlation between these two series is 0.9999536 indicating that the two types of returns move in the same manner.

Problem 13 Computing the mean and the standard deviation of the two series I get [1] "returns: mean= 0.000503; std= 0.008900" [1] "log-returns: mean= 0.000463; std= 0.008901" Notice that the mean is smaller for the log-returns while the standard deviation looks about the same.

Problem 14-15 We can use the R command t.test to compare these two series of numbers. Running that we get the following > t.test(x=rets, y=lrets, paired=TRUE) t = 15.866, df = 1175, p-value < 2.2e-16 alternative hypothesis: true difference in means is not equal to 0 95 percent confidence interval: 3.478409e-05 4.460108e-05 sample estimates: mean of the differences 3.969258e-05 This indicates that the mean difference is significant between the two samples. The assumptions of the t-test are that the random variables come from a normal distributions. While that is normally not strictly true in financial data sets the t-test is normally robust to deviations in the distribution of the inputs. There are some evidence that the log-returns will have a smaller mean than the mean of the simple returns see https://papers.ssrn.com/sol3/papers.cfm?abstract_id=1549328

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While the above t-test shows that there is a significant difference in the mean returns between the two different ways of calculating them, this difference is very small (four-tenths of a basis point) especially compared to the standard deviation of the returns (89 basis points). Thus while statistically significant the observation made by the t-test may not be practically significant.

The two returns seem close enough for small values to consider them equivalent.

Problem 16 Running the R code for this problem we run a Monte Carlo simulation and find an expected profit from playing this game to be 0.060500 which is larger than zero and one should play this game.

Problem 17 Running our Monte-Carlo simulation with these new rules gives an expected profit for playing this game of 0.243000 which is also larger than zero and one should like to play this game even more than before.

Exercises Exercise 2.1 Part (a): To have a value less than $990 means that we must have a log return less than

990 log = −0.01005034 .

1000 To find this probability we evaluate pnorm(−0.01005034, mean = 0.001, sd = 0.015) to get the value 0.2306556.

Part (b): In five trading days our log return √ will be normally distributed with a mean 5(0.001) = 0.005 and a standard deviation of 5(0.015) = 0.03354102. To be less than $990 we need to have a logarithm less than -0.01005034 (as computed above). Thus in this case we need to evaluate pnorm(−0.01005034, mean = 0.005, sd = 0.03354102) to get the value 0.3268188.

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Exercise 2.2 To have a price greater than 110 we must have a log return greater than log 0.09531018 in one years time. This will happen with a probability of

110 100

=

1 − pnorm(0.09531018, mean = 0.1, sd = 0.2) = 0.509354 .

Exercise 2.3 The log-return for two years would√be a normal random variable with a mean of by 2(0.1) = 0.2 with a standard deviation of 2(0.15) = 0.212132. To have the price in two years be larger than $90 means that the log return must be larger than log(90/80) = 0.117783. Using the theory of normal random variables this will happen with a probability of 0.6508345. This calculation is done in the following R code m = 0.2 s = sqrt(2)*0.15 lr = log(90/80) z = ( lr - m ) / s 1 - pnorm(z)

Exercise 2.4 For this problem we will need to recall the definitions of the net return Rt (k) =

Pt − Pt−k , Pt−k

(1)

and the log return rt (k) = log(1 + Rt (k)) .

Using the above formulas we have that R3 (2) =

P3 − P1 98 − 95 P3 − P3−2 = = = 0.03061224 .

P3−2 P1 95

Then r3 (2) = log(1 + 0.03061224) = 0.03015304.

Exercise 2.5 Part (a): With dividends our single period net return Rt is given by Rt =

Pt + Dt − Pt−1 , Pt−1 10

(2)

so that with the numbers for this problem we get R2 =

54 + 0.2 − 52 P 2 + D2 − P 1 = = 0.04230769 .

P1 52

Part (b): Next recall that with dividends the multiperiod gross returns 1 + Rt (k) is given by

Pt + Dt Pt−1 + Dt−1 Pt−k+1 + Dt−k+1 1 + Rt (k) = ··· .

(3) Pt−1 Pt−2 Pt−k Using the above we have

P 4 + D4 P 3 + D3 P2 + D2 1 + R4 (3) = P3 P P 2 1

53 + 0.2 54 + 0.2 59 + 0.25 = 1.147959 . = 53 54 52 Thus R4 (3) = 0.1479588.

Part (c): For this part we will use

P 3 + D3 53 + 0.2 = log = −0.01492565 . r3 = log(1 + R3 ) = log P2 54 Exercise 2.6 Part (a): Using the formula in the book for Rt (k) when we have dividends we have that

P 2 + D2 P 3 + D3 .

1 + R3 (2) = P2 P1 Using the numbers given in this exercise we compute the above to be R3 (2) = 0.01460689.

Part (b): Using the formula in the book for rt (k) when we have dividends we have that r4 (3) = log(1 + R4 ) + log(1 + R3 ) + log(1 + R2 )

P 4 + D4 P 3 + D3 P2 + D2 = log + log + log P3 P2 P

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87 + 0.125 83 + 0.1 85 + 0.1 = log + log + log .

83 85 82 Evaluating this we find r4 (3) = 0.06300449.

Exercise 2.7 Part (a): The variable rt (4) would be a normal random variable with a mean 4(0.06) = 0.24 and a variance of 4(0.47) = 1.88 (assuming that the number 0.47 quoted is the one period variance and not standard deviation).

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Part (b): We would compute this using the R command pnorm(2, mean = 0.24, sd = sqrt(1.88)) = 0.9003611 .

Part (c): For this recall that r1 (2) = r1 + r0 r2 (2) = r2 + r1 , where each of rt is i.i.d. from N(0.06, 0.47). Thus we have that Cov(r1 (2), r2 (2)) = Cov(r1 + r0 , r2 + r1 ) = Cov(r1 , r2 ) + Cov(r1 , r1 ) + Cov(r0 , r2 ) + Cov(r0 , r1 ) = 0 + σ 2 + 0 + 0 = 0.47 .

Part (d): For this note that rt (3) = rt + rt−1 + rt−2 .

Thus if we know that rt−2 was equal to 0.6 then rt (3) is made of only two random components (and a known constant) thus rt (3)|{rt−2 = 0.6} is a normal random variable with a mean 2(0.06) + 0.6 = 0.72 and a variance of 2(0.47) = 0.94.

Exercise 2.8 Part (a): For this we have

X2 > 1.3 P (X2 > 1.3X0 ) = P X0 = P (r1 + r2 > log(1.3)) = 1 − P (r1 + r2 < log(1.3))

= 1 − pnorm(log(1.3), mean = 2µ, sd =

√

2σ) .

Part (b): For this we first recall that A.4 (with X replaced with R) is given by fY (y) = fR (h(y))|h′(y)| .

For this problem these functions are

(r − µ)2 1 exp − fR (r) = √ 2σ 2 2πσ Y = g(R) = X0 eR

1 Y so h′ (Y ) = .

R = h(Y ) = log X0 Y 12

(4)

Then using Equation 4 we have

(log(Y /X0 ) − µ)2 1 exp − fY (y) = √ , 2σ 2 2πσy for the density of Y = X1 .

Part (c): As we can write Xk = X0 eR where R is a normal random variable with mean a kµ and variance kσ 2 the probability density function of the random variable Xk is derived just like the one for X1 above. In fact we have if Y = Xk we have

(log(Y /X0 ) − kµ)2 1 exp − . fY (y) = √ 2kσ 2 2πkσy Then since the transformation from R to Xk is a monotone transformation the quantiles of R transform to the quantiles of Xk using the same monotone transformation. Thus finding the 0.9 quantile of R (by using the qnorm command in R for a normal with a mean kµ and a variance kσ 2 ) which we will denote as µ0.9 . Given this we find the 0.9 quantile of Xk by computing X0 eµ0.9 .

Part (d): Now Xk2 is equal to Xk2 = X02 exp

(

2

k X

rj

j=1

)

= X02

k Y

e2rj .

j=1

The expectation of this is given by integrating the above against the density p(r1 , r2 , . . . , rk−1, rk ) =

k Y

p(rj ) .

j=1

This gives E[Xk2 ] = X02

Z ∞

−∞

k Y

2rj

e

p(rj )drj

j=1

!

= X02

k

1 (r − µ)2 dr . e √ exp − 2σ 2 2πσ −∞

Z ∞

2r

1 Lets now evaluate the integral above (dropping for now the coefficient √2πσ ) we have

Z ∞

Z ∞

(r − µ)2 dr dr = exp 2r − I= e e 2σ 2 −∞ −∞

Z ∞ r2 rµ µ2 = exp 2r − 2 + 2 − 2 dr 2σ σ 2σ −∞

Z ∞

2 r µ µ2 = exp − 2 + 2 + 2 r − 2 dr 2σ σ 2σ −∞

Z ∞

2 2 µ r − µ2 exp − 2 + 2 + 2 r dr . = e 2σ 2σ σ −∞ (r−µ) 2r − 2σ 2

2

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Working with the exponent in the above exponential we have

1 2 µ 1 2 − 2 r − 2σ 2 + 2 r = − 2 r 2 − 2 2σ 2 + µ r 2σ σ 2σ

1 = − 2 r 2 − 2(2σ 2 + µ)r + (2σ 2 + µ)2 − (2σ 2 + µ)2 2σ

2 1 2 2 2 = − 2 r − (2σ + µ) − (2σ + µ) .

2σ Putting this back into the expression for I we get

Z ∞ 2 2 (2σ 2 +µ)2 1 − µ2 2 2 I = e 2σ e 2σ dr exp − 2 r − (2σ + µ) 2σ −∞ Z ∞ 4σ 4 +4σ 2 µ 1 2 2 e− 2σ2 r dr . = e 2σ −∞

In deriving this last integral I have made the centering substitution v = r − (2σ 2 + µ) (and then dv = dr) where since 2σ 2 + µ is constant and the limits of the integrand are infinite only replaces this “center” value of 2σ 2 + µ to zero and effectively removes that shift.

To keep evaluating the expression for I we recall [3] that r Z ∞ π −αx2 . e = α −∞ Using this we have that I=

√

(5)

2

2πσe2µ+2σ .

1 Putting back the factor of √2πσ we get

2 k 2 E[Xk2 ] = X02 e2µ+2σ = X02 e2(kµ+kσ ) .

Part (e): Now we want to compute Var (Xk ) which we do using Var (Xk ) = E[Xk2 ] − E[Xk ]2 .

Thus we use this formula we need to compute E[Xk ]. Following the same procedure as used above to compute E[Xk2 ] we get

E[Xk ] = X0 e

σ2 +µ 2

k

.

Using this we get

2 2 Var (Xk ) = X02 e2kµ+2kσ − ekσ +2kµ 2

2

= X02 e2kµ ekσ (ekσ − 1) .

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Exercise 2.9 The terms Xn of a log-normal geometric random walk are generated as Xn = X0 er1 +r2 +···+rn , where each ri are iid N(µ, σ 2 ) random variables.

Part (a): The desired condition is equivalent to

X3 > ln(1.2) = P (r1 + r2 + r3 > ln(1.2)) .

P log X0 Given the distribution of rt we know that r1 + r2 + r3 is also a normal random variable with mean 3µ and a variance 3σ 2 . Using that we can compute the above and find the probability given by Z ∞

P =

N(x; 3µ; 3σ 2 )dx .

ln(1.2)

2

Here N(x; µ, σ ) is the probability density function for a normal random variable. Using the numbers given in this exercise I find this to be 0.6329619.

Part (b): We are asked to find Var

Xk k

=

1 Var (Xk ) . k2

√ k is a log-normal random variable with µ = 0.1k and σ = 0.2 k and so using formulas Now X X0 for the variance of a log-normal random variable (with these parameters) we find

Xk = e0.1k+0.04k (e0.04k − 1) .

Var X0 Using this we can compute Var (Xk ) and thus what we are asked for. We find

Xk X2 Var = 20 e0.1k+0.04k (e0.04k − 1) . k k

Part (c): Assuming Pt is a log-normal geometric random walk we know that log PP0t is normal with a mean 0.1t and a variance 0.04t. Transforming to normal random variables the condition we want to impose is to find the smallest t such that

ln(2) − 0.1t √ 1−P Z ≤ ≥ 0.9 .

0.2 t Evaluating the above for several values of t in the following R code ts = 5:20 # When is the first time this is less than 1 - 0.9 = 0.1 pnorm( ( log(2) - 0.1 * ts ) / ( 0.2 * sqrt(ts) ) ) we see that the above will be satisfied for all t ≥ 18.

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Exercise 2.10 To have the stock price greater than $100 means that the log return needs to be larger than

100 = 0.03045921 . log 97 In 20 days the log return should be a normal random variable and have a mean value of √ 20(0.0002) = 0.004 with a standard deviation of 20 × 0.03 = 0.1341641. The probability we have a return larger than the above (and a final price greater than 100) is given by 1 − pnorm(0.03045921, mean = 0.004, sd = 0.1341641) = 0.4218295 .

Exercise 2.11 To have Rt = Pt /P0 ≥ 2 means that the net return is larger than two. From the exercise description in t days the log-return should be a normal random variable and have a mean value of 0.0005t , and a variance of 0.012t .

To work this exercise we will make the assumption that Rt ∼ N(0.0005t, 0.012t) and we want to know the value of t such that

2 − 0.0005t Rt − 0.0005t √ ≥ 0.9 . ≥ √ P (Rt ≥ 2) = P 0.012t 0.012t In the above the expression Rt√−0.0005t is a standard normal random variable. From the normal 0.012t distribution we know that P (Z ≥ zc ) = 0.9 if zc = qnorm(1 − 0.9) = −1.281552 .

Thus we must find the first t such that 2 − 0.0005t √ ≤ −1.281552 .

0.012t Solving for t in the above we find that t ≥ 184.6523 days.

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1200 800

1000

price of bond

1400

1600

par= 1000, coupon payment= 40, T= 30

0.020

0.025

0.030

0.035

0.040

0.045

0.050

yield to maturity

Figure 2: A plot of the value of a bond for various yields to maturity.

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Chapter 3 (Fixed Income Securities)

R Lab See the R script Rlab.R where the problem for this chapter are worked.

Problem 1 When we run the given Rlab.R function we get the plot in Figure 2. There we see graphically that the value found by the spline function (which was 0.03239813) intersects the curve at $1200.

Problem 2 The given code with the uniroot function finds the root of the given function. In this case that is equivalent to finding the square root of 0.5.

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Problem 3 For this we would execute uniroot( function(r) bondvalue(C, T, r, par) - 1200, c(0.03,0.04) ) which gives the following output $root [1] 0.03241618 $f.root [1] -0.5497406 $iter [1] 3 $estim.prec [1] 6.103516e-05 The value of the “root” field gives the desired yield-to-maturity.

Problem 4 For this we would execute uniroot( function(r) bondvalue(280, 8, r, 10000) - 9800, c(0.01,0.04) ) which gives an output structure with a root field with a value 0.02956445.

Problem 5 Using the uniroot and the bondvalue function we find this to be r = 0.06041274.

Problem 6 Using the uniroot and the bondvalue function we find this to be C = 28.99996 dollars.

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short end

long end

0.09 yield

0.08 0.07

0.08 0.07

yield

0.09

1985−12−01 1986−01−01 1986−02−01 1986−03−01

1985−12−01 1986−01−01 1986−02−01 1986−03−01 0.0

0.5

1.0

1.5

2.0

2.5

3.0

20

maturity (monthly)

22

24

26

28

30

maturity (monthly)

Figure 3: Left: The time evolution of the short end of the yield curves for the dates given in Problem 7. Right: The time evolution of the long end of the yield curves for the dates given in Problem 7.

Problem 7 See Figure 3 (left) for the requested plot of the short end of the yield curves. There we see that the yields seemed to increase as the maturity increases. See Figure 3 (right) for the requested plot of the long end of the yield curves. There we see that the yields gets smaller as time progresses.

Problem 8 See Figure 4 for the plot of the yield curves at these four points in time. There we see that the yields seemed increase to a maximum with a duration at about 22.5 months and then decrease.

Problem 9 See Figure 5 for the plot of the numerically computed forward rates (from the yield rates) at the requested four points in time. We see the the wiggles mentioned in the comments in the book in these plots.

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0.08 0.07 0.06

yield 0.05

1986−12−01 1987−01−01 1987−02−01 1987−03−01 0

5

10

15

20

25

30

maturity (monthly)

0.09 0.07

0.08

forward rate

0.10

0.11

Figure 4: The evolution of the yield curves with duration for the dates given in Problem 8.

0.06

1986−12−01 1987−01−01 1987−02−01 1987−03−01 0

5

10

15

20

maturity (month)

Figure 5: The evolution of the yield curves with duration for the dates given in Problem 9.

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Exercises Exercise 3.1 Part (a): The yield to maturity is given by 1 yT = T

Z T

r(t)dt ,

(6)

0

thus with what we are given here we have Z 20 1 y20 = (0.028 + 0.00042t)dt 20 0

0.00042 2 1 0.028(20) + (20 ) = 0.0322 . = 20 2 Part (b): Using the same steps as in the previous part we find y15 = 0.03115 then the price is given by P (T ) = PAR exp {−T yT } .

(7) Thus for the numbers here we get P (15) = 1000e−0.3115(15) = 626.7234 .

Exercise 3.2 Part (a): The yield to maturity is given by Equation 6 thus with what we are given here we have Z 1 8 (0.04 + 0.0002t − 0.00003t2 )dt y8 = 8 0

1 0.0002 2 0.00003 3 = 0.04(8) + (8 ) − (8 ) = 0.04016 .

8 2 3 Part (b): Using the same steps as in the previous part we find y5 = 0.04025 then the price is given by Equation 7. Thus for the numbers here we get P (5) = 1000e−0.04025(5) = 817.708 .

Part (c): See Figure 6 where these two functions are plotted. As the second derivative of both functions is negative both functions are concave. The two curves differ only in the numerical value of the coefficients of the polynomial terms.

Part (d): Using arguments like the above the ten year zero-coupon bond will cost 670.32.

After a year has passed the nine year zero-coupon bond will be worth 697.1114 to give a net return of 0.03996806 or about 4%.

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0.04020 0.04000

0.04010

r(t)

0.04030

forward rate

0.0

0.5

1.0

1.5

2.0

2.5

3.0

2.0

2.5

3.0

t

0.04000 0.04005 0.04010 0.04015 0.04020

y(t)

yield curve

0.0

0.5

1.0

1.5 t

Figure 6: The forward rate and yield curve for the given function r(t).

Exercise 3.3 Part (a): As the coupon rate is larger than the current yield following the discussion in the book in the section titled “Yield to Maturity” the bond is selling above par value.

Part (b): As the bond is selling above par value, the yield to maturity is smaller than the current yield and thus is below 2.8%.

Exercise 3.4 Part (a): This would be given by y5 and is computed using Equation 6 for the forward rate function r(t) given here we find y5 = 0.03616667.

Part (b): This would be given by Equation 7 with PAR = 1 where we find P (5) = 0.8345744.

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Exercise 3.5 We are given the values of the half year spot rates r1 , r2 , r3 , r4 , T = 2, a coupon amount of C = 35 and a par value of $1000. The price of this bond will be given by C C C PAR + C + + + .

(8) 1 + r1 (1 + r2 )2 (1 + r3 )3 (1 + r4 )4 Using what we know we find this expression to be given by $1008.058.

Exercise 3.6 For this we will use the identity N X

rt =

t=1

r − r N +1 , 1−r

to write the sum on the left-hand-side as 2T +1 (r+1) 1 1 1 − (r+1) − 1+r 2T +1 1 1+r = = 1 (1 + r)t 1+r−1 1 − 1+r t=1

1 1 = 1− . r (1 + r)2T

2T X

Using this we see that the left-hand-side of the given expression becomes

1 PAR C C C 1− (1 + r)−2T , + = + PAR − r (1 + r)2T (1 + r)2T r r

(9)

the desired expression.

Exercise 3.7 Part (a): When the bond is issued the annual interest rate is 8.5% (or a semiannual rate of 4.25%) thus the semiannual coupon payment would be for 0.0425(1000) = 42.5.

Part (b): The semiannual interest rate now is 3.8%. After two coupons payments and with the new interest rates the bond is now worth 38 X 1000 42.5 + = 1089.718 , t 38 1.038 1.038 t=1 when we evaluate.

Part (c): In this case the lower limit on the first summation would start at t = 0 and we would have 38 X 1000 42.5 + = 1132.218 . t 1.038 1.03838 t=0 23

Exercise 3.8 Part (a): We are told that P (5) = 818 =

1000 , exp(5r)

which has r = 0.044017. so exp(5r) = 1000 818 Part (b): The price of the bond would now be P (4) =

1000 = 845.3538 . exp(4(0.042))

Part (c): The net return is Rt =

Pt − Pt−1 845.3538 − 818 Pt −1 = = = 0.0335 , Pt−1 Pt−1 818

or about 3.3%.

Exercise 3.9 Part (a): We are told that PAR = 1000, C = 21, and r = 0.04 (so the semiannual rate is 0.02). To compute the price of the bond we can use the right-hand-side of Equation 9 with T = 10 to get

21 21 1.02−20 = 1016.351 . + 1000 − 0.02 0.02

C 21 Part (b): The coupon rate is 1000 = 1000 = 0.021. Since the coupon rate is larger than the current yield of 0.02 this bond is selling above par value.

Exercise 3.10 Part (a): We are told that PAR = 1000 and T = 7. We find to find the value of r such that Equation 9 gives

23 C 23 C −2T (1 + r) = (1 + r)−14 . + 1000 − 1040 = + PAR − r r r r Using the bondvalue function with uniroot we get r = 0.01970275.

23 Part (b): The current yield is 1040 = 0.02211538 semiannually.

Part (c): The yield to maturity is less than the current yield since the bond is selling above par.

24

Exercise 3.11 For this we first compute the yield to maturity y15 which is given by Z 15 Z 15 1 1 y15 = r(t)dt = (0.033 + 0.0012t)dt 15 0 15 0

15 1 0.0012t2 = 0.033t + 15 2 0

1 0.0012 2 = 0.033(15) + 15 = 0.042 .

15 2 Thus with this we have the price of the zero is given by P (15) = 100 exp{−15y15 } = 53.25918 .

Exercise 3.12 The yield to maturity with this continuous forward rate r(t) is given by Z Z 1 T 1 T r(t)dt = (0.04 + 0.001t)dt yT = T 0 T 0

1 0.04T + 0.0005T 2 = 0.04 + 0.0005T . = T

Using this we find P (8) = 0.7032801 × PAR. In the same way at the later time (using the new forward rate) we find P (7.5) = 0.7698478 × PAR. This gives a return of 0.09465309 or about 9.5%.

Exercise 3.13 The yield to maturity with this continuous forward rate r(t) is given by Z 20 Z 20 1 1 y20 = r(t)dt = (0.03 + 0.001t − 0.00021(t − 10)+ )dt 20 0 20 0

Z 20 0.00021 20 0.001t2 1 − 0.03t + (t − 10)+ dt = 20 2 20 0 0

Z 0.00021 20 0.00021 (20 − 10)2 = 0.04 − = 0.039475 . (t − 10)dt = 0.04 − 20 20 2 10

25

Exercise 3.14 From the problem statement the zero spot rates are given by y1 = 0.031 y2 = 0.035 y3 = 0.04 y4 = 0.042 y5 = 0.043 .

Part (a): For the price of these zeros we will use the formula formula P (n) =

PAR .

(1 + yn /2)2n

So with the numbers given we have 1000 = 969.7061 (1 + 0.031/2)2 1000 P (3) = = 887.9714 (1 + 0.04/2)6 1000 = 808.3814 .

P (5) = (1 + 0.043/2)10 P (1) =

Part (b): This would be P (1) = 1000 since the bond pays immediately 1000 = 932.9585 P ′ (3) = P (2) = (1 + 0.035/2)4 1000 = 846.8258 .

P ′ (5) = P (4) = (1 + 0.042/2)8 Note that the yields had to change since one year has passed.

Part (c): This would be P ′ (1) = 1000 1000 = 923.8454 (1 + (0.035 + 0.005)/2)4 1000 = 830.4189 .

P ′ (5) = P (4) = (1 + (0.042 + 0.005)/2)8 P ′ (3) = P (2) =

Again the yields had to change since one year has passed and get incremented by 0.005.

Part (d): We will compute the rate of return for each of these three investments under the case where the analyst is correct and we find [1] 0.03124025 0.04040000 0.02726132 26

Thus the investment in the three year zero bond gives the largest return.

Part (e): We will compute the rate of return for each of these three investments under the case where the analyst is not correct and we find [1] 0.03100000 0.05007258 0.04700961 Thus the investment in the three year zero bond gives the largest return.

Part (f): The bond with the highest spot rate is the five year bond. From the previous part we see that when the spot rate does not change (and we sell our bond a year later) we get a rate of return of 0.04700961. This however is smaller than the three year bond which (under the same situation) would give a rate of return of 0.05007258.

Exercise 3.15 To being we recall the definition of net present value NPVi = Ci e−Ti yTi , of the weights

and of the duration

(10)

NPVi wi = PN , NPV j j=1 DUR =

N X

(11)

wi Ti .

(12)

i=1

We then get that the derivative of DUR as the spot yield changes is then N N X d X Ci Ti e−Ti yTi Ci exp (−Ti (yTi + δ))|δ=0 = − dδ i=1 i=1

=− =− =− =−

N X

i=1 N X

Ti NPVi wi

i=1

NPVj

j=1

N X

NPVj

j=1

N X

! N X

!

i=1

NPVi DUR

N X i=1

Ti

wi Ti

!

i=1

= −DUR 27

N X

Ci e−Ti yTi ,

which is the derivative we wanted to show.

P To derive the books equation 3.31 we recognize that N i=1 NPVi is the value of the bond (which we will denote as B) so that the change in the bond value (which we will call ∆B) as the spot yield changes is given by !

N d X Ci exp {−Ti (yTi + δ)|δ=0 } δy ∆B = dδ i=1 With this we have

= −DUR B δy .

∆B = −DURδy , B which is the books’ Eq. 3.31. Here δy is the change in the spot yield.

Exercise 3.16 Part (a): For the given yield curve yT = 0.04 + 0.001T , we have that y10 = 0.04 + 0.001(10) = 0.05 so the bond price is P (10) = 1000 exp(−T yT ) = 606.5307 .

Part (b): For the new yield curve we find y9 = 0.051 so the price in a year is P (9) = 631.9152 .

The net return is

P (9) − P (10) = 0.04185211 .

P (10)

Exercise 3.17 Part (a): As the coupon rate is larger than the current yield following the discussion in the book in the section titled “Yield to Maturity” the bond is selling above par value.

Part (b): As the bond is selling above par value, the yield to maturity is smaller than the current yield and thus is below 2.8%.

Exercise 3.18 Part (a): This would be given by y5 and is computed using Equation 6 for the forward rate function r(t) given here we find y5 = 0.03416667.

Part (b): This would be given by Equation 7 where we find P (5) = 0.8429621.

28

Exercise 3.19 This seems to be exactly the same problem as Exercise 3.5 above with perhaps some numbers changed slightly. It would be worked in the same way.

Exercise 3.20 Part (a): Given the prices and using the relationship P (n) =

1000 , (1 + yn )2n

we can determine yn is a semiannual rate. For example we have 1000 = 980.39 so y1/2 = 0.02000224 1 + y1/2 1000 = 957.41 so y1 = 0.02200029 (1 + y1 )2 1000 = 923.18 so y3/2 = 0.0270018 (1 + y3/2 )3 1000 = 888.489 so y2 = 0.02999943 .

(1 + y2 )4 Part (b): This would be given by C C C PAR + C + + + .

2 3 1 + y1/2 (1 + y1 ) (1 + y3/2 ) (1 + y2 )4 Using the numbers we know we find this to be 967.2278.

Exercise 3.21 We are told that PAR = 1000 and T = 4. We find to find the value of r such that Equation 9 gives

C 25 C 25 −2T 1015 = + PAR − (1 + r) = (1 + r)−8 . + 1000 − r r r r Using the bondvalue function with the uniroot we get r = 0.02291701.

Exercise 3.22 Part (a): We will first find the yield to maturity for the times to each coupon payment 1/2, 1, 3/2, 2, 5/2, 3, 7/2, 4. For example for 1/2 we have Z 1/2 1 r(t)dt = 0.02292604 . y1/2 = (1/2) 0 29

The other yields are given by [1] 0.02324167 0.02300313 0.02226667 0.02108854 0.01952500 0.01763229 0.01546667

The net present value of the first coupon payment is Ce−(1/2)y1/2 = 21e−(1/2)y1/2 = 20.76065 .

The rest of the coupon payments have net present values of [1] 20.51755 20.28776 20.08532 19.92153 19.80526 19.74321 19.74017

while the final payment of $1000 has a net present value of 940.0082. The price of the bond is the sum of these parts or 1100.87.

Part (b): The duration of the bond is given by Equation 12 which we find to be 3.741419.

The calculations for this problem are worked in this small R code yield = function(T){ ( 0.022*T + (0.005/2)*T^2 - (0.004/3)*T^3 + (0.0003/4)*T^4 ) / T } yield(1/2) times = c(1, 3/2, 2, 5/2, 3, 7/2, 4) yield( times ) pt1 = 21 * exp( -(1/2) * yield(1/2) ) pt2 = 21 * exp( -times * yield( times ) ) pt3 = 1000 * exp( -4 * yield( 4 ) ) pt1 + sum( pt2 ) + pt3 # All net present values: npvs = c( pt1, pt2, pt3 ) B = sum( npvs ) # the bond value ws = npvs / B # the weights DUR = sum( c( 1/2, times, 4 ) * ws ) # the duration

30

CAC FTSE

4000 3000

4000

5000

60001500 2000 2500 3000 3500 4000

2000

4000

SMI

6000

8000

2000

3000

DAX

5000

6000

EuStockMarkets

1992

1993

1994

1995

1996

1997

1998

Time

Figure 7: Plots of the four time series corresponding to the EuStockMarkets data.

3

Chapter 4 (Exploratory Data Analysis)

R Lab See the R script Rlab.R where the problem for this chapter are worked.

Problem 1-2 (EuStockMarkets) In Figure 7 we show the time series plots of the four indices. They don’t look stationary and the fluctuations seem to get larger towards more recent dates (i.e. increase with increasing index level). In Figure 8 we show the time series of log returns. These time series look stationary but the fluctuations appear to get larger at certain periods of time (again the more recent dates).

Problem 3 When we run the code that produces the four normal plots we get the result given in Figure 9.

All of the plots have a “sigmoidal” shape. From that plot we see that none of the indices 31

0.00 0.02 0.00 −0.04 −0.02

FTSE

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CAC

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−0.05

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0.00

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logR

1992

1993

1994

1995

1996

1997

1998

Time

Figure 8: Plots of the log returns corresponding to the four time series in the EuStockMarkets data. have returns that look normal. All of them have heaver tails than a normal distribution would predict. The FTSE index seems to be the most symmetric of the four indices with tail returns caped at around ±0.04. The other three indices appear to be skewed to the left indicating that the negative samples can be larger than the positive samples. For example, the DAX index has its most negative return of −0.1 while its most positive return is around 0.05. The Shapiro-Wilks test for the four returns gives the following data: logR[, i] W = 0.9538, p-value < 2.2e-16 data: logR[, i] W = 0.9554, p-value < 2.2e-16 data: logR[, i] W = 0.982, p-value = 1.574e-14 data: logR[, i] W = 0.9799, p-value = 1.754e-15 The Shapiro-Wilks test assumes that the input data comes from a normal distribution. The test statistics is the value of W and the p-value is the probability we get a W test statistic this extreme from random fluctuations. All of the p-values are very small indicating that we can reject the null hypothesis and conclude that the data does not come from a normal distribution.

32

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Sample Quantiles

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Sample Quantiles

Figure 9: The four normal plots corresponding to the returns of the four indices in the EuStockMarkets data.

Problem 4 1 between the The code q.grid = (1:n)/(n+1) creates a uniform grid with a spacing of n+1 nodes i.e.

> n=9; (1:n)/(n+1) [1] 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 The code qt(q.grid,df=df[j]) evaluates the quantiles of the t-distribution with df[j] degrees of freedom i.e. > qt( (1:n)/(n+1), df=4 ) [1] -1.5332063 -0.9409646 -0.5686491 -0.2707223 [8] 0.9409646 1.5332063

0.0000000

0.2707223

The paste command combines multiple strings together into a single string.

33

0.5686491

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0

0

qt(q.grid, df = df[j])

5

200 400 600

DAX , df= 4

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logR[, i]

logR[, i]

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logR[, i]

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logR[, i]

Figure 10: QQ plots for the returns of the DAX index with respect to a t-distribution with ν = 4.

Problem 5 For the DAX index the QQ plots using the t-distribution are given in Figure 10. There we see that four degrees of freedom fits the data in that the linear fit seems “visually” to be the best.

Problem 6 In Figure 11 I plot the KDE (in black) of the returns, the t model (in red), and a normal model (in blue). We can see that the t model does a good job of matching the KDE both in the central region and in the tails of the distribution.

Problem 7 Looking at the “help” for the density command we see that the bw parameter controls the bandwidth selector to be used and has a default value of nrd0. Information about this can be given by looking at the results of the online help for bw.nrd0. That help page gives a description of the procedure used and call is “Silverman’s rule of thumb”. The default kernel 34

left tail

20

15 10

Density

10

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30 −0.10

−0.05

N = 1859

0.00

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5 0

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20

both tails KDE t: df= 4 normal

−0.05

−0.03

−0.01

0.00

0.00

0.01

0.02

0.03

0.04

0.05

Bandwidth = 0.001645

Figure 11: A plot of modeling log returns of the DAX index using: a kernel density estimate (KDE), a t-distribution with four degrees of freedom, and an normal distribution. is the Gaussian kernel.

Problem 8 For the CAC index the QQ plots using t-distributions with different degrees of freedom are computed in the R code for this problem. From those plots it looks visually like six degrees of freedom fits the data best. Using this value we plot the parametric t-distribution with a KDE of the return data. These plots are given in Figure 12. Note that visually the curves agree quite well.

Problem 9-10 The price series does not appear stationary. We see a steady climb of the assets value. The return series appears more stationary than the series of prices. Notice that the variance of the returns is non-constant.

35

left tail

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−0.05

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Bandwidth = 0.001961

Figure 12: A plot of modeling log returns of the CAC index using: a kernel density estimate (KDE), a t-distribution with six degrees of freedom, and an normal distribution.

Normal Q−Q Plot

1 0

Theoretical Quantiles

−1

30

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20

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Histogram of LogRet

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Figure 13: The plots for Problem 11.

36

0.04

Problem 11 These plots are given in Figure 13. The log returns do not appear normal (the ShapiroWilks) test will also agree with this statement). From the QQ plot we see that the data has heavier tails than a normal distribution. The log returns appear to be mostly symmetrically distributed with perhaps a bit more left skew. The tails appear to have about the same weighting i.e. none appears heavier than the other.

Exercises See the R script chap 4.R where the exercises for this chapter are worked.

Exercise 4.1 Part (a): We can use the summary command to compute some of these statistics > summary(ford.s[,2]) Min. 1st Qu. -0.1810000 -0.0099070

Median 0.0000000

Mean 0.0007601

3rd Qu. 0.0112900

Max. 0.1040000

and the sd command to compute the standard deviation where we find the value of 0.01831557.

Part (b): See Figure 14 where we plot a normal QQ plot for these returns. In that plot we see the classic mismatch that normal models have with return data, that is, the normal model does not fit the tails of the distribution very well.

Part (c): The result from using the shapiro.test command is Shapiro-Wilk normality test data: ford.s[, 2] W = 0.9639, p-value < 2.2e-16 With a P -value this we can reject the null hypothesis of a normal distribution for the returns.

Part (d): See Figure 15 where we produce QQ plot for a t-distribution of the Ford returns for various degrees of freedom. From that plot we see that the line that appears to match the data best is for six degrees of freedom.

Next we remove the return that has the smallest negative absolute value from all returns (the return from black Monday) and then reproduce the QQ plots over the same values for 37

0 −3

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Theoretical Quantiles

1

2

3

Normal Q−Q Plot

−0.15

−0.10

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Sample Quantiles

Figure 14: A normal QQ plot for the returns in the ford.s data frame. the degree of freedom parameter as above. When we do that we get qualitatively the same results and the degree of freedom that appears to be the best is the still the one with six degrees of freedom.

Part (e): The formula referred to is q(1 − q) . n[f {F −1 (q)}]2

(13)

where for the median we have q = 0.5, F −1 (q) = 0.0 (using the sample median), n = 2000, and f {F −1 (q)} = 26.06007 (using the approx function and the output from density).

Using these numbers we compute the standard error of the median to be 0.0004290219. The standard error the mean is given by √σ̂n = 0.0004095486 which is slightly smaller than the standard error of the median.

Exercise 4.2 Part (a): We can use the summary command to compute some of these statistics > summary(logR) Min. 1st Qu. -0.1922000 -0.0116100

Median 0.0008147

Mean 0.0014950 38

3rd Qu. 0.0140700

Max. 0.1485000

−600

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qt(q.grid, df = df[j])

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Figure 15: QQ plots for various values of ν (the degree of freedom parameter in a tdistribution).

39

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Figure 16: A normal QQ plot for the returns from the RecentFord data frame. and the sd command to compute the standard deviation where we find the value of 0.02627306.

Part (b): See Figure 16 where we plot a normal QQ plot for these returns. In that plot we see the classic mismatch that normal models have with return data, that is, the normal model does not fit the tails of the distribution very well.

Part (c): The result from using the shapiro.test command is Shapiro-Wilk normality test data: logR W = 0.9344, p-value < 2.2e-16 With a P -value this we can reject the null hypothesis of a normal distribution for the returns.

Part (d): See Figure 17 where we produce QQ plot for a t-distribution of the Ford returns for various degrees of freedom. From that plot we see that the line that appears to match the data best is for four degrees of freedom (it was six in the previous time period).

Next we remove the return that has the smallest negative value from all returns and then reproduce the QQ plots over the same values for the degree of freedom parameter as above.

When we do that we get qualitatively the same results and the degree of freedom that appears to be the best is unchanged.

40

F, df= 4

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qt(q.grid, df = df[j])

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Figure 17: QQ plots for various values of ν (the degree of freedom parameter in a tdistribution).

I find the day of the smallest return on ford to be 20090511. A web search gives that this was during the global financial crisis and many automobile manufactures were having financial difficulties.

Part (e): The formula referred to is given in Equation 13 where for the median we have q = 0.5, F −1 (q) = 0.000814664 (using the sample median), n = 1257, and f {F −1 (q)} = 22.48967 (using the approx function and the output from density). Using these numbers we compute the standard error of the median to be 0.0006270746. The standard error the mean is given by √σ̂n = 0.0007410423 which is slightly larger than the standard error of the median.

Exercise 4.3 We plot the suggested parametric and nonparametric density estimates in Figure 18. There we see that the normal density with a mean taken as the median and the standard deviation taken to be the MAD gives a closer fit to the kernel density estimate than does the normal density with a mean and standard deviation computed from the sample.

41

density.default(x = diff_dy)

0

5000

Density

10000

15000

KDE normal(mean,sd) normal(median,mad)

−1e−04

0e+00 N = 1866

1e−04

2e−04

Bandwidth = 4.765e−06

Figure 18: Density estimation of the first differences of the variable dy in the Garch data set.

Exercise 4.4 Part (a): When the data is plotted on the y-axis a convex pattern means that the data is skewed to the right.

Part (b): When the data is plotted on the y-axis a concave pattern means that the data is skewed to the left.

Part (c): When the data is plotted on the y-axis a convex-concave pattern means that the data is more clustered in the center than a normal distribution i.e. it has lighter tails than a normal distribution.

Part (d): When the data is plotted on the y-axis a concave-convex pattern means that the data is less clustered in the center than a normal distribution i.e. it has heavier tails than a normal distribution.

Exercise 4.5 See Figure 19 where we construct the requested plots. For the plots of the data from simulated normal random variables notice that for each value of p the line touches “all” 42

1

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Figure 19: Left: The 3×2 grid of QQ plots for simulated normal random variables. Middle: The 3 × 2 grid of QQ plots for the difference in the bp variable in the Garch data frame.

Right: The 3 × 2 grid of QQ plots for the difference in the logarithm of the bp variable in the Garch data frame. of the points. When comparing the logarithm of bp vs. the direct change we see that the logarithm seems to have more points that deviate from a straight line than when we consider just the raw differences of bp. This indicates that the raw difference of bp are more normal than the differences of the logarithm of bp.

Exercise 4.6 Part (a): For this I compute 1.959964 which is the negative of the number given for Φ−1 (0.025) because the normal distribution is symmetric.

Part (b): We can compute this in R using the qnorm function with parameters specified for the mean and the standard deviation. Using this function I compute the value of 3.771808.

Exercise 4.7 From the book the qth sample quantile is approximately normally distributed with a variance given by q(1 − q) . n[f {F −1 (q)}]2

We want to find the value of q such that the above is minimized. Note that for the uniform distribution we have F −1 (q) = q so that the denominator of the above fraction simplifies to 43

n. Using calculus we find the extreme value of q(1 − q) , n given when q = 12 which turns out to be the maximum of this expression. Minimums of this expression happen when q = 0 and q = 1 where the variance is zero.

44

2 10

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1.0

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Figure 20: QQ plots of untransformed and transformed male.earnings data.

4

Chapter 5 (Modeling Univariate Distributions)

R Lab See the R script Rlab.R where the problem for this chapter are worked.

Problem 1 In the qqplots both the untransformed and the log-transformed data look concave (see Figure 20). The square-root transformed plot looks to be the straightest curve (of the three).

Note however that the left-tail of the distribution is not modeled very well in with this transformation (other transformations don’t do better however).

The box plots (see Figure 21) don’t show much qualitative difference between the methods.

All methods seem to have a large number of outliers and to be skewed to one side.

The kernel density estimate (see Figure 22) show that the untransformed and the logtransformed variables look skewed while the square-root transformed variables look to have the most “normal” (and symmetric) form.

45

sqruare−root transformed

2

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untransformed

1/4 power

1.0

1.5

1.5

2.0

2.0

2.5

3.0

3.5

2.5

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log−transformed

Figure 21: Box plots of untransformed and transformed male.earnings data. sqruare−root transformed

0.3 Density

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Bandwidth = 0.06885

N = 5956

2.0

2.5

Bandwidth = 0.03448

Figure 22: KDE plots of untransformed and transformed male.earnings data.

46

Problem 2 (the Box-cox transformation) We can automate the selection of the exponent to apply to the male.earnings data that will produce the most “normal data” by using the boxcox function from the MASS package.

The code in Rlab.R shows how to use this function on the male.earnings data.

Part (a): The logical vector ind holds a TRUE for the index in x (and y) that is the location of the maximum in the Box-Cox likelihood plot. The logical vector ind2 holds the locations where we are within the 95% confidence interval for λ.

Part (b):Running that code give the two plots shown in Figure 23. The optimal value of λ to use in the Box-Cox transformation is found to be [1] "Maxlikihood lambda value=

0.360000"

This is smaller (but on the same order of magnitude) as the value of 0.5 used in the squareroot transformation.

Part (c): A 95% confidence interval for λ is given by > range(bc$x[ind2]) [1] 0.32 0.40 Part (d): We would need to change the 0.95 to 0.99 in the qchisq call.

Problem 3 Running the given code we find that the variable fit has the following elements (partial) > fit $minimum [1] 20121.41 $estimate mean 17.322933

sd nu 7.492441 21.599882

xi 1.651652

These are the parameters in an Fernandez-Steel (F-S) skewed t-distribution. Here ξ = 1.65 > 1 which indicates right skewness of the data. The other parameters are the ones found in the standardized t-distribution.

47

density.default(x = bcPower(male.earnings, ml_lam))

Density

0.00

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95%

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−1

0

1

2

2

λ

4

6

8

N = 5956 Bandwidth = 0.1789

Figure 23: Left: The output from the boxcox command showing the location of the maximum likelihood solution. Right: A KDE of the earnings data that results when we apply the optimal value of λ to the male.earnings data.

48

0.06

density.default(x = male.earnings)

0.03 0.00

0.01

0.02

density

0.04

0.05

KDE sstd

0

10

20

30

40

50

x

Figure 24: A kernel density estimate and a F-S skewed t-distribution estimate of the male.earnings data.

Problem 4 Using the dsstd function we can evaluate the skewed t-distribution found in the previous problem. When we plot these two density estimates we get the plot in Figure 24. The two curves match quite well.

Problem 5 (fitting a skewed GED model) To fit a skewed GED model we use the function sgedFit. Using that function on the male.earnings data we get > sgedFit( male.earnings ) $par mean sd nu 17.336668 7.500945 1.770956

xi 1.654997

We can plot the resulting parametric density using the density function for this family which in this case is 49

0.06

density.default(x = male.earnings)

0.03 0.00

0.01

0.02

density

0.04

0.05

KDE sged

0

10

20

30

40

50

x

Figure 25: A kernel density estimate and a F-S skewed GED model of the male.earnings data. dsged(x, mean = 0, sd = 1, nu = 2, xi = 1.5, log = FALSE) Plotting the fitted density and the kernel density estimate gives the plot in Figure 25. Again the two curves match quite well. As it is hard to tell visually which of the models (the skewed t or the skewed GED) has the better fit we could compare them with the AIC or the BIC. As the number of parameters used in each model is the same the comparison would depend on which model had a larger log-likelihood. Each fitting procedure above minimizes the negative of the log-likelihood and at the optimal parameters we find > c( fit_sstd$minimum, fit_sged$objective ) [1] 20121.41 20120.98 Thus the skewed GED has a slightly lower objective function (slightly larger log-likelihood) and this is the better fitting model.

Problem 6 Running the required code gives the following maximum likelihood parameter estimates 50

> fit_std $par [1] 0.0007845663 0.0105790014 4.0351450626 $value [1] -5983.233 The values in the par element are the estimates of µ, σ, and ν respectively. The AIC is computed using AIC = −2 log(L(θ̂ML )) + 2p .

(14) For the fit in this problem we find this to be the value -11960.47.

Problem 7 To modify the code for a skewed t-distribution we replace the use of the function dstd with the function dsstd and have to introduce another parameter to optimize over ξ representing the amount of ”skew” to introduce into the distribution using the F-S method. We also have to specify bounds on the ξ parameter so that the optim function call with the ”L-BFGS-B” can work. When we do this we find the following parameter estimate > fit_sstd $par [1] 0.0007472667 0.0097865057 7.4064801455 1.0001388909 $value [1] -5972.539 Notice that the estimated value of ξ is very close to one indicating that skewness is not really important for this data. The AIC for this model is given by -11937.08. This is larger than the value for the model with a symmetric t-distribution also indicating that the better model is the one without skew.

Problem 8-9 In the Rlab.R we include code to compute the TKDE. We then plot the KDE and the TKDE on the same graph. This is given in Figure 26 (left). A zoom of the tail region is given in Figure 26 (right). Notice that the TKDE and the parametric estimate are smooth in the tails while the KDE estimate is not.

Problem 10 Running the code for this problem we get 51

KDE TKDE Parametric std

0

0

10

1

20

2

Density(y)

30

Density(y)

3

40

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50

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−0.10

−0.05

0.00

0.05

0.035

y

0.040

0.045

0.050

0.055

0.060

y

Figure 26: Left: The KDE and the TKDE for the DAX log returns. Right: The KDE and the TKDE for the DAX log returns for the range 0.035 < y < 0.06.

[1] "MLE=" "0.00146" "0.00962" "4.02853" "1.4982" [1] "MLE Std Err=" "0.00025" "0.00045" "0.34624" [1] "AIC= -12631.387733"

"0.08624"

Here these are the MLE estimates and standard errors of the four parameters of a F-S skewed t-distribution. The parameters correspond to µ, σ, ν, and ξ.

Problem 11 The first part of this code fits a classical t-distribution to the returns from McDonalds. We then extract the parameters found in the fit.T covert them into the parameters need to evaluate the dstd in the fGarch package. The code then plots a histogram of the returns and the fit density so that we can compare the return samples with the parametric function.

Problem 12 The mean value is found to be 0.0006401997 with a standard error of 0.0002291503 this means that there is approximately a 95% chance the mean is in the interval 52

[1] 0.0001818991 0.0010985003 As this lower limit is positive we can conclude that the mean is significantly different than zero.

Problem 13 Looking at the histogram and the t model fit it looks like the data has sharper peaks in the center (than the t-model) and also that the data might have even heavier tails than the histogram shows. The t model does seem to capture the left-hand-tail well. Even with these issues, it does look like the parametric model captures the distribution quite well.

Problem 14 The estimate of the parameter ν in the t distribution is 4.269307. The larger the value the more like the Gaussian distribution the t distribution is. This small of a value indicates that the tails are quite heavy. To have a finite kurtosis we must have ν > 4. I find the 95 confidence interval for ν given by [1] 3.171030 5.367584 As the lower limit is less than four we cannot conclude that the kurtosis is finite.

Exercises Exercise 5.1 The dimnames command gives [1] "year"

"month" "day"

"ge"

"ibm"

"mobil" "crsp"

and the requested return plot is given in Figure 27 (left). The r object has a mode of numeric and a class of ts. If we change the type of the r object to a numeric object and plot again (as suggested in the problem) we get the plot in Figure 27 (right). The remaining commands have an output given by > cov(CRSPday[,4:6]) 53

0.10 r2

0.05

0.10 0.05

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r 0.00 −0.05 −0.10

1969

1970

1971

1972

1973

1974

1975

1976

0

Time

500

1000

1500

2000

2500

Index

Figure 27: Left: A plot of the IBM returns (when r) is a object of class ts (i.e. a time series object). Right: A plot of the IBM returns (when r) is a object of class numeric.

54

ge ibm mobil ge 1.882164e-04 8.007660e-05 5.270394e-05 ibm 8.007660e-05 3.061309e-04 3.588748e-05 mobil 5.270394e-05 3.588748e-05 1.670265e-04 > apply(CRSPday[,4:6],2,mean) ge ibm mobil 0.0010713801 0.0007000767 0.0007788801 From these we can conclude that the mean of the Mobil return is 0.0007788801, the variance of the GE returns is 1.882164 10−4, the covariance between the GE and the Mobil returns is given by 5.270394 10−5. The cor function has an output given by > cor(CRSPday[,4:6]) ge ibm mobil ge 1.0000000 0.3335979 0.2972499 ibm 0.3335979 1.0000000 0.1587072 mobil 0.2972499 0.1587072 1.0000000 From this we can conclude that the correlation between the GE and the Mobil returns is given by 0.2972499.

Exercise 5.2 The likelihood (with τ = σ12 ) is given by N Y

1 (Yi − µ)2 √ f (Y1 , · · · , Yn |σ ) = exp − 2σ 2 2πσ i=1 ) ( P N 2 (Y − µ) 1 i . σ −N exp − i=1 2 = (2π)N/2 2σ 2

If we let s2 ≡ N1

PN

i=1 (Yi − µ)

2

we get

1 Ns2 −N f (Y1 , · · · , Yn |σ ) = σ exp − 2 .

(2π)N/2 2σ 2

Taking the logarithm of the above expression gives log(f (Y1 , · · · , Yn |σ 2 )) = −

Ns2 N log(2π) − N log(σ) − .

2 2σ 2

The derivative of the above with respect to σ is given by ∂ N Ns2 (−2) log(f (Y1 , · · · , Yn |σ 2 )) = − − . ∂σ σ 2σ 3 55

Setting this equal to zero gives −N +

Ns2 = 0 so σ 2 = s2 , σ2

is the ML estimate for σ 2 as we were to show.

Exercise 5.3 Eq. 5.15 in the book is ∗

f (y|ξ) =

f (yξ) y < 0 f (y/ξ) y ≥ 0

(15)

Then for this function we have Z ∞ Z 0 Z ∞ ∗ I= f (y|ξ)dy = f (yξ)dy + f (y/ξ)dy . −∞

−∞

0

In the first integral we let v = yξ so that dv = ξdy. In the second integral we let v = y/ξ so that dv = dy/ξ. With these we get 1 I= ξ

Z 0

f (v)dv + ξ

−∞

Z ∞

f (v)dv .

0

As the function f (·) is symmetric about x = 0 (and the total integral of f (·) is one) each of the above integrals is 1/2 and we get ξ 1 + , 2ξ 2 as we were to show.

Exercise 5.4 Part (a): The kurtosis is given by Kur = E

(

Y − E(Y ) σY

4 )

.

(16)

If we introduce the notation E(X) = µ and Var (X) = σ 2 we see that for any random variable X we have E(X 2 ) = σ 2 + E(X)2 = σ 2 + µ2 .

Thus if we let X be given in terms of Y by 2

Y − E(Y ) , X≡ σY 56

Then E(X 2 ) is the kurtosis of Y so using the above identity we have Kur(Y ) = E(X 2 ) = Var Now E

"

Y − E(Y ) σY

2 #

Y − E(Y ) σY

=

2 !

+E

"

Y − E(Y ) σY

2 #2

.

1 σY2 2 E[(Y − E(Y )) ] = = 1, σY2 σY2

so we get Kur(Y ) = Var

Y − E(Y ) σY

2 !

+1,

as we were to show.

Part (b): As Var (X) ≥ 0 for all random variables X by the above decomposition we have that Kur(Y ) ≥ 1 .

Exercise 5.5 Part (a): The definition of kurtosis is given in Equation 16 and for a normal random variable Y ∼ N(µ, σ 2 ) we have that E((Y − E(Y ))4 ) = 3σ 4 , for the Gaussian mixture given here we have the fourth moment given by E((X − E(X))4 ) = 0.95E((X1 − E(X1 ))4 ) + 0.05E((X2 − E(X2 ))4 ) = 0.95(3) + 0.05(3(102)) = 17.85 , and the variance given by 2 σX = 0.95(1) + 0.05(10) = 1.45 .

Thus our kurtosis is given by 17.85 E((X − E(X))4 ) = = 8.489893 .

4 σX 1.452 Part (b): Following the formulas above (but in the more general case) we have E((X − E(X))4 ) = pE((X1 − E(X1 ))4 ) + (1 − p)E((X2 − E(X2 ))4 ) = p(3) + (1 − p)(3σ 4 ) = 3(1 − p)σ 4 + 3p , and the variance of X is given by 2 σX = p(1) + (1 − p)σ 2 = (1 − p)σ 2 + p .

57

With these our kurtosis is given by Kur(X) =

3(1 − p)σ 4 + 3p . ((1 − p)σ 2 + p)2

(17)

Lets fix the value of σ 2 > 1 and look for the value of p that maximizes the expression above. Rather than maximize the expression directly we will try the equivalent procedure of maximizing the logarithm of the above expression. Taking the logarithm of Kur(X) we get log(3(1 − p)σ 4 + 3p) − 2 log((1 − p)σ 2 + p) .

Taking the derivative of the above with respect to p we get −3σ 4 + 3 2(−σ 2 + 1) − .

3(1 − p)σ 4 + 3p (1 − p)σ 2 + p

We can combine these two fractions under the same denominator to get (−3σ 4 + 3)((1 − p)σ 2 + p) + 2(σ 2 − 1)(3(1 − p)σ 4 + 3p) .

(3(1 − p)σ 4 + 3p)((1 − p)σ 2 + p)

We now want to set the numerator of the above equal to zero and then solve for p. Doing this (and skipping some of the algebra) we find σ 2 (1 − σ 2 ) , (18) p= 1 − σ4 which is a function of only σ. We will assume that this value of p gives a maximum value for the kurtosis (we could evaluate the second derivative at this value of p and show that it is negative to verify that). Taking this expression for p and putting it into Equation 17 (and simplifying) we get Kur(X) =

3(σ 4 − 1)2 σ8 = O( ) = O(σ 2 ) .

4σ 2 (σ 2 − 1)2 σ2σ4

Thus we see that by making σ large enough and then selecting p (as above) we can make Kur(X) as large as desired. For example, taking σ 2 = 20000 and then selecting p using Equation 18 we get Kur(X) = 15001.5 > 10000.0.

Exercise 5.6 While the text states to fit a F-N skewed t-distribution the code they suggest (i.e sstdFit) fits a F-S skewed t-distribution so I’ll assume the F-N was a typo. I’ll fit to the Flow1 data.

Using that code we get estimates for the parameters of a F-S skewed t-distribution given by > fit_sstd $minimum [1] 4518.062 $estimate mean sd nu xi 9.542522e+05 1.533434e+05 1.318583e+01 3.610280e-01 58

KDE sstd

1.5e−06 0.0e+00

5.0e−07

1.0e−06

density

2.0e−06

2.5e−06

3.0e−06

density.default(x = dat$Flow1)

400000

600000

800000

1000000

1200000

x

Figure 28: A kernel density estimate and a F-S skewed t-distribution model of the Flow1 data.

Plotting a KDE and the parametric fit gives the plot in Figure 28.

Exercise 5.7 In this problem we use notation slightly different than in the book. For this problem we will assume a density function given by −θx θe x≥0 p(x|θ) = 0 else .

Given the set of data, {xi }, we form the likelihood function L (assuming independence of each sample point) as n Y L(θ) = p(xi |θ) . i=1

From this we form the the log-likelihood function l(θ) (assuming the exponential likelihood form given above) of l(θ) = log(L(θ)) =

n X i=1

n X log(p(xi |θ)) = (log(θ) − θxi ) = n log(θ) − θnx̂ . i=1

59

P Where here I have defined x̂ to be the sample mean i.e. x̂ = n1 ni=1 xi . To evaluate the maximum of this expression with respect to θ; we take the derivative with respect to θ, set the resulting expression equal to zero, and solve for θ. This gives n 1 1 ∂l = 0 ⇒ − nx̂ = 0 ⇒ = x̂ ⇒ θ = . ∂θ θ θ x̂ Thus our maximum likelihood estimate of θ is given by 1 θML = 1 Pn n

i=1 xi

.

Converting back to the notation of the book we have the desired result.

Exercise 5.8 We would have MSE(θ̂) = = = = =

E[θ̂ − θ]2 E[θ̂ − E(θ̂) + E(θ̂) − θ]2 E[(θ̂ − E(θ̂))2 + 2(θ̂ − E(θ̂))(E(θ̂) − θ) + (E(θ̂) − θ)2 ] E[(θ̂ − E(θ̂))2 ] + (E(θ̂) − θ)2 Var(θ̂) + Bias(θ̂)2 .

(19)

Here we have used the fact that θ and E[θ̂] are constants with respect to the expectation so that in the middle term above we have E[(θ̂ − E(θ̂))(E(θ̂) − θ)] = E[θ̂ − E(θ̂)](E(θ̂) − θ) = (E[θ̂] − E(θ̂))(E(θ̂) − θ) = 0 .

Exercise 5.9 Note that we have E[µ̂] =

n n 1X 1X E[Xi ] = µ = µ, n i=1 n i=1

as each Xi has an expectation of µ. Thus this estimate of the mean is unbiased i.e Bias(µ̂) = 0. From elementary statistics we have that Var(µ̂) =

σ2 , n

thus

σ2 . n From the central limit theorem if Xi are samples from a population with a finite mean and a non-zero variance (i.e. like the Student t distribution considered here) then as n → ∞ the 2 distribution of the sample mean is normal with a mean µ and a variance σn . Thus the above expression for the MSE above will still hold.

MSE(µ̂) = µ2 +

60

Exercise 5.10 We know that a t-distribution has tails that look like |y|−(ν+1) for |y| → ∞. When we write our density for large y as f (y) ∼ Ay −(a+1) the value of a is called the “tail index” and thus for a t distribution a = ν. As we are told that Kurt (ν) = 3 +

6 ≈ 9. ν −4

Solving we find ν = 5.

Exercise 5.11 Part (a): Our model is that the probability of N defaults in a month is a Poisson random variable with a parameter λ that is a linear function of x or P (N = n) =

(β0 + β1 x)n e−(β0 +β1 x) . n!

Then the total log-likelihood is the product of the probabilities P (N = ni ) for each of the i = 1, 2, · · · , N − 1, N realized monthly defaults. We start our maximization of the log-likelihood at the location (β0 , β1 ) = (1, 1).

Part (b): The optim function estimated β0 = 28.0834 and β1 = 0.6884 .

Part (c): Confidence intervals for maximum likelihood estimators can be derived from the fact that the standard error for an estimator is given by 1 . sθ̂ = q I(θ̂)

(20)

Since we are given the values of the above from the Fisher information matrix we can compute 95% confidence intervals assuming normality. Namely for β0 we have > 28.0834 + 1.8098 * qnorm(1-0.05/2) * c(-1,+1) [1] 24.53626 31.63054 while for β1 we have > 0.6884 + 0.1638 * qnorm(1-0.05/2) * c(-1,+1) [1] 0.3673579 1.0094421 61

20 −10

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Figure 29: Various qq-plots for fitting symmetric t-distributions to the returns of siemens for various values of ν.

Exercise 5.12 Part (a): The function dstd is the d(ensity) of the standardized t-distribution and its in the fGarch package.

Part (b): The function solve in this case inverts the Hessian matrix.

Part (c): The estimate of ν is the third value in the par field in the mle bmw variable which is 2.9882458517.

Part (d): Taking the square root of the inverse of the Fisher information matrix we find the (3, 3) element is given by ŝν = 0.1294801353.

Exercise 5.13 Part (a): The given commands produces the plot given in Figure 29. Looking at the qq plot for ν = 2 it looks like the returns have a lighter tails than a t-distribution with ν = 2.

This is because the curve for that plot bends down (on the left side) and up (on the right side).

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Part (b): Looking for the value of ν that gives the “straightest” qq-plot we would guess a value of ν ≈ 5.

Part (c): Using the fitdistr function in the MASS package we get > fitdistr(siemens,"t") m s df 0.0002927624 0.0076637573 3.4234499635 (0.0001172093) (0.0001255543) (0.1675316915) Thus the optimal value for ν was found to be around 3.4.

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Chapter 6 (Resampling)

R Lab See the R script Rlab.R where the problem for this chapter are worked.

Problem 1 Looking at the output from the sstdFit command we have > fit_skewt $minimum [1] -17865.61 $estimate mean sd nu xi 0.0003856989 0.0161162587 2.9889166127 1.0332622264 Thus the MLE of ν is around three. To have a finite value for the skewness of a t-distribution must have ν > 3 and to have a finite value of kurtosis we must have ν > 4. Thus for the estimate of ν given above neither skewness or kurtosis for this parametric model will be finite.

Problem 2 Since we can compute exactly the quantiles of a t-distribution using the R code qt. We can then compute the quantile based kurtosis exactly for each value of ν. This is done in the function tQuantKurt. If we then plot this function for the suggested values of ν we get the plot given in Figure 30. Note that as ν decrease the value of the function tQuantKurt goes to infinity.

Problem 3 When we plot the two KDEs we get the plot given in Figure 31 (left). From the given plot we see that the model based distribution seems to be shifted to the left relative to the model free distribution. Note that the uncertainty (variance) in each distribution looks to be about the same. In Figure 31 (right) we present boxplots of samples of kurtosis under each modeling assumption.

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Problem 4 The basic percentile confidence interval on a parameter θ is given by finding the α/2-lower ∗ and α/2-upper sample quantiles of the bootstrap estimates θ̂1∗ , θ̂2∗ , . . . , θ̂B and then the basic percentile confidence interval is given by (qL , qU ) .

For the samples in the ModelFree kurt and ModelBased kurt (and for α = 0.1) we get > quantile( ModelFree_kurt, c(0.05,1-0.05) ) 5% 95% 4.064854 4.447066 > quantile( ModelBased_kurt, c(0.05,1-0.05) ) 5% 95% 3.996941 4.332154

Problem 5 Using the bcanon function we get the following output 65

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Right: Boxplots of the samples of kurtosis. The label “1” corresponds to the model free samples of kurtosis while the label “2” corresponds to the model based samples of kurtosis.

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> set.seed("1234") > bc_out = bcanon( bmwRet[,2], 5000, quantKurt, alpha=c(0.05,0.95) ) > print( bc_out$confpoints ) alpha bca point [1,] 0.05 4.104399 [2,] 0.95 4.487329 This agrees qualitatively with the results from Problem 4 above.

Problem 6 In the given R code we compute five different types of bootstrap intervals. Each of these interval types are discussed in this chapter under the topic “Bootstrap Confidence Intervals”.

Here the different bootstrap intervals are 1. boott generates a bootstrap t confidence interval 2. bcanon gives us a bias corrected bootstrap interval 3. int3 gives us the normal bootstrap confidence interval 4. int4 gives us the percentile bootstrap confidence interval 5. int5 gives us the basic bootstrap interval

Problem 7 The value of B is the number of bootstrap resamples which is the value of the variable nboot or 400.

Problem 8 We ran 500 iterations or simulations.

Problem 9 The estimate of bias comes from the error matrix. Here the mean value of that is −0.187.

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Problem 10 The estimate of the MSE can be obtained by using the error matrix and the decomposition MSE ≈ BIAS2 + s2boot , using the numbers computed in Rlab Problem.R I find the MSE given by 0.192.

Problem 11 The BCa interval comes from the bcanon function which is recorded in the second column in the correct matrix. The bootstrap-t interval comes from the boott function which is recorded in the first column in the correct matrix. We can use the values in this matrix to estimate the actual coverage probability for our resampling methods. We find > print(colMeans(correct)) [1] 0.660 0.680 0.538 0.558 0.500 This means that we would estimate the actual coverage probability of these two bootstrap methods to be 68% and 66% rather than the expected 95%.

Problem 12 We can use the values in the correct matrix (for the second column) to form a confidence interval on the actual coverage probability. We can do that with the simple R code alpha = 0.05 data = correct[, 2] n = length(data) ci = mean(data) + c(-1, +1) * qt(1-alpha/2, n-1) * ( sd(data) / sqrt(n) ) print(ci)

Running that code I get [1] 0.639 0.721 Note that these numbers are sufficiently less than 0.95 the nominal coverage probability.

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Problem 13 Taking at the columnwise statistics of the matrix widths I find > print(colMeans(width)) [1] 6.07 3.05 2.70 2.55 2.55 From this this it looks like the smallest (most accurate) confidence interval is given by the percentile and basic bootstrap intervals (which have the same widths based on how they are calculated). As the true kurtosis is four, all of these widths are a very large percentage of the value of the parameter and would not be considered very accurate.

Problem 14 Running the code I get that this simulation took about 6.3 minutes to run

Problem 15 In that R Code I change the function Kurtosis to quantKurt (my implementation of the function mentioned) and run the code again where I get print(colMeans(width)) [1] 2.81 2.89 2.79 2.70 2.70 Again compared with the true value of four none of these intervals are very accurate. This code took about 24 minutes to run.

Exercises Exercise 6.1 The α/2 (when α = 0.1) quantile of the samples sb,boot would be 10000(0.05) = 500 thus to compute a percentile confidence interval we would need to find the s500,boot and s9500,boot bootstrap samples. These are given to us in the problem where we have 0.71 ≤

sb,boot ≤ 1.67 . s

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The bootstrap approximation means that under enough sampling we can imagine sb,boot → s and s → σ so that we would have s 0.71 ≤ ≤ 1.67 , σ with a probability of approximately 0.9. We can solve the above for σ and find σ 1 1 ≤ ≤ .

1.67 s 0.71 This means that 0.1856287 ≤ σ ≤ 0.4366197 .

is the percentile confidence interval for σ.

Exercise 6.2 Part (a): The bias is estimated using BIASboot (θ̂) = θ̂∗ − θ̂ = 0.68431 − 0.69119 = −0.00688 .

Part (b): An estimate of the standard deviation of the sample correlation is given by v u B u 1 X t (θ̂∗ − θ̂∗ )2 = 0.11293 , sboot (θ) = B − 1 b=1 b from the numbers given in the text.

Part (c): An estimate of the MSE for the sample correlation is given by MSEboot (θ) = BIAS2boot (θ̂) + s2boot (θ̂) = (−0.00688)2 + 0.112932 = 0.01280052 .

Part (d): This would be (−0.00688)2 BIAS2boot (θ̂) = = 0.00369785 .

MSEboot (θ) 0.01280052 This is a very small fraction of the MSE and it is probably worth more to reduce the variance of our estimator than to spend time trying to reduce the bias.

Exercise 6.3 Part (a): The bias is estimated using BIASboot (θ̂) = θ̂∗ − θ̂ = 1.283 − 1.323 = −0.04 .

Part (b): An estimate of the MSE for the sample standard deviation is given by MSEboot (θ) = BIAS2boot (θ̂) + s2boot (θ̂) = (−0.04)2 + 0.23862 = 0.05852996 .

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Chapter 7 (Multivariate Statistical Models)

R Lab See the R script Rlab.R where the problem for this chapter are worked.

Problem 1 (Equity Returns) In Figure 32 we show the scatter plot produced from the pairs function for the four suggested returns. Calling the cov function on the data frame of returns gives > cov(Berndt) CITCRP CONED CONTIL DATGEN CITCRP 0.006556442 0.0010933247 0.006052187 0.0055091027 CONED 0.001093325 0.0025272604 0.000795463 0.0006124665 CONTIL 0.006052187 0.0007954630 0.022710259 0.0050502504 DATGEN 0.005509103 0.0006124665 0.005050250 0.0162664367 Then to compute (estimate) the variance of the given linear combination we need to compute w T Σw where Σ is the matrix above and w is a vector of portfolio weights w T =

0.5 0.3 0.2 0 . When we compute the above we get the value 0.004408865 71

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Problem 2 (A multivariate t distribution) In Figure 33 we plot the loglikelihood of the return data as a function of the degrees of freedom parameter ν. To determine the 90% confidence interval for this value of ν we will use the fact that the standard errors of our maximum likelihood estimates are given by q sθ̂i = {I(θ̂)−1 }ii , where I(θ̂) is the Fisher information matrix. Rather than take the required expectation we will follow the suggestion in the book and approximate this use the value based on the sample of data we have. This means that we have to estimate the second derivative of the loglikelihood at the optimal location. There are many way to do this but we will do it by approximating the second derivative using central finite differences [1] y ′′ (x) ≈

y(x + h) − 2y(x) + y(x − h) . h2

Using this approximation we find −

∂ 2 log(L) ≈ 1.028893 . ∂ν 2

Thus we find our standard error given by sν̂ = 0.9858593 .

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Figure 34: Examples of samples drawn from four different multivariate t distributions a-d from left to right. where zα/2 are the critical points of a standard normal and α = 0.1. In Figure 33 we show these two values as black vertical lines. As a check to the validity of using the finite difference approximation I also used the fdHess function in the nlme library to estimate the Hessian at the maximum likelihood value of ν. The two methods gave results that were very close.

Problems 3-5 (example multivariate t distributions) See Figure 34 where the simulations of these t-distributions are presented.

• Then the sample with independent variates is (c) since the covariance matrix has zero off diagonal elements and there is no linkage between the two variables from the w parameter. In fact here two different vectors w1 and w2 are used.

• The sample with correlated variates without tail dependence is (b) since in that case the covariance matrix has nonzero off diagonal elements but there is no linkage between the two variables with the w parameter.

• The sample with uncorrelated variates but with tail dependence is (d) since in that case the covariance matrix has zero off diagonal elements but linkage between the two variables with a common w parameter.

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Problem 6 Part (a): From the discussion in the book entitled “Using the t-distribution in portfolio analysis” the random variable R will also have a t-distribution but with a mean of w T µ and ν w T Λw = w T Σw. Thus in this case R will have a t-distribution with ν = 5, a variance of ν−2 a mean value of

0.001 1/2 1/2 = 0.0015 , 0.002 and a variance of

1/2 1/2

Thus our scale “matrix” is Λ=

0.1 0.03 0.03 0.15

1/2 1/2

= 0.0775 .

ν−2 3 Σ= 0.0775 = 0.0465 . ν 5

Given all of these things we can state that R is given by a tν (µ, Λ) = t5 (0.0015, 0.0465) distribution.

X Part (b): This is done in the R code Rlab.R. In that code we generate raw data from Y the correct multivariate t-distribution using the code rmt and then for each sample compute X wT to get a sample for R. Running that code for a very large number of samples we Y find [1] "Using N= [1] "Using N= [1] "Using N=

10000 has a mean of upper 100000 has a mean of upper 1000000 has a mean of upper

0.990000 quantile of= 0.990000 quantile of= 0.990000 quantile of=

0.733422" 0.740097" 0.743422"

Problem 7 Part (a): The object Y holds returns of IBM and CRSP. The call to cov creates a covariance matrix and the call to chol compute the Cholesky factorization of this matrix. This is an upper triangular matrix U that when “squared” as U T U gives back the covariance matrix for Y. This step makes the initial matrix used for the “scale matrix” equal to the sample covariance matrix.

Part (b): The call from optim gives > fit_mvt$par [1] 0.0003789236 0.0008317070 0.0126907363 0.0026859493 0.0051011198 [6] 4.2618395311

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for the ML estimate of the variables in θ.

Part (c-d): The member hessian of the variable fit mvt gives the second derivative of the loglikelihood when viewed as a function of the parameters. Since the R optimization routine optim looks for a minimum, the function loglik that returns the loglikelihood was multiplied by −1 to find the maximum of the loglikelihood. This means that to get an approximation to the Fisher information matrix I(θ) we can just report the value of the member hessian. We need to invert this, extract the diagonal elements, and then take the square root of these to get the standard errors of the components of θ. When we do that we find > sqrt( diag(inv_fisher) ) [1] 2.886679e-04 1.310061e-04 2.506014e-04 1.292280e-04 9.372555e-05 [6] 2.570202e-01 Part (e): For this we need to extract the needed components of the parameter vector θ and form the covariance matrix. When we do this we find [,1] [,2] [1,] 1.610548e-04 3.408667e-05 [2,] 3.408667e-05 3.323575e-05 Part (f): Using the above covariance matrix we can construct the correlation matrix to get > cor_matrix = DInv %*% ( cov_matrix %*% DInv ) > cor_matrix [,1] [,2] [1,] 1.0000000 0.4659025 [2,] 0.4659025 1.0000000

Exercises Exercise 7.1 Part (a): These would be E(0.2X + 0.8Y ) = 0.2E(X) + 0.8E(Y ) = 0.2(1) + 0.8(1.5) = 1.4 Var (0.2X + 0.8Y ) = 0.22 Var (X) + 0.82 Var (Y ) + 2(0.2)(0.8)Cov (X, Y ) = 0.22 (2) + 0.82 (2.7) + 2(0.2)(0.8)(0.8) = 2.064 .

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Part (b): We can expand the given expression to get Var (wX + (1 − w)Y ) = w 2 Var (X) + (1 − w)2 Var (Y ) + 2w(1 − w)Cov (X, Y ) = 2w 2 + 2.7(1 − w)2 + 1.6w(1 − w) .

Taking the derivative of this expression with respect to w, setting the result equal to zero and solving for w we get 6.2w − 3.8 = 0 so w = 0.6129032 .

This procedure is finding the minimum variance combination of the two random variables X and Y .

Exercise 7.2 Part (a): From the definition of the covariance we have Cov (X1 + X2 , Y1 + Y2 ) = E[(X1 + X2 − E[X1 + X2 ])(Y1 + Y2 − E[Y1 + Y2 ])] = E[((X1 − E[X1 ]) + (X2 − E[X2 ]))((Y1 − E[Y1 ]) + (Y2 − E[Y2 ]))] .

Then expanding the multiplication of terms in the outermost expectation, taking the expectation, and recognizing pairwise covariance we have the desired result.

Part (b): An arbitrary number of terms would be expanded in the same way as above.

That is the outer expectation is of the expression ! !

X X XX (Xi − E[Xi ]) (Yj − E[Yj ]) = (Xi − E[Xi ])(Yj − E[Yj ]) , i

j

i

j

which is the same as the double sum of pairwise covariances.

Exercise 7.3 We want to derive equivalent expressions for Cov (X, Y ) = σX,Y . From the definition of the covariance we have Cov (X, Y ) = E[(X − E[X])(Y − E[Y ])] = E[XY − XE[Y ] − Y E[X] + E[X]E[Y ]] = E[XY ] − E[X]E[Y ] − E[Y ]E[X] + E[X]E[Y ] = E[XY ] − E[X]E[Y ] .

(21)

We can also evaluate the definition of covariance as follows Cov (X, Y ) = E[(X − E[X])(Y − E[Y ])] = E[X(Y − E[Y ]) − E[X](Y − E[Y ])] = E[X(Y − E[Y ])] − E[X]E[Y − E[Y ]] = E[X(Y − E[Y ])] .

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Since E[Y − E[Y ]] = 0. Since the covariance expression is symmetric in X and Y interchanging the roll of X and Y gives Cov (X, Y ) = E[Y (X − E[X])] .

(23)

Finally from Equation 21 we see that if either E[X] or E[Y ] is zero we have Cov (X, Y ) = E[XY ] .

(24)

Exercise 7.4 Part (a): Since E[X] is a constant (not a random variable) its expectation is itself. Thus using the linearity of the expectation we have E[X − E[X]] = E[X] − E[X] = 0 .

Part (b): If two random variables are independent then E[XY ] = E[X]E[Y ] , so that Equation 21 then gives Cov (X, Y ) = corr(X, Y ) = 0 and X and Y are uncorrelated.

Exercise 7.5 We have

Cov (X, Y ) = Cov X, X 2 = E[X 3 ] − E[X]E[X 2 ] = 0 − 0 = 0 ,

since E[X] = 0 (in fact E[X p ] = 0 for any odd power p). This shows that X and Y are uncorrelated. To show that X and Y are not independent we want to see if p(x, y) = p(x)p(y).

To evaluate the right-hand-side of this expression we need to know p(y). For Y the cdf is given by √ √ FY (y) = P {Y ≤ y} = P {X 2 ≤ y} = P {− y ≤ X ≤ + y} Z +√ y √ y dx = √ = . a − y 2a Thus taking the derivative we have pY (y) =

1 √ .

2a y

Thus the product of the marginal distributions is

1 1 1 = 2√ . p(x)p(y) = √ 2a y 2a 4a y Next to evaluate p(x, y) we will use p(x, y) = p(y|x)p(x) = p(y|x) . Now p(y|x) is a delta 2a 2 function with the only nonzero value when y = x so we have δ(y − x2 ) .

2a As this does not equal p(x)p(y) we have that X and Y are not independent. p(x, y) =

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Exercise 7.6 We write w T X =

PN

i=1 wi Xi and then use the linearity of the expectation to get

T

E[w X] =

N X

wi E[Xi ] = w T E[X] .

i=1

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N X i=1

w i Xi ,

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wi wj Cov (Xi , Xj ) .

i=1 j=1

This is equivalent to the expression for w T COV(X)w when we expand.

Exercise 7.7

Using the fact that Var w T Y = w T Var (Y ) w ≥ 0 we can show that a cannot be zero. To do this assume that a = 0 and then find the eigenvalues/eigenvectors of the given covariance matrix. We do this in the R file chap 7.R. There we see that one eigenvalue would be negative with a value of -0.2727922 (which can’t be true for a covariance matrix). That eigenvalue has an eigenvector given by [1]

0.5000000 -0.7071068

0.5000000

thus if did have a = and consider the random variable w T Y with w given by the eigenvector

above we would have Var w T Y < 0 which cannot happen. Thus a cannot be zero and have a valid covariance matrix.

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7

Chapter 8 (Copulas)

Notes On The Text The derivation of the equation 8.6 in the book Following the text the authors came to the statement that the CDF of Xj is FXj (xj ) = FYj {gj−1(xj )} .

From this we will now derive the inverse CDF of Xj which I found hard to “see” directly.

To do that we let u = FXj (xj ) , (25) then from the above we have gj−1 (xj ) = FY−1 (u) , j so taking one more inverse (this time of the function gj−1) we end with xj = gj {FY−1 (u)} . j Recalling how xj was related to u in Equation 25 we get FX−1j (u) = gj {FY−1 (u)} , j

(26)

which is the book’s Eq. 8.6.

R Lab What was difficult about doing the Rlab for this problem was installing the copula library in R. After much searching on the web, the following steps worked for me on a recent version of Ubuntu (mileage may vary). The first step is to make sure that you have all of the auxiliary programs that copula will need. This can be done with the following installs: # These were all needed to install copula in R: sudo apt-get install -y libgsl0-dev # for GSL gnu scientific library sudo apt-get install -y r-cran-rcpparmadillo r-cran-rgl sudo apt-get install -y libgmp3-dev sudo apt-get install -y libboost-all-dev build-essential sudo apt-get install -y libmpfr-dev The “-y” flag stops the apt-get program from asking the user if they want to install the packages or not. Next in R one has to perform 79

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Once this is done one should be able to execute the R script Rlab.R where the problem for this chapter are worked.

Problem 1 When we run the given Rlab.R function we get the pairs plot in Figure 35. From the definition of the copula we have a t-copula with one degree of freedom. The vector given in the call to tCopula is used to construct the correlation matrix which can be obtained by using the getSigma function on the constructed copula as > getSigma(cop_t_dim3) 80

[,1] [,2] [,3] [1,] 1.00 -0.6 0.75 [2,] -0.60 1.0 0.00 [3,] 0.75 0.0 1.00 The sample size is 500.

Problem 2 See the scatter plot matrix in Figure 35.

Part (a): It looks like when Var 3 is very small Var 2 can be either be very small or very large. The same seems to hold true if Var 3 is very large. Thus these two variables don’t seem independent.

Part (b-c): Note that when Var 1 is large Var 2 is small and when Var 2 is small Var 1 is large thus the two variables show tail dependence but in opposite directions (the correlation between these two variables is negative). In the same way if Var 3 is large/small then Var 1 is large/small thus the tails of these two distributions seem to move together (the correlation between these two variables is positive). Between Var 2 and Var 3 we see that when Var 2 is small Var 3 is equally likely to be small/large (here the correlation is zero).

Part (d): It seems that this is not due to random variations since the suggested confidence interval is given by > cor.test( rand_t_cop[,1], rand_t_cop[,3] ) Pearson’s product-moment correlation data: rand_t_cop[, 1] and rand_t_cop[, 3] t = 22.314, df = 498, p-value < 2.2e-16 alternative hypothesis: true correlation is not equal to 0 95 percent confidence interval: 0.6603249 0.7483624 sample estimates: cor 0.707073 Here we see that the value of 0.75 is not in the confidence interval.

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Figure 36: Left: The scatter plot matrix for data generated by a t-copula. Right: KDE of the marginal distributions.

Problem 3 The pairs plot for this is given in Figure 36 (left) and the KDE marginals are estimated in Figure 36 (right).

Part (a): The marginals are exponential that have rates 2, 3, and 4. These translate into mean values of 1/2, 1/3, and 1/4.

Part (b): The correlation is specified to be zero between these two variables.

Problem 4 In the book we have

2 arcsin(Ωi,j ) , (27) π where ρτ is Kendall’s tau correlation. Since we want the value of Ωi,j we can invert the above equation to get

π ρτ (Yi , Yj ) . Ωi,j = sin 2 When we compute that on the given data we find the value of omega given by 0.7018346. ρτ (Yi , Yj ) =

82

Problem 5 Part (a): The fitCopula function has an input that should be the cumulative density of the marginal variables. This can be done by using the pstd which assumes that the marginals are from a t-distribution or by using the empirical cumulative distribution. These two versions are the what are used for fitting the two values for ft1 and ft2.

Part (b): The two estimates are given by > print(ft1) Call: fitCopula(copula, data = data, method = "ml") Fit based on "maximum likelihood" and 2516 2-dimensional observations.

Copula: tCopula rho.1 df 0.7022 2.9834 The maximized loglikelihood is 967.2 Optimization converged > print(ft2) Call: fitCopula(copula, data = data, method = "ml") Fit based on "maximum likelihood" and 2516 2-dimensional observations.

Copula: tCopula rho.1 df 0.7031 3.0222 The maximized loglikelihood is 964.6 Optimization converged Note that the two estimates of the parameters are very close to each other and not practically different.

Problem 6 Part (a-b): The output from the optim command gives > fit_cop $par [1] 0.06504702 1.37982785 3.35792635 0.07422142 1.80751172 2.33415925 0.70421613 [8] 2.96934965 $value [1] 6828.219 The first three numbers in par are the (µ, s, ν) parameters of the marginal t-distribution for the ibm returns, the next three numbers are the (µ, s, ν) parameters of the marginal 83

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Part (c): This is maximum likelihood estimation as we are estimating the parameters of the marginal distributions with the parameters of the t-distribution copula at the same time.

If we had frozen the parameters of the marginal (at their individual maximum likelihood values) and only optimized over the parameters in the copula we would have been using pseudo-maximum likelihood estimation.

Part (d): From the text the coefficient of lower tail dependence is defined as λl and for a t-copula with ν degrees of freedom and a correlation ρ we have ( s ) (ν + 1)(1 − ρ) λl = 2Ft,ν+1 − .

(28) 1+ρ Using the numbers estimated for ρ and ν we find this to be 0.4535334.

Problem 7 The contour plots for this problem are given in Figure 37. Many of them look quite similar to the empirical copula. Perhaps the t-copula is best but its hard to tell visually.

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Problem 8 The contour plots for this problem are given in Figure 38. In this case the t-distribution copula looks to match the empirical copula the closest.

Problem 9 Using the log-likelihood the AIC is given by −2log-likelihood + 2p .

For the given copulas we get the following values for the AIC AICs name 1 -1930.338 T 5 -1733.932 Gumbel 2 -1664.648 Normal 3 -1560.383 Frank 4 -1458.880 Clayton 6 -1377.822 Joe 85

Seeking the smallest value of AIC would point to the t model as the best one. This is equivalent to the conclusion made above.

Exercises Exercise 8.1 We are told that ρτ (X, Y ) = E[sign{(X − X ′ )(Y − Y ′ )}] = 0.55 .

If we flip Y to 1/Y we get

1 1 1 ′ ρτ X, = E sign (X − X ) − ′ Y Y Y

′

Y −Y ′ = E sign (X − X ) YY′ = −E [sign {(X − X ′ )(Y − Y ′ )}] = −0.55 .

If we now flip X to 1/X we get

1 1 1 1 1 1 = E sign , − − ′ ρτ X Y X X′ Y Y ′

′ Y −Y X −X = E sign XX ′ YY′ = E [sign {(X − X ′ )(Y − Y ′ )}] = 0.55 .

Exercise 8.2 (Spearman, Kendall, and Pearson correlations) We have that X ∼ U[0, 1] with a second random variable Y defined as Y = X 2 . Using these two variables I would calculate Spearman’s rank correlation, Kendall’s tau, and Pearson’s correlation.

To calculate Kendall’s tau we have ρτ (X, Y ) = ρτ (X, X 2 ) = P {(X − X ∗ )(X 2 − X ∗ 2 ) > 0} − P {(X − X ∗ )(X 2 − X ∗ 2 ) < 0} = P {(X − X ∗ )2 (X + X ∗ ) > 0} − P {(X − X ∗ )2 (X + X ∗ ) < 0} = 1−0 = 1.

Next to compute Spearman’s rank coefficient we have ρS (X, X 2 ) = corr{FX (X), FX 2 (X 2 )} .

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To evaluate this, note that as X is uniform its distribution function is linear FX (x) = x.

Next we need to compute the distribution function for X 2 when X is uniform. In that case note that Z √v √ 2 dξ = v .

FX 2 (v) = P {X < v} = ξ=0

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From this we see that FX 2 (X ) = X and we have ρS (X, X 2 ) = corr{X, X} = 1 .

This proves the two statements given in the problem.

Now to compute Pearson’s correlation coefficient recall that it is the ratio of the covariance of X and X 2 to the product of σX and σX 2 . Using moments of the uniform distribution we have

1 1 1 1 2 3 2 = Cov X, X = E(X ) − E(X)E(X ) = − 4 2 3 12 1 1 1 Var (X) = E(X 2 ) − E(X)2 = − = 3 4 12 2

1 1 4 Var X 2 = E(X 4 ) − E(X 2 )2 = − . = 5 3 45

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The reason “why” has to do with what each correlation metric is trying to measure. In Kendall’s tau we are considering pairs of samples (X, X 2 ) and (X ∗ , X ∗ 2 ) and looking to see when X < X ∗ how often is X 2 < X ∗ 2 (and vice versa with the opposite inequality). From how X and X 2 are related this will always be true giving a Kendall tau of one. The same argument holds for the Spearman rank correlation. For Pearson’s correlation we are looking to determine how well one variable is predicted from the other variable in a linear regression.

The only way to get ρP = 1 is if the two variables can be perfectly predicted from each other.

For X and X 2 that is not possible due to the non-linearity and we must have ρP < 1.

Exercise 8.3 (Archimedean copula with generator function φ(u) = − log(u)) For the given function φ(u) = − log(u) note that the inverse of this function is φ−1 (u) = e−u .

Then if we consider the book’s Eq. 8.9 we have C(u1, . . . , ud ) = φ−1 (φ(u1 ) + φ(u2) + · · · + φ(ud)) = φ−1 (− log(u1 u2 · · · ud )) = u1 u2 · · · ud , which is the independence copula. If φ(u) = − log10 (u) then the same steps above can be carried out and we would still get the independence copula in this case.

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Exercise 8.4 Researching this exercise leads to the fact that the non-strict generator φ(u) = 1 − u (which has φ−1 (u) = 1 − u) will “generate” the counter-monotonicity copula C− . At least in that we will have φ−1 (φ(u1 ) + φ(u2)) = 1 − (1 − u1 + 1 − u2 ) = u1 + u2 − 1 .

Note that this is the counter-monotonicity copula (without the maximization). If anyone knows how to impose the maximization needed please contact me.

Exercise 8.5 For the generator of a Frank copula −θu

e −1 Fr φ (u) = − log e−θ − 1

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− ∞ < θ < +∞ .

(29)

Note that φ is continuous as all the functions in its construction are continuous.

The first derivative of φFr with respect to u is given by

−θ −θe−θu θe−θu e −1 dφFr = = − −θu . du e −1 e−θ − 1 e−θu − 1 If θ > 0 and u ∈ [0, 1] then −uθ < 0 so e−uθ < 1 and e−uθ − 1 < 0. In addition θe−θu > 0 so we have θe−θu < 0, e−θu − 1 Fr

thus φdu < 0 showing that φFr is decreasing in the case θ > 0. In the same way if θ < 0 and u ∈ [0, 1] then −uθ > 0 so e−uθ > 1 so e−uθ − 1 > 0. In addition we have θe−θu < 0 so again φFr < 0 and we have that φFr (u) is decreasing both θ > 0 and θ < 0. du Now the second derivative of φFr is given by θ(−θe−θu ) θe−θu (−θeθu ) d2 φFr = − du2 e−θu − 1 (e−θu − 1)2 −θ2 e−θu (e−θu − 1) + θ2 e−2θu = (e−θu − 1)2 θ2 e−2θu = −θu > 0, (e − 1)2 for all θ and u. Therefore we have that φFr is convex function on u ∈ [0, 1].

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Next we consider the value of φFr when u = 0. Using much of the same arguments above if θ < 0 we observe that as u → 0+ we have e−θu − 1 → 0+ , eθ − 1

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lim φFr (u) = − log(0+ ) = ∞ .

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so that again we have

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Exercise 8.6 For this problem we want to show that when u1 and u2 have a domain of [0, 1] we have lim C Fr (u1 , u2 ) = min(u1 , u2) .

θ→∞

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(e−θu1 − 1)(e−θu2 − 1) 1 .

C (u1 , u2 ) = − log 1 + θ e−θ − 1 Fr

Now as θ → +∞ the argument of the logarithm goes to 1+

(−1)(−1) = 0. (−1)

Thus the entire function goes to −∞ log(0) → , θ ∞ which is indeterminate. Thus to directly evaluate this limit we would have to use something like L’Hospital’s rule. That calculation is somewhat complicated to do directly so we will argue this result heuristically. To do this lets write the function C Fr (u1, u2 ) as −θ

e − 1 + (e−θ(u1 +u2 ) − e−θu1 − e−θu2 + 1) 1 Fr log C (u1 , u2 ) = − θ e−θ − 1

−θ(u1 +u2 )

1 e − e−θu1 − e−θu2 + e−θ =− log . θ e−θ − 1 89

Lets now assume that u1 < u2 . Now since u1 + u2 > u2 > u1 we have that e−θ(u1 +u2 ) < e−θu2 < e−θu1 .

Since u1 < 1 we also have that

e−θ < e−θu1 .

Thus in the numerator when θ is large we have e−θ(u1 +u2 ) − e−θu1 − e−θu2 + e−θ ∼ e−θu1 , meaning that the first, third, and fourth term become much smaller than the second terms as θ → ∞. If we use this approximation then we have that −θu1 e 1 Fr .

C (u1 , u2) ∼ − log −θ θ e −1 Taking the limit of θ → ∞ directly still gives an indeterminate expression but using L’Hospital’s rule on this functional form is much easier to apply. Using it we get −θ

e −1 u1 e−θu1 e−θu1 e−θ Fr lim C (u1, u2 ) = − lim − −θ + θ→∞ θ→∞ e−θu1 e − 1 (e−θ − 1)2

1 − e−θ −u1 e−θ(u1 +1) + u1 e−θu1 + e−θ(1+u1 ) = lim θ→∞ e−θu1 (e−θ − 1)2

u1 e−θu1 1 = u1 . = e−θu1 12 Now if we had started with u2 < u1 we would find this limit equal to u2 . Thus we see that in either case the limit is min(u1 , u2 ).

Exercise 8.7 From the construction of φ(u) we know that it is • continuous • decreasing • convex • have φ(0) = ∞ • have φ(1) = 0 thus φ(u) is a strict generator. For the case k = 2 I’m not sure how to compute the copula function due to the difficulty in determining φ−1 (u) in the general case. If anyone sees how this should be done please contact me.

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Exercise 8.8 The given expression for C(u1 , u2|θ) is the Archimedean copula with the suggested generating function φ. We now need to show the suggested limits.

Recall that 0 ≤ ui ≤ 1 so that 0 ≤ 1 − ui ≤ 1 so at as θ → ∞ the smaller of 1 − u1 or 1 − u2 when raised to the θ power will shrink to zero before the other. This means that (1 − u1 )θ + (1 − u2 )θ → max(1 − u1 , 1 − u2 )θ , as θ → ∞. Thus our copula limits to C → max(0, 1 − max(1 − u1 , 1 − u2 )) .

If 1 − u1 > 1 − u2 then u1 < u2 and in that case we get C → max(0, 1 − (1 − u1 )) = u1 . In the same way if 1 − u2 > 1 − u1 then u2 < u1 and we get C → max(0, 1 − (1 − u2 )) = u2 .

Thus we see that C → min(u1 , u2 ) in this limit.

If θ → 1 we have C → max(0, 1 − (1 − u1 + 1 − u2 )) = max(0, u1 + u2 − 1) , as we were to show.

Exercise 8.9 I think the answer is yes. For a convex combination of copulas would have the expression k X

αi Ci (Y1 , Y2 , . . . , Yd ) .

i=1

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Exercise 8.10 (Y with a meta-Gaussian distribution and ρij = 0) If Y has a meta-Gaussian distribution then Y has a Gaussian copula, C Gauss (·|Ω), which only depends on Ω the correlation matrix for the random vector Y . This means that 91

CYGauss (u1 , . . . , ud |Ω) is the cumulative distribution function of a d dimensional normal random variable with correlation matrix Ω. An expression for C Gauss (·|Ω) for an arbitrary correlation matrix can be found in several places1 . From those references, if Ω = I then we see that cGauss (·|I) = 1 (note the lowercase c), that is we have the independent copula. This statement is also made in the book: “A d dimensional Gaussian copula whose correlation matrix is the identity is the d-dimensional independence copula”. If we use this fact in the book’s Eq. 8.4 which is fY (y1 , . . . , yd ) = cY {FY1 (y1 ), . . . , FYd (yd )}fY1 (y1 ) · · · fYd (yd ) ,

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we get fY (y1 , . . . , yd ) = fY1 (y1 ) · · · fYd (yd ) .

This is the statement that the joint pdf is equal to the product of the marginals which tells us that Y1 , . . . Yd are independent the statement we wanted to show.

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For example the wikipedia page on copulas.

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8

Chapter 9 (Regression: Basics)

R Lab See the R script Rlab.R for this chapter. We plot a pairwise scatter plot of the variables of interest in Figure 39 (left) From that plot we see that it looks like the strongest linear relationship exists between consumption and dpi and unemp. The variables cpi and government don’t seem to be as linearly related to consumption. There seem to be some small outliers in several variables namely: cpi (for large values), government (large values), and unemp (large values). There does not seem to be too much correlation between the variable in that none of the scatter plots seem to look strongly linear and thus there does not look to be collinearity problems.

If we fit a linear model on all four variables we get Call: lm(formula = consumption ~ dpi + cpi + government + unemp, data = MacroDiff) Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 14.752317 2.520168 5.854 1.97e-08 *** dpi 0.353044 0.047982 7.358 4.87e-12 *** cpi 0.726576 0.678754 1.070 0.286 government -0.002158 0.118142 -0.018 0.985 unemp -16.304368 3.855214 -4.229 3.58e-05 *** Residual standard error: 20.39 on 198 degrees of freedom Multiple R-squared: 0.3385, Adjusted R-squared: 0.3252 F-statistic: 25.33 on 4 and 198 DF, p-value: < 2.2e-16 The two variables suggested to be the most important above namely dpi and unemp have the most significant regression coefficients. The anova command gives the following > anova(fitLm1) Analysis of Variance Table Response: consumption Df Sum Sq Mean Sq F value Pr(>F) dpi 1 34258 34258 82.4294 < 2.2e-16 *** cpi 1 253 253 0.6089 0.4361 government 1 171 171 0.4110 0.5222 unemp 1 7434 7434 17.8859 3.582e-05 *** Residuals 198 82290 416

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The anova table emphasizes the facts that when we add cpi and government to the regression of consumption on dpi we don’t reduce the regression sum of square significantly enough to make a difference in the modeling. Since two of the variables don’t look promising in the modeling of consumption we will consider dropping them using stepAIC in the MASS library. The stepAIC suggests that we should first drop government and then cpi from the regression. Comparing the AIC for the two models gives that the reduction in AIC is 2.827648 starting with an AIC of 1807.064. This does not seem like a huge change.

The two different vif give > vif(fitLm1) dpi cpi government 1.100321 1.005814 1.024822 > vif(fitLm2) dpi unemp 1.095699 1.095699

unemp 1.127610

Note that after removing the two “noise” variables the variance inflation factors of the remaining two variables decreases (as it should) since now we can determine the coefficients with more precision.

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Problem 7 The outputs from the crPlot are given in Figure 39 (right). Looking at these plots we see at the least-squares lines for government and cpi are almost horizontal indicating that these variables don’t add much predictive power when the other ones are in the model. This agrees with our earlier arguments. In the plots for dpi and unemp the least squares lines are close the lowess curves indicating a linear model is appropriate.

Exercises Exercise 9.1 (the distributions in regression) Part (a): Since Xi = 1 we have Yi ∼ N(1.4 + 1.7, 0.3) = N(3.1, 0.3) .

Thus the conditional mean of Y is 3.1 and the conditional standard deviation of Y is 0.5477226.

√

0.3 =

To compute P (Yi ≤ 3|Xi = 1) in R this would be pnorm( 3, mean=3.1, sd=sqrt(0.3) ) to find 0.4275661.

Part (b): We can compute the density of P (Yi = y) as Z ∞ P (Yi = y) = P (Yi = y|X)P (X)dX −∞

Z ∞ (y − β0 − β1 x)2 (x − 1)2 1 1 √ √ exp − exp − dx = 2σ12 2σ22 2πσ1 2πσ2 −∞

2

Z ∞ 1 x − 2x + 1 (y − β0 )2 − 2β1 (y − β0 )x + β12 x2 = exp − dx exp − 2πσ1 σ2 −∞ 2σ12 2σ22

1 (y − β0 )2 1 − 2 exp − = 2πσ1 σ2 2σ12 2σ2

2

Z ∞ 1 β1 (y − β0 ) 1 1 β1 2 + x + + 2 x dx . × exp − 2 σ12 σ22 σ12 σ2 −∞ √ √ Here σ1 = 0.3 and σ2 = 0.7. To evaluate this we now define 1 β12 σ22 + σ12 β12 + = σ12 σ22 σ12 σ22 1 β1 σ22 (y − β0 ) + σ12 β1 (y − β0 ) + = , b= σ12 σ22 σ12 σ22

a=

and so the integral above we want to evaluate is given by Z ∞

1 1 (y − β0 )2 − 12 ax2 +bx P (Yi = y) = − e exp − dx .

2πσ1 σ2 2σ12 2σ22 −∞ 95

Using Mathematica (or integral tables) we find that r Z ∞ 1

2

e− 2 ax +bx dx =

−∞

2π b2 e 2a . a

From the definition a and b we can compute the argument of the exponential above b2 (β1 σ22 (y − β0 ) + σ12 )2 = .

2a 2σ12 σ22 (β12 σ22 + σ12 ) Thus we find P (Yi = y) = √

1 p

2π β12 σ22 + σ12

(y − β0 )2 1 (β1 σ22 (y − β0 ) + σ12 )2 × exp − − 2+ .

2σ12 2σ2 2σ12 σ22 (β12 σ22 + σ12 )

We will further evaluate the argument of the exponent which we denote as A (for argument).

If we define v as v 2 ≡ β12 σ22 + σ12 we then have

1 2 2 2 2 2 2 4 2 2 2 4 (y − β ) σ v + σ v − β σ (y − β ) + 2σ σ β (y − β ) + σ 0 0 1 0 2 1 1 1 1 2 1 2σ12 σ22 v 2

1 = − 2 2 2 (σ22 v 2 − β12 σ24 )(y − β0 )2 − 2σ12 σ22 β1 (y − β0 ) + (σ12 v 2 − σ14 ) .

2σ1 σ2 v

A=−

From the definition of v 2 we have

σ22 v 2 − β12 σ24 = σ22 (β12 σ22 + σ12 ) − β12 σ24 = σ12 − σ22 σ12 v 2 − σ14 = σ12 (β12 σ22 + σ12 ) − σ14 = β12 σ12 σ22 ,

and

so the above becomes A=−

1 2σ12 σ22 v 2

2 2

σ1 σ2 (y − β0 )2 − 2σ12 σ22 β1 (y − β0 ) + β12 σ12 σ22

1 (y − β0 )2 − 2β1 (y − β0 ) + β12 2 2v

1 = − 2 (y − β0 − β1 )2 .

2v

=−

Thus we have finally shown

(y − β0 − β1 )2 p P (Yi = y) = √ exp − 2(β12 σ22 + σ12 ) 2π β12 σ22 + σ12

1

,

Note that this density is a normal density with a mean of β0 + β1 and a variance of β12 σ22 + σ12 .

We can evaluate the requested probability using the cumulative normal density function.

For the numbers given in this exercise that means that E(Yi ) = β0 + β1 = 3.1 Var (Yi ) = β12 σ22 + σ12 = 1.72 (0.7) + 0.3 = 2.323 P (Yi ≤ 3) = pnorm(3, 3.1, 2.323) = 0.4828317 .

96

Exercise 9.2 (least squares is the same as maximum likelihood) Maximum likelihood estimation would seek parameters β0 and β1 to maximize the loglikelihood of the parameters given the data. For the assumptions in this problem this becomes !

N Y X LL = log p(Yi |Xi ) = log p(Yi |Xi ) i=1

(yi − β0 − β1 xi )2 1 exp − = log √ 2σǫ2 2πσǫ X 1 = a constant − 2 (yi − β0 − β1 xi )2 .

2σǫ i X

This later summation expression is what we are minimizing when we perform least-squares minimization.

Exercise 9.3 Using β̂1 =

n X

wi Yi ,

i=1

with Yi iid we have that β̂1 =

n X

wi2Var (Yi ) =

i=1

Now as

we see that

n X

wi2 σǫ2 = σǫ2

i=1

n X

wi2 .

i=1

n X

n X 1 1 2 , wi = Pn (Xi − X̄)2 = Pn 2 2 2 i=1 (Xi − X̄) i=1 i=1 i=1 (Xi − X̄)

as we were to show.

σǫ2 σǫ2 , = 2 2 (n − 1)SX i=1 (Xi − X̄)

Var β̂1 = Pn

Exercise 9.4 (the VIF for centered variables) In the R code chap 9.R we perform the requested experiment and if we denote the variable X − X̄ as V we find [1] "cor(X,X^2)= 0.974" [1] "cor(V,V^2)= 0.000" [1] "VIF for X and X^2= 19.389" [1] "VIF for (X-barX) and (X-barX)^2= 1.000" 97

Thus we get a very large reduction in the variance inflation factor when we center our variable.

Exercise 9.5 (the definitions of some terms in linear regression) In this problem we are told that n = 40 and that the empirical correlation r(Y, Ŷ ) = 0.65.

Using these facts and the definitions provided in the text we can compute the requested expressions.

Part (a): R2 = rY2 Ŷ = (0.65)2 = 0.4225 Part (b): From the definition of R2 we can write R2 = 1 −

residual error SS . total SS

(31)

Since we know the value of R2 and that the total sum of squares, given by, X total SS = (Yi − Ȳ )2 , i

is 100 we can solve Equation 31 for the residual sum of square. We find we have a residual error sum of squares given by 57.75.

Part (c): Since we can decompose the total sum of squares into the regression and residual sum of squares as total SS = regression SS + residual SS , (32) and we know the values of the total sum of squares and the residual sum of squares we can solve for the regression sum of squares, in that 100 = regression SS + 57.75 .

Thus regression SS = 42.25.

Part (d): We can compute s2 as s2 =

57.75 57.75 residual error SS = = = 1.604167 . residual degrees of freedom n−1−p 40 − 1 − 3

Exercise 9.6 (model selection with R2 and Cp ) For this problem we are told that n = 66 and p = 5. We will compute several metrics used to select which of the models (the value of the number of predictors or p) one should use in

98

the final regression. The metrics we will consider include residual error SS total SS (n − p − 1)−1 residual error SS Adjusted R2 = 1 − (n − 1)−1 total SS SSE(p) − n + 2(p + 1) where Cp = 2 σ̂ǫ,M n X SSE(p) = (Yi − Ŷi )2 and R2 = 1 −

(33) (34) (35)

i=1

2 σ̂ǫ,M =

M X 1 (Yi − Ŷi )2 . n − 1 − M i=1

2 Here σ̂ǫ,M is the estimated residual variance using all of the M = 5 predictors, and SSE(p) is computed using values for Ŷi produced under the model with p < M predictors. From the numbers given we compute it to be 0.1666667. Given the above when we compute the three model selection metrics we find

> print( rbind( R2, Adjusted_R2, Cp ) ) [,1] [,2] [,3] R2 0.7458333 0.7895833 0.7916667 Adjusted_R2 0.7335349 0.7757855 0.7743056 Cp 15.2000000 4.6000000 6.0000000 To use these metrics in model selection we would want to maximize R2 and the adjusted R2 and minimize Cp . Thus the R2 metric would select p = 5, the adjusted R2 metric would select p = 4, and the Cp metric would select p = 4.

Exercise 9.7 (high p-values) The p-values reported by R are computed under the assumption that the other predictors are still in the model. Thus the large p-values indicate that given X is in the model X 2 does not seem to help much and vice versa. One would need to study the model with either X or X 2 as the predictors. Since X and X 2 are highly correlated one might do better modeling if we subtract the mean of X from all samples i.e. take as predictors (X − X̄) and (X − X̄)2 rather than X and X 2 .

Exercise 9.8 (regression through the origin) The least square estimator for β1 is obtained by finding the value of β̂1 such that the given RSS(β1 ) is minimized. Taking the derivative of the given expression for RSS(β̂1 ) with respect 99

to β̂1 and setting the resulting expression equal to zero we find d dβ̂1

RSS(β̂1 ) = 2

or −

X

X (Yi − β̂1 Xi )(−Xi ) = 0 ,

Yi Xi + β̂1

Solving this expression for β̂1 we find

X

Xi2 = 0 .

P Xi Y i β̂1 = P 2 .

Xi

(36)

To study the bias introduced by this estimator of β1 we compute P P 2 Xi E(Yi ) X P 2 E(β̂1 ) = = β1 P i2 = β1 , Xi Xi

showing that this estimator is unbiased. To study the variance of this estimator we compute Var(β̂1 ) = =

1 (

P

Xi2 )

2

2

σ P 2 2 ( Xi )

X i

X i

X 1 Var(Xi Yi ) = P 2 2 Xi2 Var(Yi ) ( Xi ) i σ2 Xi2 = P 2 , i Xi

(37)

the requested expression. An estimate of σ̂ is given by the usual σ̂ 2 =

RSS , n−1

which has n − 1 degrees of freedom.

Exercise 9.9 (filling in the values in an ANOVA table) To solve this problem we will use the given information to fill in the values for the unknown values. As the total degrees of freedom is 15 the number of points (not really needed) must be one more than this or 16. Since our model has two slopes the degrees of freedom of the regression is 2. Since the degrees of freedom of the regression (2) and the error must add to the total degrees of freedom (15) the degrees of freedom of the error must be 15 − 2 = 13.

The remaining entries in this table are computed in the R code chap 9.R.

Exercise 9.10 (least squares with a t-distribution) For this problem in the R code chap 9.R we generate data according to a model where y is linearly related to x with an error distribution that is t-distributed (rather than the classical 100

normal distribution). Given this working code we can observe its performance and match the outputs with the outputs given in the problem statement. We find Part (a): This is the second number in the mle$par vector or 1.042.

Part (b): Since the degrees-of-freedom parameter is the fourth one the standard-error of it is given by the fourth number from the output from the sqrt(diag(FishInfo)) or 0.93492.

Part (c): This would be given by combining the mean and the standard error for the standard deviation estimate or 0.152 ± 1.96(0.01209) = (0.1283036, 0.1756964) .

Part (d): Since mle$convergence had the value of 0 the optimization converged.

101

(b)

1.0

abs(resid)

0 −3

0.0

−2

0.5

−1

resid

1

1.5

2

3

2.0

(a)

500

1000

0

fitLm$fit

(c)

(d)

1000

resid

−2

0

2

40 30 20 0

−4

10

Sample Quantiles

500

fitLm$fit

4

0

−4

−2

0

2

4

0

Theoretical Quantiles

5

10

15

education

0 −3

−2

−1

resid

1

2

3

(e)

0

10

20

30

40

50

60

experience

Figure 40: Several regression diagnostics plots for the CPS1988 dataset.

9

Chapter 10 (Regression: Troubleshooting)

R Lab See the R script Rlab.R for this chapter. To make the plots more visible I had to change the y limits of the suggested plots. When these limits are changed we get the sequence of plots shown in Figure 40. The plots (in the order in which they are coded to plot) are given by • The externally studentized residuals as a function of the fitted values which is used to look for bias in the model. Note that for small and large values of the fitted values the red curve deviates from the blue curve indicating that the bias maybe present.

• The absolute value of the externally studentized residuals as a function of the fitted values which is used to look for heteroscedasticity.

• The qqplot is used to look for error distributions that are skewed or significantly 102

non-normal. This might suggest applying a log or square root transformation to the response Y to try and make the distribution of residuals more Gaussian.

• Plots of the externally studentized residuals as a function of the variable education which can be used to look for nonlinear regression affects in the given variable.

• Plots of the externally studentized residuals as a function of the variable experience which can be used to look for nonlinear regression affects in the given variable. Note the concave downward shape of the red curve indicating that there is a slight nonlinear affect from this variable.

There are a couple of things of note from this plot. The most striking item in the plots presented is in the qqplot. The right limit of the qqplot has a large deviation from a straight line. This indicates that the residuals are not normally distributed and perhaps a transformation of the response will correct this.

We choose to apply a log transformation to the response wage and not to use ethnicity as a predictor (as was done in the previous part of this problem). When we plot the same diagnostic plots as earlier (under this new model) we get the plots shown in Figure 41. The qqplot in this case looks “more” normal (at least both tails of the residual distribution are more symmetric). The distribution of residuals still has heavy tails but certainly not as severe as they were before (without the log transformation). After looking at the plots in Figure 41 we see that there are still non-normal residuals. We also see that it looks like there is a small nonlinear affect in the variable experience. We could fit a model that includes this term. We can try a model of log(wage) with a quadratic term. When we do that, and then reconsider the diagnostic plots presented so far we get the plots shown in Figure 42.

We can then add in the variable ethnicity and reproduce the same plots be have been presenting previously. These plots look much like the last ones presented.

Problem 5 In Figure 43 we present this plot. There we see two examples with very large Cook’s D. That Those samples do not seem to be in error and should probably be kept.

Exercises Exercise 10.1 Some notes on the diagnostic plots are • From Plot (a) there should be a nonlinear term in x added to the regression.

103

(b)

1.0

abs(resid)

0 −3

0.0

−2

0.5

−1

resid

1

1.5

2

3

2.0

(a)

5.5

6.0

6.5

7.0

5.0

5.5

6.0

6.5

fitLm$fit

fitLm$fit

(c)

(d)

7.0

2 −2

0

resid

2 0 −2

−4

−4

Sample Quantiles

4

4

6

5.0

−4

−2

0

2

4

0

Theoretical Quantiles

5

10

15

education

0 −3

−2

−1

resid

1

2

3

(e)

0

10

20

30

40

50

60

experience

Figure 41: Several regression diagnostic plots for the CPS1988 dataset where we apply a log transformation to the response.

104

(b)

1.0

abs(resid)

0 −3

0.0

−2

0.5

−1

resid

1

1.5

2

3

2.0

(a)

4.5

5.0

5.5

6.0

6.5

7.0

4.0

4.5

5.0

5.5

fitLm$fit

fitLm$fit

(c)

(d)

6.0

6.5

7.0

resid

0

2

4 2

−2

0 −4

−4

−2

Sample Quantiles

6

4

8

4.0

−4

−2

0

2

4

0

Theoretical Quantiles

5

10

15

education

0 −3

−2

−1

resid

1

2

3

(e)

0

10

20

30

40

50

60

experience

Figure 42: Several regression diagnostic plots for the CPS1988 dataset where we apply a log transformation to the response and model with a quadratic term in experience (as well as a linear term).

105

5000

15000 Index

25000

sqrt(cooks.distance(fitLm4))

0

5000

15000

25000

0.10

0.15 Index

0.05

106

0

1

2

3 Half−normal quantiles

Figure 43: Looking for high leverage points with Cook’s D.

0

Sorted Data 0.00

0.15 0.05

0.10 0.00

0.0030 0.0025 0.0020 0.0015 0.0010 0.0005 0.0000

hatvalues(fitLm4)

15387

17228

4

• From Plot (b) we have some heteroscedasticity in that it looks like we have different values of variance for small and larger values of ŷ.

• From Plot (c) there might be some heavy tails and or some outliers.

• From Plot (d) it looks like we have autocorrelated errors.

• From Plot (f) we might have some outliers (samples 1 and 100).

Exercise 10.2 Most of these plots seem to emphasis an outlier (the sample with index 58). This sample should be investigated and most likely removed.

Exercise 10.3 Some notes on the diagnostic plots are • From Plot (a) there should be a nonlinear term in x added to the regression.

• From Plot (b) we don’t have much heteroscedasticity i.e. the residual variance looks uniform.

• From Plot (d) it looks like we have autocorrelated residual errors.

Exercise 10.4 Some notes on the diagnostic plots are • From Plot (a) there is perhaps a small nonlinear term in x that could be added to the regression.

• From Plot (c) we see that the distribution of the residuals have very large tails. Thus we might want to consider taking a logarithmic or a square root transformation of the response Y .

• From Plot (f) it looks like there are two samples (89 and 95) that could be outliers.

107

Exercise 10.5 Cook’s D is a measure of how sensitive are the predictions to the removal of one sample. To have a large Cook’s D a sample needs to have high leverage and a Yi value that is an outlier.

If Yi is not an outlier it might not result in a large Cook’s D.

108

10

Chapter 11 (Regression: Advanced Topics)

Notes on the Text The maximum likelihood estimation of σ 2 To evaluate what σ is once β has been computed, we take the derivative of LGAUSS with respect to σ, set the result equal to zero, and then solve for the value of σ. For the first derivative of LGAUSS we have

n ∂LGAUSS Yi − xTi β n X Yi − xTi β . = − 2 ∂σ σ σ σ i=1

Setting this expression equal to zero (and multiply by σ) we get n

Solving for σ then gives

1 X n− 2 (Yi − xTi β)2 = 0 . σ i=1 n

1X σ = (Yi − xTi β)2 . n i=1 2

Notes on the Transform-Both-Sides Method (Poisson responses) To use the “transform-both-sides” method we are told to pick a function h(y) such that Var(Yi |Xi ) = [h(1) {f (Xi ; β̂)}]−2 .

(38)

holds.

As an example of this, in Poisson regression the variance of Yi is equal to the mean of Yi or Var(Yi |Xi ) = f (Xi ; β̂) .

Using this in the above gives [h(1) {f (Xi ; β̂)}]−2 = f (Xi ; β̂) .

To simplify notation if we take X ≡ f (Xi ; β̂) this is

[h(1) {X}]−2 = X .

This means that

1 h(1) (X) = √ .

X

If we integrate this we have X 1/2 = 2X 1/2 .

1/2 This means that we should take α = 21 in the h(y) = y α transformation. h(X) =

109

Notes on the best linear prediction If we desire to estimate Y with the linear combination β0 + β1 X then to compute β0 and β1 we seek to minimize E((Y − (β0 − β1 X))2 ). This can be expanded to produce a polynomial in these two variables as E((Y − (β0 − β1 X))2 ) = E(Y 2 − 2Y (β0 + β1 X) + (β0 + β1 X)2 )

= E(Y 2 − 2β0 Y − 2β1 XY + β02 + 2β0 β1 X + β12 X 2 )

= E(Y 2 ) − 2β0 E(Y ) − 2β1 E(XY ) + β02 + 2β0 β1 E(X) + β12 E(X 2 ) .

(39)

Take the β0 and β1 derivatives of this result, and then setting them equal to zero gives 0 = −2E(Y ) + 2β0 + 2β1 E(X) 0 = −2E(XY ) + 2β0 E(X) + 2β1 E(X 2 ) ,

(40) (41)

as the two equations we must solve for β0 and β1 to evaluate the minimum of our expectation.

Writing the above system in matrix notation gives

E(Y ) β0 1 E(X) . = E(XY ) β1 E(X) E(X 2 ) Using Cramer’s rule we find

β0 β1

1 E(X 2 ) −E(X) E(Y ) = 1 E(XY ) E(X 2 ) − E(X)2 −E(X)

2 1 E(X )E(Y ) − E(X)E(XY ) = −E(X)E(Y ) + E(XY ) Var (X)

1 E(X 2 )E(Y ) − E(X)E(XY ) = .

Cov (X, Y ) Var (X)

Thus we see that β1 =

Cov (X, Y ) σXY = 2 .

Var (X) σX

(42)

From Equation 40 we have β0 = E(Y ) − β1 E(X) = E(Y ) −

σXY 2 σX

E(X) .

(43)

These are the equations presented in the text.

Notes on the error of the best linear prediction Once we have specified β0 and β1 we can evaluate the expected error in using these values for our parameters. With Ŷ = β̂0 + β̂1 X and the expressions we computed for β0 and β1

110

when we use Equation 39 we have

σXY σXY 2 2 E((Y − Ŷ ) ) = E(Y ) − 2 E(Y ) − 2 E(X) E(Y ) − 2 E(XY ) 2 σX σX

2

2

σXY σXY σXY σXY E(X) + + E(Y ) − 2 E(X) + 2 E(Y ) − 2 E(X) E(X 2 ) 2 2 σX σX σX σX

σXY σXY 2 2 E(XY ) = E(Y ) − 2E(Y ) + 2 2 E(X)E(Y ) − 2 2 σX σX σXY σ2 + E(Y )2 − 2 2 E(X)E(Y ) + XY E(X)2 4 σX σX

2 2

σXY σXY σXY 2 + 2 2 E(X)E(Y ) − 2 E(X) + E(X 2 ) 2 2 σX σX σX 2 σ2 σXY σXY 2 E(X) + E(X 2 ) = E(Y 2 ) − E(Y )2 + 2 2 (E(X)E(Y ) − E(XY )) − XY 4 4 σX σX σX σXY σ2 = Var (Y ) − 2 2 Cov (X, Y ) + XY (E(X 2 ) − E(X)2 ) 4 σX σX σXY σ2 σXY 2 = Var (Y ) − 2 2 Cov (X, Y ) + XY σX = σY2 − 2 4 σX σX σX

2 σ = σY2 1 − 2XY2 . σX σY

Since σXY = σX σY ρXY we can write the above as

E((Y − Ŷ )2 ) = σY2 1 − ρ2XY ,

(44)

which is the equation presented in the book. Next we evaluate the expectation of (Y − c)2 for a constant c. We find E((Y − c)2 ) = E((Y − E(Y ) + E(Y ) − c)2 ) = E((Y − E(Y ))2 + 2(Y − E(Y ))(E(Y ) − c) + (E(Y ) − c)2 ) = E((Y − E(Y ))2 ) + 0 + E((E(Y ) − c)2 )) = Var (Y ) + (E(Y ) − c)2 ,

(45)

since this last term is a constant (independent of the random variable Y ).

R Lab See the R script Rlab.R for this chapter where these problems are worked.

Nonlinear Regression The R command help(Irates) tells us that the r1 column from the Irates dataframe is a ts object of interest rates sampled each month from Dec 1946 until Feb 1991 from the United-States. These rates are expressed as a percentage per year.

111

(b)

0

−4

5

−2

r1

delta_r1

10

0

2

15

(a)

1960

1970

1980

1990

1950

1960

1970

Time

Time

(c)

(d)

1980

1990

0 delta_r1 −2

10 0

−4

5

delta_r1^2

15

2

20

1950

1950

1960

1970

1980

1990

0

Time

5

10

15

lag_r1

10 0

5

delta_r1^2

15

20

(e)

0

5

10

15

lag_r1

Figure 44: The plots for estimating the short rate models.

When the above R code is run we get the plots shown in Figure 44. These plots are used in building the models for µ(t, r) and σ(t, r). From the plot labeled (d) we see that ∆rt seems (on average) to be relatively constant at least for small values of rt−1 i.e. less than 10. For values greater than that we have fewer samples and it is harder to say if a constant would be the best fitting function. From the plot labeled (b) it looks like there are times when ∆rt is larger than others (namely around 1980s). This would perhaps argue for a time dependent µ function. There does not seem to be a strong trend. From the summary command we see that a and θ are estimated as > summary( nlmod_CKLS ) Formula: delta_r1 ~ a * (theta - lag_r1) Parameters: Estimate Std. Error t value Pr(>|t|) theta 5.32754 1.33971 3.977 7.96e-05 *** a 0.01984 0.00822 2.414 0.0161 * 112

Running the R code for this problem I also get estimates for A and B which I can translate into estimates for σ and γ. I then get that the 95% confidence intervals for these parameter are given by [1] "sigma ci:" [1] 0.00000000 0.05458282 [1] "gamma ci:" [1] 1.23251 1.74067 The unweighted estimate of θ shown as the red line in panel (a) is larger than the weighted estimate shown in panel (c). They differ because the weights used (the fitted values from the first model) are so much larger in the second half of the dataset than in the first half of the dataset.

Response Transformations The boxcox function returns x which is a list of the values of α tried and y the value of the loglikelihood for each of these values of α. We want to pick a value of α that maximizes the loglikelihood. Finding the maximum of the loglikelihood we see that it is achieved at a value of α = 0.1414141. The new model with Y transformed using the box-cox transform has a much smaller value of the AIC [1] "AIC(fit1)= 12094.187991, AIC(fit3)= 1583.144759" This is a significant reduction in AIC. Plots of the residuals of the box-cox model as a function of the fitted values indicate that there is not a problem of heteroscedasticity. The residuals of this box-cox fit appear to be autocorrelated but since this is not time series data this behavior is probably spurious (not likely to repeat out of sample).

Who Owns an Air Conditioner?

Computing a linear model using all of the variables gives that several of the coefficients are not estimated well (given the others in the model). We find > summary(fit1) Call: glm(formula = aircon ~ ., family = "binomial", data = HousePrices) Deviance Residuals: 113

Min -2.9183

1Q -0.7235

Median -0.5104

3Q 0.6578

Max 3.2650

Coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -3.576e+00 5.967e-01 -5.992 2.07e-09 *** price 5.450e-05 8.011e-06 6.803 1.02e-11 *** lotsize -4.482e-05 6.232e-05 -0.719 0.472060 bedrooms -6.732e-02 1.746e-01 -0.385 0.699887 bathrooms -5.587e-01 2.705e-01 -2.065 0.038907 * stories 3.155e-01 1.540e-01 2.048 0.040520 * drivewayyes -4.089e-01 3.550e-01 -1.152 0.249366 recreationyes 1.052e-01 2.967e-01 0.355 0.722905 fullbaseyes 1.777e-02 2.608e-01 0.068 0.945675 gasheatyes -3.929e+00 1.121e+00 -3.506 0.000454 *** garage 6.893e-02 1.374e-01 0.502 0.615841 preferyes -3.294e-01 2.743e-01 -1.201 0.229886 We can use the stepAIC in the MASS library to sequentially remove predictors. The final step from the stepAIC command gives Step: AIC=539.36 aircon ~ price + bathrooms + stories + gasheat Df Deviance AIC <none> 529.36 539.36 - bathrooms 1 532.87 540.87 - stories 1 535.46 543.46 - gasheat 1 554.74 562.74 - price 1 615.25 623.25 The summary command on the resulting linear model gives > summary(fit2) Call: glm(formula = aircon ~ price + bathrooms + stories + gasheat, family = "binomial", data = HousePrices) Deviance Residuals: Min 1Q Median -2.8433 -0.7278 -0.5121

3Q 0.6876

Max 3.0753

Coefficients: Estimate Std. Error z value Pr(>|z|) 114

(Intercept) -4.045e+00 price 4.782e-05 bathrooms -4.723e-01 stories 3.224e-01 gasheatyes -3.657e+00

4.050e-01 6.008e-06 2.576e-01 1.317e-01 1.082e+00

-9.987 < 2e-16 *** 7.959 1.73e-15 *** -1.833 0.066786 .

2.449 0.011334 * -3.378 0.000729 ***

Looking at the signs of the coefficients estimated we first see that as price and stories increases the probability of air conditioning increases which seams reasonable. From the same table, increasing bathrooms and gasheatyes should decrease the probability that we have air conditioning. One would not expect that having more bathrooms should decrease our probability of air conditioning. The same might be said for the gasheatyes predictor. The difference in AIC between the model suggested and the one when we remove the predictor bathrooms is not very large indicating that removing it does not give a very different model.

As the sample we are told to look at seems to be the same as the first element in the training set we can just extract that sample and use the predict function to evaluate the given model. When we do this (using the first model) we get 0.1191283.

Exercises Exercise 11.1 (computing β0 and β1 ) See the notes on Page 110.

Exercise 11.2 (hedging) The combined portfolio is F20 P20 − F10 P10 − F30 P30 .

Lets now consider how this portfolio changes as the yield curve changes. From the book we would have that the change in the total portfolio is given by −F20 P20 DUR20 ∆y20 + F10 P10 DUR10 ∆y10 + F30 P30 DUR30 ∆y30 .

We are told that we have modeled ∆y20 as ∆y20 = β̂1 ∆y10 + β̂2 ∆y30 .

When we put this expression for ∆y20 into the above (and then group by ∆y10 and ∆y30 ) we can write the above as (−F20 P20 DUR20 β̂1 + F10 P10 DUR10 )∆y10 + (−F20 P20 DUR20 β̂2 + F30 P30 DUR30 )∆y30 .

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We will then take F10 and F30 to be the values that would make the coefficients of ∆y10 and ∆y30 both zero. These would be

P20 DUR20 F10 = F20 β̂1 P10 DUR10

P20 DUR20 F30 = F20 β̂2 .

P30 DUR30 Exercise 11.3 (fitting a yield curve) We are given the short rate r(t; θ), which we need to integrate to get the yield yt (θ). For the Nelson-Siegel model for r(t; θ) this integration is presented in the book on page 383. Then given the yield the price is given by Pi = exp(−Ti yTi (θ)) + ǫi .

I found it hard to fit the model “all at once”. In order to fit the model I had to estimate each parameter θi in a sequential fashion. See the R code chap 11.R for the fitting procedure used. When that code is run we get estimate of the four θ parameters given by theta0 theta1 theta2 theta3 0.009863576 0.049477242 0.002103376 0.056459908 When we reconstruct the yield curve with these numbers we get the plot shown in Figure 45.

Exercise 11.4 Using the estimate and the standard error I compute a 95% confidence interval in r of [1] 0.05557965 0.06142035 As the true value of r = 0.6 is inside this interval I would assume that this result (that r̂ 6= 0.6) is not due to a biased estimate but due to sampling error.

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Figure 45: The model (in red) and the discrete bond prices (in black) for Exercise 11.3.

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11

Chapter 12 (Time Series Models: Basics)

Notes on the text: ARMA(1, 1) Process From the ARMA(1, 1) expression Yt = φYt−1 + θǫt−1 + ǫt , if we square this we get 2 Yt2 = φ2 Yt−1 + θ2 ǫ2t−1 + ǫ2t + 2φθYt−1 ǫt−1 + 2φYt−1 ǫt + 2θǫt−1 ǫt .

Then taking expectations of both sides and using the facts that 2 E[Yt2 ] = E[Yt−1 ] = γ(0) 2 2 E[ǫt ] = σǫ E[Yt−1 ǫt−1 ] = σǫ2 E[Yt−1 ǫt ] = 0 E[ǫt−1 ǫt ] = 0 ,

we get γ(0) = φ2 γ(0) + θ2 σǫ2 + σǫ2 + 2φθσǫ2 , which we can solve for γ(0) to get the variance of an ARMA(1, 1) process.

R Lab See the R script Rlab.R for my implementations of the R Lab for this chapter.

Note that some of the codes use the forecast package. A dependency of that package is RcppArmadillo which is no longer supported for some versions of R. If I installed an older version of RcppArmadillo I was able to download all required packages and work the problems successfully. To install an older version of RcppArmadillo I used the following R code link = "https://cran.r-project.org/src/contrib/Archive/chpArmadillo/chpArmadillo_0.6.100.0.0.tar.gz" install.packages(link, repos=NULL, type="source")

Problem 1 (stationarity of the raw variables) Part (a): To start this analysis see Figure 46 (left) for plots of time series for the variables r, y, and π. From that plot, the variables r and y seem to be moving in the same direction for extended periods of time (indicating that they may not be stationary time series). The plot 118

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Figure 46: Left: The time series plots of the data in the Tbrate dataset. Right: The autocorrelation plots for the data in the Tbrate dataset. of π seems to be more stationary. Next see Figure 46 (right) for plots of the auto-correlation of the same three variables r, y, and π (on the diagonal) and plots of the cross-correlation functions (on the off-diagonal). The very slow decay of the autocorrelation functions along the diagonal indicate that differencing will probably be needed to obtain stationary time series that we can model with ARMA models.

Part (b): We next run the adf.test on each of the given time series before any transformations are taken. Note that the null hypothesis for the adf.test is that the time series has a unit-root i.e. is nonstationary. The alternative hypothesis is that the time series is actually stationary. Thus “large” and negative t-values (or small p-values) indicate that this time series probably does not have a unit root and might already be stationary. On the other hand “small” t values (or large p-values) indicates that the given time series probably does have a unit root. If a time series has a unit root taking the first difference of it may result in a time series that is more easily modeled. For the variable in this dataframe we find • For the variable r we get Dickey-Fuller = -2.5622, Lag order = 6, p-value = 0.3395 Indicating that most likely have a unit-root in this data.

• For the variable y we get 119

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Right: The autocorrelation plots of first difference of the data in the Tbrate dataset.

Dickey-Fuller = -1.3474, Lag order = 6, p-value = 0.8502 Indicating that it is very likely that we have a unit-root in this data.

• For the variable π we get Dickey-Fuller = -3.1444, Lag order = 6, p-value = 0.09794 Indicating that there is less chance that we have a unit-root in this data.

Looking back at plots of the time series of the variables using the plot(Tbrate) we see that the first two time series seem to be steadily growing while the one for π seems to more stationary. That is what the numbers above state in that y and r seem to very strongly have a unit root while for π there is less evidence of that. The above p-values for all of the series are quite large and we could conclude from this that none of the time series are stationary.

Problem 2 (stationarity of the first differences) Next we take the first difference of this data and look if the result is more stationary. We first plot the time series of this first difference in Figure 47 (left). There we see that the 120

series are much more mean reverting in this case. Next we plot the ACF functions for these first differences in Figure 47 (right). All of these plots show series that look much more stationary than before.

Next we consider the adf.test looking for a unit root for these first differences. For these three time series we get • For the variable ∆r we get Dickey-Fuller = -6.3425, Lag order = 6, p-value = 0.01 Indicating that there is most likely not a unit-root in this difference and that the time series is stationary.

• For the variable ∆y we get Dickey-Fuller = -5.6647, Lag order = 6, p-value = 0.01 Also indicating that there is most likely not a unit-root in this difference and that the time series is stationary.

• For the variable ∆π we get Dickey-Fuller = -7.9521, Lag order = 6, p-value = 0.01 Again indicating that there is most likely not a unit-root in this difference and that the time series is stationary.

We conclude that all three series appear stationary after taking first differences.

Problem 3 (seasonality differences) Next we look to see if the mean level for ∆r depends on the quarter. In Figure 48 we construct a box plot of the samples of ∆r that fall in each quarter. From that plot it does not look like there is much dependence on the mean level with the quarter.

Problem 4 (fitting an ARIMA model) The output from the auto.arima function call is

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Figure 48: A box plot of ∆r grouped by the quarter 1, 2, 3 or 4 the measurement was taken in. > auto.arima( Tbrate[,1], max.P=0, max.Q=0, ic="aic" ) Series: Tbrate[, 1] ARIMA(2,1,3) Coefficients: ar1 ar2 0.1368 -0.2225 s.e. 0.1655 0.1483

ma1 -0.2597 0.1539

ma2 0.1041 0.1483

ma3 0.3364 0.0649

sigma^2 estimated as 0.758: log likelihood=-298.6 AIC=609.19 AICc=609.56 BIC=629.95 Which indicates that when we use the AIC information criterion we choose to take the first difference d = 1 and then to model ∆r with a ARMA(2, 3) model. Changing the ic argument in the auto.arima call to bic selects a ARIMA(2, 1, 2) model. This model is ∆rt = rt − rt−1 = φ1 ∆rt−1 + φ2 ∆rt−2 + ǫt + θ1 ǫt−1 + θ2 ǫt−2 , where the values of the parameters φi and θi are given in the above R output.

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Problem 5 (residual autocorrelation) The plot of the autocorrelation function of the ARIMA(2, 1, 2) model residuals show no spikes reaching far outside of the confidence intervals plotted on the graph. There are some spikes that are just outside these limits but these are probability due to statistical fluctuations rather than to a real phenomena that we should try to model. The output from the Box.test command on these residuals is > Box.test( residuals(fit1), lag=12, type="Ljung", fitdf=2+2 ) X-squared = 23.33, df = 8, p-value = 0.002966 Which indicates that if we accept the hypothesis of randomly ordered data there is a probability of around 0.03 of getting results as “extreme” as the ones we have. We would like the p-value returned from this test to be as large as possible so as to conclude that the values of the autocorrelation function at reasonable lags are all zero. This value of the p-value is perhaps not as large as we would like to conclude that there is no further autocorrelation.

We could perhaps reduce this autocorrlation by using seasonal models or a long-memory process.

Problem 6 (GARCH effects) The plot of the autocorrelation function for the residuals squares is given in Figure 49. There we see two large spikes above the 2σ horizontal lines. The output from the Box.test also indicates the presence of non-randomness > Box.test(resid2, lag=12, type="Ljung") X-squared = 69.368, df = 12, p-value = 4.204e-10 A p-value this small means that it is very unlikely that this data is purely random and we see evidence of GARCH effects.

Problem 7 (forecasting) The output from the auto.arima call is > auto.arima( pi, max.P=0, max.Q=0, ic="bic" ) Series: pi ARIMA(0,1,1) Coefficients: 123

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s.e.

ma1 -0.7882 0.0451

sigma^2 estimated as 0.5708: log likelihood=-267.56 AIC=539.12 AICc=539.17 BIC=546.04 Which indicates that we should model this with an ARIMA(0,1,1) model.

We plot the forecast and the standard errors in Figure 50. In a non rigorous way, the prediction intervals widen since at every timestep our ∆Yt build from terms of the form ǫt + θ1 ǫt−1 (along with other terms). These two terms are random and have an associated variance and as we predict further in the future we have to add more and more of these random variables to estimate ∆Yt . The more random variables we add together the larger the variance of the resulting sum.

Exercises See the R code chap 12.R where these problems are worked.

Exercise 12.1 Plots of the output from the acf function for each of the two calls are given in Figure 51.

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Part (a): Note that variable crsp is a ts (time series) object and as such has some auxiliary data associated with it. If we try to display the variable crsp in the command window we get the following (partial output) > crsp Time Series: Start = c(1969, 1) End = c(1975, 338) Frequency = 365 [1] -0.007619 0.013016 [8] 0.004027 0.001736

0.002815 0.003064 0.001504 -0.001778

0.001633 -0.001991 0.004671 0.008950 0.001920 -0.000024

When we plot the acf of the ts object the lag is in units of time since the time series object has a notion of how much time is represented between each data point. For the output above we see that the frequency is 365 (corresponding to the number of days in a year) thus a lag 1 of one day corresponds to 365 = 0.002739726 yearly time units which is the spacing between the acf values in the leftmost plot. If we look at the rightmost acf plot in Figure 51 we see 7 a small downward spike at lag seven. In units of time this is located at 365 = 0.01917808 in the leftmost plot.

Part (b): We see that the three significant autocorrelations (with the smallest lags) are found at lags 1, 7, and 16. I would expect the autocorrelation at lag 1 to be statistically significant but the others are most likely due to chance. Autocorrelations with even larger lags that seem “significant” (in that they are above the 2σ standard error bars) are also probably due to chance.

Exercise 12.2 Part (a): From the autocorrelation plots for the crsp data it looks like a better model to fit would be a MA(1) model rather than any AR(p) model (as suggested in this exercise).

In any case, if we are to fit AR(1) and AR(2) model the output from the two arima calls is > arima(crsp,order=c(1,0,0)) Series: crsp ARIMA(1,0,0) with non-zero mean Coefficients: ar1 intercept 0.0853 7e-04 s.e. 0.01128 2e-04 sigma^2 estimated as 5.973e-05: AIC=-17406.37 AICc=-17406.36

log likelihood=8706.18 BIC=-17388.86 126

> arima(crsp,order=c(2,0,0)) Series: crsp ARIMA(2,0,0) with non-zero mean Coefficients: ar1 ar2 0.0865 -0.0141 s.e. 0.0199 0.0199

intercept 7e-04 2e-04

sigma^2 estimated as 5.972e-05: AIC=-17404.87 AICc=-17404.85

log likelihood=8706.43 BIC=-17381.53

In comparing these two models we would select the one that has the lower value of the Akaike information criterion (aic) or the Bayesian information criterion (bic). For either of these two metrics using the above results the AR(1) model is preferred to the AR(2) model.

Part (b): Using the data given for the AR(1) fit we have a 95% confidence interval for the value of φ given by alpha = 0.05 0.0853 + 0.0198 * qnorm( 1 - 0.5 * alpha ) * c(-1,+1) [1] 0.04649271 0.12410729

Exercise 12.3 (an AR(1) model) Part (a): An AR(1) model looks like Yt − µ = φ(Yt−1 − µ) + ǫt .

To have a stationary AR(1) model we must have |φ| < 1. Since for this problem we have φ = −0.55 we do have |φ| < 1 and this AR model is stationary.

Part (b): We can write the above expression for an AR(1) model in the form Yt = (1 − φ)µ + φYt−1 + ǫt .

With φ = −0.55 and equating this to the model we were given in this problem we have µ=

5 5 = = 3.225806 .

1−φ 1 + 0.55

Part (c): From the book we have that γ(0) = Var (Yt ) =

σǫ2 1.2 = = 1.72043 .

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Part (d): From the book we have that γ(h) = Cov(Yt , Yt+h ) =

σǫ2 φ|h| = 1.72043(−0.55)|h| .

1 − φ2

Exercise 12.4 (another AR(1) model) Part (a): For an AR(1) model we know Var (Y1 ) =

1.2 σǫ2 = = 1.6 .

1 − φ2 1 − 0.52

Part (b): Again from the discussion in the book on AR(1) models we have σǫ2 φ|1| 1.2(0.5) = = 0.8 2 1−φ 1 − 0.52 1.2(0.5)2 σ 2 φ|2| = 0.4 .

Cov(Y1 , Y3) = ǫ 2 = 1−φ 1 − 0.52

Cov(Y1 , Y2) =

Part (c): We can compute this expression as follows

Y1 + Y2 + Y3 1 Var = Var(Y1 + Y2 + Y3 ) 2 4 " 3 # X 1 X = Var (Yi ) + 2 Cov(Yi , Yj ) 4 i=1 i<j

1 [3Var (Y1 ) + 2 (Cov(Y1 , Y2 ) + Cov(Y1 , Y3 ) + Cov(Y2 , Y3 ))] 4 1 = [3(1.6) + 2 (0.8 + 0.4 + 0.8)] = 2.2 .

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Exercise 12.5 (an AR(3) model) An AR(3) model takes the form Yt − µ = φ1 (Yt−1 − µ) + φ2 (Yt−2 − µ) + φ3 (Yt−3 − µ) + ǫt .

To predict the value of Yt+1 we increment the t index by one to get Yt+1 − µ = φ1 (Yt − µ) + φ2 (Yt−1 − µ) + φ3 (Yt−2 − µ) + ǫt+1 , and then take expectations (conditioned data that has arrived by the time t) to get Ŷt+1 = µ̂ + φ̂1 (Yt − µ̂) + φ̂2 (Yt−1 − µ̂) + φ̂3 (Yt−2 − µ̂) = 104 + 0.4(99 − 104) + 0.25(103 − 104) + 0.1(102 − 104) = 101.55 .

For the prediction of Yt+2 we use Equation 46 again (evaluated at t = t + 2) to get Ŷt+2 = µ̂ + φ̂1 (Ŷt+1 − µ̂) + φ̂2 (Yt − µ̂) + φ̂3 (Yt−1 − µ̂) = 104 + 0.4(101.55 − 104) + 0.25(99 − 104) + 0.1(103 − 104) = 101.67 .

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Exercise 12.6 (the autocovariance of a MA(2) model) For the MA(2) model Yt = µ + ǫt + θ1 ǫt−1 + θ2 ǫt−2 , note that E[Yt ] = µ .

Thus we can compute γ(h) = Cov(Yt , Yt+h ) = E[(Yt − µ)(Yt+h − µ)] = E[(ǫt + θ1 ǫt−1 + θ2 ǫt−2 )(ǫt+h + θ1 ǫt+h−1 + θ2 ǫt+h−2 )] = E[ǫt ǫt+h ] + θ1 E[ǫt ǫt+h−1 ] + θ2 E[ǫt ǫt+h−2 ] + θ1 E[ǫt−1 ǫt+h ] + θ12 E[ǫt−1 ǫt+h−1 ] + θ1 θ2 E[ǫt−1 ǫt+h−2 ] + θ2 E[ǫt−2 ǫt+h ] + θ1 θ2 E[ǫt−2 ǫt+h−1 ] + θ22 E[ǫt−2 ǫt+h−2 ] .

In the above if h = 0 then we get γ(0) = (1 + θ12 + θ22 )σǫ2 .

In the above if h = ±1 then we get γ(1) = (θ1 + θ1 θ2 )σǫ2 .

In the above if h = ±2 then we get γ(2) = θ2 σǫ2 .

In the above we get γ(h) = 0 if |h| > 2. Using these results we have for the autocorrelation of Yt   1 h=0   θ1 +θ1 θ2  h = ±1 γ(h) 1+θ12 +θ22 ρ(h) = = θ2 γ(0)   1+θ12 +θ22 h = ±2   0 |h| > 2 Exercise 12.7 An AR(2) process has the form Yt − µ = φ1 (Yt−1 − µ) + φ2 (Yt−2 − µ) + ǫt .

Part (a): Now given this process we can compute the autocovariance function as γ(k) = E[(Yt − µ)(Yt−k − µ)] = E[(φ1 (Yt−1 − µ) + φ2 (Yt−2 − µ) + ǫt )(Yt−k − µ)] = φ1 E[(Yt−1 − µ)(Yt−k − µ)] + φ2 E[(Yt−2 − µ)(Yt−k − µ)] + 0 = φ1 γ(k − 1) + φ2 γ(k − 2) .

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Note that the last term in the equation above the last is zero since ǫt and Yt−k are independent and E[ǫt ] = 0. If we now divide both sides by γ(0) we get ρ(k) = φ1 ρ(k − 1) + φ2 ρ(k − 2) , the expression we were trying to show.

Part (b): If we let k = 1 in the above equation then we get ρ(1) = φ1 (1) + φ2 ρ(−1) = φ1 + φ2 ρ(1) .

If we let k = 2 in the above equation we get ρ(2) = φ1 ρ(1) + φ2 (1) .

Putting these two equations together gives the system

φ1 1 ρ(1) ρ(1) . = φ2 ρ(1) 1 ρ(2) Thus if we think that a time series is generated from an AR(2) model we could measure ρ(1) and ρ(2) numerically and then invert the above system to compute values for φ1 and φ2 that would be estimates of the parameters in this AR(2) model.

Part (c): These values would give rise to the system

0.4 1 0.4 φ1 = , 0.2 0.4 1 φ2 which if we invert gives φ1 = 0.38095238 and φ2 = 0.04761905. Using these parameter estimates we can compute ρ(3) as ρ(3) = φ1 ρ(2) + φ2 ρ(1) = 0.38095238(0.2) + 0.04761905(0.4) = 0.0952381 .

Exercise 12.8 The left-hand-side of the book’s equation 12.12 is ! ∞ X ∞ ∞ ∞ X X X φi φj Cov(ǫt−i , ǫt+h−j ) .

Cov ǫt−i φi , ǫt+h−j φj = i=0

i=0 j=0

j=0

Now Cov(ǫt−i , ǫt+h−j ) = 0 unless t − i = t + h − j or j = h + i. Thus our double summation above simplifies to a single summation to give

∞ ∞ X X 1 2i 2 h i h+i 2 h . φ = σǫ φ φ φ Cov(ǫt−i , ǫt−i ) = σǫ φ 1 − φ2 i=0 i=0 Since the autocovariance function is symmetric in h we need to take the absolute value of h in the above to get φ|h| which gives the desired expression.

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Exercise 12.9 The expression for wt is wt = wt0 + Yt0 + Yt0 +1 + · · · + Yt ,

so that when we compute ∆wt we get

∆wt = wt0 + Yt0 + Yt0 +1 + · · · + Yt − (wt0 + Yt0 + Yt0 +1 + · · · + Yt−1 ) = Yt , as we were to show.

Exercise 12.10 Strictly stationary means that all aspects of its statistical behavior are unchanged by shifts in time. Weakly stationary means only that the mean, the variance, and the pairwise covariance are unchanged by shifts in time. A weakly stationary time series could have “higher order moments” change as the time origin changes.

Exercise 12.11 For a large enough lengths of time maybe. One could get some evidence if an I(2) model would fit a given time series of prices pt by looking at the results of the Ljung-Box test on the second difference sequence ∆2 pt .

Exercise 12.12 Let t = n + 1 and from the form of Yt we see that Yn+1 is given by Yn+1 = µ + ǫn+1 + θ1 ǫn + θ2 ǫn−1 , so from this we see that our prediction is given by Ŷn+1 = µ̂ + 0 + θ̂1 ǫ̂n + θ̂2 ǫ̂n−1 = 45 + 0.3(1.5) − 0.15(−4.3) = 46.095 .

For Yn+2 from the form of Yt we see that we have Yn+2 = µ + ǫn+2 + θ1 ǫn+1 + θ2 ǫn .

Then using this our prediction Ŷn+2 is given by Ŷn+2 = µ̂ + 0 + 0 + θ̂2 ǫ̂n = 45 − 0.15(1.5) = 44.775 .

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Exercise 12.13 For the given time series model with t = n + 1 we have Yn+1 = µ + φ1 Yn + ǫn+1 + θ1 ǫn + θ2 ǫn−1 , so our prediction of Yn+1 is given by Ŷn+1 = µ̂ + φ̂1 Yn + 0 + θ̂1 ǫ̂n + θ̂2 ǫ̂n−1 = 103 + 0.2(118.3) + 0.4(2.6) − 0.25(−2.3) = 128.275 .

Next using the given time series model with t = n + 2 we have Yn+2 = µ + φ1 Yn+1 + ǫn+2 + θ1 ǫn+1 + θ2 ǫn , so our prediction of Yn+2 is given by Ŷn+2 = µ̂ + φ̂1 Ŷn+1 + 0 + 0 + θ̂2 ǫ̂n = 103 + 0.2(128.275) − 0.25(2.6) = 128.005 .

Exercise 12.14 The autocorrelation plot of ∆yt shows a function that is effectively zero and thus the simplest model would be to model the time series of yt as ∆yt = ǫt (i.e. d = 1).

Exercise 12.15 Part (a): We plot the time series in Figure 52. There we see trending that indicates that we do not have a stationary time series. In Figure 53 (left) we plot the autocorrelation function of the raw time series. The very slow decay of the values indicates that we need to take a forward different of this time series to proceed to try to model it with an ARMA model.

In Figure 53 (right) we plot the autocorrelation function of the first difference of the time series. There it looks like we can model this first difference with a MA(1) model. Well it looks like there are several spikes in the autocorrelation plot that are above the 2σ horizontal lines and that these seem to be spaced periodically in time. Thus we would probably need a seasonal MA model to fully model this process.

Part (b): The output from the auto.arima command using the AIC option is > auto.arima( tb1, max.P=0, max.Q=0, ic="aic" ) Series: tb1 ARIMA(5,1,3)

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Figure 52: A plot of the tb1 time series.

Coefficients: ar1 ar2 0.7659 -0.7175 s.e. 0.0578 0.0690

ar3 0.6176 0.0697

ar4 0.2595 0.0576

ar5 -0.1117 0.0480

ma1 -0.9432 0.0387

ma2 0.9001 0.0513

ma3 -0.8921 0.0389

sigma^2 estimated as 0.0199: log likelihood=263.81 AIC=-509.62 AICc=-509.25 BIC=-471.87 This does not look like a very parsimonious model (given the large number of parameters that used in fitting it). The output from the auto.arima command using the BIC option is much cleaner > auto.arima( tb1, max.P=0, max.Q=0, ic="bic" ) Series: tb1 ARIMA(0,1,1) Coefficients: ma1 -0.1692 s.e.

0.0448 sigma^2 estimated as 0.02158:

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Right: A plot of the autocorrelation function of the first difference of the tb1 time series.

AIC=-484.95

AICc=-484.93

BIC=-476.57

Which is a much simpler model.

Part (c): We plot the autocorrelation function of the residuals of the model fit above using the AIC in Figure 54 (left). There we still see a significant spike at the integer lag 21 which might be problematic. The autocorrelation function of the model fit above using the BIC is given in Figure 54 (right). There we see several spikes that are outside of the 2σ horizontal lines. Both of these results indicate that we have not modeled this series completely and we should probably add a seasonal component.

Exercise 12.16 An AR(2) time series model looks like Yt = µ + φ1 (Yt−1 − µ) + φ2 (Yt−2 − µ) + ǫt , so letting t = n + 1 we get Yn+1 = µ + φ1 (Yn − µ) + φ2 (Yn−1 − µ) + ǫn+1 .

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Figure 54: Left: A plot of the autocorrelation function for the residuals of the tb1 time series when we model this as an ARIMA(5,1,3) series. Right: A plot of the autocorrelation function for the residuals of the tb1 time series when we model this as an ARIMA(0,1,1) series.

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Using this, our prediction of Yn+1 is then given by Ŷn+1 = µ̂ + φ̂1 (Yn − µ̂) + φ̂2 (Yn−1 − µ̂) + 0 = 100.1 + 0.5(102.3 − 100.1) + 0.1(99.5 − 100.1) = 101.14 .

Next letting t = n + 2 in our time series model we get Yn+2 = µ + φ1 (Yn+1 − µ) + φ2 (Yn − µ) + ǫn+2 .

Using this, our prediction of Yn+2 is then given by Ŷn+2 = µ̂ + φ̂1 (Ŷn+1 − µ̂) + φ̂2 (Yn − µ̂) + 0 = 100.1 + 0.5(101.14 − 100.1) + 0.1(102.3 − 100.1) = 100.84 .

Next letting t = n + 3 in our time series model we get Yn+3 = µ + φ1 (Yn+2 − µ) + φ2 (Yn+1 − µ) + ǫn+3 .

Using this, our prediction of Yn+3 is then given by Ŷn+3 = µ̂ + φ̂1 (Ŷn+2 − µ̂) + φ̂2 (Ŷn+1 − µ̂) + 0 = 100.1 + 0.5(100.84 − 100.1) + 0.1(101.14 − 100.1) = 100.574 .

Exercise 12.17 We will just make arguments along these if Yt has an mth degree polynomial Plines. First k β t .

Note that the expectation of the first time trend then we can write E(Yt ) = m k=0 k difference of Yt is then given by E(∆Yt ) =

m X

βk ∆tk .

k=0

Lets looks at what a few forward differences look like. Note that the forward difference of 1, t, and t2 are given by ∆1 = 1 − 1 = 0 ∆t = t − (t − 1) = 1 ∆t2 = t2 − (t − 1)2 = t2 − (t2 − 2t + 1) = 2t − 1 .

From these it looks like the first difference of a monomial is a polynomial of one less degree.

We can prove this in general using the binomial theorem as follows k X k l k k k k t (−1)k−l ∆t = t − (t − 1) = t − l l=0 !

k−1 k−1 X X k k l l k−l k k t (−1)k−l , =− t (−1) =t − t + l l l=0 l=0

which is a polynomial of degree k − 1 (one less than we started with). Thus for each forward difference we take we reduce the order of the polynomial time trend by one. If we take d differences we will end with a polynomial time trend of degree m − d. If d is larger than m then E(∆d Yt ) = 0.

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Figure 55: The time series plot of the consumption data.

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Chapter 13 (Time Series Models: Further Topics)

R Lab See the R script Rlab.R where the problem for this chapter are worked.

Problem 1 (seasonal ARIMA models) From the plot of consumption given in Figure 55 we see that the time series is not stationary and will need to take differences to make it stationary. In addition, noticing the periodic “up ticks” that seem to happen in the third quarter of every year imply that we need seasonal differencing to make this time series stationary. This also gives the indication that the order of the seasonal difference should be four.

In Figure 56 we plot the autocorrelation functions (ACF) for various differences of the consumption time series data.

• The leftmost plot is the ACF for the untransformed consumption data.

• The left-middle plot is the ACF for the first difference of the consumption data.

• The right-middle plot is the ACF for the seasonal difference (of order four) of the consumption data.

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Figure 56: Plots of the autocorrelation functions for the consumption time series.

• The rightmost plot is the ACF for the combined nonseasonal and seasonal difference (of order four) of the consumption data.

Taking a logarithmic transform of the data gives an ACF that look very much the same as these presented here. In each case it looks like one seasonal difference combined with one nonseasonal difference can be modeled with a seasonal moving average (MA) model with a single coefficient. Fitting this model to both the raw consumption time series and the logarithm of consumption we will compare which is a better fit by looking at the Ljung-Box test on each models residuals. When we do that we get > consumption_ts_model = arima( consumption, order=c(0,1,0), seasonal=list(order=c(0,1,1), period=4) ) > Box.test( residuals(consumption_ts_model), lag=10, type="Ljung" ) Box-Ljung test X-squared = 6.5425, df = 10, p-value = 0.7678 > log_consumption_ts_model = arima( lconsumption, order=c(0,1,0), seasonal=list(order=c(0,1,1), period=4) ) > Box.test( residuals(log_consumption_ts_model), lag=10, type="Ljung" ) Box-Ljung test X-squared = 4.1145, df = 10, p-value = 0.942

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Intuitively, larger P-values in the Box.text indicate that the time series looks like it is a sequence of independent events. Since the logarithm of consumption has a larger P-value we might argue that taking the logarithm results in a better fit since the model then produces more independent residuals.

Problem 2 (fitting a seasonal ARIMA model) Based on the above discussion, we propose a ARIMA{(0, 1, 0) × (0, 1, 1)4} model for the logarithm of consumption. Fitting such a model gives > arima( lconsumption, order=c(0,1,0), seasonal=list(order=c(0,1,1), period=4) ) ARIMA(0,1,0)(0,1,1)[4] Coefficients: sma1 -0.5348 s.e.

0.1164 sigma^2 estimated as 0.0002923: log likelihood=139.77 AIC=-275.55 AICc=-275.31 BIC=-271.61

Problem 3 (the residuals of the fit) A plot of the autocorrelation function of the residuals (not shown) for the model above show no statistically significant autocorrelations.

Problem 4 (using auto.arima) The model found by auto.arima is the same model as suggested from the acf plots. Running auto.arima gives > auto.arima( lconsumption, ic="bic" ) ARIMA(0,1,0)(0,1,1)[4] Coefficients: sma1 -0.5348 s.e.

0.1164 sigma^2 estimated as 0.0002923: log likelihood=139.77 AIC=-275.55 AICc=-275.31 BIC=-271.61 139

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Figure 57: Predictions for eight quarters ahead using two models for the consumption data set.

Problem 5 (forecasting) In Figure 57 I plot the predictions (the solid lines) and two-sigma error bars (the dashed lines) on these predictions for the next eight quarters. The red curves are for the consumption data directly and the green curves are for the logarithm model. Notice that the log model predicts larger numbers with a larger confidence interval.

Problems 6-8 (Regression with HAC Standard Errors) Running the given code we see that the plot of the autocorrelation function indicates that there is a significant autocorrelation at lag one and lag twelve (some other lags also appear significant).

The output from lm gives > print(round(summary(fit)$coef, 4)) Estimate Std. Error t value Pr(>|t|) (Intercept) 0.0000 0.0207 0.0001 0.9999 tb3_dif 0.9447 0.0370 25.5559 0.0000 140

while the output from the HC standard errors is > print(round(coef(fit)/sqrt(diag(NeweyWest(fit, lag=0, prewhite=F))), 4)) # the t-values (Intercept) tb3_dif 0.0001 11.7840

Note that the t-values of the slope are different but the conclusion of a significant slope is not changed (the t-values are either still very small or very large).

For the t-values at different lags I’m finding [1] 1 (Intercept) 0.0002 [1] 2 (Intercept) 0.0002 [1] 3 (Intercept) 0.0002

tb3_dif 12.4873 tb3_dif 12.8494 tb3_dif 13.0995

There is not much change in the t-values as a function of lag.

Problems 9-11 (Regression with ARMA Noise) When the above R code is run it computes the two requested AIC values [1] "AIC(arima fit)=

86.852; AIC(lm fit)=

138.882"

and also generates Figure 58. Note that both of these diagnostics indicate that the model that considers autocorrelation of residuals is the preferred model i.e. this would mean the model computed using the R command arima. I was not exactly sure how to compute the BIC directly but since it is related to the AIC (which the output of the arima command gives us) I will compute it using BIC = AIC + {log(n) − 2}p .

Using this we find [1] "BIC(arima fit)=

96.792; BIC(lm fit)=

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Figure 58: Left: Autocorrelation function for the residuals for the arima fit of the MacroDiff dataset. Right: Autocorrelation function for the residuals for the lm fit of the MacroDiff dataset. Note the significant autocorrelation present at lag one. which again specifies that the first model is better. We can fit several ARIMA models and compare the BIC of each. When we do that we find [1] "BIC(arima(2,0,0))=

99.532; BIC(arima(1,0,1))=

100.451; "

Thus we see that according to the BIC criterion the ARIMA(1,0,0) model is still the best.

Problem 12 (fitting VAR models) Part (a): For this part we have p = 3 and the matrix Φ1 is given by $ar , , 1 r y pi r -0.1455106 16.2063 -0.0981919 y 0.0005167 0.2923 0.0009493 pi 0.1820831 5.0277 -0.8057485 Estimates of Φ2 and Φ3 are given in the corresponding R code.

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Figure 59: The output from the acf on the residuals of the VAR(3) model fit to the time series in the Tbrate data set. Note that all auto and cross-correlations are quite small.

Part (b): The estimate of the covariance of the ǫt is given by $var.pred r y pi r 0.737581 2.087e-03 0.1086986 y 0.002087 6.764e-05 0.0004953 pi 0.108699 4.953e-04 0.4617927 Part (c): The output from the acf function is given in Figure 59. In that plot we see very few spikes above the two sigma standard error lines and the few that are could be due to statistical fluctuations and not real effects.

Problem 13-18 (predicting with VAR models) To predict with the model that we extracted from the ar function we need to convert the given data into a ts object to pass to the predict function. I’ll assume that the time stamps to associate with the new data occur after the given training data set. Running the predict function on this sample gives the following

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$pred r y pi 2014 Q1 0.004251706 0.007801101 -0.0027438936 2014 Q2 -0.003960056 0.007652677 0.0002221654 2014 Q3 -0.005299672 0.007601372 -0.0027603079 2014 Q4 -0.005283604 0.007582847 -0.0015794426 2015 Q1 -0.005726351 0.007578012 -0.0021177241 for the next five predictions. We then compute the ratio of these predictions to var1$x.mean doing this we find r y pi 2014 Q1 -0.7456349 1.029853 1.3725708 2014 Q2 0.6944874 1.010259 -0.1111332 2014 Q3 0.9294200 1.003486 1.3807817 2014 Q4 0.9266021 1.001040 0.7900805 2015 Q1 1.0042481 1.000402 1.0593436 Notice that the different components seem to revert to the value of var1$x.mean (i.e. they all end with a ratio value near one).

Printing the matrix Phi hat we get r y pi r -0.132891458 12.9957312 -0.1692522981 y 0.001047693 0.2994738 0.0005891462 pi 0.209316080 -0.9506641 -0.4735984075 The signs of the above numbers indicate which direction a change in r, y, and π will make to the next change in a variable. For example from the first row in the above matrix we see that a negative change in r, a positive change in y, and a negative change in π will give rise to a predicted positive change in r. The same can be said for the other rows. The second row for example will predict a positive change in y if the change in all three variables is positive.

Printing the eigenvalues, I get that each one is around on third in value. This process is then stationary and will have sequences that revert to the mean (as seen in a previous part of the problem).

Running the ar code with aic=TRUE I’m finding the code selects a tenth order model! This certainly does not seem very parsimonious. In the R code associated with this lab I then plot the AIC and the BIC from this fit. Looking at this plot it looks like the AIC is minimized at the 10th lag(the AIC at p = 5 is the next smallest) and the BIC is minimized at the fifth lag. Given this I would think that we should fit a fifth order model (and not a 10th order model).

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Figure 60: Left: A plot of the time series DiffSqrtCpi. Right: A plot of the autocorrelation function for the time series DiffSqrtCpi. Problem 19 (long term memory DiffSqrtCpi) A plot of the time series and the autocorrelation function (ACF) for the suggested transformations of the Consumer Price Index (CPI) data is given in Figure 60. Notice that the ACF decays very slowly and its value is not strictly decreasing (there are periods where it increases as we move from left to right). This is characteristic of perhaps an AR(p) model or the need for fractional differencing (i.e. a long term memory process).

Problem 20 (fractional differencing) Applying the code fracdiff to estimate the amount of fractional differencing we get d = 0.4134047 applying this amount of differencing and then plotting the ACF we get Figure 61.

Notice that there are no significant spikes above the two sigma horizontal lines indicating that we have found a reasonable fit.

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Problem 21 (ARIMA model on fdiff) If we run auto.arima on the time series fdiff we see that this routine will choose to model this with an ARIMA(0,1,1) model. > auto.arima(fdiff,max.P=0,max.Q=0,ic="aic") Series: fdiff ARIMA(0,1,1) Coefficients: ma1 -0.9857 s.e.

0.0093 sigma^2 estimated as 0.0001029: log likelihood=1549.4 AIC=-3094.79 AICc=-3094.77 BIC=-3086.41 Using BIC does not give a different result. This is a bit strange since looking at the ACF of the raw time series it looks like there are no significant autocorrelations. If we look at the ACF for the first difference of the fdiff time series we do indeed see a very significant spike at lag one. What is strange is that forcing auto.arima to not consider any differences we get a result that has lower AIC and BIC metrics and should be the preferable model. > auto.arima(fdiff,d=0,max.P=0,max.Q=0,ic="aic") Series: fdiff ARIMA(0,0,0) with zero mean sigma^2 estimated as 0.0001028: log likelihood=1554.56 AIC=-3107.11 AICc=-3107.1 BIC=-3102.92 This is a case of spurious differencing (more differencing than is needed). Given these results it seems prudent to model fdiff using the simplest model (that of no additional differencing).

Problem 22 (selecting the correct ARIMA model) Running the given code gives an ARIMA(2,1,0) model (as also stated in the text) > auto.arima(price,ic="bic") Series: price ARIMA(2,1,0) 146

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Coefficients: ar1 ar2 0.2825 0.0570 s.e. 0.0407 0.0408 sigma^2 estimated as 9.989: AIC=3146.23 AICc=3146.27

log likelihood=-1570.11 BIC=3159.47

Running the given bootstrap code in the text we find that of the 10 runs we only get an ARIMA(2,1,0) once. Even in that single case the estimate of φ2 is quite different than that used to generate the data (0.1759 vs. 0.0570). Thus it is safe to say that none of the bootstrap samples found the correct model.

Problem 23 (estimating the correct parameters) We next want to determine how well we can estimate the given AR(2) parameters under the assumption that we know the correct model i.e. that our data is coming from an ARIMA(2,1,0) model. When we run the suggested R code from the book, for the biases, standard deviations, and MSE we get

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[,1] [,2] bias 0.0001361095 0.005447920 std_devs 0.0427685740 0.044955387 mses 0.0027049221 0.002843228 The first column has statistics of the estimate for φ1 while the second column holds statistics for the estimates of φ2 . We see that the bias in the estimate of φ2 is larger than that in the estimate of φ1 . We commented on how different the estimate of φ2 was in the single case where auto.arima estimated the correct model in the previous problem. Note that the standard error of the two estimate is about the same. This leads one to conclude that the estimate of φ2 (because the bias is so large) will in general not be correct.

Exercises See the R script chap 13.R for the implementation of the exercises for this chapter.

Exercise 13.1 From the given picture is is clear that we need both a seasonal and a nonseasonal difference to make this data stationary. Both differences gives autocorrelations with only a few large spikes beyond the two-sigma error lines.

Exercise 13.2 In this case the seasonal difference gives an ACF that has only a few (two) spikes beyond the horizontal two-sigma lines. In this case we could probably fit the seasonal difference of the time series using a MA(2) model.

Exercise 13.3 In this case a single nonseasonal difference gives an ACF that has only a few (two) spikes beyond the two-sigma lines. Thus we could model the first nonseasonal difference using a MA(2) model.

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Figure 62: A plot of the income time series.

Exercise 13.4 A two-dimensional AR(1) model for the variables ∆CPI and ∆IP is written in the form 1

∆CPIt−1 µ1 ǫt ∆CPIt µ1 . = Φ1 − + − ǫ2t µ2 ∆IPt−1 µ2 ∆IPt Using the given numbers in the book (in gory detail) our prediction of the vector at t + 1 becomes

∆CPIt+1 ∆IPt+1

=

0.00518 0.00215

+

0.767 0.0112 −0.33 0.3014

0.00173 0.00591

−

0.00518 0.00215

=

0.002575962 0.004421764

.

The next prediction is done the same way but using the prediction we just made for the state vector. That is

∆CPIt+2 ∆IPt+2

=

0.00518 0.00215

+

0.767 0.0112 −0.33 0.3014

0.002575962 0.004421764

−

0.00518 0.00215

=

0.003208147 0.003694042

.

If we compare this with the results from figure 13.8 in the book we see that the numbers computed here agree with the ones plotted on that graph.

Exercise 13.5 In the R code chap 13.R we first plot this time series. This plot can be seen in Figure 62.

From that plot we see a steady increase in the value of income. Next we look at the ACF 149

of various differences of this time series. In Figure 63 we plot several of these (for various amounts of differencing). From that plot it looks like either a single seasonal difference or a seasonal difference combined with a single nonseasonal difference would be the two models to consider.

• If we consider the single seasonal difference it looks like the resulting time series would best be modeled with an AR(1) model or at least the ACF has a decaying that looks like it might be exponential (consistent with AR models).

• If we consider the combined seasonal and nonseasonal differencing then the resulting series would best be modeled perhaps with a MA(2) model (or no additional model beyond differencing at all).

From these plots it looks like the most parsimonious model (to me) is given by seasonal differencing followed by nonseasonal differencing. It is interesting to compare this hypothesis with what the auto.arima would give. One run of that function gives > auto.arima( income, ic="aic" ) Series: income ARIMA(2,1,2)(0,1,0)[4] Coefficients: ar1 -0.1315 s.e.

0.2091

ar2 -0.3016 0.1571

ma1 0.0743 0.1338

ma2 0.9129 0.1183

sigma^2 estimated as 334423: log likelihood=-413.54 AIC=837.08 AICc=838.36 BIC=846.93 Which arrives at what seems like a relatively complicated model. This model is to be contrasted with the one we get when we take the ic option to the auto.arima function to the value bic. In that case we get > auto.arima( income, ic="bic" ) Series: income ARIMA(0,1,0)(0,1,0)[4] sigma^2 estimated as 486420: log likelihood=-422.22 AIC=846.43 AICc=846.51 BIC=848.4 This states that no other modeling (other than taking the differences) is needed. This result seems to have only a marginal increase in the AIC and BIC metrics but the benefit of being quite a bit simpler. It is also the model we argued for above based on the autocorrelation plots.

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Figure 63: A plot of the ACF for various differences of the income time series. From left to right we have (in terms of the amount of differencing): none, nonseasonal, seasonal (4), a seasonal (4) and a nonseasonal difference. Notice that all of the spikes in the fourth plot are within the two sigma horizontal error bars indicating that no additional modeling (other than taking these differences) maybe needed.

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Exercise 13.6 Part (a): For this time series each of the differenced autocorrelation function does not look uniformly “flat”. This makes modeling this time series difficult since it does not seem to fit nicely into a particular case. Rather than puzzle over the ACF functions in an attempt to figure out what model fits best we will use a data driven approach and let the auto.arima function in the forecast package give us an estimated model. Running this command on the unemp data gives > auto.arima(unemp,ic="aic") Series: unemp ARIMA(1,1,0)(0,0,2)[4] Coefficients: ar1 sma1 0.6611 -0.4199 s.e. 0.0544 0.0675

sma2 -0.2623 0.0660

sigma^2 estimated as 0.07992: log likelihood=-32.85 AIC=73.69 AICc=73.89 BIC=86.94 As predicted, this gives a somewhat “mixed” model in that it includes effects from several different sources (both seasonal and nonseasonal). We can verify that the residuals show no autocorrelation when using this model by calling the acf on the model residuals.

Part (b): For this part we will use the simulate.Arima function from the forecast library to generate Monte-Carlo samples of the seasonal ARIMA model we fit above. The model we found above (in R notation) is given by ARIMA(1,1,0)(0,0,2)[4] Looking at the eight bootstrap samples we see that the correct values of d and D where selected in none of the samples. To study this with more Monte-Carlo runs I considered 130 samples (rather than eight). In this case the results were not much better. In all 100 cases the correct model was never selected.

Exercise 13.7 For the problem mentioned if we look at Figure 59 (left) we see the ACF plots for all the variables in the Tbrate dataframe. From that plot there seem to be a few spikes that look like they might be significant at spaced at seasonal lags. See for example the variable r which looks like it might have a spike downwards at around lag “two”. Most of these 152

spikes are just on the border between significant and non-significant and as such unless there is strong compelling evidence to include seasonal effects the model is more parsimonious without them. As an example of this in a previous chapter, we forecasted the variable pi (the inflation rate) using a nonseasonal ARIMA (see Page 123) where use of auto.arima (restricted to nonseasonal models) gave a ARIMA(1,1,1) model with AIC = 772.24 and BIC = 781.94. Running auto.arima unrestricted with the two options for ic gives (results have been truncated for clarity) > auto.arima( pi, ic="aic" ) ARIMA(1,1,1)(2,0,1)[4] sigma^2 estimated as 3.432: log likelihood=-380.97 AIC=773.95 AICc=774.41 BIC=793.33 > auto.arima( pi, ic="bic" ) ARIMA(1,1,1)(2,0,0)[4] sigma^2 estimated as 3.438: log likelihood=-381.13 AIC=772.26 AICc=772.59 BIC=788.41 In this case the nonseasonal model gives smaller values for AIC and BIC and is the more parsimonious model. Thus it seems prudent to not include seasonal effects in modeling pi.

One could do a similar study for the other two variables.

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Figure 64: The Tbill time series and ACF plot.

13

Chapter 14 (GARCH Models)

R Lab See the R script Rlab.R where the problem for this chapter are worked.

Problem 1 In Figure 64 we plot the time series and the ACF for the Tbill data. In that plot we do not see a stationary time series. The ADF and KPSS test for this series give Augmented Dickey-Fuller Test data: Tbill Dickey-Fuller = -2.5622, Lag order = 6, p-value = 0.3395 alternative hypothesis: stationary KPSS Test for Level Stationarity data: Tbill KPSS Level = 1.3192, Truncation lag parameter = 3, p-value = 0.01 154

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Figure 65: The time series and ACF plot of the first difference of the Tbill data. with a warning that the P-value is smaller than the printed 0.01. In the ADF test the null hypothesis is that the time series has a unit root thus a “large” P-value indicates that it does in fact have one. In the KPSS test has a null hypothesis of stationarity (level in the default case) so a small P-value indicates that there is “alot” of evidence that the time series is in fact not stationary. Thus from each of these tests we conclude that our time series is not stationary.

Now we do the same thing with the first difference of Tbill. In Figure 65 we plot the time series and the ACF for the first difference of theTbill data. In that plot we see that this series looks more stationary. There are still some spikes at larger lags in the ACF but the series is much more stationary. Results for the ADF and KPSS test for this series give Augmented Dickey-Fuller Test data: Del.Tbill Dickey-Fuller = -6.3425, Lag order = 6, p-value = 0.01 alternative hypothesis: stationary KPSS Test for Level Stationarity data: Del.Tbill KPSS Level = 0.15565, Truncation lag parameter = 3, p-value = 0.1 with a warning from the ADF test that the P-value is smaller than the printed 0.01 and a 155

warning from the KPSS test that the P-value is larger than 0.1. Following the discussion above if a time series is stationary then the P-value from the ADF test should be “small” (yes) and the P-value from the KPSS test should be “large” (yes). Thus both of these tests are telling us that the differenced time series is stationary.

Finally, from Figure 65 (left) we see osculations of different magnitudes especially large osculations around 1980 and smaller ones around 1965. This change in the magnitude of the series is an observation of heteroscedasticity.

Problem 2 Part (a-b): In the previous problem we argued that Del.Tbill is stationary and should be further modeled with an ARMA/GARCH model. From the ugarchspec parameters specified in the R code we are using a AR(1) model for the conditional mean or yt − µ = φ(yt−1 − µ) + at , and a GARCH(1,1) model for the “noise” term at of at = σt ǫt q 2 . σt = ω + α1 a2t−1 + β1 σt−1

When we run the given R code and look at the output (partial) from the show command we get Optimal Parameters -----------------------------------Estimate Std. Error t value Pr(>|t|) mu 0.018556 0.032524 0.57054 0.568312 ar1 0.139618 0.069092 2.02074 0.043306 omega 0.016961 0.010715 1.58296 0.113430 alpha1 0.383655 0.065991 5.81378 0.000000 beta1 0.615345 0.041021 15.00089 0.000000 The numerical values of the model parameters are given in the “estimate” column. Thus the model fit is yt − 0.018556 = 0.139618(yt−1 − 0.018556) + at , where at is modeled as

at = σt ǫt q 2 σt = 0.016961 + 0.383655a2t−1 + 0.615345σt−1 .

Here ǫt ∼ N(0, 1). In fitting the first model we see some model missmatch. The model fit to the logarithm fits much better.

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Figure 66: Autocorrelation plots for the residuals and the residual squared for the Del.Tbill time series. The two ACF for the standardized residuals look like the model is fitting well.

Problem 3 Part (a): See the upper middle plot in Figure 66. There we see that the model for the mean of the Del.Tbill series is modeled well.

Part (b): See the upper right most plot in Figure 66. There we see that the model for the variance of the time series has some spikes above the horizontal 2σ lines that indicate that there is conditional heteroscedasity and the need for a GARCH model. Since we are actually fitting a GARCH model we should look at the standardized residuals to see how well our model is fitting.

Part (c): See the bottom right most plot in Figure 66. There we see that the model for the standardized residuals squared has no significant spikes indicating that our GARCH model is fitting well.

Part (d): See the bottom left most plot in Figure 66. There we see an almost uniform noise like pattern with no noticeable volatility clustering. This is an indicator that our model is fitting well. There do seem to be very large moves which indicate that maybe the noise has heavier tails than Gaussian.

157

Problem 4 For this problem we try to fit an ARMA/GARCH model to the series Del.Log.Tbill defined as in the text. We perform the same model fitting (using ugarchfit) that we did for Del.Tbill. For comparison, the show command on the fitted model of Del.Tbill has the following partial results Information Criteria -----------------------------------Akaike 1.9475 Bayes 2.0211 Shibata 1.9466 Hannan-Quinn 1.9772 the same thing for the GARCH model of the difference of the logarithm gives Information Criteria -----------------------------------Akaike -0.40010 Bayes -0.32649 Shibata -0.40098 Hannan-Quinn -0.37042 In minimizing any information criterion metric, the model of the difference in the logarithm is a much better fit. This can also be seen in the residual plots for this model (see Figure 67).

In particular these are much nicer plots than when we modeled Del.Tbill with a GARCH process. See Figure 66 where the ACF for the residuals squared still have some significant spikes that are not found in Figure 67.

Problem 5 From the output of the show command on the fitted model we have Optimal Parameters -----------------------------------Estimate Std. Error t value Pr(>|t|) mu -0.102829 0.030827 -3.3357 0.000851 archm 2.096790 0.598834 3.5015 0.000463 omega 0.000963 0.000514 1.8733 0.061033 alpha1 0.000000 0.044867 0.0000 1.000000 beta1 0.613558 0.215919 2.8416 0.004489 158

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Figure 67: Plots of standardized residuals, the ACF of the standardized residuals, and the ACF of the standardized residuals squared when fitting an AR(1)/GARCH(1,1) model to the first difference in the logarithm of Tbill. This is the best fitting model for this time series seen thus far.

159

This means that the model we have found is explicitly given by Yt = −0.102829 + 2.096784σt + at at = σt ǫt 2 σt2 = 0.000963 + 0.613525σt−1 .

Here we are assuming α = 0 (as indicated in the “Optimal Parameters” output above) in the model.

Problem 6 From the output of the show command (above) we see that the P-value for the two-sided hypothesis on δ is 0.000463 which indicates that the one sided P-value would be half off this value or 0.0002315. This is small enough that we can reasonably conclude that δ > 0.

Problem 7-10 Running the mLjungBox command we find K Q(K) d.f. p-value 1 8 132.57 32 0 this indicates that there is significant series correlation in the first K = 8 lags indicating the series exhibits conditional heteroscedasticity.

The estimated value of λ is given by 0.8665812. Notice that the ts plots have large values for σ̂1 and σ̂2 at the same points in time where the two original time series (from the plot(TbGdpPi) command) have their largest variations. We also see a nonzero estimate for σ̂12 .

Computing the residuals under the EWMA model and then running the mLjungBox command on their square we get K Q(K) d.f. p-value 1 8 62.07 32 0.001 The small P-value indicates that the EWMA model is not adequate as there still seems to be conditional heteroscedasticity in the residuals.

Running the DOC.test command we get 160

> DOC.test(TbGdpPi.diff^2, 8) m Q(m) d.f. p-value 1 0 0.955 1 0.328 2 1 3.285 3 0.350 3 2 3.882 5 0.566 4 3 3.986 7 0.781 5 4 4.523 9 0.874 6 5 4.573 11 0.950 7 6 6.302 13 0.934 8 7 6.343 15 0.973 9 8 7.447 17 0.977 In the section of the book on “Dynamic Orthogonal Component (DOC) Models” we see that this test computes the statistic Q0d (s2 ; K) and the null hypothesis consists of the fact that the correlation of s2i and s2j−h is zero for all i 6= j and h = 0, . . . , K. The large P-values indicate that there is not enough evidence to reject the null and we should conclude that there is no significant autocorrelation between the components of s2i and s2j−h . This means that we don’t need to model the cross correlations as is done with the EWMA model and we should model the conditional heteroscedasticity in this pairwise time series separately.

We could then fit GARCH models to each time series individually to remove the conditional heteroscedasticity.

Exercises See the R script chap 14.R where the exercises for this chapter are worked.

Exercise 14.1 Following the hint we have r Z ∞ r

Z ∞ 2 2 2 −z 2 /2 ∞ d −z 2 /2 −z 2 /2 ze dz = E(|Z|) = √ dz = − e − e 0 π 0 dz π 2π 0 r r 2 2 (0 − 1) = , =− π π which is the expression we were to show.

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Exercise 14.2 To show fX (x) integrates to one we have Z +1 Z ∞ Z ∞ Z −1 1 dx dx + dx + fX (x)dx = 2 2 −1 4 +1 4x −∞ −∞ 4x 1 1 1 −1 ∞ = −x−1 −∞ + (2) + −x−1 +1 4 4 4 1 1 1 = − (−1 − 0) + − (0 − 1) = 1 , 4 2 4 as we were to show. So show that X does not have an expectation consider the integral over negative x which is given by Z 0 Z 0 Z −1 x dx + dx xfX (x)dx = −1 4 −∞ −∞ 4x 0

1 1 x2 = log(|x|)|−1 + −∞ 4 4 2 −1 1 1 = (log(1) − log(+∞)) + (0 − 1) = −∞ .

4 8

Next the integral over positive x which is given by Z +∞ Z +∞ Z 1 dx x dx + xfX (x)dx = 4x 1 0 0 4 +1

1 x2 1 + log(|x|)|∞ 1 4 2 0 4 1 1 = (1 − 0) + (log(| + ∞|) − 0) = +∞ .

8 4

=

Exercise 14.3 This is an AR(1)/ARCH(1) model i.e. it takes the form yt − µ = φ(yt−1 − µ) + at , with at = σt ǫt =

q

(47)

ω + α1 a2t−1 ǫt .

Part (a): Taking the expectation of the given expression with the numbers as given we have µ = 3 + 0.72µ or µ = 10.71429 .

Part (b): The unconditional variance of yt is given by γu (0) =

1 ω = = 1.538462 .

1 − α1 1 − 0.35 162

The conditional variance of yt is given by σt2 = 1 + 0.35a2t−1 .

Part (c): This is the same as for an AR(1) model or ρu (h) = φ|h| = 0.72|h| for all h. |h|

Part (d): The process a2t has an ACF given by ρa2 (h) = α1 = 0.35|h| for all h.

Exercise 14.4 Part (a): As E(at ) = 0 for all t this is given by E(u2 |u1 = 1, u0 = 0.2) = µ + φ(u1 − µ) + 0 = 0.4 + 0.45(1 − 0.4) = 0.67 .

Part (b): Evaluate the model for the mean (given by Equation 47) at t = 1 where we get u1 − µ = φ(u0 − µ) + a1 .

We know everything in the above except a1 . Putting in everything we know and solving we get a1 = 0.69. For the variance we then get σt2 = ω + α1 a2t−1

so σ22 = ω + α1 a21 = 1 + 0.3(0.692) = 1.14283 .

Exercise 14.5 Note that this is an AR(1)/ARCH(1) model for Yt with φ = 0.67, ω = 7, and α1 = 0.5.

Part (a): Taking the expectation of the given expression for Yt we get µ = 2 + 0.67µ so µ = 6.060606 .

Part (b): Since Yt is a AR(1)/ARCH(1) model the ACF for Yt is the same as an AR(1) model or ρY (h) = 0.67|h| , for all h.

Part (c): This is discussed in the section entitled “ARCH(1) Processes” where it is stated that ρa (h) = 0 for all h.

Part (d): This is discussed in the section entitled “ARCH(1) Processes” where it is stated |h| that ρa2 (h) = α1 = 0.5|h| for all h.

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Exercise 14.6 Part (a): For the models of the type we have been discussion in this chapter we always take at = σt ǫt so for the given expression here this means that q σt = 1 + 0.5a2t−1 .

p When at−1 = 0.6 we can evaluate the above to get σt = 1 + 0.5(0.62 ) = 1.086278. Now E(at ) = 0 so we know all of the terms in the expectation we are to compute. We have E(Yt |Xt−1 = 0.1 and at−1 = 0.6) = β0 + β1 (0.1) + δ(1.086278) + 0 = 0.06 + 0.35(0.1) + 0.22(1.086278) = 0.3339812 .

Part (b): For this we have Var(Yt |Xt = 0.1 and at−1 = 0.6) = Var(at |Xt = 0.1 and at−1 = 0.6) = 1 + 0.5(0.62 ) = 1.18 .

Part (c): Yes because once Xt and at−1 are known we can compute σt exactly (using the above formula). Thus all of the terms β0 + β1 Xt + δσt are known exactly. The noise term at takes the form at = σt ǫt which is a multiple of a normal random variable and is therefore itself a normal random variable. Adding all of these pieces together we get that the conditional distribution of Yt is normal.

Part (d): I would say not since unconditionally we have the model q Yt = β0 + β1 Xt + δ 1 + 0.5a2t−1 + at ,

which has two random variables at−1 and at added together in a nonlinear way.

Exercise 14.7 Part (a): at is a ARCH(1) model.

Part (b): ut is a AR(1)/ARCH(1) model.

Part (c): at is not Gaussian and has heavier tails than a Gaussian distribution.

Part (d): The ACF of at is φa (h) = 0 for all h. |h|

Part (e): The ACF of a2t is φa2 (h) = α1 = 0.3|h| for all h.

Part (f): The ACF of ut is φu (h) = φ|h| = 0.6|h for all h.

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Figure 68: Daily returns leading up to (and including) black Monday.

Exercise 14.8 We run the given code and plot the daily returns in Figure 68. Notice the very large negative return on the right side of the graph.

Part (a): The output from the ugarchforecast command is 0-roll forecast [T0=1971-05-21 20:00:00]: Series Sigma T+1 -0.003325 0.01203 I then compute the “z-score” for the return observed on black Monday −0.2280063 − (−0.003324955) = −18.68138 .

0.01202702 Then I use this value and the pt function to compute the probability we would get a sample this small (or smaller) from a t-distribution with the estimated degrees of freedom (which was 3.304503). Doing this I get the probability of 9.115669 10−5.

Part (b): We plot the standardized residuals in Figure 69. There we see that the suggested AR(1)/GARCH(1,1) model appears to fit quite well.

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Figure 69: Plots of standardized residuals, the ACF of the standardized residuals, and the ACF of the standardized residuals squared.

Part (c): We can refit with the suggested AR(1)+ARCH(1) model and then look at diagnostic plots (like residuals plots). Doing this we get plots that look similar to the ones in the previous part and show no problems with their fit. To compare the two models in the first case show(fit) we get (partial) Information Criteria -----------------------------------Akaike -6.5205 Bayes -6.4703 Shibata -6.5207 Hannan-Quinn -6.5008 For the AR(1)/ARCH(1) model considered in this part the same pieces from the show(results c) command look like Information Criteria -----------------------------------Akaike

-6.5177 166

Bayes -6.4759 Shibata -6.5179 Hannan-Quinn -6.5013 Since there are 506 points in the data vector x and the information criterion numbers above are normalized by this length we need to multiply by 506 to get the absolute information criterion numbers. For the first models the AIC is -3299.373 while for the second model the AIC is -3297.956. The better model will be the one that minimizes the AIC and thus in this case we would select the first model (although the difference is very slight).

Part (d): The (partial) output from the arima function is sigma^2 estimated as 9.451e-05:

log likelihood = 1626.52,

aic = -3247.03

The value of the AIC here (-3247.03) is larger than that from either of the two models above.

This gives an indication that we would prefer one of the previous two models. In Figure 70 we plot the standardized residuals and the ACF’s for the residuals and the residuals squared.

There we see that the ACF of the squared residuals has a second lag that is significant and the the residuals seem to show some volatility clustering. These are indications that the pure AR(1) model is not fully adequate.

Exercise 14.9 Part (a): We are fitting an AR(1)/GARCH(1,1) model to the differences in rt . A model of this type has (defining ut ≡ ∆rt ) the following form ut − µ = φ1 (ut−1 − µ) + at , and a GARCH(1,1) model for the “noise” term at of at = σt ǫt q 2 σt = ω + α1 a2t−1 + β1 σt−1 .

Part (b): The output from the show command gives

Optimal Parameters -----------------------------------Estimate Std. Error t value Pr(>|t|) mu 0.010084 0.002587 3.8979 0.000097 ar1 0.095981 0.043276 2.2179 0.026564 omega 0.000503 0.000236 2.1312 0.033076 alpha1 0.254352 0.080236 3.1701 0.001524 beta1 0.744648 0.063585 11.7111 0.000000 shape 3.102618 0.404827 7.6640 0.000000 167

0

100

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0.8 0.6 ACF 0.4 0.2 0.0

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Series res_std^2

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10

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Figure 70: Plots of standardized residuals, the ACF of the standardized residuals, and the ACF of the standardized residuals squared when fitting an AR(1) model. for the estimated model parameters.

Part (c): Based on the AR(1)/GARCH(1,1) model specified the ACF for ∆rt is the same |h| as for an AR(1) model so ρ∆r (h) = φ1 for all h.

Part (d): Based on the AR(1)/GARCH(1,1) model specified the ACF for at will be zero.

Part (e): Based on the AR(1)/GARCH(1,1) model specified the ACF for a2t will be the same as that discussed in the text namely α1 (1 − α1 β1 − β12 ) 1 − 2α1 β1 − β12 ρa2 (h) = (α1 + β1 )h−1 ρa2 (1) for h ≥ 2 . ρa2 (1) =

We would put in the estimated values for α1 and β1 in the above.

Exercise 14.10 Part (a): Using the acf function we see a few autocorrelations outside of the two sigma bounds. Running the Box-Ljung test on the residuals gives 168

Box-Ljung test data: y X-squared = 52.272, df = 6, p-value = 1.644e-09 which indicates that there is significant autocorrelation.

Part (b): The same two things on the residual squared we see that there are ARCH effects also.

Part (c): For this part of the problem, the model we will fit is yt − µ = φ1 (yt−1 − µ) + at at = σt ǫt σt2 = ω + α1 a2t−1 .

In the notation from the book this means that we will fit an ARMA(1, 0) + GARCH(0, 1) model to the log returns of the S&P 500 index. The parameters for this model are found using the show command on the output of the ugarchfit command and found to be Optimal Parameters -----------------------------------Estimate Std. Error t value Pr(>|t|) mu 0.000567 0.000254 2.2335 0.025513 ar1 -0.073614 0.025509 -2.8859 0.003903 omega 0.000000 0.000000 1.3318 0.182938 beta1 0.999000 0.000012 82771.2814 0.000000 Note that the AR(1) coefficient φ1 is estimated negative which is the correct sign given the acf plot.

Part (d): Our fitted ARCH model will be stationary if β1 < 1 which is barley true (assuming the estimate above is correct).

Part (e): This model is specified in Problem 2 above with the same determination of the meaning of the parameters. Using ugarchfit I find estimates of the parameters given by Optimal Parameters -----------------------------------Estimate Std. Error t value Pr(>|t|) mu 0.000808 0.000202 4.0094 0.000061 ar1 -0.038525 0.028270 -1.3627 0.172966 omega 0.000003 0.000001 1.8534 0.063827 alpha1 0.121811 0.017929 6.7941 0.000000 beta1 0.857884 0.018593 46.1409 0.000000 169

Part (f): The Box.test gives Box-Ljung test data: res_std X-squared = 16.501, df = 20, p-value = 0.6851 for the residuals and Box-Ljung test data: res_std^2 X-squared = 25.354, df = 20, p-value = 0.1882 for the residuals squared. The first result indicates that there is no autocorrelation of the residuals (since the P-value is relatively large). The second result (with the much smaller Pvalue) indicates that there may be autocorrelation of the residuals squared and that perhaps we have not fully modeled the variance of this process.

Part (g): For our GARCH process to be stationary we must have α1 + β1 < 1. Using the estimates above we have α1 + β1 = 0.121811 + 0.857884 = 0.979695 < 1 . which would be stationary.

Part (h): Running the R code for this problem gives the QQ-plot. A Gaussian distribution does not look like the appropriate model for the residuals as we have heavy tails.

Part (i): Looking at a plot of σ̂t we see a process that does not look stationary.

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14

Chapter 15 (Cointegration)

R Lab See the R script Rlab.R where the problem for this chapter are worked.

Problem 1 The values of the test statistic output from the ca.jo command is given by test 10pct 5pct 1pct r <= 9 | 0.60 6.5 8.18 11.7 r <= 8 | 6.31 12.9 14.90 19.2 r <= 7 | 9.20 18.9 21.07 25.8 r <= 6 | 11.79 24.8 27.14 32.1 r <= 5 | 15.32 30.8 33.32 38.8 r <= 4 | 19.89 36.2 39.43 44.6 r <= 3 | 21.65 42.1 44.91 51.3 r <= 2 | 22.98 48.4 51.07 57.1 r <= 1 | 35.62 54.0 57.00 63.4 r = 0 | 58.39 59.0 62.42 68.6 To read the output from this command we start at the top and can reject a hypothesis r ≤ d if the value in the “test” column is larger than the various confidence given in the 10%, 5%, and 1% columns. For this example none of the values in the “test” column is larger than the other values in the same row. Thus we can not reject the hypothesis that r ≤ d for each of the d’s given. We are forced to conclude that r = 0 or that there are no cointegration vectors.

Problem 2 The maturities for the bonds considered is stored in the mk.maturity dataframe. Looking at the maturities of the bonds we are considering (in months) for cointegration we find > print(mk.maturity$Maturity[2:11] * 12) [1] 2 3 4 5 6 7 8 9 10 11 Thus these are short-term maturities.

171

Problem 3 The values of the test statistic output from the ca.jo command on this data is given by test 10pct 5pct 1pct r <= 9 | 1.43 6.5 8.18 11.7 r <= 8 | 8.03 12.9 14.90 19.2 r <= 7 | 18.48 18.9 21.07 25.8 r <= 6 | 27.87 24.8 27.14 32.1 r <= 5 | 36.12 30.8 33.32 38.8 r <= 4 | 40.96 36.2 39.43 44.6 r <= 3 | 51.87 42.1 44.91 51.3 r <= 2 | 59.25 48.4 51.07 57.1 r <= 1 | 70.71 54.0 57.00 63.4 r = 0 | 90.57 59.0 62.42 68.6 Looking at the 1% level we can reject r ≤ 3 (since 51.87 > 51.3) but not r ≤ 4 (since 40.96 6> 44.6). Thus we must conclude that r = 4. Thus we have four cointegration relationships.

Problem 4-6 (Cointegration Analysis of Coke and Pepsi) The time series plots do seem to move together but there are periods where the moves are quite different in magnitude. A plot of the difference appears more stationary in that it seems to oscillate back and forth around a constant mean value. Output from the ca.jo command is test 10pct 5pct 1pct r <= 1 | 0.69 6.5 8.18 11.7 r = 0 | 8.48 12.9 14.90 19.2 We cannot reject any rows and must conclude that these two time series are not stationary.

Problem 7-8 (Cointegration Analysis of Daily Stock Prices) The output from the ca.jo command is

r <= 9 | r <= 8 |

test 10pct 5pct 1pct 1.76 6.5 8.18 11.7 5.53 12.9 14.90 19.2 172

r <= 7 | 14.04 r <= 6 | 24.98 r <= 5 | 30.92 r <= 4 | 36.70 r <= 3 | 38.66 r <= 2 | 45.22 r <= 1 | 71.57 r = 0 | 88.40

18.9 21.07 25.8 24.8 27.14 32.1 30.8 33.32 38.8 36.2 39.43 44.6 42.1 44.91 51.3 48.4 51.07 57.1 54.0 57.00 63.4 59.0 62.42 68.6

Thus we can reject r = 0 and r ≤ 1 at the 1% level but not r ≤ 2. Thus there are r = 2 cointegrating relationships. The rank of the Π = αβ ′ matrix would be r = 2. The cointegrating vectors are the first two columns from the “Eigenvector” matrix. In this case these are GM_AC.l2 F_AC.l2 UTX_AC.l2 CAT_AC.l2 MRK_AC.l2 PFE_AC.l2 IBM_AC.l2 MSFT_AC.l2 C_AC.l2 XOM_AC.l2

1.0000 -0.9403 -0.4511 0.0203 0.1477 0.1098 0.8553 -2.4203 -1.8426 1.8555

1.000 2.199 -2.189 1.614 1.425 -5.859 0.595 -0.916 3.540 -1.381

If we take K = 8 then the 10 stocks are still cointegrated and we still have r = 2.

Problem 9-12 (Simulation) Using the simulation written in Rlab.R we find that the expected profit is 13238, the probability that we have to liquidate for a loss is given by 0.0966, the expected waiting time is 44.1 days, and the expected yearly return is 5.267559.

Exercises Exercise 15.1 Now Eq. 15.4 from the book is given by ∆(Y1,t − λY2,t ) = (φ1 − λφ2 )(Y1,t−1 − λY2,t−1 ) + (ǫ1,t − λǫ2,t ) 173

(48)

When we use the definition of the forward difference ∆ we get that the left-hand-side of the above is equal to (Y1,t − λY2,t ) − (Y1,t−1 − λY2,t−1 ) .

When we move Y1,t−1 − λY2,t−1 in the above to the right-hand-side of Equation 48 we have the equation Y1,t − λY2,t = (1 + φ1 − λφ2 )(Y1,t−1 − λY2,t−1 ) + (ǫ1,t − λǫ2,t ) , which shows that Y1,t−1 − λY2,t−1 is an AR(1) process with a coefficient 1 + φ1 − λφ2 .

Exercise 15.2 Now Eq. 15.2 from the book is given by ∆Y1,t = φ1 (Y1,t−1 − λY2,t−1 ) + ǫ1,t ,

(49)

and Eq. 15.3 from the book is given by ∆Y2,t = φ2 (Y1,t−1 − λY2,t−1 ) + ǫ2,t .

(50)

If we add constants to them we get the equations ∆Y1,t = φ1 (Y1,t−1 − λY2,t−1 ) + µ1 + ǫ1,t ∆Y2,t = φ2 (Y1,t−1 − λY2,t−1 ) + µ2 + ǫ2,t .

From these we find that ∆(Y1,t − λY2,t ) is given by ∆(Y1,t − λY2,t ) = (φ1 − λφ2 )(Y1,t−1 − λY2,t−1 ) + (µ1 − λµ2 ) + (ǫ1,t − λǫ2,t ) , or that Y1,t − λY2,t is given by Y1,t − λY2,t = (1 + φ1 − λφ2 )(Y1,t−1 − λY2,t−1 ) + (µ1 − λµ2 ) + (ǫ1,t − λǫ2,t ) , This is an AR(1) model for Y1,t − λY2,t but it is not yet written in the “standard form” for an AR(1) models (with a mean m) which takes the form Zt − m = γ(Zt−1 − m) + ǫt

or Zt = γZt−1 + m(1 − γ) + ǫt .

To find the mean of the AR(1) process for Y1,t − λY2,t using the second equation for Zt above we see that the m mean must satisfy m(1 − γ) = m[1 − (1 + φ1 − λφ2 )] = µ1 − λµ2 .

The solution for m is given by m=−

µ1 − λµ2 . φ1 − λφ2

The point of this is that the process Y1,t − λY2,t now has a nonzero mean m (assuming the numerator is nonzero).

174

Exercise 15.3 In Example 15.2 we were told to take φ1 = 0.5, φ2 = 0.55 and λ = 1. From Exercise 1 above we know that Y1,t − λY2,t is an AR(1) process with a parameter that has a magnitude of |1 + φ1 − λφ2 | = |1 + 0.5 − 0.55| = 0.95 .

Since this is less than one we have that Y1,t − λY2,t is stationary.

Exercise 15.4 Eq. 15.2 and 15.3 together are the vector model

Y1,t−1 Y1,t φ1 ǫ1,t 1 −λ ∆ = + .

Y2,t φ2 Y2,t−1 ǫ2,t

Y1,t When we write this as a VAR(1) model for we get Y2,t

ǫ1,t Y1,t φ1 + 1 −λφ1 Y1,t , + = ǫ2,t Y2,t φ2 −λφ2 + 1 Y2,t To have the vector AR process for Y be stationary means that the coefficient matrix Φ above has all eigenvalues less than one in magnitude. For the numbers we were to use for this problem φ1 = 0.5, φ2 = 0.55 and λ = 1 the coefficient matrix Φ is given by

1.5 −0.5 . Φ= 0.55 0.45 This matrix has eigenvalues given by 1 and 0.95. Since on of the eigenvalues has a magnitude of one this process is not technically stationary.

175

15

Chapter 16 (Portfolio Theory)

Notes on the Text: The derivation of a quadratic expected utility From the class of utility functions considered here U(x; λ) = 1 − e−λx ,

(51)

we find E[U(X; λ)] = E[U(X0 (1 + R); λ)] = E[1 − e−λ(X0 (1+R)) ] = 1 − e−λX0 E[e−λX0 R ] .

Now if we assume that R is a normal random variable with mean µ and variance σ 2 then the random variable −λX0 R is also normal with mean −λX0 µ and variance λ2 X02 σ 2 . This means that e−λX0 R is a log-normal random variable. From the properties of a log-normal random variable we have that 1

2

2 2

E[e−λX0 R ] = e−λX0 µ+ 2 λ X0 σ .

Using this result we get that the expectation of our utility U(X; λ) can be written as 1

2

E[U(X; λ)] = 1 − e−λX0 [e−λX0 (µ− 2 (λX0 )σ ) ] .

Now to maximize E[U(X; λ)] we will minimize −E[U(X; λ)] which means that we want to minimize the argument of the exponential or to maximize the expression 1 1 µ − (λX0 )σ 2 = E(R) − (λX0 )Var (R) , 2 2 which is the expression stated in the book.

R Lab See the R script Rlab.R for the R Lab solutions for this chapter.

Problem 1 When the above R code is run it produces the plot given in Figure 71. In that figure we show the requested items. The data that went into this plot was produced by the function efficient frontier. Note that to have the constraints −0.1 ≤ wj ≤ 0.5 on the portfolio weights enforced in our quadratic program we need to write them in the form ATneq w ≥ bneq .

To do this we write the above constraints as the two conditions wj ≥ −0.1 and −wj ≥ −0.5.

176

0.08

Problem 1

CAT

*

UTX

MRK

0.06

+

F

0.04

GM

0.02

muP

TBM

0.00

* 0.0

0.5

1.0

1.5

2.0

sdP

Figure 71: The efficient frontier for the requested problem.

Problem 2 We will combine the tangency portfolio with the risk free portfolio in such a way that we achieve the desired portfolio return of 0.07%. To do this we will let Rp be the return of the portfolio that allocates a fraction ω of the investment to the tangency portfolio and 1 − ω to the risk-free asset. Then the return (and standard deviation) of the total portfolio is given by Rp = ωE(RT ) + (1 − ω)µf = µf + ω{E(RT ) − µf } and σRp = ωσT .

If we desire Rp = 0.0007 (unitless) then using the properties of the tangency portfolio (namely E(RT ) found earlier) we find that ω must satisfy (in terms of percent) 0.07 = 0.008219178 + ω(0.073411544 − 0.08219178) .

This gives ω = 0.947679 to be invested in the tangency portfolio and with 1−ω = 0.05233042 to be invested in the risk-free asset. Combining this with the stock weights found in the computation of the tangency portfolio we get the final weights of > print( omega*result$weights[ind,] ) [1] -0.086866909 -0.002917908 0.318313977 [6] 0.052288168

0.364024462

0.302827793

These represent the fraction of the portfolio to buy of each asset. We would also need to buy 177

a fraction 1 − ω = 0.05233042 of our portfolio of the risk-free asset. As an explicit example, if we have 100000 dollars to invest we will • Short −0.086866909(100000) = −8686.6909 dollars worth of GM • Short −0.002917908(100000) = −291.7908 dollars worth of F • Go long 0.318313977(100000) = 31831.3977 dollars worth of CAT • Go long 0.364024462(100000) = 36402.4462 dollars worth of UTX • Go long 0.302827793(100000) = 30282.7793 dollars worth of MRK • Go long 0.052288168(100000) = 5228.8168 dollars worth of IBM • Go long 0.05233042(100000) = 5233.042 dollars worth of the risk-free asset Summing all of the numbers on the right-hand-side of the above equations gives exactly our starting dollar value of 100000.

Problem 3 Black Monday is October 19th 1987. If we look at the return data we see that that data point is included and the huge drop in prices that happened over that date. > dat[201:202, c(’Date’, ’GM_AC’, ’F_AC’, ’CAT_AC’, ’UTX_AC’, ’MRK_AC’, ’IBM_AC’)] Date GM_AC F_AC CAT_AC UTX_AC MRK_AC IBM_AC 201 16-Oct-87 11.80 3.46 5.49 3.89 5.93 23.11 202 19-Oct-87 9.32 2.83 4.30 3.28 5.16 17.67

Problem 4 When we run the above R code it produces the plot given in Figure 72. It also outputs [1] "max sharpe (daily)= 0.085318" [1] "max sharpe (yearly)= 1.354379" [1] "Tangency portfolio weights:" [1] 0.0619 0.5000 0.1132 0.3248

178

0.07

Problem 4

0.06

XOM

*

0.05

+

MCD

muP

0.04

TGT

0.00

0.01

0.02

0.03

AAPL

* 0.0

0.5

1.0

1.5

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Figure 72: The efficient frontier for the requested problem.

0.0510 0.0508 E(return) 0.0506 0.0504

0.5800

0.0504

0.5802

0.0506

E(return)

SD(return)

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8

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Figure 73: The portfolio mean, portfolio standard deviation, and efficient frontier for the requested problem.

179

Problem 5 This procedure is implemented in the R function maximize expected utility. I tested this code in the code Rlab.R and used it to duplicate the book’s Figure 16.8. Unfortunately when I ran this code on the data suggested for this problem I don’t find that the efficient portfolio matched the one from the previous problem. I also didn’t find that the value µP = 0.046 was a feasible solution over all ranges I specified for log-λ. Thus I changed the problem slightly so that I was able to work it with the numbers I was getting. When we run the associated R code we get the plot in Figure 73. I choose to take µP = 0.0506 and this target portfolio return is drawn as a red horizontal line in the left-most plot in Figure 73. Running the given R code I get [1] "For mu_P= 0.050600 take lambda= 2.328247" If anyone sees anything wrong with what I did please contact me.

Problem 6-7 This procedure is implemented in the R function possible expected returns. We can test that the code a bit by showing that a long only portfolio has it minimum and maximum return bounded by the minimum and maximum return of the individual stocks over that time frame. Running this test we get [1] 0.04089730 0.04786051 0.07989876 0.07284620 0.06240313 0.04616853 [1] "Long only portfolio return bounds: minRet= 0.040897; maxRet= 0.079899"

If I specify the desired bounds on the weights wi i.e. such that −B2 ≤ wi ≤ B1 and run the code in Rlab.R I’m getting [1] "Problem 6: minRet= 0.043924; maxRet= 0.070196" To notice that we cannot have the constrains that |wi | ≤ 0.15 and produce a feasible solution recall that in this example the number of stocks is M = 6 and that the bounds above are ”too small” because M M X X wi ≤ |wi | ≤ 0.15M = 0.9 , i=1

which indicates that the constraint

i=1

PM

i=1 wi = 1 can never hold.

180

Exercises Exercise 16.1 We are told the values of the parameters µA , µB , σA , σB , and ρAB .

Part (a): Following the book, by letting ω be the fraction of the portfolio to invest in the A asset we have E(Rp ) = ωµA + (1 − ω)µB .

If we desire to have E(RP ) = 3 then we need to have ωµA + (1 − ω)µB = 3 which has a solution ω = 0.6818182 (so that 1−ω = 0.3181818). This is the fraction of the total portfolio we invest in asset A and 1 − ω is the fraction of the total portfolio we invest in asset B. Part (b): Since the variance of our return satisfies σR2 = ω 2 σA2 + (1 − ω)2σB2 + 2ω(1 − ω)ρAB σA σB .

(52)

Then using the numbers given in this problem (in percents) this is σR2 = 6ω 2 + 11(1 − ω)2 + 2.762173ω(1 − ω) .

As we want σR2 = 5.5 the above becomes a quadratic equation in ω with solutions 0.4107773 and 0.9403999. A portfolio with these weights will have a mean return given by ωµA + (1 − ω)µB .

Using the two values for ω above we find these two numbers to be [1] "omega_1 portfolio return= 3.596290 percent" [1] "omega_2 portfolio return= 2.431120 percent" As the first allocation has the larger return it would be the preferred portfolio.

Exercise 16.2 In the notation of this chapter we are told that E(RT ) = 5, σT = 7, µf = 1.5, ωC = 0.65, and ωD = 0.35. When we combine this tangency portfolio with the risk-free one we get E(Rp ) = µf + ω{E(RT ) − µF } and σR = ωσT .

Since we want σR = 5 this means that ω = 75 = 0.7142857. Thus we should invest • ωC ω = 0.4642857 fraction of our portfolio in C • ωD ω = 0.25 fraction of our portfolio in D • 1 − ω = 0.2857143 in the risk free asset.

181

Exercise 16.3 Part (a): The value of this portfolio is 75(300) + 115(100) = 34000 , so that the weights of A and B are 75(300) = 0.6617647 34000 115(100) = 0.3382353 .

1−ω = 34000 ω=

Part (b): We would have nj Pj wj = PN i=1 ni Pi

for 1 ≤ j ≤ N .

Exercise 16.4 (linearity of returns) We recall that the gross return2 which we will denote as Rt is defined as Rt ≡

Pt , Pt−1

(53)

and the net return3 which we will denote as rt is defined as rt ≡

Pt − Pt−1 = Rt − 1 , Pt−1

(54)

This notation is consistent with that in [2]. Now consider a portfolio of N assets and assume that we have Xi0 dollars worth of each i = 1, 2, . . . , N − 1, N. The total initial value of the investment is N X X0 = Xi0 , i=1

and let wi be the fraction of investment i in the initial portfolio so wi ≡

Xi0 .

X0

Then if each asset experiences the gross return of Ri then the new value of the asset is Xi = Ri Xi0 , 2 3

sometimes this is called the total return sometimes this is called the rate of return

182

and the portfolio now has value of N X

Ri Xi0 .

i=1

Thus the portfolios gross return is given by PN

i=1 Ri Xi0

X0

N X

N Xi0 X = Ri = wi Ri .

X0 i=1 i=1

Showing that the portfolios gross return is a weighted sum of its components gross returns.

Since the net return is the gross return minus one we have that the portfolios net return is r=

N X i=1

wi (ri + 1) − 1 =

N X

w i ri +

i=1

N X i=1

wi − 1 =

N X

w i ri ,

i=1

P using N i=1 wi = 1. Showing that the portfolios net return is also a weighted sum of its components net returns. This expression is not true if the returns are log returns but it is a good approximation in that case.

Exercise 16.5 Part (a): From the resampled numbers we would compute a value of ω such that E(Rp ) = ωµ1 + (1 − ω)µ2 = ω0.0047 + (1 − ω)0.0065 = 0.005 , or ω = 0.8333333.

Part (b): Since we are given the covariance and the variances we can compute the correlation to find ρ12 = 0.3420653. Using this, the portfolio variance we seek would be given by σR2 p = ω 2 σ12 + (1 − ω)2 σ22 + 2ω(1 − ω)ρ12 σ1 σ2 .

Putting in the numbers given we get σR2 p = 0.01093056 so that σRp = 0.1045493.

Part (c): For the weight given in Part (a) we find that the true portfolio returns and variance is given by Rp = 0.8333333µ1 + 0.1666667µ2 = 0.0039 σR2 p = 0.01413889 so σRp = 0.1189071 .

Notice that the true (realized) portfolio return is actually much less than the target of 0.005 and the true portfolio variance is larger than what is estimated in Part (b). These values were computed with the following simple R code # These are bootstrap resamples: mu_1 = 0.0047

183

mu_2 = 0.0065 sigma_1 = sqrt(0.0125) sigma_2 = sqrt(0.023) rho_12 = 0.0058/(sigma_1 * sigma_2) # compute the correlation from the covariance # The portfolio weights: omega = ( 0.005 - mu_2 ) / ( mu_1 - mu_2 ) # The estimated variance using the bootstrap estimates: s2Rp = omega^2 * sigma_1^2 + (1-omega)^2 * sigma_2^2 + 2 * omega * ( 1 - omega ) * rho_12 * sigma_1 * sigma_2 sqrt(s2Rp) # These are the population parameters: mu_1 = 0.0032 mu_2 = 0.0074 sigma_1 = sqrt(0.017) sigma_2 = sqrt(0.025) rho_12 = 0.0059/(sigma_1 * sigma_2) omega * mu_1 + (1-omega) * mu_2 s2Rp = omega^2 * sigma_1^2 + (1-omega)^2 * sigma_2^2 + 2 * omega * ( 1 - omega ) * rho_12 * sigma_1 * sigma_2 sqrt(s2Rp)

Exercise 16.6 From the numbers given in this problem the weights of the two stocks in this portfolio are given by 100(200) = 0.6153846 100(200) + 125(100) 125(100) = 0.3846154 . ωB = 100(200) + 125(100) ωA =

The covariance matrix for these two stocks is given by

σA2 ρAB σA σB 0.00090 0.00042 Σ= = . ρAB σA σB σB2 0.00042 0.00160 The mean return of the portfolio is given by E(Rp ) = ωA µA + ωB µB = 0.001192308 , with a variance of σR2 p =

ωA ωB

Σ

ωA ωB

= 0.0007763314 .

Using the codes written for another chapter (namely VaR norm which assumes normally distributed return and S = 100000) we compute the one-week VaR(0.05) to be 4463.779.

184

16

Chapter 17 (The Capital Asset Pricing Model)

R Lab See the R script Rlab.R where the Rlab problem from this chapter are worked.

Problem 1 We can look at the summary command output for each regression to see if any of the Pvalues for the constant (in R this is denoted as the Intercept coefficient) of the simple linear regression is sufficiently small. At an intuitive level a very small P-value indicates that a nonzero result is likely to “actually” be nonzero. Scanning the results we find that the smallest P-value of the Intercept terms is for F AC and is the value 0.14. Typical values (for a significant result) need to be at least below 0.05. Thus we can conclude that none of the α terms are nonzero.

Problem 2 Eq. 17.19 from the book is given by Rj,t = µf,t + βj (RM,t − µf,t ) + ǫj,t .

(55)

Thus the expected excess return of the jth stock is given in terms of the expected excess return of the market by E[Rj,t − µf,t ] = βj E[RM,t − µf,t ] .

I took this problem to mean that we will compute the expected excess return of the j stock to be equal to the product of the estimate of βj (computed earlier) and the excess return of the market (see the previous equation). When we do this we get the following > mean(market) * betas GM_AC F_AC UTX_AC CAT_AC MRK_AC PFE_AC IBM_AC 0.0233 0.0240 0.0190 0.0274 0.0146 0.0184 0.0159 These estimates are to be compared against just computing the values for Rj,t − µf,t and then taking the average of these points. Computing the average in this way gives > apply( stockExRet, 2, mean ) GM_AC F_AC UTX_AC CAT_AC -0.0452 -0.0736 0.0485 0.0842

MRK_AC PFE_AC IBM_AC 0.0071 -0.0218 -0.0163

These results are very different from each other.

185

Problem 3 Using the cor command we get these to be > cor(res) GM_AC F_AC UTX_AC CAT_AC MRK_AC PFE_AC IBM_AC GM_AC 1.00000 0.50911 0.03967 0.0202 -0.0472 -0.0188 0.00785 F_AC 0.50911 1.00000 -0.00714 0.0289 0.0128 0.0114 0.03575 UTX_AC 0.03967 -0.00714 1.00000 0.1498 -0.0154 -0.1110 -0.06949 CAT_AC 0.02023 0.02895 0.14977 1.0000 -0.0757 -0.0650 -0.08342 MRK_AC -0.04715 0.01279 -0.01540 -0.0757 1.0000 0.2833 -0.07817 PFE_AC -0.01877 0.01142 -0.11103 -0.0650 0.2833 1.0000 -0.04606 IBM_AC 0.00785 0.03575 -0.06949 -0.0834 -0.0782 -0.0461 1.00000 From the above correlation matrix it looks like some of the tickers have larger correlations with certain other tickers. Some examples are F and GM (both automobile companies) and MRK and PFE (both pharmaceutical companies). The companies UTX and CAT are both in the “industrial goods” sector and have a relatively large correlation also.

Problem 4 Once we have estimated βj and σǫj for each security then Eq. 17.19 implies that the covariance matrix for the returns between securities has 2 σj2 = βj2 σM + σǫ2j ,

for its diagonal elements and 2 , σjj ′ = βj βj ′ σM

for its off diagonal elements. Computing these elements from the given data we get > as.matrix(betas) %*% t(as.matrix(betas)) * sigma2_M + diag( sigmas2 ) GM_AC F_AC UTX_AC CAT_AC MRK_AC PFE_AC IBM_AC GM_AC 5.519 0.686 0.541 0.781 0.416 0.526 0.455 F_AC 0.686 3.677 0.557 0.804 0.428 0.542 0.468 UTX_AC 0.541 0.557 1.231 0.634 0.338 0.427 0.370 CAT_AC 0.781 0.804 0.634 2.443 0.487 0.617 0.534 MRK_AC 0.416 0.428 0.338 0.487 3.334 0.328 0.284 PFE_AC 0.526 0.542 0.427 0.617 0.328 2.007 0.359 IBM_AC 0.455 0.468 0.370 0.534 0.284 0.359 0.994 This is to be compared with just computing the covariance matrix directly from the excess returns which gives 186

> cov(stockExRet) GM_AC F_AC UTX_AC CAT_AC MRK_AC PFE_AC IBM_AC GM_AC 5.512 2.616 0.619 0.836 0.234 0.474 0.469 F_AC 2.616 3.673 0.546 0.866 0.466 0.566 0.519 UTX_AC 0.619 0.546 1.230 0.799 0.314 0.303 0.319 CAT_AC 0.836 0.866 0.799 2.441 0.323 0.516 0.448 MRK_AC 0.234 0.466 0.314 0.323 3.329 0.954 0.171 PFE_AC 0.474 0.566 0.303 0.516 0.954 2.004 0.311 IBM_AC 0.469 0.519 0.319 0.448 0.171 0.311 0.992 These two results are quite similar.

Problem 5 This is the R2 of the CAPM regression performed on the UTX symbol. From the R code for this problem we find a summary of this regression given by > summary_fit_reg[3] Response UTX_AC : Call: lm(formula = UTX_AC ~ market) Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 0.0296 0.0343 0.86 0.39 market 0.9766 0.0506 19.31 <2e-16 *** Residual standard error: 0.89 on 670 degrees of freedom Multiple R-squared: 0.357, Adjusted R-squared: 0.356 F-statistic: 373 on 1 and 670 DF, p-value: <2e-16 Thus we have an R2 of 0.357 indicating 35.7% of the variance in the returns of UTX explained by the variance in the market.

Problem 6 The expected excess return from each stock will be given by its beta times the expected excess return of the market. Thus we would get > betas * ex_ret_M_predicted 187

GM_AC F_AC UTX_AC CAT_AC MRK_AC PFE_AC IBM_AC 0.0481 0.0495 0.0391 0.0564 0.0300 0.0380 0.0328 For the excess return for our stocks.

Problem 7 The alpha for the portfolio is given by αP =

N X

w j αj ,

j=1

which is a linear objective function of the weights wj . The linear programming problem we want to solve is then to maximize αP subject to the constraints that βP =

N X

wj βj = 0 ,

j=1

and that

−0.25 ≤ wj ≤ +0.25 .

This is set up and solved in the code Rlab.R using the linprog library. When we run that code we compute the portfolio weights (in the variable w), the value of αP , and the value of βP . Running that code we find [1] "alpha_F (max)= 0.245266; beta_P= 0.000000"

Problem 8 I’m actually getting a solution using the above code restricted to the first twenty stocks. In fact I find [1] "alpha_F (max)= 0.085899; beta_P= -0.000000" If anyone sees anything wrong with what I have done please let me know.

Exercises Exercise 17.1 Using the expression Rj,t = µf,t + βj (RM,t − µf,t ) + ǫj,t , 188

by taking expectations we get E[RP ] = µf + β(E[RM ] − µf ) .

Solving for β we get β=

0.16 − 0.055 E[RP ] − µf = = 1.909091 .

E[RM ] − µf 0.11 − 0.055

Exercise 17.2 We are told that µf = 0.03 and E[RM ] = 0.14 with σM = 0.12.

Part (a): If we want E[RP ] = 0.11 then following CAPM we would compute w=

E[RP ] − µf 0.11 − 0.03 = = 0.7272727 .

E[RM ] − µf 0.14 − 0.03

We would then invest w percent of our dollars (capital) in a fund that tracks the market and invest 1 − w = 0.2727273 percent of our capital in risk-free treasury bills.

Part (b): The Capital Market Line (CML) states that the returns and risk must be related as µM − µf µP − µf = , σP σM so solving for σP we get

µP − µf 0.11 − 0.03 σP = σM = 0.12 = 0.08727273 . µM − µf 0.14 − 0.03 Note that σP < σM as it must be.

Exercise 17.3 Part (a): If σM = 0.12 then the CML gives 0.1 − 0.023 µP − 0.023 = so µP = 0.05508333 .

0.05 0.12 Part (b): We have σA,M = 0.004 so σA,M 0.004 βA = 2 = = 0.2777778 . σM 0.122 Part (c): For this part we are told that βB = 1.5 βC = 1.8 σǫB = 0.08 σǫC = 0.1 .

189

Part (i): With our portfolio of stocks B and C our return will be 1 1 RP = RB + RC , 2 2 thus the expected return is given by 1 1 E[RP ] = µB + µC .

2 2 But in terms of the market portfolio we have µB = µf + βB (µM − µf ) and similarly for µC .

Using what we know we get µB = 0.1385 and µC = 0.1616 so that E[RP ] = 0.15005.

Part (ii): For this we have 2 2 1 1 1 1 2 2 2 σP = σǫ,B + σǫ,C = 0.082 + 0.12 = 0.0041 .

2 2 4 4

thus σP = 0.06403124.

Exercise 17.4 Equation 7.8 from the book is Cov(w1T Y, w2T Y ) = w1T Cov(Y )w2 .

(56)

To use this to derive the books equation 17.15 we let w1 be a vector of all zeros but with a single one at the jth location (denoted by the vector ej ) and let w2 have elements given by wi,M i.e. the vector of weights (denoted by the vector w). Then if we define !

N X wi,M Ri,t , σj,M ≡ Cov Rj,t , i=1

we see that this is equal to

Cov eTj R, w T R ,

with R a random vector of returns. Then using Equation 56 to evaluate it since

eTj Cov(R) = σj,1 σj,2 · · · σj,N −1 σj,N ,

we get

eTj Cov(R)w =

N X

σj,i wi,M =

i=1

N X

wi,M σi,j ,

i=1

as we should. Note that we used the symmetry of Cov(R) as σj,i = σi,j in the above.

Exercise 17.5 This is True. In order to get a larger return rate for µP the risk σP must increase to keep µP − µf µM − µf = , σP σM constant.

190

Exercise 17.6 Part (a): The value for beta for stock A is given by βA =

σA,M 0.0165 = 1.363636 . = 2 σM 0.112

Part (b): For this we have µA − µf = βA (µM − µf ) so µA = 0.1490909 .

Part (c): For this we would have σA2 =

220 = 0.022 .

10000

2 The market risk variance is βA2 σM = 0.0225 so the fraction of the variance of A due to the market is 2 0.0225 βA2 σM = = 1.022727 .

2 σA 0.022 Warning: Something must be wrong with this since the above fraction is larger than one.

If anyone sees anything wrong with what I have done please contact me.

Exercise 17.7 For a portfolio of N assets the returns under the model Rj = βj RM + ǫj are given by !

N N N X X X RP = wj Rj = wj ǫj = βP RM + ǫP . wj βj RM + j=1

j=1

j=1

Part (a): This would be given by βP =

N X

1 1 1 wj βj = β1 + β2 + β3 = 0.8666667 .

3 3 3 j=1

Part (b): This would be given by Var (RP ) or Var (RP ) = βP2 Var (RM ) + Var (ǫP ) = βP2 Var (RM ) +

N X

2 wj2σǫ,j = 0.0145 .

j=1

Part (c): For asset number one this would be 2 2 β12 σM β12 σM = = 0.531 .

2 2 σ12 β12 σM + σǫ,1

We might also ask what proportion of the total risk (in the portfolio) is due to market risk.

This would be given by 2 βP2 σM = 0.724 . σP2 191

Exercise 17.8 (two risky assets) Part (a): This would be given by µT = wC µC + wD µD = 0.6(0.04) + 0.4(0.06) = 0.048 .

Part (b): This would be given by q 2 2 σD + 2wC wD σC σD ρC,D σT = wC2 σC2 + wD p = 0.62 (0.1)2 + 0.42 (0.18)2 + 2(0.6)(0.4)(0.1)(0.18)(0.5) = 0.1144727 .

Part (c): We will combine the tangency portfolio with the risk free portfolio in such a way that we achieve the desired portfolio risk of 3%. To do this we will let RP be the return of the portfolio that allocates a fraction ω of the investment to the tangency portfolio and 1 − ω to the risk-free asset. Then the return (and standard deviation) of the total portfolio is given by RP = ωE(RT ) + (1 − ω)µf = µf + ω{E(RT ) − µf } and σRP = ωσT .

If we desire σRp = 0.03 (unitless) then using the properties of the tangency portfolio found earlier we find that ω = 0.2620712 to be invested in the tangency portfolio and with 1 − ω = 0.7379288 to be invested in the risk-free asset.

Part (d): In this case we want to have RP = 0.07 = µf + ω(µT − µf ) ,

this means that ω = 1.6111111 so that 1 − ω = −0.6111111. This means that we should invest ωwC = 0.9666667 , in asset C and ωwD = 0.6444444 , in asset D and in the risk free asset.

1 − ω = −0.6111111 ,

Note that since we desire to have E[RP ] = 0.07 which is larger than both µC and µD we need to borrow money (loosing return at the risk free rate µf ) to purchase more of the equities.

The above numbers mean that if we have V0 dollars to invest we need to borrow an additional 0.6111111V0 dollars. This will give us V0 + 0.6111111V0 = 1.6111111V0 dollars to invest into C and D, From this 1.6111111V0 dollars we invest wC of it into C and wD of it into D. This gives an allocation of • 0.6111111V0 dollars borrowed • 1.6111111V0wC = 0.9666667V0 dollars invested into asset C • 1.6111111V0wD = 0.6444444V0 dollars invested into asset D Note the sum of these allocations is V0 as it must be.

192

Exercise 17.9 For this we will use E[RP ] − µf = β(E[RM ] − µf ) .

Putting in the numbers given in the problem we have

0.15 − 0.06 = β(0.12 − 0.06) so β = 1.5 .

Exercise 17.10 Part (a): We will want to allocate 0.11 − 0.07 µP − µf = = 0.5714286 , w= µM − µf 0.14 − 0.07

percent of our investment dollars in the market and then allocate 1 − w = 0.4285714 percent of our investment dollars in the risk-free asset.

Part (b): The risk standard deviation of the portfolio in the above must be on the Capital Market Line (CML) or must satisfy µM − µf µP − µf = . σP σM Solving for σP we get 0.06857143.

Exercise 17.11 This problem is just like Problem 7 above but with different numbers.

Part (a): This would be given by β=

N X

1 1 1 wj βj = β1 + β2 + β3 = 0.7 .

3 3 3 j=1

Part (b): This would be given by Var (RP ) or Var (RP ) = βP2 Var (RM ) + Var (ǫP ) = βP2 Var (RM ) +

N X

2 wj2σǫ,j = 0.0109 .

j=1

Part (c): For asset number one this would be 2 2 β12 σM β12 σM = 2 2 = 0.407 .

2 σ12 β1 σM + σǫ,1

We might also ask what proportion of the total risk (in the portfolio) is due to market risk.

This would be given by 2 βP2 σM = 0.632 . σP2 193

Exercise 17.12 We would prefer the more diversified portfolio since the expectations is that it would have a much lower variance going forward.

194

17

Chapter 18 (Factor Models and Principal Components)

Notes on the Book Notes on Estimating Expectations and Covariances using Factors Given the expression for Rj,t we can evaluate the covariance between two difference asset returns as follows Cov (Rj,t , Rj ′ ,t ) = Cov (β0,j + β1,j F1,t + β2,j F2,t + ǫj,t , β0,j ′ + β1,j ′ F1,t + β2,j ′ F2,t + ǫj ′ ,t ) = Cov (β1,j F1,t , β0,j ′ + β1,j ′ F1,t + β2,j ′ F2,t + ǫj ′ ,t ) + Cov (β2,j F2,t , β0,j ′ + β1,j ′ F1,t + β2,j ′ F2,t + ǫj ′ ,t ) + Cov (ǫj,t , β0,j ′ + β1,j ′ F1,t + β2,j ′ F2,t + ǫj ′ ,t ) = β1,j β1,j ′ Var (F1,t ) + β1,j β2,j ′ Cov (F1,t , F2,t ) + β2,j β1,j ′ Cov (F2,t , F1,t ) + β2,j β2,j ′ Var (F2,t ) = β1,j β1,j ′ Var (F1,t ) + β2,j β2,j ′ Var (F2,t ) + (β1,j β2,j ′ + β1,j ′ β2,j )Cov (F1,t , F2,t ) , which is the same as the books equation 18.6.

R Lab See the R script Rlab.R for this chapter. We first duplicate the bar plot of the eigenvalues and eigenvectors of the covariance matrix of the dataframe yielddat. These are shown in Figure 74.

Problem 1-2 (for fixed maturity are the yields stationary?)

See Figure 75 for a plot of the first four columns of the yield data (the first four maturities).

These plots do not look stationary. This is especially true for index values from 1000 to 1400 where all yield curves seem to trend upwards.

As suggested in the book we can also use the augmented Dickey-Fuller test to test for stationarity. When we do this for each possible maturity we get [1] "column index= [1] "column index= [1] "column index= [1] "column index=

1; p_value= 2; p_value= 3; p_value= 4; p_value=

0.924927" 0.543508" 0.410602" 0.382128" 195

0.6

0.6

0.4 0.2 0.0

eig$vector[, 2]

−0.6

−0.6

−0.4

−0.2

0.2 0.0 −0.4

−0.2

eig$vector[, 1]

0.4

3.5 3.0 2.5 2.0

2

4

6

8

10

2

4

6

8

10

8

10

Index

0.6 0.4 0.2 0.0

eig$vector[, 4]

−0.6

−0.4

−0.2

0.2 0.0

eig$vector[, 3]

0.0

−0.6

−0.4

0.5

−0.2

1.0

0.4

0.6

1.5

Index

2

4

6 Index

8

10

2

4

6 Index

Figure 74: Left: The distribution of the eigenvalues of the yield data. Right: Plots of the first four eigenvectors of the yield data.

[1] "column index= 5; p_value= [1] "column index= 6; p_value= [1] "column index= 7; p_value= [1] "column index= 8; p_value= [1] "column index= 9; p_value= [1] "column index= 10; p_value= [1] "column index= 11; p_value=

0.382183" 0.386320" 0.391729" 0.437045" 0.461692" 0.460651" 0.486028"

As all of these p values are “large” (none of them are less than 0.05) we can conclude that the raw yield curve data is not stationary.

Problem 3 (for fixed maturity are the difference in yields stationary?)

See Figure 76 for a plot of the first difference of each of the four columns of the yield data (the first difference of the first four maturities). These plots now do look stationary. Using the augmented Dickey-Fuller test we can show that the time series of yield differences are stationarity. Using the same code as before we get [1] "column index=

1; p_value=

0.010000" 196

8 4

5

6

yieldDat[, 2]

7

8 7 6 4

5

yieldDat[, 1]

0

200

400

600

800

1000

1400

0

200

400

800

1000

1400

1000

1400

8 7 4

5

6

yieldDat[, 4]

7 6 5 4

yieldDat[, 3]

600

Index

8

Index

0

200

400

600

800

1000

1400

0

Index

200

400

600

800

Index

Figure 75: Plots time series of the first four yield maturities.

197

0

0

200

200

400

400

600

800

800

Index

600

1000

1000

1400

1400

198

0

0

200

200

400

400

600

800

800

Index

600

Index

1000

1000

1400

1400

Figure 76: Plots the time series of the first difference of the first four interest rate yield maturities.

Index

0.2 0.1 0.0 −0.1 −0.2 −0.3 0.3 0.2 0.1 0.0 −0.1 −0.2 −0.3

delta_yield[, 2] delta_yield[, 4]

0.6 0.4 0.2 0.0 −0.2 −0.4 0.3 0.2 0.1 0.0 −0.1 −0.2 −0.3

delta_yield[, 1] delta_yield[, 3]

[1] "column index= 2; p_value= 0.010000" [1] "column index= 3; p_value= 0.010000" [1] "column index= 4; p_value= 0.010000" [1] "column index= 5; p_value= 0.010000" [1] "column index= 6; p_value= 0.010000" [1] "column index= 7; p_value= 0.010000" [1] "column index= 8; p_value= 0.010000" [1] "column index= 9; p_value= 0.010000" [1] "column index= 10; p_value= 0.010000" [1] "column index= 11; p_value= 0.010000" There were 11 warnings (use warnings() to see them) As all of these p values are “small” (they are all less than 0.01) we can conclude that the first differences of a yield at a fixed maturity is stationary. The warnings indicates that the adf.test command could not actually compute the correct p-value and that the true p-values are actually smaller than the ones printed above.

Problem 4 (PCA on differences between the yield curves) Part (a): The variable sdev holds the standard deviations of each principal components, these are also the square root of the eigenvalues of the covariance matrix. The variable loadings hold the eigenvectors of the covariance matrix. The variable center hold the means that were subtracted in each feature dimension in computing the covariance matrix.

The variable scores holds a matrix of each vector variable projected into all its principle components. We can check that this is so by comparing the two outputs t(as.matrix(pca_del$loadings[,])) %*% t( delta_yield[1,] - pca_del$center ) 2 Comp.1 0.1905643953 Comp.2 0.0375662026 Comp.3 0.0438591813 Comp.4 -0.0179855611 Comp.5 0.0002473111 Comp.6 0.0002924385 Comp.7 0.0101975886 Comp.8 -0.0093768514 Comp.9 -0.0036798653 Comp.10 0.0004287954 Comp.11 -0.0005602180 with pca_del$scores[1,] 199

Comp.1 0.1905643953 Comp.6 0.0002924385 Comp.11 -0.0005602180

Comp.2 Comp.3 Comp.4 0.0375662026 0.0438591813 -0.0179855611 Comp.7 Comp.8 Comp.9 0.0101975886 -0.0093768514 -0.0036798653

Comp.5 0.0002473111 Comp.10 0.0004287954

These two outputs are exactly the same (as they should be).

Part (b): Squaring the first two values of the sdev output we get > pca_del$sdev[1:2]^2 Comp.1 Comp.2 0.031287874 0.002844532 Part (c): The eigenvector corresponding to the largest eigenvalue is the first one and has values given by > pca_del$loadings[,1] X1mon X2mon X3mon X4mon X5mon X5.5mon -0.06464327 -0.21518811 -0.29722014 -0.32199492 -0.33497517 -0.33411403 X6.5mon X7.5mon X8.5mon X9.5mon NA. -0.33220277 -0.33383143 -0.32985565 -0.32056039 -0.31668346 Part (d): Using the output from the summary(pca_del) which in a truncated form is given by Importance of components: Comp.1 Comp.2 Comp.3 Comp.4 Comp.5 Standard deviation 0.1768838 0.05333415 0.03200475 0.014442572 0.011029556 Proportion of Variance 0.8762330 0.07966257 0.02868616 0.005841611 0.003406902 Cumulative Proportion 0.8762330 0.95589559 0.98458175 0.990423362 0.993830264 we see from the Cumulative Proportion row above that to obtain 95% of the variance we must have at least 2 components. Taking three components gives more than 98% of the variance.

Problem 5 (zero intercepts in CAPM?) The output of the lm gives the fitted coefficients and their standard errors, capturing the partial output of the summary command we get the following 200

Response GM : Estimate Std. Error t value Pr(>|t|) (Intercept) -0.0103747 0.0008924 -11.626 <2e-16 *** FF_data$Mkt.RF 0.0124748 0.0013140 9.494 <2e-16 *** Response Ford : Estimate Std. Error t value Pr(>|t|) (Intercept) -0.0099192 0.0007054 -14.06 <2e-16 *** FF_data$Mkt.RF 0.0131701 0.0010386 12.68 <2e-16 *** Response UTX : Estimate Std. Error t value Pr(>|t|) (Intercept) -0.0080626 0.0004199 -19.20 <2e-16 *** FF_data$Mkt.RF 0.0091681 0.0006183 14.83 <2e-16 *** Response Merck : Estimate Std. Error t value Pr(>|t|) (Intercept) -0.0089728 0.0009305 -9.643 < 2e-16 *** FF_data$Mkt.RF 0.0062294 0.0013702 4.546 6.85e-06 *** Notice that the p-value of all intercepts are smaller than the given value of α i.e. 5%. Thus we cannot accept the hypothesis that the coefficient β0 is zero.

Problem 6 We can use the cor command to compute the correlation of the residuals of each of the CAPM models which gives > cor( fit1$residuals ) GM Ford UTX Merck GM 1.00000000 0.55410840 0.09020925 -0.04331890 Ford 0.55410840 1.00000000 0.09110409 0.03647845 UTX 0.09020925 0.09110409 1.00000000 0.05171316 Merck -0.04331890 0.03647845 0.05171316 1.00000000 The correlation between GM and Ford is quite large. To get confidence intervals for each correlation coefficient we will use the command cor.test to compute the 95% confidence intervals. We find [1] "Correlation between Ford and [1] "Correlation between UTX and [1] "Correlation between UTX and [1] "Correlation between Merck and

GM; ( 0.490439, GM; ( 0.002803, Ford; ( 0.003705, GM; ( -0.130254,

201

0.554108, 0.090209, 0.091104, -0.043319,

0.611894)" 0.176248)" 0.177122)" 0.044277)"

[1] "Correlation between Merck and [1] "Correlation between Merck and

Ford; ( -0.051113, UTX; ( -0.035878,

0.036478, 0.051713,

0.123513)" 0.138515)"

From the above output only the correlations between Merck and GM, Ford, and UTX seem to be zero. The others don’t seem to be zero.

Problem 7 (comparing covariances) The sample covariance or ΣR can be given by using the cov command. Using the factor returns the covariance matrix ΣR can be written as ΣR = β T ΣF β + Σǫ ,

(57)

where β is the row vector of each stocks CAPM beta value. In the R code Rlab.R we compute both ΣR and the right-hand-side of Equation 57 (which we denote as Σ̂). If we plot these two matrices sequentially we get the following > Sigma_R GM Ford UTX Merck GM 4.705901e-04 2.504410e-04 6.966363e-05 1.781501e-05 Ford 2.504410e-04 3.291703e-04 6.918793e-05 4.982034e-05 UTX 6.966363e-05 6.918793e-05 1.270578e-04 3.645322e-05 Merck 1.781501e-05 4.982034e-05 3.645322e-05 4.515822e-04 > Sigma_R_hat GM Ford UTX Merck GM 4.705901e-04 7.575403e-05 5.273459e-05 3.583123e-05 Ford 7.575403e-05 3.291703e-04 5.567370e-05 3.782825e-05 UTX 5.273459e-05 5.567370e-05 1.270578e-04 2.633334e-05 Merck 3.583123e-05 3.782825e-05 2.633334e-05 4.515822e-04 The errors between these two matrices are primarily in the off diagonal elements. We expect the pairs that have their residual correlation non-zero to have the largest discrepancy. If we consider the absolute value of the difference of these two matrices we get > abs( Sigma_R - Sigma_R_hat ) GM Ford UTX Merck GM 1.084202e-19 1.746870e-04 1.692904e-05 1.801622e-05 Ford 1.746870e-04 2.168404e-19 1.351424e-05 1.199209e-05 UTX 1.692904e-05 1.351424e-05 1.084202e-19 1.011988e-05 Merck 1.801622e-05 1.199209e-05 1.011988e-05 0.000000e+00 The largest difference is between GM and Ford which are also the two stocks that had the largest residual correlations under the CAPM model.

202

Problem 8 (the beta of SMB and HML) If we look at the p-values of the fitted model on each stock we are getting results like the following > sfit2$’Response GM’$coefficients Estimate Std. Error t value Pr(>|t|) (Intercept) -0.010607689 0.000892357 -11.887271 7.243666e-29 FF_data$Mkt.RF 0.013862140 0.001565213 8.856390 1.451297e-17 FF_data$SMB -0.002425130 0.002308093 -1.050708 2.939015e-01 FF_data$HML 0.006373645 0.002727395 2.336899 1.983913e-02 > sfit2$’Response GM’$coefficients Estimate Std. Error t value Pr(>|t|) (Intercept) -0.010607689 0.000892357 -11.887271 7.243666e-29 FF_data$Mkt.RF 0.013862140 0.001565213 8.856390 1.451297e-17 FF_data$SMB -0.002425130 0.002308093 -1.050708 2.939015e-01 FF_data$HML 0.006373645 0.002727395 2.336899 1.983913e-02 > sfit2$’Response Ford’$coefficients Estimate Std. Error t value Pr(>|t|) (Intercept) -1.004705e-02 0.0007082909 -14.18492403 1.296101e-38 FF_data$Mkt.RF 1.348451e-02 0.0012423574 10.85396920 8.752040e-25 FF_data$SMB -7.779018e-05 0.0018320033 -0.04246181 9.661475e-01 FF_data$HML 3.780222e-03 0.0021648160 1.74620926 8.138996e-02 > sfit2$’Response UTX’$coefficients Estimate Std. Error t value Pr(>|t|) (Intercept) -0.0080963376 0.0004199014 -19.2815220 2.544599e-62 FF_data$Mkt.RF 0.0102591816 0.0007365160 13.9293389 1.721546e-37 FF_data$SMB -0.0028475161 0.0010860802 -2.6218286 9.013048e-03 FF_data$HML 0.0003584478 0.0012833841 0.2792989 7.801311e-01 > sfit2$’Response Merck’$coefficients Estimate Std. Error t value Pr(>|t|) (Intercept) -0.008694614 0.000926005 -9.389381 2.142386e-19 FF_data$Mkt.RF 0.007065701 0.001624233 4.350178 1.650293e-05 FF_data$SMB -0.004094797 0.002395124 -1.709639 8.795427e-02 FF_data$HML -0.009191144 0.002830236 -3.247483 1.242661e-03

In the fits above we see that the slope of the SMB and HML for different stocks have significance at the 2% - 8% level. For example, the HML slope for GM is significant at the 1.9% level. Based on this we cannot accept the null hypothesis of zero value for slopes.

Problem 9 (correlation of the residuals in the Fama-French model) If we look at the 95% confidence interval under the Fama-French model we get [1] "Correlation between Ford and [1] "Correlation between UTX and [1] "Correlation between UTX and [1] "Correlation between Merck and [1] "Correlation between Merck and

GM; ( 0.487024, GM; ( -0.004525, Ford; ( 0.002240, GM; ( -0.119609, Ford; ( -0.039887,

203

0.550991, 0.082936, 0.089651, -0.032521, 0.047708,

0.609079)" 0.169138)" 0.175702)" 0.055064)" 0.134575)"

[1] "Correlation between Merck and

UTX; ( -0.040087,

0.047508,

0.134378)"

Now the correlation between UTX and GM is zero (it was not in the CAPM). We still have a significant correlation between Ford and GM and between UTX and Ford (but it is now smaller).

Problem 10 (model fitting) The AIC and BIC between the two models is given by [1] "AIC(fit1)= -10659.895869; AIC(fit2)= -10688.779045" [1] "BIC(fit1)= -10651.454689; BIC(fit2)= -10671.896684" The smaller value in each case comes from the second fit or the Fama-French model.

Problem 11 (matching covariance) The two covariance matrices are now > Sigma_R GM Ford UTX Merck GM 4.705901e-04 2.504410e-04 6.966363e-05 1.781501e-05 Ford 2.504410e-04 3.291703e-04 6.918793e-05 4.982034e-05 UTX 6.966363e-05 6.918793e-05 1.270578e-04 3.645322e-05 Merck 1.781501e-05 4.982034e-05 3.645322e-05 4.515822e-04 > Sigma_R_hat GM Ford UTX Merck GM 4.705901e-04 7.853015e-05 5.432317e-05 3.108052e-05 Ford 7.853015e-05 3.291703e-04 5.602592e-05 3.437406e-05 UTX 5.432317e-05 5.602592e-05 1.270578e-04 2.733456e-05 Merck 3.108052e-05 3.437406e-05 2.733456e-05 4.515822e-04 The difference between these two matrices are smaller than in the CAPM model.

Problem 12 (Fama-French betas to excess returns covariance) We will use the formula ΣR = β T ΣF βj + Σǫ .

Here we have already calculated the value of ΣF in the R code Rlab.R. We had found 204

(58)

Mkt.RF SMB HML Mkt.RF 0.46108683 0.17229574 -0.03480511 SMB 0.17229574 0.21464312 -0.02904749 HML -0.03480511 -0.02904749 0.11023817 This factor covariance matrix will not change if the stock we are considering changes.

Part (a-c): Given the factor loadings for each of the two stocks and their residual variances we can compute the right-hand-side of Equation 58 and find [,1] [,2] [1,] 23.2254396 0.1799701 [2,] 0.1799701 37.2205144 Thus we compute that the variance of the excess return of Stock 1 is 23.2254396, the variance of the excess return of Stock 2 is 37.2205144 and the covariance between the excess return of Stock 1 and Stock 2 is 0.1799701.

Problem 13 Using the factanal command we see that the factor loadings are given by Factor1 Factor2 GM_AC 0.693 -0.348 F_AC 0.692 -0.313 UTX_AC 0.531 -0.028 CAT_AC 0.529 -0.079 MRK_AC 0.551 0.421 PFE_AC 0.574 0.511 IBM_AC 0.490 MSFT_AC 0.499 0.044 The variance of the unique risks for Ford and GM are the values that are found in the “Uniquenesses” list which we found is given by GM_AC 0.399

F_AC 0.423

UTX_AC 0.718

CAT_AC 0.714

MRK_AC 0.519

PFE_AC 0.410

IBM_AC MSFT_AC 0.760 0.749

Thus the two numbers we are looking for are 0.423 and 0.399.

205

Problem 14 The p-value for the factanal command is very small (less than 10−10 ) indicating that we should reject the null hypothesis and try a larger number of factors. Using four factors (the largest that we can use with eight inputs) gives a larger p-value 0.86.

Problem 15 For statistical factor models the covariance between the log returns is given by ΣR = β̂ T β̂ + Σ̂ǫ , where the β̂ and Σ̂ǫ are the estimated loadings and uniqueness found using the factanal command. When we do that we get an approximate value for ΣR given by [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [1,] 1.0000 0.5883 0.3774 0.3939 0.2357 0.2197 0.3383 0.3305 [2,] 0.5883 1.0000 0.3759 0.3906 0.2495 0.2366 0.3379 0.3314 [3,] 0.3774 0.3759 1.0000 0.2829 0.2807 0.2899 0.2600 0.2636 [4,] 0.3939 0.3906 0.2829 1.0000 0.2587 0.2633 0.2591 0.2606 [5,] 0.2357 0.2495 0.2807 0.2587 1.0000 0.5314 0.2722 0.2937 [6,] 0.2197 0.2366 0.2899 0.2633 0.5314 1.0000 0.2834 0.3088 [7,] 0.3383 0.3379 0.2600 0.2591 0.2722 0.2834 1.0000 0.2449 [8,] 0.3305 0.3314 0.2636 0.2606 0.2937 0.3088 0.2449 1.0000 As Ford is located at index 2 and IBM is located at index 7 we want to look at the (2, 7)th or (7, 2)th element of the above matrix where we find the value 0.3379.

Exercises Exercise 18.1-2 These are very similar to the Rlab for this chapter.

Exercise 18.3 See the notes on Page 195.

206

18

Chapter 19 (Risk Management)

R Lab See the R script Rlab.R where the problem for this chapter are worked.

Problem 1 The expression qt(alpha, df=nu) gives the α quantile (often denoted as tα ) of the tdistribution that has its parameters fit to the stock returns of Coca-Cola. The expression with dt verifies that the value output from qt has an α fraction of the area of the t-distribution “in front” of it.

Problem 2 Running the R code for this problem I compute [1] "VaR_t= 75.305683; ES_t= 122.114367"

Problem 3 From the ugarchspec parameters specified in the R code we are using a ARMA(0, 0) model for the conditional mean of yt (the return of KO) which is yt − µ = at , and a GARCH(1,1) model for the “noise” term at of at = σt ǫt q 2 σt = ω + α1 a2t−1 + β1 σt−1 .

When we run the given R code and look at the output (partial) from the show command we get Optimal Parameters -----------------------------------Estimate Std. Error t value Pr(>|t|) mu 0.000635 0.000240 2.6486 0.008082 omega 0.000003 0.000004 0.6981 0.485113 207

alpha1 beta1 shape

0.093609 0.892841 5.995587

0.039389 0.043317 1.192118

2.3766 0.017475 20.6118 0.000000 5.0294 0.000000

The numerical values of the model parameters are given in the “estimate” column. Thus the model fit is yt − 0.000635 = at ,

where at is modeled as

at = σt ǫt q 2 . σt = 0.000003 + 0.093609a2t−1 + 0.892841σt−1

Here ǫt ∼ t6 . Here I have rounded the degree of freedom estimate above to six.

Problem 4 Running the R code for this problem we get > pred = ugarchforecast(KO.garch.t, data=returns, n.ahead=1); pred *------------------------------------* * GARCH Model Forecast * *------------------------------------* Model: sGARCH Horizon: 1 Roll Steps: 0 Out of Sample: 0 0-roll forecast [T0=1974-01-15 20:00:00]: Series Sigma T+1 0.0006349 0.01039 The predicted one-step conditional mean and standard deviation are given in the Series and Sigma columns in the above output.

Problem 5 Following the similar example in this chapter of the book and running the R associated with this problem we get [1] "VaR_t= 63.404352; ES_t= 89.461279" 208

Problem 6 We specify a range of values for ν and then use cov.trob to estimate the parameters for a multidimensional t-distribution. Once we have the parameters we can use the function dmt to evaluate the loglikelihood of the data under the given parameter setting. When we run this code we get the following > result$fit $cov DATGEN DEC DATGEN 0.009993897 0.004546750 DEC 0.004546750 0.005406026 $center DATGEN DEC 0.005422298 0.021636475 with an estimate of ν̂ = 4.3. The label of the output from the cov.trob is the string cov but I believe that is actually an approximation to the scale matrix Λ. This is based on usage since its value is passed directly into the function dmt as the argument S which (according to the documentation) is a scale matrix. This hypothesis is further validated when we compute the covariance matrix from the scale matrix using Σ=

ν Λ. ν−2

For the approximation above we get > result$fit$cov * ( 4.3 / (4.3-2) ) DATGEN DEC DATGEN 0.018684242 0.008500445 DEC 0.008500445 0.010106918 This is a much better match for the covariance of the data which is given by > cov(Berndt) DATGEN DEC DATGEN 0.016266437 0.007283029 DEC 0.007283029 0.009829483 than is result$fit$cov see above. If anyone knows that this logic is incorrect please contact me.

209

Problem 7 Part (a): Following the discussion in Chapter 7 on multivariate distributions in the section titled ”Using the t-distribution in Portfolio Analysis” the portfolio as specified will have returns given by rp = 0.3rDATGEN + 0.7rDEC .

I see two ways to work this problem. The first is to take the given fractional distribution of each asset and the raw returns for the two assets and combine them to compute the portfolio return (using the above formula). We can then use the R code fitdistr to fit a t-distribution to this data. When we do that we get the following parameters of the t-distribution > params m s df 0.015713110 0.077582356 5.389583510 (0.008109552) (0.008802226) (2.692582265) As discussed in Chapter 5 in the q book fitdistr fits the classical t-distribution and returns which is called s in the output. This is the parameter the standard deviation times ν−2 ν called that is also called the ”scale parameter” and is denoted λ.

A second method to work this problem is to recall the fact that linear projections of random variables that are distributed as multivariate t-distribution will have a univariate tν w T Λw = w T Σw, and the same degrees of distribution with a mean w T µ, a variance ν−2 freedom ν. This gives the parameter estimate > list( m=mu_2, s=lambda_2, df=nu_2 ) $m [1] 0.01677222 $s [,1] [1,] 0.07387854 $df [1] 4.3 These numerical values match well with the ones we obtained using fitdistr.

Part (b): Given parameters for a t-distribution we can use the parametric estimate of the value-at-risk and the expected shortfall that apply when we assume that we have a t-distribution. In this case, the formula for the value-at-risk is given by n o t ˆ VaR (α) = −S µ̂ + qα,t (ν̂)λ̂ , (59) and that for the expected shortfall is given by the books Eq. 19.8. Using these two equations (and the two ways we estimated the parameters of the t-distribution) we get 210

[1] "VaR_1= 13815.711768; VaR_2= 13762.077715" [1] "ES_1= 20227.220528; ES_2= 21130.813757"

Problem 8 I used the boot command in the library of the same name to generate the bootstrap samples.

When that command finishes one has the bootstrap samples in the member field t. In addition, when we display the output from the boot command we get > boot_results ORDINARY NONPARAMETRIC BOOTSTRAP Call: boot(data = r_port, statistic = boot_fn, R = 250) Bootstrap Statistics : original bias std. error t1* 1.570073e-02 -0.0002939243 6.366081e-03 t2* 7.786412e-02 -0.0000742150 6.593052e-03 t3* 5.512324e+00 1.0626691845 4.214329e+00 t4* 1.380347e+04 -0.3143866541 1.146888e+03 t5* 2.013453e+04 94.8409343322 1.892534e+03 As the bootstrap gives the standard error and we assume that the distribution of values of VaR(0.05) are normally distributed using the numbers above the 90% confidence interval for VaR(0.05) is given by 1.380347e+04 + c(-1,+1) * qnorm(1-0.1/2) * 1.146888e+03 [1] 11917.01 15689.93 We plot the KDE of the bootstrap samples of VaR(0.05) in Figure 77. From that plot we see that the distribution of VaR is somewhat symmetric and somewhat Gaussian. Using the shapiro.test on this data gives > shapiro.test(VaR_boots) Shapiro-Wilk normality test data: VaR_boots W = 0.9957, p-value = 0.714 The large p-value indicates that this data could be normal. By contract in Figure 78 we present the KDE of the bootstrap samples of the parameter ν (the degree of freedom in the univariate t-distribution fit). Here we see a very skewed distribution that does not look normal at all. The shapiro.test on this data gives 211

0.00015 0.00000

0.00005

0.00010

Density

0.00020

0.00025

0.00030

0.00035

VaR (kde)

10000

12000

14000 N = 250

16000

18000

Bandwidth = 342.1

Figure 77: The KDE of the bootstrap samples of VaR(0.05).

0.00

0.05

0.10

Density

0.15

0.20

0.25

df (kde)

0

10

20 N = 250

30

40

Bandwidth = 0.5901

Figure 78: The KDE of the bootstrap samples of the degrees of freedom parameter ν in the t-distribution.

212

0

1

a

2

3

4

a^{Hill} estimate

0

10

20

30

40

50

nc

Figure 79: Hill plots of the estimated tail index a. > shapiro.test(df_boots) Shapiro-Wilk normality test data: df_boots W = 0.557, p-value < 2.2e-16 Indicating a very non-normal distribution.

Problem 9 (the Hill tail index estimate) When computing a tail index we are assuming that the density for y < 0 takes the functional form f (y) = A|y|−(a+1) .

Then a > 0 is the tail index. The Hill estimate of the left tail index is given by n(c) , Y(i) n(c) log i=1 c

âHill (c) = P

(60)

where n(c) = #{Y(i) ≤ c} and c = Y(n(c)) . We will vary the value of n(c) from the smallest it can be (of one) to the largest it can be (the number of samples in the dataset) and then plot the estimate of âHill (c) as a function of n(c). This is done in Figure 79 Notice that for 213

a range of n(c) between 10 and 30 the estimate of a is constant around the value of 2.25.

When n(c) = 20 the estimated value of a is 2.18973.

Exercises See the R script chap 19.R where the exercises for this chapter are worked.

Exercise 19.1 Part (a): For the nonparametric estimates we will use the formulas VaRnp = −S q̂(α) , and ES For this data these give

np

Pn i=1 Ri I{Ri < q̂(α)} . = −S P n i=1 I{Ri < q̂(α)}

(61) (62)

> as.vector( c( VaR_np, ES_np ) ) [1] 40.79757 56.49151 Part (b): In this case we get > c( VaR_norm(alpha,m,s,S), ES_norm(alpha,m,s,S) ) [1] 33.98577 38.98592 Part (c): In this case we get [1] 42.00555 64.88889 Part (d): The Shapiro-Wilk test on a sample of this data gives > shapiro.test( rets[sample.int(length(rets),size=2500)] ) Shapiro-Wilk normality test data: rets[sample.int(length(rets), size = 2500)] W = 0.918, p-value < 2.2e-16 Showing that these returns are not very normally distributed. In fact a histogram indicates that they have very long tails. Thus we would expect the t-distribution to fit better and give more realistic risk metrics. The estimated value of ν in the t-distribution was 2.9968678708.

A value this small further indicates non-normality.

214

Exercise 19.2 When a density has a polynomial tail with a left tail index a we can show that the value-atrisk for two different values of α is related as α 1/a 0 .

(63) VaR(α) = VaR(α0 ) α

Using what we know we then have that

0.05 Var(0.005) = VaR(0.05) 0.005

1/3.1

= 529.6405 .

Exercise 19.3 I downloaded prices for “IBM” and took the last 1000 prices. I also assumed that S = 10000.

If you download prices at a different point in time your results might be different than these.

Part (a): For this we compute > VaR_t_pt_a [1] 212.2063 Part (b): For this we compute > VaR_np 2.5% 233.4143 Part (c): For this we compute a normal QQ plot and a t QQ plot. Each of these is displayed in Figure 80. The t distribution is a better fit (especially along the right tail).

Neither model does a great job modeling the negative returns. Normality of the returns does not seem reasonable.

Part (d): For this we use the Hill plot given in Figure 81. Notice for a large range of value for n(c) the estimate â is very stable. Taking the mean value of the estimate of a over the range from 50 to 125 we get â = 2.110415. Then using Equation 63 we compute VaR(0.0025) = 631.8278 .

215

QQ plot for IBM : df= 5.27822575300788

0

qt(q.grid, df = df)

−2

0.00

−6

−4

−0.05

−3

−2

−1

0

1

2

3

−0.05

0.00

Theoretical Quantiles

0.05

rets

Figure 80: Left: A normal QQ plot. Right: A t QQ plot.

2

3

4

a^{Hill} estimate

0

1

a

Sample Quantiles

2

0.05

4

6

Normal Q−Q Plot

0

50

100 nc

Figure 81: A Hill plot.

216

150

200

ES (kde)

Density 0.005

0.015

0.000

0.000

0.005

0.010

Density

0.020

0.010

0.025

0.030

0.015

VaR (kde)

80

100

120

140

160

100

N = 1000 Bandwidth = 2.613

150

200

250

N = 1000 Bandwidth = 5.299

Figure 82: Left: A KDE estimate of VaR. Right: A KDE estimate of ES. Exercise 19.4 In Figure 82 I plot KDE of the bootstrap densities of VaR (left) and ES (right). We can directly use the results from the output of the boot command (like we did above) or the function boot.ci to derive confidence intervals. Based on the location of the result we are interested in we have > boot.ci(boot_results, conf=0.95, type="norm", index=4) Intervals : Level Normal 95% ( 92.6, 137.9 ) > boot.ci(boot_results, conf=0.95, type="norm", index=5) Intervals : Level Normal 95% (111.2, 203.1 )

217

Exercise 19.5 (subadditivity of the VaR metric) There is a problem in the second edition of this textbook that is something like the following.

Suppose the risk measure R is VaR(α) for some α. Let P1 and P2 be two portfolios whose returns have a joint normal distribution with means µ1 and µ2 , standard deviations σ1 and σ2 , and correlation ρ. Suppose the initial investments are S1 and S2 . Show that R(P1 +P2 ) ≤ R(P1 ) + R(P2 ) under joint normality.

To show this we will recall that when we have normally distributed returns the value-of-risk is given by

VaRnorm (α) = −S µ + Φ−1 (α)σ .

(64)

When we invest S1 and S2 into each asset we combine the returns with weights S1 S1 + S2 S2 x2 = , S1 + S2

x1 =

to get statistics of the portfolio of µP = x1 µ1 + x2 µ2 σP2 = x1 σ12 + x1 σ12 + 2x1 x2 σ1 σ2 ρ .

(65) (66)

We can show that σP < x1 σ1 + x2 σ2 , as long as ρ < 1 (which it must be for a correlation), x1 > 0, and x2 > 0. To show this is true we can assume it is true, square both sides, and use the expression in Equation 66 to show that it will be true if the given requirements are meet. For “normal” values of α (say less than 0.5) we will have Φ−1 (α) < 0. Thus multiplying both sides of the previous inequity by Φ−1 (α) we have Φ−1 (α)σP > x1 Φ−1 (α)σ1 + x2 Φ−1 (α)σ2 .

Adding the portfolio mean to both sides of this (and using Equation 65) we get µP + Φ−1 (α)σP > x1 µ1 + x1 Φ−1 (α)σ1 + x2 µ2 + x2 Φ−1 (α)σ2 = x1 (µ1 + Φ−1 (α)σ1 ) + x2 (µ2 + Φ−1 (α)σ2 ) .

If we multiply both sides by −S = −(S1 + S2 ) this becomes −S{µP + Φ−1 (α)σP } < −Sx1 (µ1 + Φ−1 (α)σ1 ) − Sx2 (µ2 + Φ−1 (α)σ2 ) = −S1 (µ1 + Φ−1 (α)σ1 ) − S2 (µ2 + Φ−1 (α)σ2 ) .

Using Equation 64 we see that this equation is norm VaRnorm (α) < VaRnorm P P1 (α) + VaRP2 (α) ,

as we were to show.

218

Exercise 19.6 (the five-day VaR) Part (a): The estimated vector of daily mean returns µ, for the selected stocks is given by > apply( rets, 2, mean ) GM_AC F_AC UTX_AC CAT_AC MRK_AC 9.801282e-04 1.556664e-03 5.400236e-05 1.231639e-03 8.568635e-04 The estimated covariance matrix Σ, for the selected stocks is given by > cov( rets ) GM_AC F_AC UTX_AC CAT_AC MRK_AC GM_AC 0.0004081672 0.0002913652 0.0001877597 0.0002314422 0.0001858492 F_AC 0.0002913652 0.0004493332 0.0001973972 0.0002771111 0.0002188128 UTX_AC 0.0001877597 0.0001973972 0.0003744431 0.0002231245 0.0001698379 CAT_AC 0.0002314422 0.0002771111 0.0002231245 0.0005037975 0.0002291694 MRK_AC 0.0001858492 0.0002188128 0.0001698379 0.0002291694 0.0003123000 Part (b): Since there are five stocks in this portfolio each stock would get 50/5 = 10 million dollars. Dividing this number by the price-per-share gives GM_AC F_AC UTX_AC CAT_AC MRK_AC 594530.3 2232142.9 2923976.6 1821493.6 1754386.0 for the number of shares of each stock to purchase to get the desired portfolio.

Part (c): For this portfolio, the weight vector is a five element vector with 51 in every component. This means that our portfolio return will be a normal random variable with a mean given by w T µ and a variance of w T Σw .

Evaluating these on the data given, we get a portfolio mean of w T µ = 0.0009358594 and a variance of w T Σw = 0.0002588712 (which implies a standard deviation of return of 0.01608947). We can then have everything we need to use the formula for the value-atrisk when we have normally distributed returns and given in the function VaR norm. Calling this function with S = 100000 and the above estimates we get the value of 1968.363.

Part (d): Over five days (rather than one) the mean return will be five times the one day mean return and the covariance will be five times the one day covariance. Assuming that the correlations between stocks can be taken to be zero we can zero out the off-diagonal elements in Σ to obtain a matrix Σ̂ and then again compute the one day variance using w T Σ̂w. Doing this, for the data in this problem we get a five day mean return of 0.004679297 and a five day variance of 0.0004096082. With these parameters we find the value-at-risk (again using the function VaR norm) given by 2125.774.

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19

Chapter 20 (Bayesian Data Analysis and MCMC)

R Lab One of the hard parts (for me) about running the code from this chapter was getting the needed BUGS program installed. As I run Ubuntu Linux I choose to install OpenBUGS vs.

WinBUGS. To do that I needed to first # to install 32 bit headers on 64 bit machine needed to install OpenBUGS sudo apt-get install g++-multilib Then one needs to download and install the OpenBUGS software from http://www.openbugs.net/w/Downloads Then in R I needed to do the following install.packages( "R2WinBUGS", dependencies=TRUE ) install.packages( "BRugs", dependencies=TRUE ) After these steps there were still a couple of small changes I needed to make to the given codes to make them run. See the R script Rlab.R for this chapter.

Problem 1 Part (a-b): The output from the print(univt.mcmc) command is > print(univt.mcmc,digits=4) Inference for Bugs model at "Tbrate_t.bug", fit using OpenBUGS, 3 chains, each with 2600 iterations (first 100 discarded) n.sims = 7500 iterations saved mean sd 2.5% 25% 50% 75% 97.5% mu 0.0095 0.0035 0.0025 0.0072 0.0096 0.0118 0.0164 tau 237.7120 22.5085 198.6151 221.8916 236.1194 250.9109 288.5372 nu 27.6553 11.4585 8.9672 18.1431 26.9800 37.0714 48.4221 sigma 0.0651 0.0030 0.0589 0.0631 0.0651 0.0671 0.0710 deviance -916.2988 2.1460 -919.0373 -917.7026 -916.8050 -915.4008 -910.7087 Rhat n.eff mu 1.0014 3500 220

sigma

0.055

0.000

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6000

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iteration

2000

4000

6000

iteration

10

20

30

40

50

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0

2000

4000

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iteration

Figure 83: Plots of the MCMC chains for mu, sigma, and nu. tau 1.0024 nu 1.0058 sigma 1.0024 deviance 1.0008

2400 480 2400 7500

For each parameter, n.eff is a crude measure of effective sample size, and Rhat is the potential scale reduction factor (at convergence, Rhat=1).

DIC info (using the rule, pD = Dbar-Dhat) pD = 2.0 and DIC = -914.3 DIC is an estimate of expected predictive error (lower deviance is better).

The parameter n.eff can be thought of as “the number of effective samples” and larger values are better (meaning more information is extracted). The parameter Rhat is a measure of chain convergence and here smaller values are better. Using that from the above output the deviance parameter mixes best while the parameter nu mixes the worst.

Part (c): A 95% confidence interval for nu can be found by reading off the 2.5% and 97.5% quantiles in the above row corresponding to the nu parameter. From the above output we find that to be the values (8.9672, 48.4221). Note that this is a very large range. This matches our observation above about nu having the worst mixing properties.

221

sigma

ACF 0.0

0.0

0.2

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Lag

0.0

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0.4

ACF

0.6

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df

0

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30

Lag

Figure 84: The ACF for the MCMC chains for mu, sigma, and nu.

Problem 2 Part (a): See Figure 83 where we see that based on the time series plots in order of best mixing to worst mixing we have mu, sigma, and nu. The time series plot for nu does not look stationary.

Part (b): See Figure 84 where we plot the ACF for the three variables mu, sigma, and nu.

There we again see the ordering from best to worst mixing is the same as that found in Part (a).

Part (c): We can use the extracted matrix variable nu to determine the skewness and kurtosis where we find > library("fGarch") # for skewness and kurtosis > skewness(nu,method="moment") [1] 0.7075442 > kurtosis(nu,method="moment") [1] 2.59612

222

sigma

mu

2000

120 100 80

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0.005 0.010 0.015 0.020 0.025 N = 7500 Bandwidth = 0.000528

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0.075

N = 7500 Bandwidth = 0.0004512

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df

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nu

10

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40

50

N = 7500 Bandwidth = 1.731

Figure 85: Left: Histograms of the MCMC chains for mu, sigma, and nu. Right: KDE of the MCMC samples of the posterior for mu, sigma, and nu.

Problem 3 See Figure 85 for plots of the histograms (left) and the kernel density estimates (right) for the mu, sigma, and nu variables. From this figure we see that the variable nu appears to be the most skewed. The sample skewness for all three variables is given by > c( skewness(mu), skewness(sigma), skewness(nu) ) [1] -0.01635687 -0.13948345 0.70754418 From which we see that nu has the largest value (in magnitude).

Problem 4 We have samples of the posterior for ν and then using the formula given 3(ν−2) we can ν−4 compute the kurtosis when ν > 4. If ν ≤ 4 we take the kurtosis to be +∞.

Part (a): We find these to be > quantile( posterior_kurtosis, c( 0.01, 0.05, 0.25, 0.5, 0.75, 0.95, 0.99 ) ) 223

phi ACF

ACF 0.0

0.0

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Figure 86: Time series plots and the ACF for the phi variable.

1% 5% 25% 50% 75% 95% 99% 3.132109 3.140067 3.181426 3.261097 3.424235 3.946165 4.763263 Part (c): We did a similar problem in Chapter 6. Here we will just use the model free bootstrap (see Chapter 6 for an example where we use the model based bootstrap method).

Using the boot library we find the requested quantiles to be > quantile( boot.kurtosis$t, c( 0.01, 0.05, 0.25, 0.5, 0.75, 0.95, 0.99 ) ) 1% 5% 25% 50% 75% 95% 99% 2.577871 2.824477 3.571534 4.660616 5.477031 6.468237 7.109982 Part (d): The confidence interval for the samples from the posterior is (3.140067, 3.946165), while that from the model free bootstrap is (2.824477, 6.468237). The confidence interval for the samples from the posterior is smaller.

Problem 5 Part (a): Plots of the time series and the ACF for the phi variable are given in Figure 86 while plots of the time series and the ACF for the sigma variable are given in Figure 87.

224

sigma ACF

0.0

0.010

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4000

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Lag

Figure 87: Time series plots and the ACF for the sigma variable.

There we see that for both parameters the samples of the posterior seem to be coming from a stationary distribution. This is an indication that the parameters chosen (the number of chains, the number of iterations, and the burn-in) are adequate.

Part (b): The output from the print statement is given by > print(ar1.mcmc,digits=3) Inference for Bugs model at "ar1.bug", fit using OpenBUGS, 3 chains, each with 2600 iterations (first 100 discarded) n.sims = 7500 iterations saved mean sd 2.5% 25% 50% 75% 97.5% phi 0.219 0.072 0.079 0.170 0.219 0.267 0.361 sigma 0.012 0.001 0.011 0.011 0.012 0.012 0.013 deviance -1129.076 1.965 -1131.020 -1130.493 -1129.655 -1128.314 -1123.843 Rhat n.eff phi 1.001 7500 sigma 1.001 7500 deviance 1.001 3100 For each parameter, n.eff is a crude measure of effective sample size, and Rhat is the potential scale reduction factor (at convergence, Rhat=1).

DIC info (using the rule, pD = Dbar-Dhat) 225

pD = 2.0 and DIC = -1127.0 DIC is an estimate of expected predictive error (lower deviance is better).

While the output from the arima call which computes the MLE estimate of the parameters in our AR(1) model is > arima(y,order=c(1,0,0)) Call: arima(x = y, order = c(1, 0, 0)) Coefficients: ar1 intercept 0.2181 0.0000 s.e. 0.0713 0.0011 sigma^2 estimated as 0.0001336:

log likelihood = 568.72,

aic = -1131.44

There we see that the Bayesian and the MLE estimates of phi are quite close to each other.

The Bayesian estimate of sigma squared (or 0.0122 = 0.000144) matches the MLE estimate of sigma squared (of 0.0001336).

Exercises Exercise 20.1 Part (a): We need to find the MAP estimator for the θ parameter in the density π(θ|5) = 56θ6 (1 − θ) .

To find the MAP estimator we need to maximize the above expression as a function of θ.

Taking the derivative and setting it equal to zero we get 56(6)θ5 (1 − θ) + 56θ6 (−1) = 56θ5 (6 − 7θ) = 0 .

Solving the above for θ we get θ = 76 . Taking the second derivative of π we get d2 π = 56(5)θ4 (6 − 7θ) + 56θ5 (−7) dθ2 = 56θ4 (30 − 35θ − 7θ) = 56θ4 (30 − 42θ) .

Note that

d2 π dθ2

6 θ= 7

showing that θ = 67 is a maximum.

4 6 = −6(56) < 0, 7

226

Exercise 20.2 To show the desired expression consider the likelihood n Y

(Yi − µ)2 1 √ exp − f (Y1 , · · · , Yn |τ ) = 2σ 2 2πσ i=1

n Y 1 1/2 (Yi − µ)2 √ τ exp − τ in terms of τ = 2 2π i=1 ( )

n τX 1 n/2 n/2 2 2 ∝ τ exp − (Yi − µ) = τ exp − nτ s , 2 i=1 2 which is the desired expression.

227

20

Chapter 21 (Nonparametric Regression and Splines)

Notes on the Text Notes on polynomial splines When we force the linear fits on each side of the knot x = t to be continuous we have that a + bt = c + dt and this gives c = a + (b − d)t. When we use this fact we can simplify the formula for s(x) when x > t as s(x) = c + dx = a + (b − d)t + dx = a + bt + d(x − t) = a + bx − bx + bt + d(x − t) = a + bx − b(x − t) + d(x − t) = a + bx + (d − b)(x − t) , which is the books equation 21.11.

R Lab See the R script Rlab.R for this chapter.

R lab: An Additive Model for Wages, Education, and Experience When we enter and then run the given R code we see that the summary command gives that β0 = 6.189742 and β1 = −0.241280 .

Plots from the fitGam are duplicated in Figure 88. Notice that s1 is very linear (meaning wages increase with education). Next notice that s2 has more curvature to it meaning that the relationship between wages and experience is more complicated.

R lab: An Extended CKLS model for the Short Rate We are using 10 knots. The outer function here takes the outer difference of the values in t with those in knots. The statement that computes X2 then computes the value of the “plus” functions for the various knots i.e. evaluates (t − k)+ where t is the time variable and k is a knot. Then X3 holds the linear spline basis function i.e. the total spline we are using to predict µ(t, r) is given by 10 X θi (t − ki )+ . α0 + α1 t + i=1

228

s(education,7.65)

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s(experience,8.91) −1.0 −1.5 −2.0

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Figure 88: Plots for the splines CPS1988 dataset.

5

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Figure 89: The three plots for the short rate example.

Here α0 and α1 are the coefficients of the initial linear fit, and θi are the jumps in the first derivatives at each of the ki knots. The first column of X3 is the a column of all ones (for the constant α0 term), the second column of X3 is a column of time relative to 1946. The rest of the columns of X3 are samples of the spline basis “plus” functions i.e. (t − ki )+ for 1 ≤ i ≤ 10. When we run the given R code we generate the plot in Figure 89. Note that X3[,1:2]%*%a is a linear function in t but because of the way that X3 is constructed (its last 10 columns) X3%*%theta is the evaluation of a spline.

The interpretation of a time-varying θ(t) is that interest rates r revert to a mean that is time varying.

230

Our estimates of the coefficients of α0 and α1 are not significant. The call summary( nlmod_CKLS_ext )$parameters gives

a1 a2

Estimate Std. Error 0.082619604 0.087780955 0.002423339 0.002706119

t value Pr(>|t|) 0.94120192 3.470418e-01 0.89550342 3.709356e-01

The large p-values indicate that these coefficients are not well approximated and might not be real effects.

Exercises Exercise 21.1 (manipulations with splines) Our expression for s(t) is given by s(t) = β0 + β1 x + b1 (x − 1)+ + b2 (x − 2)+ + b3 (x − 3)+ .

Where the “plus function” is defined by (x − t)+ =

0 x<t . x−t x ≥t

From the given values of x and s(x) we can compute s(0) = 1 = β0 s(1) = 1.3 = 1 + β1 so β1 = 0.3 s(2) = 5.5 = 1 + 0.3(2) + b1 (1) so b1 = 3.9 s(4) = 6 = 1 + 0.3(4) + 3.9(3) + b2 (2) + b3 (1) s(5) = 6 = 1 + 0.3(5) + 3.9(4) + b2 (3) + b3 (2) .

Solving these two equations gives b2 = −3.7 and b3 = −0.5. Thus we have s(x) given by s(x) = 1 + 0.3x + 3.9(x − 1)+ − 3.7(x − 2)+ − 0.5(x − 3)+ .

Part (a): We would find s(0.5) = 1 + 0.3(0.5) = 1.15 .

231

Part (b): We would find s(3) = 1 + 0.3(3) + 3.9(2) − 3.7(1) = 6 .

Part (c): We would evaluate Z 4 Z 4 Z 4 s(t)dt = (1 + 0.3t + 3.9(t − 1) − 3.7(t − 2))dt − 0.5 (t − 3)dt , 2

2

3

each term could then be evaluated.

Exercise 21.2 The model 21.1 in the book is ∆rt = rt − rt−1 = µ(rt−1 ) + σ(rt−1 )ǫt .

(67)

In this problem we are told functional forms for µ(rt−1 ) and σ(rt−1 ).

Part (a): Since ǫt has a mean of zero we have that E[rt |rt−1 = 0.04] = rt−1 + µ(rt−1 ) = 0.04 + 0.1(0.035 − 0.04) = 0.0395 .

Part (b): Since ǫt has a variance of one we have that Var (rt |rt−1 = 0.02) = σ 2 (rt−1 ) = 2.32 rt−1 2 = 2.32 (0.02)2 = 0.002116 .

Exercise 21.3 Part (a): This function s(x) defined as s(x) = (x)+ − 3(x − 1)+ + (x − 2)+ , takes the following form x<0 0<x<1 1<x<2 2<x

s(x) = 0 s(x) = x s(x) = x − 3(x − 1) = 3 − 2x s(x) = 3 − 2x + x − 2 = 1 − x .

Note that in this form we have limx→2 s(x) = −1 < 0 and thus this cannot be a pdf or a cdf. Because of this I think that there is a typo in the book and the function s(x) should be written s(x) = (x)+ − 2(x − 1)+ + (x − 2)+ .

232

In that case we see that x<0 0<x<1 1<x<2 2<x

s(x) = 0 s(x) = x s(x) = x − 2(x − 1) = 2 − x s(x) = 2 − x + (x − 2) = 0 .

Thus s(x) is a piece-wise linear function that increases to the location (1, 1) and then decreases to the location (2, 0). Since when x > 1 the term −2(x − 1)+ is causes the value of s(x) to decrease, thus s(x) cannot be a cumulative distribution function (which must be non-decreasing). Note also that the area under this curve (because it is a triangle) is 1 (2)(1) = 1 .

2 Thus this function is a pdf.

Part (b): Given the above pdf we now compute the cdf. Recall that Z x S(x) = s(t)dt .

0

Thus if 0 < x < 1 this means S(x) =

Z x

tdt =

0

If 1 < x < 2 then S(x) =

Z 1 0

tdt +

Z x 1

(2 − t)dt =

x2 .

2

1 1 1 − ((2 − x)2 − 1) = 1 − (2 − x)2 .

2 2 2

Then to find the 90% percentile of X we need to solve S(x) = 0.9 .

Doing this I get x=2±

√

0.2 .

We need √ to take the minus to have x < 2 and we find that the 90% percentile of X is given by 2 − 0.2 = 1.552786.

Exercise 21.4 For the given spline we have s′ (x) = 0.65 + 2x + 2(x − 1)+ + 1.2(x − 2)+ s′′ (x) = 2 + 2(x − 1)0+ + 1.2(x − 2)0+ .

Part (a): We have s(1.5) = 1 + 0.65(1.5) + 1.52 + 0.52 = 4.475 .

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and s′ (1.5) = 0.65 + 3 + 2(0.5) + 1.2(0) = 4.65 .

Part (b): We have s′′ (2.2) = 2 + 2(1) + 1.2(1) = 5.2 .

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References [1] J. Kiusalaas. Numerical methods in engineering with Python; 2nd ed. Cambridge Univ.

Press, New York, NY, 2010.

[2] D. G. Luenberger. Investment science. Oxford university press, New York, 1998.

[3] S. Mahajan. Street-fighting mathematics: the art of educated guessing and opportunistic problem solving. The MIT Press, Cambridge, 2010.

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