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程序代写案例-ETC3460
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Page 1 of 14
Semester One 2021
Exam - Alternative Assessment Task
UNIT CODE: ETC3460-ETC5346
UNIT TITLE: Financial Econometrics
ASSESSMENT DURATION: 2 hours 40 minutes (includes reading,
downloading, and uploading time)
This is an individual assessment task.
All responses must be included in ONE separate PDF document. Clearly
state your ID number, surname and given name at the top of this
document.
Clearly state the question numbers corresponding to your answers.
Students are required to answer ALL questions.
This assessment accounts for 60% of the total in the unit and has a hurdle
requirement of 45% to pass the unit.
Upon completion of this assessment task, please upload your PDF answer
document to Moodle using the assignment submission link.
You will sign the Student Statement shown on the next page
automatically when using the assignment submission link.
Your submission must occur within 2 hours 40 minutes of the official
commencement of this assessment task (Australian Eastern Standard
Time).
Please read the next page carefully (Student Statement) before
commencing the assessment task.
Page 2 of 14
Intentional plagiarism or collusion amounts to cheating under Part 7 of the Monash
University (Council) Regulations
Plagiarism: Plagiarism means taking and using another person’s ideas or manner of
expressing them and passing them off as one’s own. For example, by failing to give
appropriate acknowledgment. The material used can be from any source (staff, students or
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written, oral or practical work and includes paying another person to complete all or part of
the work.
Where there are reasonable grounds for believing that intentional plagiarism or collusion has
occurred, this will be reported to the Associate Dean (Education) or delegate, who may
disallow the work concerned by prohibiting assessment or refer the matter to the Faculty
Discipline Panel for a hearing.
Student Statement:
I have read the university’s Student Academic Integrity Policy and Procedures.
I understand the consequences of engaging in plagiarism and collusion as described in
Part 7 of the Monash University (Council) Regulations
https://www.monash.edu/legal/legislation/current-statute-regulations-and-related-
resolutions
I have taken proper care to safeguard this work and made all reasonable efforts to ensure
it could not be copied.
I have not used any unauthorised materials in the completion of this assessment task.
No part of this assessment has been previously submitted as part of another unit/course.
I acknowledge and agree that the assessor of this assessment task may for the purposes
of assessment, reproduce the assessment and:
i. provide to another member of faculty and any external marker; and/or
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Date:
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Page 3 of 14
MARKS ALLOCATED TO QUESTIONS WITHIN THIS ASSESSMENT TASK
Question 1 2 3 4 5 6 7 8 9 TOTAL
Allocated Marks 17 16 17 10 60
Office Use Only
Mark received
Second marking
INSTRUCTIONS TO STUDENTS
When testing a hypothesis, to obtain full marks you need to specify the null and
the alternative hypotheses, the test statistic and its distribution under the null,
and then perform the test and state your conclusion.
If a question does not specify the level of signi cance of a hypothesis test explicitly,
use 5%.
Statistical tables are provided after Question 4.
Question 1 (17 marks)
1. Suppose that Daniel is interested in making an investment in assets A and B over the next
year. The expected returns on assets A and B are A = 0:1 and B = 0:35 and their volatilities
are equal to A = 0:14 and B = 0:27, respectively. The two asset returns have correlation
equal to A;B = 0:2.
(a) Provide the formulae and numerical values for the optimal (risk minimizing) weights for
assets A and B. [You are not required to solve the optimization problem]. (1pt)
(b) Write the regression equation that you would use to obtain the same weights. Explain
which parameter refers to the optimal weight(s) and analytically show why. (4pts)
2. Consider the following Fama-French four factor model
rt rft = 0 + 1(rmt rft) + 2SMBt + 3HMLt + 4MOMt + t; (1)
where rt denote monthly returns on the Fama-French hitec portfolio, (rmt rft) is the excess
market return (the market factor), rft is the risk free interest rate, SMB is the size factor,
HML is the value factor and MOM is the momentum factor. OLS output from estimating (1)
over the period January 1990 to January 2021 is shown in Figure 1.
(a) Test whether the hitec portfolio provides adequate reward for the assumed risk. (2pts)
(b) Test whether the hitec portfolio tracks the market versus that it is aggressive, at 1%
signi cance level. Based on your conclusion interpret the relevant coe¢ cient estimate.
(2pts)
(c) Test for the joint signi cance of the market and momentum factors using the information
provided in Figures 1 and 2. What conclusion do you draw? Is this a sensible conclusion
given your knowledge of models for asset pricing? (3pts)
3. In the regression results of Figure 1, the standard errors have been adjusted for presence of
heteroskedasticity.
(a) What is the source of the variation in error variance in (1) implied by this adjustment?
Write the corresponding auxiliary regression that you would use to test for existence of
this type of heteroskedasticity. (2pts)
(b) What other type of variation in error variance in (1) would you consider testing for?
Write the corresponding auxiliary regression that you would use to test for existence of
this type of heteroskedasticity. (2pts)
4. The results shown in Figure 3 correspond to the Breusch-Godfrey test for serial correlation
in residuals of (1). How do you relate these results to one or more of the stylized facts for
nancial returns that you know? Briey explain. (1pt)
Page 4 of 14
Figure 1
Figure 2
Figure 3
Page 5 of 14
Question 2 (16 marks)
Assume that time series fytgTt=1 is de ned by equations
yt = xt + "t (2)
xt = xt1 + t; (3)
where jj < 1 and "t WN
0; 2"
and t WN
0; 2
are independent stationary white noise
processes.
1. Compute the unconditional mean and variance of yt and nd its autocorrelation function.
(5pts)
2. Assume that xt are returns on ANZ stock and are normally distributed. We run OLS on (3)
over the period 5 Jan 2000 - 3 Jan 2018 and the tted regression is written as:
x^t = 0:04
(0:015)
xt1; t = 2; 3; : : : ; 4528:
Given the information shown in Table 1, derive the formula and numerical value for the one-
step ahead point forecast for xt and its 95% prediction interval. (3pts)
3. Analytically show that the forecast variance of xt converges to the unconditional variance of
xt as the forecast horizon, h, tends to in nity (h!1). (4pts)
4. Now assume that tjFt1 N
0; 2;t
. We re-run (3) over the period 5 Jan 2000 - 3 Jan 2018
and obtain the following additional conditional variance equation:
^2;t = 0:0004+
(0:000)
0:304
(0:017)
^2t1; t = 2; 3; : : : ; 4528:
Using the information shown in Table 1, compute the new 95% prediction interval for the one-
step ahead forecast of xt. Compare your results with those in Q2.2 and justify any di¤erences.
(4pts)
Table 1
T xT ^T ^
4528
(3 Jan 2018)
0.0039 0.0051 0.0243
Page 6 of 14
Question 3 (17 marks)
A researcher is interested in modelling the daily log returns for Honda, denoted by rH;t, over the
period 5 January 2004 to 30 April 2021.
1. Speci cally, he ts an ARMA(1,1) model to (rH;t rH)2, where rH = 1T
PT
t=1 rH;t and T is
the number of observations in the sample.
(a) What stylized fact of asset returns is the researcher interested in capturing by tting this
model? (1pt)
(b) What other model speci cation could he have used instead to obtain the same results?
Analytically prove the equivalence of these two models. (3pts)
2. The researcher settled for the speci cation implied by the results of Figure 4 in order to model
the Honda returns, rH;t.
(a) What observations did he make about the properties of the Honda returns to arrive at
this choice of model? (1pt)
(b) The News Impact Curve (NIC) associated with this model is shown in Figure 5. Provide
the expression of the NIC based on the estimation results of Figure 4 and explain which
parameters determine its shape. (3pts)
3. Figure 6 provides results for an ARCH(5) speci cation for the variance equation of Honda
returns.
(a) Verify that the parameter estimates in Figure 6 comply to the conditions of this model.
Why are these conditions necessary? (2pts)
(b) Formally compare the ARCH(5) model with the model implied by the results of Figure 4,
and verify that the researcher made the right choice in terms of model selection. Explain
the metrics that you used and the steps in your testing procedure. (3pts)
4. The researcher fears that he is not taking into consideration potential co-movements of the
Honda returns with returns on Toyota. He considers using a DCC model.
(a) Provide the formula for his conditional covariance matrix and explain the steps that the
researcher needs to go through in order to model this matrix. (3pts)
(b) Is he correct in opting for a DCC model over a BEKK model for this purpose? Explain
why. (1pt)
Page 7 of 14
Figure 4
Figure 5
News Impact Curve
Figure 6
Page 8 of 14
Question 4 (10 marks)
Consider the following model for non-syncronous trading. Denote the virtual returns for security
i by rit. In each period t, there is a probability i that security i does not trade, which is independent
from the virtual return rit:
it =
1 (no trade) with probability i
0 (trade) with probability 1 i ; it iid:
In each period t, the observed return r0it depends on whether or not security i trades in that period
or not. The observed returns are de ned as:
r0it =
1X
k=0
Xit (k) ri;tk;
where
Xit (k) = (1 it) i;t1i;t2 : : : i;tk; k > 0;
Xit (0) (1 it) :
Under these assumptions, consider running a linear regression of virtual returns for security i on an
intercept and the market factor, namely:
rit = i + ifmkt;t + "it;
where fmkt;t is the market factor which is a white noise stationary process with zero mean and
variance equal to 2f , and "it iid
0; 2"
.
1. Compute the expression for Cov (rit; fmkt;t). (2pts)
2. Compute the expression for Cov
r0it; fmkt;t
. (5pts)
3. Quantify how your population slope coe¢ cient i would change if observed returns for security
i were regressed on an intercept and the market factor. Interpret your results. (3pts)
END OF EXAMINATION
Page 9 of 14
TABLE G.2 Critical Values of the tDistribution
1-Tailed: .10 .05
2-Tailed: .20 .10
1 3.078 6.314
2 1.886 2.920
3 1.638 2.353
4 1.533 2.132
5 1.476 2.015
6 1.440 1.943
7 1.415 1.895
8 1.397 1.860
9 1.383 1.833
10 1.372 1.812
11 1.363 1.796
12 1.356 1.782 e
g 13 1.350 1.771
r 14 1.345 1.761
e 15 1.341 1.753
e
s 16 1.337 1.746
17 1.333 1.740
0 18 1.330 1.734
19 1.328 1.729
F 20 1.325 1.725
r 21 1.323 1.721
e
e 22 1.321 1.717
d 23 1.319 1.714
0 24 1.318 1.711
25 1.316 1.708
26 1.315 1.706
27 1.314 1.703
28 1.313 1.701
29 1.311 1.699
30 1.310 1.697
40 1.303 1.684
60 1.296 1.671
90 1.291 1.662
120 1.289 1.658
00 1.282 1.645
Significance Level
.025 .01 .005
.05 .02 .01
12.706 31.821 63.657
4.303 6.965 9.925
3.182 4.541 5.841
2.776 3.747 4.604
2.571 3.365 4.032
2.447 3.143 3.707
2.365 2.998 3.499
2.306 2.896 3.355
2.262 2.821 3.250
2.228 2.764 3.169
2.201 2.718 3.106
2.179 2.681 3.055
2.160 2.650 3.012
2.145 2.624 2.977
2.131 2.602 2.947
2.120 2.583 2.921
2.110 2.567 2.898
2.101 2.552 2.878
2.093 2.539 2.861
2.086 2.528 2.845
2.080 2.518 2.831
2.074 2.508 2.819
2.069 2.500 2.807
2.064 2.492 2.797
2.060 2.485 2.787
2.056 2.479 2.779
2.052 2.473 2.771
2.048 2.467 2.763
2.045 2.462 2.756
2.042 2.457 2.750
2.021 2.423 2.704
2.000 2.390 2.660
1.987 2.368 2.632
1.980 2.358 2.617
1.960 2.326 2.576
Examples: The 1 % critical value for a one-tailed test with 25 df is 2.485. The 5% critical value for a two-tailed test with large
(> 120) dfis 1.96.
Source: This table was generated using the Stata® function invttail.
STATISTICAL TABLES
Page 10 of 14
TABLE G.3a 10% Critical Values of the F Distribution
Numerator Degrees of Freedom
1 2 3 4 5 6
10 3.29 2.92 2.73 2.61 2.52 2.46
D
11 3.23 2.86 2.66 2.54 2.45 2.39
e 12 3.18 2.81 2.61 2.48 2.39 2.33
n 13 3.14 2.76 2.56 2.43 2.35 2.28
0 14 3.10 2.73 2.52 2.39 2.31 2.24
m
i 15 3.07 2.70 2.49 2.36 2.27 2.21
n 16 3.05 2.67 2.46 2.33 2.24 2.18
a 17 3.03 2.64 2.44 2.31 2.22 2.15
0 18 3.01 2.62 2.42 2.29 2.20 2.13
r 19 2.99 2.61 2.40 2.27 2.18 2.11
20 2.97 2.59 2.38 2.25 2.16 2.09
D .
e 21 2.96 2.57 2.36 2.23 2.14 2.08
g 22 2.95 2.56 2.35 2.22 2.13 2.06
r 23 2.94 2.55 2.34 2.21 2.11 2.05
e
24 e 2.93 2.54 2.33 2.19 2.10 2.04
s 25 2.92 2.53 2.32 2.18 2.09 2.02
26 2.91 2.52 2.31 2.17 2.08 2.01
0 27 2.90 2.51 2.30 2.17 2.07 2.00
28 2.89 2.50 2.29 2.16 2.06 2.00
F 29 2.89 2.50 2.28 2.15 2.06 1.99
r 30 2.88 2.49 2.28 2.14 2.05 1.98
e
e 40 2.84 2.44 2.23 2.09 2.00 1.93
d 60 2.79 2.39 2.18 2.04 1.95 1.87
0 90 2.76 2.36 2.15 2.01 1.91 1.84
m
120 2.75 2.35 2.13 1.99 1.90 1.82
00 2.71 2.30 2.08 1.94 1.85 1.77
Example: The 10% critical value for numerator df = 2 and denominator df = 40 is 2.44.
Source: This table was generated using the Stata® function invFtail.
7
2.41
2.34
2.28
2.23
2.19
2.16
2.13
2.10
2.08
2.06
2.04
2.02
2.01
1.99
1.98
1.97
1.96
1.95
1.94
1.93
1.93
1.87
1.82
1.78
1.77
1.72
8 9 10
2.38 2.35 2.32
2.30 2.27 2.25
2.24 2.21 2.19
2.20 2.16 2.14
2.15 2.12 2.10
2.12 2.09 2.06
2.09 2.06 2.03
2.06 2.03 2.00
2.04 2.00 1.98
2.02 1.98 1.96
2.00 1.96 1.94
1.98 1.95 1.92
1.97 1.93 1.90
1.95 1.92 1.89
1.94 1.91 1.88
1.93 1.89 1.87
1.92 1.88 1.86
1.91 1.87 1.85
1.90 1.87 1.84
1.89 1.86 1.83
1.88 1.85 1.82
1.83 1.79 1.76
1.77 1.74 1.71
1.74 1.70 1.67
1.72 1.68 1.65
1.67 1.63 1.60
Page 11 of 14
TABLE G.Jb 5% Critical Values of the f Distribution
Numerator Degrees of Freedom
1 2 3 4 5 6 7
D 10 4.96 4.10 3.71 3.48 3.33 3.22 3.14
e 11 4.84 3.98 3.59 3.36 3.20 3.09 3.01
n 12 4.75 3.89 3.49 3.26 3.11 3.00 2.91
0 13 4.67 3.81 3.41 3.18 3.03 2.92 2.83
m 14 4.60 3.74 3.34 3.11 2.96 2.85 2.76
15 4.54 3.68 3.29 3.06 2.90 2.79 2.71
n
a 16 4.49 3.63 3.24 3.01 2.85 2.74 2.66
t 17 4.45 3.59 3.20 2.96 2.81 2.70 2.61
0 18 4.41 3.55 3.16 2.93 2.77 2.66 2.58
r 19 4.38 3.52 3.13 2.90 2.74 2.63 2.54
20 4.35 3.49 3.10 2.87 2.71 2.60 2.51
D
21 4.32 3.47 3.07 2.84 2.68 2.57 2.49
g 22 4.30 3.44 3.05 2.82 2.66 2.55 2.46
r 23 4.28 3.42 . 3.03 2.80 2.64 2.53 2.44
e 24 4.26 3.40 3.01 2.78 2.62 2.51 2.42
e 25 4.24 3.39 2.99 2.76 2.60 2.49 2.40
26 4.23 3.37 2.98 2.74 2.59 2.47 2.39
0 27 4.21 3.35 2.96 2.73 2.57 2.46 2.37
f 28 4.20 3.34 2.95 2.71 2.56 2.45 2.36
29 4.18 3.33 2.93 2.70 2.55 2.43 2.35
F 30 4.17 3.32 2.92 2.69 2.53 2.42 2.33
r 40 4.08 3.23 2.84 2.61 2.45 2.34 2.25
e
60 4.00 3.15 e 2.76 2.53 2.37 2.25 2.17
d 90 3.95 3.10 2.71 2.47 2.32 2.20 2.11
0 120 3.92 3.07 2.68 2.45 2.29 2.17 2.09
m 00 3.84 3.00 2.60 2.37 2.21 2.10 2.01
Example: The 5% critical value for numerator df = 4 and large denominator df( oo) is 2.37.
Source: This table was generated using the Stata® function invFtail.
8 9 10
3.07 3.02 2.98
2.95 2.90 2.85
2.85 2.80 2.75
2.77 2.71 2.67
2.70 2.65 2.60
2.64 2.59 2.54
2.59 2.54 2.49
2.55 2.49 2.45
2.51 2.46 2.41
2.48 2.42 2.38
2.45 2.39 2.35
2.42 2.37 2.32
2.40 2.34 2.30
2.37 2.32 2.27
2.36 2.30 2.25
2.34 2.28 2.24
2.32 2.27 2.22
2.31 2.25 2.20
2.29 2.24 2.19
2.28 2.22 2.18
2.27 2.21 2.16
2.18 2.12 2.08
2.10 2.04 1.99
2.04 1.99 1.94
2.02 1.96 1.91
1.94 1.88 1.83
Page 12 of 14
TABLE G.3c 1 % Critical Values of the F Distribution
Numerator Degrees of Freedom
1 2 3 4 5 6
10 10.04 7.56 6.55 5.99 5.64 5.39
D 11 9.65 7.21 6.22 5.67 5.32 5.07
e 12 9.33 6.93 5.95 5.41 5.06 4.82
n
13 9.07 6.70 5.74 5.21 4.86 4.62
0
m 14 8.86 6.51 5.56 5.04 4.69 4.46
i 15 8.68 6.36 5.42 4.89 4.56 4.32
n 16 8.53 6.23 5.29 4.77 4.44 4.20
17 8.40 6.11 5.18 4.67 4.34 4.10 t
0 18 8.29 6.01 5.09 4.58 4.25 4.01
r 19 8.18 5.93 5.01 4.50 4.17 3.94
20 8.10 5.85 4.94 4.43 4.10 3.87
D
21 8.02 5.78 4.87 4.37 4.04 3.81 e
. g 22 7.95 5.72 4.82 4.31 3.99 3.76
r 23 7.88 5.66 4.76 ·4.26 3.94 3.71
e 24 7.82 5.61 4.72 4.22 3.90 3.67
e
25 7.77 5.57 4.68 4.18 3.85 3.63 s
26 7.72 5.53 4.64 4.14 3.82 3.59
0 27 7.68 5.49 4.60 4.11 3.78 3.56
f 28 7.64 5.45 4.57 4.07 3.75 3.53
F 29 7.60 5.42 4.54 4.04 3.73 3.50
r 30 7.56 5.39 4.51 4.02 3.70 3.47
e 40 7.31 5.18 4.31 3.83 3.51 3.29
e 60 7.08 4.98 4.13 3.65 3.34 3.12
0
90 6.93 4.85 4.01 3.54 3.23 3.01
m 120 6.85 4.79 3.95 3.48 3.17 2.96
00 6.63 4.61 3.78 3.32 3.02 2.80
Example: The I% critical value for numerator df = 3 and denominator df = 60 is 4.13.
Source: This table was generated using the Stata® function invFtail.
7
5.20
4.89
4.64
4.44
4.28
4.14
4.03
3.93
3.84
3.77
3.70
3.64
3.59
3.54
3.50
3.46
3.42
3.39
3.36
3.33
3.30
3.12
2.95
2.84
2.79
2.64
8 9 10
5.06 4.94 4.85
4.74 4.63 4.54
4.50 4.39 4.30
4.30 4.19 4.10
4.14 4.03 3.94
4.00 3.89 3.80
3.89 3.78 3.69
3.79 3.68 3.59
3.71 3.60 3.51
3.63 3.52 3.43
3.56 3.46 3.37
3.51 3.40 3.31
3.45 3.35 3.26
3.41 3.30 3.21
3.36 3.26 3.17
3.32 3.22 3.13
3.29 3.18 3.09
3.26 3.15 3.06
3.23 3.12 3.03
3.20 3.09 3.00
3.17 3.07 2.98
2.99 2.89 2.80
2.82 2.72 2.63
2.72 2.61 2.52
2.66 2.56 2.47
2.51 2.41 2.32
Page 13 of 14
TABLE G.4 Critical Values of the Chi-Square Distribution
Significance Level
.10 .05 .01
1 2.71 3.84 6.63
2 4.61 5.99 9.21
3 6.25 7.81 11.34
4 7.78 9.49 13.28
5 9.24 11.07 15.09
6 10.64 12.59 16.81
D 7 12.02 14.07 18.48
e 8 13.36 15.51 20.09
g 9 14.68 16.92 21.67
r 10 15.99 18.31 23.21
e 11 17.28 19.68 24.72
e 12 18.55 21.03 26.22
s 13 19.81 22.36 27.69
0
14 21.06 23.68 29.14
f 15 22.31 25.00 30.58
16 23.54 26.30 32.00
F 17 24.77 27.59 33.41
r 18 25.99 28.87 34.81
e 19 27.20 30.14 36.19
e 20 28.41 31.41 37.57
21 29.62 32.67 38.93
0
22 30.81 33.92 40.29
23 32.01 35.17 41.64
24 33.20 36.42 42.98
25 34.38 37.65 44.31
26 35.56 38.89 45.64
27 36.74 40.11 46.96
28 37.92 41.34 48.28
29 39.09 42.56 49.59
30 40.26 43.77 50.89
Example: The 5% critical value with df = 8 is 15.51.
Source: This table was generated using the Stata® function invchi2tail.
Page 14 of 14
学霸联盟
Semester One 2021
Exam - Alternative Assessment Task
UNIT CODE: ETC3460-ETC5346
UNIT TITLE: Financial Econometrics
ASSESSMENT DURATION: 2 hours 40 minutes (includes reading,
downloading, and uploading time)
This is an individual assessment task.
All responses must be included in ONE separate PDF document. Clearly
state your ID number, surname and given name at the top of this
document.
Clearly state the question numbers corresponding to your answers.
Students are required to answer ALL questions.
This assessment accounts for 60% of the total in the unit and has a hurdle
requirement of 45% to pass the unit.
Upon completion of this assessment task, please upload your PDF answer
document to Moodle using the assignment submission link.
You will sign the Student Statement shown on the next page
automatically when using the assignment submission link.
Your submission must occur within 2 hours 40 minutes of the official
commencement of this assessment task (Australian Eastern Standard
Time).
Please read the next page carefully (Student Statement) before
commencing the assessment task.
Page 2 of 14
Intentional plagiarism or collusion amounts to cheating under Part 7 of the Monash
University (Council) Regulations
Plagiarism: Plagiarism means taking and using another person’s ideas or manner of
expressing them and passing them off as one’s own. For example, by failing to give
appropriate acknowledgment. The material used can be from any source (staff, students or
the internet, published and unpublished works).
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Page 3 of 14
MARKS ALLOCATED TO QUESTIONS WITHIN THIS ASSESSMENT TASK
Question 1 2 3 4 5 6 7 8 9 TOTAL
Allocated Marks 17 16 17 10 60
Office Use Only
Mark received
Second marking
INSTRUCTIONS TO STUDENTS
When testing a hypothesis, to obtain full marks you need to specify the null and
the alternative hypotheses, the test statistic and its distribution under the null,
and then perform the test and state your conclusion.
If a question does not specify the level of signi cance of a hypothesis test explicitly,
use 5%.
Statistical tables are provided after Question 4.
Question 1 (17 marks)
1. Suppose that Daniel is interested in making an investment in assets A and B over the next
year. The expected returns on assets A and B are A = 0:1 and B = 0:35 and their volatilities
are equal to A = 0:14 and B = 0:27, respectively. The two asset returns have correlation
equal to A;B = 0:2.
(a) Provide the formulae and numerical values for the optimal (risk minimizing) weights for
assets A and B. [You are not required to solve the optimization problem]. (1pt)
(b) Write the regression equation that you would use to obtain the same weights. Explain
which parameter refers to the optimal weight(s) and analytically show why. (4pts)
2. Consider the following Fama-French four factor model
rt rft = 0 + 1(rmt rft) + 2SMBt + 3HMLt + 4MOMt + t; (1)
where rt denote monthly returns on the Fama-French hitec portfolio, (rmt rft) is the excess
market return (the market factor), rft is the risk free interest rate, SMB is the size factor,
HML is the value factor and MOM is the momentum factor. OLS output from estimating (1)
over the period January 1990 to January 2021 is shown in Figure 1.
(a) Test whether the hitec portfolio provides adequate reward for the assumed risk. (2pts)
(b) Test whether the hitec portfolio tracks the market versus that it is aggressive, at 1%
signi cance level. Based on your conclusion interpret the relevant coe¢ cient estimate.
(2pts)
(c) Test for the joint signi cance of the market and momentum factors using the information
provided in Figures 1 and 2. What conclusion do you draw? Is this a sensible conclusion
given your knowledge of models for asset pricing? (3pts)
3. In the regression results of Figure 1, the standard errors have been adjusted for presence of
heteroskedasticity.
(a) What is the source of the variation in error variance in (1) implied by this adjustment?
Write the corresponding auxiliary regression that you would use to test for existence of
this type of heteroskedasticity. (2pts)
(b) What other type of variation in error variance in (1) would you consider testing for?
Write the corresponding auxiliary regression that you would use to test for existence of
this type of heteroskedasticity. (2pts)
4. The results shown in Figure 3 correspond to the Breusch-Godfrey test for serial correlation
in residuals of (1). How do you relate these results to one or more of the stylized facts for
nancial returns that you know? Briey explain. (1pt)
Page 4 of 14
Figure 1
Figure 2
Figure 3
Page 5 of 14
Question 2 (16 marks)
Assume that time series fytgTt=1 is de ned by equations
yt = xt + "t (2)
xt = xt1 + t; (3)
where jj < 1 and "t WN
0; 2"
and t WN
0; 2
are independent stationary white noise
processes.
1. Compute the unconditional mean and variance of yt and nd its autocorrelation function.
(5pts)
2. Assume that xt are returns on ANZ stock and are normally distributed. We run OLS on (3)
over the period 5 Jan 2000 - 3 Jan 2018 and the tted regression is written as:
x^t = 0:04
(0:015)
xt1; t = 2; 3; : : : ; 4528:
Given the information shown in Table 1, derive the formula and numerical value for the one-
step ahead point forecast for xt and its 95% prediction interval. (3pts)
3. Analytically show that the forecast variance of xt converges to the unconditional variance of
xt as the forecast horizon, h, tends to in nity (h!1). (4pts)
4. Now assume that tjFt1 N
0; 2;t
. We re-run (3) over the period 5 Jan 2000 - 3 Jan 2018
and obtain the following additional conditional variance equation:
^2;t = 0:0004+
(0:000)
0:304
(0:017)
^2t1; t = 2; 3; : : : ; 4528:
Using the information shown in Table 1, compute the new 95% prediction interval for the one-
step ahead forecast of xt. Compare your results with those in Q2.2 and justify any di¤erences.
(4pts)
Table 1
T xT ^T ^
4528
(3 Jan 2018)
0.0039 0.0051 0.0243
Page 6 of 14
Question 3 (17 marks)
A researcher is interested in modelling the daily log returns for Honda, denoted by rH;t, over the
period 5 January 2004 to 30 April 2021.
1. Speci cally, he ts an ARMA(1,1) model to (rH;t rH)2, where rH = 1T
PT
t=1 rH;t and T is
the number of observations in the sample.
(a) What stylized fact of asset returns is the researcher interested in capturing by tting this
model? (1pt)
(b) What other model speci cation could he have used instead to obtain the same results?
Analytically prove the equivalence of these two models. (3pts)
2. The researcher settled for the speci cation implied by the results of Figure 4 in order to model
the Honda returns, rH;t.
(a) What observations did he make about the properties of the Honda returns to arrive at
this choice of model? (1pt)
(b) The News Impact Curve (NIC) associated with this model is shown in Figure 5. Provide
the expression of the NIC based on the estimation results of Figure 4 and explain which
parameters determine its shape. (3pts)
3. Figure 6 provides results for an ARCH(5) speci cation for the variance equation of Honda
returns.
(a) Verify that the parameter estimates in Figure 6 comply to the conditions of this model.
Why are these conditions necessary? (2pts)
(b) Formally compare the ARCH(5) model with the model implied by the results of Figure 4,
and verify that the researcher made the right choice in terms of model selection. Explain
the metrics that you used and the steps in your testing procedure. (3pts)
4. The researcher fears that he is not taking into consideration potential co-movements of the
Honda returns with returns on Toyota. He considers using a DCC model.
(a) Provide the formula for his conditional covariance matrix and explain the steps that the
researcher needs to go through in order to model this matrix. (3pts)
(b) Is he correct in opting for a DCC model over a BEKK model for this purpose? Explain
why. (1pt)
Page 7 of 14
Figure 4
Figure 5
News Impact Curve
Figure 6
Page 8 of 14
Question 4 (10 marks)
Consider the following model for non-syncronous trading. Denote the virtual returns for security
i by rit. In each period t, there is a probability i that security i does not trade, which is independent
from the virtual return rit:
it =
1 (no trade) with probability i
0 (trade) with probability 1 i ; it iid:
In each period t, the observed return r0it depends on whether or not security i trades in that period
or not. The observed returns are de ned as:
r0it =
1X
k=0
Xit (k) ri;tk;
where
Xit (k) = (1 it) i;t1i;t2 : : : i;tk; k > 0;
Xit (0) (1 it) :
Under these assumptions, consider running a linear regression of virtual returns for security i on an
intercept and the market factor, namely:
rit = i + ifmkt;t + "it;
where fmkt;t is the market factor which is a white noise stationary process with zero mean and
variance equal to 2f , and "it iid
0; 2"
.
1. Compute the expression for Cov (rit; fmkt;t). (2pts)
2. Compute the expression for Cov
r0it; fmkt;t
. (5pts)
3. Quantify how your population slope coe¢ cient i would change if observed returns for security
i were regressed on an intercept and the market factor. Interpret your results. (3pts)
END OF EXAMINATION
Page 9 of 14
TABLE G.2 Critical Values of the tDistribution
1-Tailed: .10 .05
2-Tailed: .20 .10
1 3.078 6.314
2 1.886 2.920
3 1.638 2.353
4 1.533 2.132
5 1.476 2.015
6 1.440 1.943
7 1.415 1.895
8 1.397 1.860
9 1.383 1.833
10 1.372 1.812
11 1.363 1.796
12 1.356 1.782 e
g 13 1.350 1.771
r 14 1.345 1.761
e 15 1.341 1.753
e
s 16 1.337 1.746
17 1.333 1.740
0 18 1.330 1.734
19 1.328 1.729
F 20 1.325 1.725
r 21 1.323 1.721
e
e 22 1.321 1.717
d 23 1.319 1.714
0 24 1.318 1.711
25 1.316 1.708
26 1.315 1.706
27 1.314 1.703
28 1.313 1.701
29 1.311 1.699
30 1.310 1.697
40 1.303 1.684
60 1.296 1.671
90 1.291 1.662
120 1.289 1.658
00 1.282 1.645
Significance Level
.025 .01 .005
.05 .02 .01
12.706 31.821 63.657
4.303 6.965 9.925
3.182 4.541 5.841
2.776 3.747 4.604
2.571 3.365 4.032
2.447 3.143 3.707
2.365 2.998 3.499
2.306 2.896 3.355
2.262 2.821 3.250
2.228 2.764 3.169
2.201 2.718 3.106
2.179 2.681 3.055
2.160 2.650 3.012
2.145 2.624 2.977
2.131 2.602 2.947
2.120 2.583 2.921
2.110 2.567 2.898
2.101 2.552 2.878
2.093 2.539 2.861
2.086 2.528 2.845
2.080 2.518 2.831
2.074 2.508 2.819
2.069 2.500 2.807
2.064 2.492 2.797
2.060 2.485 2.787
2.056 2.479 2.779
2.052 2.473 2.771
2.048 2.467 2.763
2.045 2.462 2.756
2.042 2.457 2.750
2.021 2.423 2.704
2.000 2.390 2.660
1.987 2.368 2.632
1.980 2.358 2.617
1.960 2.326 2.576
Examples: The 1 % critical value for a one-tailed test with 25 df is 2.485. The 5% critical value for a two-tailed test with large
(> 120) dfis 1.96.
Source: This table was generated using the Stata® function invttail.
STATISTICAL TABLES
Page 10 of 14
TABLE G.3a 10% Critical Values of the F Distribution
Numerator Degrees of Freedom
1 2 3 4 5 6
10 3.29 2.92 2.73 2.61 2.52 2.46
D
11 3.23 2.86 2.66 2.54 2.45 2.39
e 12 3.18 2.81 2.61 2.48 2.39 2.33
n 13 3.14 2.76 2.56 2.43 2.35 2.28
0 14 3.10 2.73 2.52 2.39 2.31 2.24
m
i 15 3.07 2.70 2.49 2.36 2.27 2.21
n 16 3.05 2.67 2.46 2.33 2.24 2.18
a 17 3.03 2.64 2.44 2.31 2.22 2.15
0 18 3.01 2.62 2.42 2.29 2.20 2.13
r 19 2.99 2.61 2.40 2.27 2.18 2.11
20 2.97 2.59 2.38 2.25 2.16 2.09
D .
e 21 2.96 2.57 2.36 2.23 2.14 2.08
g 22 2.95 2.56 2.35 2.22 2.13 2.06
r 23 2.94 2.55 2.34 2.21 2.11 2.05
e
24 e 2.93 2.54 2.33 2.19 2.10 2.04
s 25 2.92 2.53 2.32 2.18 2.09 2.02
26 2.91 2.52 2.31 2.17 2.08 2.01
0 27 2.90 2.51 2.30 2.17 2.07 2.00
28 2.89 2.50 2.29 2.16 2.06 2.00
F 29 2.89 2.50 2.28 2.15 2.06 1.99
r 30 2.88 2.49 2.28 2.14 2.05 1.98
e
e 40 2.84 2.44 2.23 2.09 2.00 1.93
d 60 2.79 2.39 2.18 2.04 1.95 1.87
0 90 2.76 2.36 2.15 2.01 1.91 1.84
m
120 2.75 2.35 2.13 1.99 1.90 1.82
00 2.71 2.30 2.08 1.94 1.85 1.77
Example: The 10% critical value for numerator df = 2 and denominator df = 40 is 2.44.
Source: This table was generated using the Stata® function invFtail.
7
2.41
2.34
2.28
2.23
2.19
2.16
2.13
2.10
2.08
2.06
2.04
2.02
2.01
1.99
1.98
1.97
1.96
1.95
1.94
1.93
1.93
1.87
1.82
1.78
1.77
1.72
8 9 10
2.38 2.35 2.32
2.30 2.27 2.25
2.24 2.21 2.19
2.20 2.16 2.14
2.15 2.12 2.10
2.12 2.09 2.06
2.09 2.06 2.03
2.06 2.03 2.00
2.04 2.00 1.98
2.02 1.98 1.96
2.00 1.96 1.94
1.98 1.95 1.92
1.97 1.93 1.90
1.95 1.92 1.89
1.94 1.91 1.88
1.93 1.89 1.87
1.92 1.88 1.86
1.91 1.87 1.85
1.90 1.87 1.84
1.89 1.86 1.83
1.88 1.85 1.82
1.83 1.79 1.76
1.77 1.74 1.71
1.74 1.70 1.67
1.72 1.68 1.65
1.67 1.63 1.60
Page 11 of 14
TABLE G.Jb 5% Critical Values of the f Distribution
Numerator Degrees of Freedom
1 2 3 4 5 6 7
D 10 4.96 4.10 3.71 3.48 3.33 3.22 3.14
e 11 4.84 3.98 3.59 3.36 3.20 3.09 3.01
n 12 4.75 3.89 3.49 3.26 3.11 3.00 2.91
0 13 4.67 3.81 3.41 3.18 3.03 2.92 2.83
m 14 4.60 3.74 3.34 3.11 2.96 2.85 2.76
15 4.54 3.68 3.29 3.06 2.90 2.79 2.71
n
a 16 4.49 3.63 3.24 3.01 2.85 2.74 2.66
t 17 4.45 3.59 3.20 2.96 2.81 2.70 2.61
0 18 4.41 3.55 3.16 2.93 2.77 2.66 2.58
r 19 4.38 3.52 3.13 2.90 2.74 2.63 2.54
20 4.35 3.49 3.10 2.87 2.71 2.60 2.51
D
21 4.32 3.47 3.07 2.84 2.68 2.57 2.49
g 22 4.30 3.44 3.05 2.82 2.66 2.55 2.46
r 23 4.28 3.42 . 3.03 2.80 2.64 2.53 2.44
e 24 4.26 3.40 3.01 2.78 2.62 2.51 2.42
e 25 4.24 3.39 2.99 2.76 2.60 2.49 2.40
26 4.23 3.37 2.98 2.74 2.59 2.47 2.39
0 27 4.21 3.35 2.96 2.73 2.57 2.46 2.37
f 28 4.20 3.34 2.95 2.71 2.56 2.45 2.36
29 4.18 3.33 2.93 2.70 2.55 2.43 2.35
F 30 4.17 3.32 2.92 2.69 2.53 2.42 2.33
r 40 4.08 3.23 2.84 2.61 2.45 2.34 2.25
e
60 4.00 3.15 e 2.76 2.53 2.37 2.25 2.17
d 90 3.95 3.10 2.71 2.47 2.32 2.20 2.11
0 120 3.92 3.07 2.68 2.45 2.29 2.17 2.09
m 00 3.84 3.00 2.60 2.37 2.21 2.10 2.01
Example: The 5% critical value for numerator df = 4 and large denominator df( oo) is 2.37.
Source: This table was generated using the Stata® function invFtail.
8 9 10
3.07 3.02 2.98
2.95 2.90 2.85
2.85 2.80 2.75
2.77 2.71 2.67
2.70 2.65 2.60
2.64 2.59 2.54
2.59 2.54 2.49
2.55 2.49 2.45
2.51 2.46 2.41
2.48 2.42 2.38
2.45 2.39 2.35
2.42 2.37 2.32
2.40 2.34 2.30
2.37 2.32 2.27
2.36 2.30 2.25
2.34 2.28 2.24
2.32 2.27 2.22
2.31 2.25 2.20
2.29 2.24 2.19
2.28 2.22 2.18
2.27 2.21 2.16
2.18 2.12 2.08
2.10 2.04 1.99
2.04 1.99 1.94
2.02 1.96 1.91
1.94 1.88 1.83
Page 12 of 14
TABLE G.3c 1 % Critical Values of the F Distribution
Numerator Degrees of Freedom
1 2 3 4 5 6
10 10.04 7.56 6.55 5.99 5.64 5.39
D 11 9.65 7.21 6.22 5.67 5.32 5.07
e 12 9.33 6.93 5.95 5.41 5.06 4.82
n
13 9.07 6.70 5.74 5.21 4.86 4.62
0
m 14 8.86 6.51 5.56 5.04 4.69 4.46
i 15 8.68 6.36 5.42 4.89 4.56 4.32
n 16 8.53 6.23 5.29 4.77 4.44 4.20
17 8.40 6.11 5.18 4.67 4.34 4.10 t
0 18 8.29 6.01 5.09 4.58 4.25 4.01
r 19 8.18 5.93 5.01 4.50 4.17 3.94
20 8.10 5.85 4.94 4.43 4.10 3.87
D
21 8.02 5.78 4.87 4.37 4.04 3.81 e
. g 22 7.95 5.72 4.82 4.31 3.99 3.76
r 23 7.88 5.66 4.76 ·4.26 3.94 3.71
e 24 7.82 5.61 4.72 4.22 3.90 3.67
e
25 7.77 5.57 4.68 4.18 3.85 3.63 s
26 7.72 5.53 4.64 4.14 3.82 3.59
0 27 7.68 5.49 4.60 4.11 3.78 3.56
f 28 7.64 5.45 4.57 4.07 3.75 3.53
F 29 7.60 5.42 4.54 4.04 3.73 3.50
r 30 7.56 5.39 4.51 4.02 3.70 3.47
e 40 7.31 5.18 4.31 3.83 3.51 3.29
e 60 7.08 4.98 4.13 3.65 3.34 3.12
0
90 6.93 4.85 4.01 3.54 3.23 3.01
m 120 6.85 4.79 3.95 3.48 3.17 2.96
00 6.63 4.61 3.78 3.32 3.02 2.80
Example: The I% critical value for numerator df = 3 and denominator df = 60 is 4.13.
Source: This table was generated using the Stata® function invFtail.
7
5.20
4.89
4.64
4.44
4.28
4.14
4.03
3.93
3.84
3.77
3.70
3.64
3.59
3.54
3.50
3.46
3.42
3.39
3.36
3.33
3.30
3.12
2.95
2.84
2.79
2.64
8 9 10
5.06 4.94 4.85
4.74 4.63 4.54
4.50 4.39 4.30
4.30 4.19 4.10
4.14 4.03 3.94
4.00 3.89 3.80
3.89 3.78 3.69
3.79 3.68 3.59
3.71 3.60 3.51
3.63 3.52 3.43
3.56 3.46 3.37
3.51 3.40 3.31
3.45 3.35 3.26
3.41 3.30 3.21
3.36 3.26 3.17
3.32 3.22 3.13
3.29 3.18 3.09
3.26 3.15 3.06
3.23 3.12 3.03
3.20 3.09 3.00
3.17 3.07 2.98
2.99 2.89 2.80
2.82 2.72 2.63
2.72 2.61 2.52
2.66 2.56 2.47
2.51 2.41 2.32
Page 13 of 14
TABLE G.4 Critical Values of the Chi-Square Distribution
Significance Level
.10 .05 .01
1 2.71 3.84 6.63
2 4.61 5.99 9.21
3 6.25 7.81 11.34
4 7.78 9.49 13.28
5 9.24 11.07 15.09
6 10.64 12.59 16.81
D 7 12.02 14.07 18.48
e 8 13.36 15.51 20.09
g 9 14.68 16.92 21.67
r 10 15.99 18.31 23.21
e 11 17.28 19.68 24.72
e 12 18.55 21.03 26.22
s 13 19.81 22.36 27.69
0
14 21.06 23.68 29.14
f 15 22.31 25.00 30.58
16 23.54 26.30 32.00
F 17 24.77 27.59 33.41
r 18 25.99 28.87 34.81
e 19 27.20 30.14 36.19
e 20 28.41 31.41 37.57
21 29.62 32.67 38.93
0
22 30.81 33.92 40.29
23 32.01 35.17 41.64
24 33.20 36.42 42.98
25 34.38 37.65 44.31
26 35.56 38.89 45.64
27 36.74 40.11 46.96
28 37.92 41.34 48.28
29 39.09 42.56 49.59
30 40.26 43.77 50.89
Example: The 5% critical value with df = 8 is 15.51.
Source: This table was generated using the Stata® function invchi2tail.
Page 14 of 14
学霸联盟
