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Semester One 2020
Exam - Alternative Assessment Task



UNIT CODE: ETC3460-BEX3460-ETC5346

UNIT TITLE: Financial Econometrics

ASSESSMENT DURATION: 3 hours 30 minutes (includes reading,
downloading, and uploading time)


 This is an individual assessment task.

 All responses must be included in ONE separate 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 40% to pass the unit.

 Upon completion of this assessment task, please upload this 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 3 hours 30 minutes of the official
commencement of this assessment task (Australian Eastern Daylight
Time).

Please read the next page carefully (Student Statement) before
commencing the assessment task.

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University (Council) Regulations

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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
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 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
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collaboration when preparing this assessment.

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Date:

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MARKS ALLOCATED TO QUESTIONS WITHIN THIS ASSESSMENT TASK
Question 1 2 3 4 5 6 7 8 9 TOTAL
Allocated Marks 15 15 15 15 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 (15 marks)
This question uses monthly returns on the Fama-French manufacturing portfolio, denoted by rt.
Consider the following Fama-French three factor model
rt rft = + 1(rmt rft) + 2SMBt + 3HMLt + vt; (1)
where (rmt rft) is the market factor, rft is the risk free interest rate, SMB is the size factor, and
HML is the value factor.
1. The expression on the left hand side of (1), (rt rft), is referred to as excess return on the
manufacturing portfolio at time t.
a. Provide a de…nition of the term ‘excess return’on a …nancial asset. (1pt)
b. Explain how this quantity relates to parameter in (1). (1pt)
2. When estimating (1) over the period January 1990 to January 2020, we obtain the results
shown in Figure 1.
a. Provide the …tted regression equation. Neatly report the coe¢ cient estimate of the size
factor and test for its statistical signi…cance at 1% signi…cance level. Based on your
conclusion interpret this coe¢ cient estimate. (3pts)
b. How would you test for joint signi…cance of the size and value factors? Provide all required
steps. If you found that the size and value factors were jointly not statistically signi…cant
what conclusion would you draw in terms of (1)? (3pts)
3. Use the residuals obtained from regression (1), v^t.
a. Explain how you would check whether the E¢ cient Market Hypothesis (EMH) holds and
why. Provide the required steps. (2pts)
b. Assume that the R2 of the relevant auxiliary regression amounts to 0.56. What conclusion
would you draw when comparing your results with the appropriate distribution with 5
degrees of freedom? (1pt)
Page 4 of 14
4. Given the results in Q1.3 we re-run (1) and results are shown in Figure 2.
a. Explain what change was made to the regression speci…cations and why. (1pt)
b. For what other reason would you make the same change? Explain how you would go
about testing this property using the residuals in (1). [Hint: we have discussed 3 such
tests. Either one of these is acceptable]. (2pts)
c. Do the results in Figure 2 change your conclusions in Q1.2.a on the coe¢ cient estimate
of the size factor? (1pt)
Figure 1
Figure 2
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Question 2 (15 marks)
Assume a time series fytgTt=1. Its corresponding correlogram is displayed in Figure 3.
1. Provide a de…nition for stationarity of a time series. Based on the information in Figure 3 do
you consider yt to be a stationary or non-stationary time series? Brie‡y explain why. (3pts)
2. Based on the information in Figure 3, a researcher decides to use the following model to …t yt:
yt = 0 + 1yt1 + 2yt2 + "t;
where "t WN(0; 2"). Justify his choice. (2pts)
3. Let Ft1 denote all information available at time t 1. Derive the conditional mean of yt
given Ft1 and derive the unconditional mean of yt. What is the condition for this process to
have a constant unconditional mean? (3pts)
4. Derive the expression for the 3-step ahead point forecast based on this model. (4pts)
5. Derive the unconditional autocovariance function, k = Cov (yt; ytk), for k = 0 and k 6= 0.
Here we assume that the unconditional mean of yt derived in Q2.3 is equal to zero. (3pts)
Figure 3
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Question 3 (15 marks)
A researcher is interested in modelling the Nikkei index daily log returns over the period 5
January 2000 to 8 May 2020. He uses a constant mean equation and initially considers an ARCH(5)
speci…cation for the variance equation.
1. Which stylised feature(s) of the Nikkei log returns does the researcher attempt to address by
using an ARCH model? Give a brief explanation of the feature(s). (1pt)
2. The estimation results from this ARCH(5) model are shown in Figure 4. Write the theoretical
model that is implied by the regression output in Figure 4. Explain what conditions are needed
on the slope parameters of this model and why. Given the information in Figure 4, are these
conditions satis…ed in this case? (3pts)
3. Next, the researcher considers a GARCH speci…cation and re-estimates his model. The results
are shown in Figure 5.
a. What is his motivation for considering this new model for the variance equation? Derive
the relationship between this new model and an ARCH model. [Hint: start with the
GARCH equation and substitute recursively the lagged conditional variance component]
(4pts)
b. Using the information in Figures 4 and 5 evaluate which out of the GARCH and ARCH
models you would opt for. Explain your reasoning and metrics that you used. (2pts)
4. Next, the researcher considers a GJR-GARCH speci…cation and re-estimates his model. The
results are shown in Figure 6.
a. What is his motivation for considering this new model for the variance equation? Write
the theoretical model speci…cation that is implied by the regression output in Figure 6
and interpret the coe¢ cient estimate of the threshold parameter. Is its sign reasonable?
Brie‡y explain. (2pts)
b. Using the information in Figures 5 and 6, evaluate which out of the GARCH and GJR-
GARCH models you would opt for. Explain your reasoning and metrics that you used.
Provide all steps in your testing procedure to obtain full credit. (3pts)
Figure 4
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Figure 5
Figure 6
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Question 4 (15 marks)
Let Pt be a random walk process:
Pt = Pt1 + "t;
where "t WN

0; 2

.
1. Show that for an initial condition P0 = 1, we have that:
Pt = P0 +
tX
i=1
"i;
and compute E (Pt). (2pts)
2. Show that k = Cov (Pt; Ptk) = (t k)2. Brie‡y comment on this result. (3pts)
3. Assume the corresponding log-return is distributed as follows:
rt N

r;
2
r

:
a. Show that the simple net return has a log-normal distribution with:
E (Rt) = e
r+0:5
2
r 1:
(4pts)
b. Show that the simple net return has a log-normal distribution with:
V (Rt) = e
2r+
2
r

e
2
r 1

:
(6pts)
[Hint: Recall that rt = ln (1 +Rt) and start with deriving lnE (1 +Rt).]
[Hint: Use the fact that lnE (a) = E [ln (a)] + 0:5V [ln (a)].]
[Hint: Recall that V (Rt) = E

R2t
E (Rt)2.]
— — — — — END OF EXAMINATION — — — — —
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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
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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

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