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Financial Econometrics

Part 1a
...

ii
...

iv
...


Part 1b
...
Write the estimated fitted model; comment on the coefficients’ significance at 1% and at 5%
...


Perform the t-test for
(a) H0 : β2=0 vs
...
5 vs
...
5 at 5%
...
NB: assume known any inputs needed!

Part 2
...
Are the returns normally distributed?
(Hint: you can use graphical procedures, moments & normality tests)

• Use log-returns on your asset (Rt) and on the index (Rmt) in order to estimate the market
model regression:
Rt = a + bRmt + ut
Discuss and interpret the regression output (coefficients, standard errors, t-stats, p-values) and
basic diagnostics (R2, F-stat)
...
Presentation (20%)

School of Humanities, Social Sciences and Economics MSc in Banking & Finance
MSc in International Accounting, Auditing, and Financial Management

Financial Econometrics
Coursework

1

Contents
1
...
Theoretical Analysis

3

2
...
Theoretical Analysis

8

3
...
Data Analysis

20

Stock Quant Report - Model Estimation and Testing

20

4
...
Presentation

30

5
...
Part 1a
...
0) where the disturbance
is ut ~ iid N(0,σ2)

i
...
We are interested in testing stationarity for the model; in
other words, to test if the statistical properties of the process do not change over time
...
1)
and the corresponding AR characteristic Equation:
φ(x) = 1 - φ1x - φ2x2 = 0 (1
...
We can also
say that the model (1
...

This statement can be generalized to the pth-order case without change
...

This will be true if, and only if, three conditions are satisfied:
𝜑𝜑1 + 𝜑𝜑2 < 1,

𝜑𝜑2 − 𝜑𝜑1 < 1,

𝑎𝑎𝑎𝑎𝑎𝑎 |𝜑𝜑2 | < 1

Proof: Let the reciprocals of the roots be denoted G1 and G2:
G1 =

2𝜑𝜑2

−𝜑𝜑1 −�𝜑𝜑12 +4𝜑𝜑2

Similarly, G2 =

=

2𝜑𝜑2

�

−𝜑𝜑1 +�𝜑𝜑12 +4𝜑𝜑2

−𝜑𝜑1 −�𝜑𝜑12 +4𝜑𝜑2 −𝜑𝜑1 +�𝜑𝜑12 +4𝜑𝜑2

�=

2𝜑𝜑2 (−𝜑𝜑1 +�𝜑𝜑12 +4𝜑𝜑2 )
𝜑𝜑12 −(𝜑𝜑12 +4𝜑𝜑2 )

=

𝜑𝜑1 −�𝜑𝜑12 +4𝜑𝜑2
2

𝜑𝜑1 +�𝜑𝜑12 +4𝜑𝜑2

2

3

We now divide the proof into two cases corresponding to real and complex roots
...

a
...
Now -2 < 𝜑𝜑1 - �𝜑𝜑12 + 4𝜑𝜑2 if, and only if,
�𝜑𝜑12 + 4𝜑𝜑2 <𝜑𝜑1 + 2 if, and only if, 𝜑𝜑12 + 4𝜑𝜑2 <𝜑𝜑12 + 4𝜑𝜑1+4, if and only if, 𝜑𝜑2 < 𝜑𝜑1 +1,

or 𝝋𝝋𝟐𝟐 − 𝝋𝝋𝟏𝟏 < 𝟏𝟏
...


These equations together with 𝝋𝝋𝟐𝟐𝟏𝟏 + 𝟒𝟒𝝋𝝋𝟐𝟐 ≥ 0 define the stationarity for the real
root case
...

The stationarity region for the AR(2) model is displayed in the following diagram
(Figure 1), where stationarity requires that the roots have to be inside the red,
blue and green triangle:

Figure 1: The real roots of the AR(2) model

4

b
...
Here G1 and G2 will be complex conjugates and |𝐺𝐺1 | = |𝐺𝐺2 | <
1 if and only if |𝐺𝐺1 |2 < 1
...
This together with the

inequality
+ 4𝜑𝜑2 < 0 defines the part of the stationarity region for complex
roots
...


Figure 2: The complex unit circle

ii
...
bauer
...
edu/rsusmel/4397/fec-8
...
0) is stationary and the roots are outside the unit circle,
the mean of the AR(2) model is2:
Ε[yt] = E[φ(L)ϵt] = μ/ (1 − 𝜑𝜑1 − 𝜑𝜑2) = 0

given that 𝜑𝜑1 + 𝜑𝜑2 ≠ 1
...

iii
...
0) model's Variance, it should be stated that the Variance of an AR(p)
model is only finite if it is stationary
...
0) and multiply both sides by Yt-k
...
3) for k=1,2,3,…
...

In all cases, the magnitude of γk dies out exponentially fast as the lag k increases
...
0) yields
γ0 = (φ12+ φ22) γ0 + 2𝜑𝜑1 𝜑𝜑2 +𝜎𝜎2𝛼𝛼 (1
...
3) gives a second linear equation γ0 and γ1, γ1= 𝜑𝜑1 γ0 + 𝜑𝜑2 γ1,
which can be solved simultaneously with Equation (1
...
bauer
...
edu/rsusmel/4397/fec-8
...
econ
...
ac
...
pdf
https://bookdown
...
html

6

iv
...

It is used to describe specific time-varying processes in economics
...

Taking into account the previous analysis, we can come to a conclusion that, regarding
stationarity, mean, and Variance, the AR(2) model follows the same conditions as the
AR(1)
...
The same conditions apply both for AR(1) and AR(2) models
...

The mean for AR(1) model has been proved to be zero 3
...
The same applies for
the AR(2) model if it has proved to be stationary
...
As with the
AR(1) process, the variance of an AR(2) process is only finite if it is stationary
...
Therefore, as it was proved in
(iii), the AR(2) model follows also the conditions of the AR(1) model for the Variance
...
sfu
...
pdf

3

https://bookdown
...
html#ar1-process

7

https://stats
...
com/questions/118019/a-proof-for-the-stationarity-of-an-ar2
http://www2
...
osaka-u
...
jp/~tanizaki/class/2014/model_analysis1/08
...
bauer
...
edu/rsusmel/phd/ec2-3
...
ucsd
...
pdf
https://www
...
com/ar2_process
...
divms
...
edu/~kchan/s156
...
pdf
https://en
...
org/wiki/Autoregressive_model#Explicit_mean/difference_form_of_A
R(1)_process
https://www
...
uh
...
pdf
http://www2
...
osaka-u
...
jp/~tanizaki/class/2014/model_analysis1/08
...
org/gary_a_napier/time_series_lecture_notes/ChapterThree
...
Part 1b
...
Write the estimated fitted model; comment on the coefficients’ significance at 1% and
at 5%
...
𝟑𝟑𝟑𝟑 𝑸𝑸 + 𝒖𝒖

The regression coefficients β0, β1, β2, β3, β4 are known as partial regression or partial
regression coefficients
...
Τhe same applies for the coefficients β2, β3,

8

β4
...

To test the "significance of the correlation coefficient," we perform a hypothesis test
...
We choose this based on the
sample correlation coefficient and the sample size T 4
...

Coefficients' significance for a significance level of 1% (α = 0
...
24−0
1
...
733

Test statistics derived in this way can be shown to follow a t- distribution with T-2
degrees of freedom
...

From the t-table of the critical values, we use the value for 300 degrees of freedom,
as it is closer to 298 degrees of freedom for our model
...

Given a significance level of 1%, α = 0
...
005)
...
lumenlearning
...
01 and for 2-sided test
...

The critical values are -2
...
5923
...
Therefore, we cannot reject the null hypothesis
...
(Method 1)
•

H0: β1 = 0
H1: β1 ≠ 0
The first step is to calculate the test statistic, where 𝛽𝛽 ∗ is the β enter the null
hypothesis:
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =

�1 − 𝛽𝛽∗
𝛽𝛽
�1 )
𝑠𝑠𝑠𝑠(𝛽𝛽

=

0
...
08

= 2
...
The degrees of freedom for the model are T – 2 = 300 – 2 = 298
...
01, we can determine a rejection region and a
non-rejection region for a 2-sided test
...
The critical values are -2
...
5923
...
Therefore, we cannot reject the null
hypothesis
...

•

H0: β2 = 0
H1: β2 ≠ 0

10

The first step again is to calculate the test statistic, where 𝛽𝛽 ∗ is the β enter the null
hypothesis:
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =

�2 − 𝛽𝛽∗
𝛽𝛽
�2 )
𝑠𝑠𝑠𝑠(𝛽𝛽

=

−0
...
001

= -2
...
Consequently, the degrees of freedom for the model are 298
...
01, we can determine a rejection region and a
non-rejection region for a 2-sided test (α/2 = 0
...
We use a t-table to obtain a
critical value with which we can compare the test statistic
...
5923 and +2
...
The test statistic lies in the non-rejection region
...

•

H0: β3 = 0
H0: β3 ≠ 0
We calculate the test statistic, where 𝛽𝛽 ∗ equals to the β enter the null hypothesis:
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =

�3 − 𝛽𝛽∗
𝛽𝛽
�3 )
𝑠𝑠𝑠𝑠(𝛽𝛽

=

0
...
03

= 2
...
So, the degrees of freedom remain 298
...
01, we can determine a rejection region and a
non-rejection region for a 2-sided test (α/2 = 0
...
We use a t-table to obtain a
critical value with which we can compare the test statistic
...
5923 and +2
...
The test statistic lies in the non-rejection region
...

•

H0: β4 = 0
H0: β4 ≠ 0
The coefficient's β4 p-value is 0
...
According to the theory, if the p-value is less than
the significance level (α = 0,01), we can reject the null hypothesis
...
In conclusion, we reject the null hypothesis and support that there is a
11

significant linear relationship between Q and Y because the correlation coefficient is
significantly different from zero
...

Finally, it is important to be mentioned, that according to the previous result we
expect that β4 will be statistically significant in the significance level of 5% and
generally in every other lower significance level
...
05):
•

H0: β0 = 0
H1: β0 ≠ 0
The first step again is to calculate the test statistic, where 𝛽𝛽 ∗ is the β enter the null
hypothesis:
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =

�0 − 𝛽𝛽∗
𝛽𝛽
�0 )
𝑠𝑠𝑠𝑠(𝛽𝛽

=

3
...
87

= 1
...
The degrees of freedom for the model are for another time T – 2
= 300 – 2 = 298
...
05, we can determine a rejection region and a
non-rejection region for a 2-sided test (α/2 = 0
...


Figure 4: Rejection and non-rejection region for α = 0
...


12

We use a t-table to obtain a critical value with which we can compare the test statistic
...
9679 and +1
...
The test statistic lies in the non-rejection
region
...

•

H0: β1 = 0
H1: β1 ≠ 0
We firstly calculate the test statistic, where 𝛽𝛽 ∗ is the β enter the null hypothesis:
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =

�1 − 𝛽𝛽∗
𝛽𝛽
�1 )
𝑠𝑠𝑠𝑠(𝛽𝛽

=

0
...
08

= 2
...
The degrees of freedom for the model is 298
...
05, we can determine a rejection region and a
non-rejection region for a 2-sided test (α/2 = 0
...
We use as always a t-table to
obtain a critical value with which we can compare the test statistic
...
9679 and +1
...
The test statistic lies in the rejection region
...

•

H0: β2 = 0
H1: β2 ≠ 0
The first step for another time includes the calculation of the test statistic, where 𝛽𝛽 ∗
is the β enter the null hypothesis:
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =

�2 − 𝛽𝛽∗
𝛽𝛽
�2 )
𝑠𝑠𝑠𝑠(𝛽𝛽

=

−0
...
001

= -2
...
The degrees of freedom for the model are 298
...
05, we can determine a rejection region and a
non-rejection region for a 2-sided test (α/2 = 0
...
We use again a t-table to obtain
a critical value with which we can compare the test statistic
...
9679 and +1
...
The test statistic lies in the rejection region
...

•

H0: β3 = 0
H1: β3 ≠ 0
We firstly calculate for another time the test statistic, where 𝛽𝛽 ∗ is the β enter the
null hypothesis:
𝑡𝑡𝑡𝑡𝑡𝑡𝑡𝑡 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =

�3 − 𝛽𝛽∗
𝛽𝛽
�3 )
𝑠𝑠𝑠𝑠(𝛽𝛽

=

0
...
03

= 2
...
The degrees of freedom for the model are 298
...
05, we can determine a rejection region and a
non-rejection region for a 2-sided test
...
The critical values are -1
...
9679
...
To conclude, we can reject the
null hypothesis, which means that there is a significant linear relationship between P
and Y because the correlation coefficient is significantly different from zero
...
According to the theory, if the p-value is less than
the significance level (α = 0,05), we reject the null hypothesis
...
Therefore,
we can reject the null hypothesis and conclude that there is a significant linear
relationship between Q and Y because the correlation coefficient is significantly
different from zero, something which was expected since it is evident previously that
β4 was highly statistically significant (in significance level of 1%)
...
500
- 2
...
330
1
...
002
-

Significant at 5%
YES
YES
YES
YES
NO

Significant at 1%
NO
NO
NO
YES
NO
14

Table 1: Summary of the significance of the coefficients at α=0
...
05

ii
...
H0: β2 = 0 vs H1: β2 < 0 at 5%; what about at 1% ?
To test if the β2 is statistically significant, we use the t-test to test single hypotheses, i
...

hypotheses involving only one coefficient
...

So:
H0: β2 = 0
H1: β2 < 0
It is followed exactly the same procedure (Method 1) which we used before
...
0023−0
0
...
3

Test statistics derived in this way can be shown to follow a t- distribution with T-2 degrees
of freedom
...
From the ttable of the critical values, we use the value for 300 degrees of freedom, as it is closer
to 298 degrees of freedom for our model
...
05, we can determine a rejection region and a nonrejection region, but for a 1-sided test this time
...
05 and for 1-sided test
...
The critical value is - 1
...
The test statistic lies in the rejection region
...

On the other hand, given a significance level of 1%, α = 0
...
3388
...
Consequently, we cannot reject the null
hypothesis that the β2 coefficient is 0 at a significance level of 1%, which means that we
can consider β2 as statistically insignificant at this level
...
H0: β1 = 0
...
5 at 5%
...
5, we use the t-test to test single hypotheses, i
...

hypotheses involving only one coefficient
...
20−0
...
08

= -3
...
05, we can determine a rejection region and a nonrejection region for a 2-sided test
...
The critical values are -1
...
9679
...
Consequently, we can reject the null hypothesis that
the β1 coefficient is equal to 0
...


c
...


H0: β1+β2=0

In many cases, we are interested in testing a hypothesis involving more than one of
the population parameters
...
The initial given
model is:
𝒀𝒀 = 𝜷𝜷𝟎𝟎 + 𝜷𝜷𝟏𝟏 𝑿𝑿 + 𝜷𝜷𝟐𝟐 𝑿𝑿𝟐𝟐 + 𝜷𝜷𝟑𝟑 𝑷𝑷 + 𝜷𝜷𝟒𝟒 𝑸𝑸 + 𝒖𝒖

To present the theory for this kind of a test, firstly, we should define its hypotheses
...
Consequently, the t statistic
should now be based on whether the estimated sum 𝛽𝛽̂1 + 𝛽𝛽̂2 = 0 is significantly
different from 0 in order to be able to reject H0
...
5 The first one based on the covariance matrix of the estimators, in
contrast to the second one where the model is reparametrized by introducing a new
parameter:
1st Procedure: Using Covariance Matrix of Estimators
Taking into account, firstly, the sampling error in our estimators, we must standardize
this sum by dividing by its standard error:

We also have:

Where:

𝑡𝑡𝛽𝛽�1+𝛽𝛽�2 =

�1 + 𝛽𝛽
�2
𝛽𝛽
�1 + 𝛽𝛽
�2 )
𝑠𝑠𝑠𝑠(𝛽𝛽

�
�1 + 𝛽𝛽
�2 � = �𝑣𝑣𝑣𝑣𝑣𝑣 (𝛽𝛽
̂1 + 𝛽𝛽̂2 )
𝑠𝑠𝑠𝑠�𝛽𝛽
�+ 𝛽𝛽�)= 𝑣𝑣𝑣𝑣𝑣𝑣(𝛽𝛽
��) + 𝑣𝑣𝑣𝑣𝑣𝑣(𝛽𝛽
��) + 2 ∗ 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐(𝛽𝛽
�
�
�
�
𝑣𝑣𝑣𝑣𝑣𝑣(𝛽𝛽
1
2
1
2
1 , 𝛽𝛽2 )

�1 + 𝛽𝛽
�2 �, we need information on the estimated covariance of
Hence, to calculate 𝑠𝑠𝑠𝑠�𝛽𝛽
estimators
...
If we have the covariance, we can calculate the t
statistic
...

Finally, we have to notice that in this case the suitable type of test in order to exam
the estimated sum 𝛽𝛽̂1 + 𝛽𝛽̂2 is a two-sided t-test
...
coursehero
...

Firstly, we define:
θ = β1 + β2
thus, the null hypothesis now is equivalent to H0: θ = 0
...
Substituting β2 in the original equation, we have:

Hence,

𝒀𝒀 = 𝜷𝜷𝟎𝟎 + 𝜷𝜷𝟏𝟏 𝑿𝑿 + (𝜽𝜽 − 𝜷𝜷𝟏𝟏 ) 𝑿𝑿𝟐𝟐 + 𝜷𝜷𝟑𝟑 𝑷𝑷 + 𝜷𝜷𝟒𝟒 𝑸𝑸 + 𝒖𝒖
𝒀𝒀 = 𝜷𝜷𝟎𝟎 + 𝜷𝜷𝟏𝟏 (𝑿𝑿 − 𝑿𝑿𝟐𝟐 ) + 𝜽𝜽 𝑿𝑿𝟐𝟐 + 𝜷𝜷𝟑𝟑 𝑷𝑷 + 𝜷𝜷𝟒𝟒 𝑸𝑸 + 𝒖𝒖

Therefore, to test whether β1 + β2 = 0 is equivalent to carrying out a significance test
on the coefficient of X2 in the above model
...

Finally, we estimate the output for the reparametrized model, and we run a two-sided
t-test:
𝑡𝑡𝜃𝜃� =

𝜃𝜃�

𝑠𝑠𝑠𝑠(𝜃𝜃�)

If the test statistic lies in the rejection area, we can reject the null hypothesis that β1
+ β2 is equal to zero, while if this does not happened, we cannot
...


The joint linear restrictions of the hypothesis H0: β1+β2=0 and β3=0
...
A test of multiple restrictions is
called a joint hypothesis test
...
The unrestricted model is the initial model, while the restricted model is
obtained by imposing H0 on the original model
...
The F – statistic (or F – ratio) is defined
by:

Where:






Test statistic =

𝑅𝑅𝑅𝑅𝑅𝑅𝑅𝑅−𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈
𝑈𝑈𝑈𝑈𝑈𝑈𝑈𝑈

×

𝑇𝑇−𝑘𝑘
𝑚𝑚

T is the number of the observations (T= 300 for our model)
m is the number of restrictions (m= 2 for our model)
k is the number of regressors in unrestricted regression (k= 5 for our model)
URSS is the residual sum of squares from unrestricted regression
RRSS is the residual sum of squares from restricted regression

The test statistic follows an F-distribution, which has 2 d
...
parameters
...
The appropriate
critical value will be in column m, row T-k
...
We therefore only reject the null hypothesis if
the test statistic is higher than the critical F-value
...
It is still true that:
RRSS ≥ URSS
Because OLS are chosen to minimize the sum of squared residuals, the RSS never
decreases when specific restrictions are introduced into the model
...
If we obtain a large increase, this is evidence against H0, and the
hypothesis will be rejected and it will be concluded that the restrictions were
imposed, were not supported by the data
...
Generally:
19

If RRSS ≈ URSS , then the F-statistic will be "small" and we cannot reject the H0
...


3
...
Data Analysis
Stock Quant Report - Model Estimation and Testing
Given an asset and an index series undertake a graphical and descriptive statistical
analysis of your logarithmic stock returns
...
There are certain tools available
to us in order to determine if a dataset comes from a normal distribution or not
...
The tools that we will use for our purpose are the following:
a
...
The graph (Figure 6) of the
logarithmic stock returns of the given asset and index is the following:
0,5
0,4
0,3
0,2
0,1
log_asset

•

-0,3

-0,25

-0,2

-0,15

-0,1

-0,05

0
-0,1

0

0,05

0,1

0,15

0,2

0,25

-0,2
-0,3
-0,4
-0,5
-0,6
log_index
Figure 6: Graph of the logarithmic stock returns of the given asset and index

20

In order to test for normality, we should use the residuals of the model regression
...
One
of the assumptions of linear regression is that the residuals are drawn from a normal
distribution; another way of saying this would be that the plot a histogram would be
normal-like
...

Still, the histogram can be deceptive since changing the number of bins alter the shape of
the distribution, and this may lead to some confusion
...


21

Figure 9: Histogram of the residuals

Figure 9 above shows a bell-shaped distribution of the residuals
...
The residuals are
generally not normally distributed as their mean is much higher than the normal
distribution, while the residuals on the tails are not evenly distributed in order to follow
a normal curve
...

This statement can also be proved by the following graph (Figure 10)
...

b
...
If our data comes from a normal distribution, the points on the graph will form
a line
...
Therefore, it could be
concluded that the data are not normally distributed
...
Quantile-quantile plot
An alternative to using a histogram is to use a quantile-quantile plot
...
The y-coordinate of a qq plot is the
dataset values, the x coordinates are values from the normal distribution
...
If the
majority of points fall onto this line, then we assume that the data represents a normal
distribution
...
As we can observe in Figure 12 only the data in the
middle of the graph fall onto the line, while residuals tend to devia

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