EC388: FINANCIAL ECONOMETRICS
Mark scheme/answers for final exam May 2013
Question 1
a)
For calculating the determinant (4,500) and the inverse matrix 25 marks. A further 10
marks for correct coefficient estimates.
(X X) 1
b)
 0.04 - 0.02 - 0.02
- 0.02 0.06 0.01 


- 0.02 0.01 0.11 

b

- 1.0
 2.0 


 5.0 
The cov-var matrix is (all steps must be shown in its calculation for 35 marks):
est.V(b) 
 0.22 - 0.11 - 0.11
- 0.11 0.31 0.05 


- 0.11 0.05 0.55 
Note: RSS = 135
c)
A correct statement on ‘BLUE’ estimates is required for full marks
d)
i)
t = -3.4641
t27 * = -2.052 (reject null)
ii)
F = 26.25
F2,27* = 3.354 (reject null)
Question 2
a)
An accurate account of the Ramsey RESET test, that specifies the null (correct FF)
and alternative hypotheses under test and test statistic used (t-test) is required where it
is assumed that the square of predicted values will improve the fit of the model if a
linear functional form is inappropriate. General remedies: change the form of the
model (use logs) or include missing variables (including dynamics, trends dummies
for qualitative changes etc.)
b)
A number of tests that can be used here – but generally the students are expected to
show knowledge of a Chow (F) Test and describe the following procedure.
1) Identify the point of the structural break.
2) Split sample into two about the break in the data
3) Estimate two equations one for each sub-sample and compute the unrestricted
RSSU
4) Estimate the pooled equation for the entire period and obtain the restricted RSSR
5) Use the F-test F =
RSS R  RSS U  k  ~ F
k ,n n 2k k = number of parameters.
RSS U n1  n 2  2k 
1
2
6) If greater than the critical value, reject H0: coefficients are stable.
Students are expected to acknowledge that the use of dummy variables to capture the
break in question as a possible cure, and their significance can be determined by a
simple t-test.
c)
Students should derive the following expression where Z is the missing variable and
b1 is the coefficient estimate
(b1) = β1  β 2
 Y1 Z 2
2
 Y1
Students are expected to show how this formula is derived and acknowledge the
following: b1 is only unbiased if (a)  0 or (b) there is zero correlation between Y1
and Z2 i.e. Y1 Z2= 0. (a) is less of a problem than (b). The direction of bias when
estimating 1 depends on the sign of 2 and the sign of Y1 Z2.
Question 3
a)
Heteroscedasticity is likely to be found in cross sectional data due to wide variations
in the in specific explanatory variables across units. In this case any of the variables
on the RHS may vary widely between the included countries. Students may give
specific examples using the expression in the question.
b)
Students should recognise that the standard error will be biased in validating the use
of t- and F tests. Students are expected to note that the coefficient estimates although
inefficient will not be biased and are consistent. So the problem resides with the covvar matrix.
c)
i) This requires an accurate account of the Goldfeld-Quandt test which requires
knowledge of the variable causing the heteroscedasticity and the form of
heteroscedasticity (variance increasing/decreasing in the variable thought to be
causing it). The test uses an F-testing framework and the null is homoscedasticity.
The procedure should be correctly described for full marks e.g. reordering of the
variable thought to be causing the problem.
ii) An accurate account of the White’s procedure is required for full marks (although
some may quote the Breusch-Pagan test). In particular students should describe and
specify the auxiliary equation used in the test, recognise that the test is a chi-squared
(LM) test. The test statistics is created using the unadjusted R2 from the auxiliary
regression multiplied by the number of observations. A large chi-square generally
rejects the null of homoscedasticity. Student expected to comment correctly for full
marks
d)
A general account of the following is required. Divide the error term by the variable
causing the heteroscedasticity say X, thus the error is weighted u/X. Then square the
last expression and take the expected value to get: σ2/X2 (1). If the variance is
increasing in X then σ2 = K2X2 where K2 is a constant thus X2 = σ2/K2 and when
substituted into expression (1) gives a constant variance.
Question 4
a)
A comment on the signs and the significance of the estimated coefficients (up to 10
marks) and their impact noting that since the model is a log-log model they represent
elasticities (up to 5 more marks). Any comment on model fit (F-stat, R2 etc) a further
5 marks.
b)
ρ = 1-(0.991629/2) = 0.504  moderate correlation between error terms may imply
autocorrelation
c)
Students are expected to use the Durbin-Watson AR(1) test to answer this question.
Given N=31 and k = 2 the upper and lower critical values from the DW tables
provided are U = 1.57 and L = 1.30. The test suggests that the model suffers from 1st
degree positive autocorrelation. A correct description of the procedure is required
together with the null (no auto) and alternative hypothesis is also required. Students
should also note that t- and F-test are invalidated.
d)
Since the model includes a LDV the Durbin-h test must be used. An accurate
description is required and students should note that this test can only account for
AR(1) processes. The formula for the test is provides and the test statistic is
computed:
h = 0.504*(30/1-(30*0.0075) = 3.14
Since the test statistic is distributed h ~ N(0,σ2) the critical value for the test is 1.64
therefore the null of no positive autocorrelation is rejected.
e)
Any two relevant causes of autocorrelation is required for full marks e.g. missing
variables, structural breaks, wrong functional form etc is required for full marks)
Question 5
a)
Students are expected to note that time series data are often found to be non-stationary
and are expected to note that a time series {Yt} is non-stationary if its mean and
variance and the joint distribution of Yt and Yt+ is dependent of time. Student should
be also be familiar with the definition of a ‘weakly stationary process’ i.e. E{yt} and
V{Yt} are both finite constant and cov{Yt ,Yt+} only depends on difference in the
time index and an adaptation of this definition is also acceptable.
b)
i) is a pure random walk
ii) is a random walk with drift
iii) is a random walk with trend and drift (it is usual to test if both the trend and drift
terms are need using an F-test)
A correct definition of these models and their properties is need for full marks (10
marks for each)
c)
i) We would expect to see a slow decay of the autocorrelation function over many
lags.
ii) A full description of the DF and ADF procedure is required together with a correct
statement of the null and alternative hypothesis. Students should be aware that the ttest is not standard and the Mckinnon critical values are used. Very good answers will
also refer to the SIC and AIC as helpful tools in determining the lag order in the ADF
procedure.
iii) Take first difference of non-stationary variables.
Question 6
a)
Students should recognise that both models model conditional heteroscedasticity and
take account of volatility clustering that is generally present in high frequency
financial time series data. Each model notes that information in the recent past might
influence the conditional variance (volatility), but each models the conditional
variance in a different way. For general recognition of these points (10 marks).
Model 1 is an ARCH (p) (2 marks). The conditional variance is given under the
conditional mean equation for Model 1. Students should state that the conditional
variance has two components: a weighted average of the long term average variance
(the constant term α0) (3 marks) and previous news about volatility measured by
lag(s) of the squared error (the ARCH terms) (4 marks).
Model 2 is a GARCH (1,1) (2 marks) and models the conditional heteroscedasticity
in a different way. The conditional variance has 3 three components and is given
under the conditional mean equation: a constant (as above) (2 marks); information
about the volatility observed in the last period (the ARCH term) (3 marks); and the
fitted variance (the lagged dependent variable) from the model during the previous
period (the GARCH term) (4 marks).
b)
A brief description of how the disadvantages of ARCH are overcome by GARCH.
Students should mention and develop the issues surrounding the following points. The
major disadvantage of ARCH(p) is deciding on the appropriate number of lags (7
marks), which could be large and the non negativity constraint is more likely to be
violated under an ARCH specification (7 marks). Other issues concerning
multicollinearity and degrees of freedom may also be alluded to (6 marks). Students
are expected to realise that the GARCH(1,1) goes some way to overcome this
problem as it results in a more parsimonious model.
c)
Students should note that the unconditional error has a zero mean and constant
variance, and E(utus) = 0 (6 marks). This implies that OLS is still efficient and
consistent (2 marks). However, as both models model conditional heteroscedasticity
one arrives at more efficient parameter estimates using maximum likelihood
techniques (no discussion of this procedure is expected or required) (2 marks).
d)
Given that all the estimated coefficients in the conditional variance(mean) equation
are significant at conventional levels of significance and the estimated coefficients in
the conditional variance sum to 0.98 approx (i.e. α1 + γ = 0.97619) there does appear
to be some degree of volatility clustering in the data (5 marks). Students should
develop this point along the following line: shocks to the conditional variance are
highly persistent and given the size of the estimated coefficients large positive or
negative returns will lead to a future variance to also be high for a protracted period
(10 marks).
e)
i) 0.0068 with comment (5 marks)
ii) 1.63×10-6/(1-(0.09839 + 0.8778)) ≈ 6.846×10-5 + comment (10 marks).
iii) Annualised volatility = √(6.846 x10 -5) × √252 ≈ 13.13% + comment (10 marks).