Solutions to the Review Questions at the End of Chapter 4
1. In the same way as we make assumptions about the true value of beta and
not the estimated values, we make assumptions about the true unobservable
disturbance terms rather than their estimated counterparts, the residuals.
We know the exact value of the residuals, since they are defined by
uˆ t  y t  yˆ t . So we do not need to make any assumptions about the residuals
since we already know their value. We make assumptions about the
unobservable error terms since it is always the true value of the population
disturbances that we are really interested in, although we never actually know
what these are.
2. We would like to see no pattern in the residual plot! If there is a pattern in
the residual plot, this is an indication that there is still some “action” or
variability left in yt that has not been explained by our model. This indicates
that potentially it may be possible to form a better model, perhaps using
additional or completely different explanatory variables, or by using lags of
either the dependent or of one or more of the explanatory variables. Recall
that the two plots shown on pages 157 and 159, where the residuals followed a
cyclical pattern, and when they followed an alternating pattern are used as
indications that the residuals are positively and negatively autocorrelated
respectively.
Another problem if there is a “pattern” in the residuals is that, if it does
indicate the presence of autocorrelation, then this may suggest that our
standard error estimates for the coefficients could be wrong and hence any
inferences we make about the coefficients could be misleading.
3. The t-ratios for the coefficients in this model are given in the third row after
the standard errors. They are calculated by dividing the individual coefficients
by their standard errors.
R 2  0.96,R 2  0.89
0.638 + 0.402 x2t - 0.891 x3t
(0.436) (0.291)
(0.763)
t-ratios
1.46
1.38
-1.17
ŷt =
The problem appears to be that the regression parameters are all individually
insignificant (i.e. not significantly different from zero), although the value of
R2 and its adjusted version are both very high, so that the regression taken as a
whole seems to indicate a good fit. This looks like a classic example of what we
term near multicollinearity. This is where the individual regressors are very
closely related, so that it becomes difficult to disentangle the effect of each
individual variable upon the dependent variable.
The solution to near multicollinearity that is usually suggested is that since the
problem is really one of insufficient information in the sample to determine
each of the coefficients, then one should go out and get more data. In other
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words, we should switch to a higher frequency of data for analysis (e.g. weekly
instead of monthly, monthly instead of quarterly etc.). An alternative is also to
get more data by using a longer sample period (i.e. one going further back in
time), or to combine the two independent variables in a ratio (e.g. x2t / x3t ).
Other, more ad hoc methods for dealing with the possible existence of near
multicollinearity were discussed in Chapter 4:
-
Ignore it: if the model is otherwise adequate, i.e. statistically and in terms
of each coefficient being of a plausible magnitude and having an
appropriate sign. Sometimes, the existence of multicollinearity does not
reduce the t-ratios on variables that would have been significant without
the multicollinearity sufficiently to make them insignificant. It is worth
stating that the presence of near multicollinearity does not affect the BLUE
properties of the OLS estimator – i.e. it will still be consistent, unbiased
and efficient since the presence of near multicollinearity does not violate
any of the CLRM assumptions 1-4. However, in the presence of near
multicollinearity, it will be hard to obtain small standard errors. This will
not matter if the aim of the model-building exercise is to produce forecasts
from the estimated model, since the forecasts will be unaffected by the
presence of near multicollinearity so long as this relationship between the
explanatory variables continues to hold over the forecasted sample.
-
Drop one of the collinear variables - so that the problem disappears.
However, this may be unacceptable to the researcher if there were strong a
priori theoretical reasons for including both variables in the model. Also, if
the removed variable was relevant in the data generating process for y, an
omitted variable bias would result.
-
Transform the highly correlated variables into a ratio and include only the
ratio and not the individual variables in the regression. Again, this may be
unacceptable if financial theory suggests that changes in the dependent
variable should occur following changes in the individual explanatory
variables, and not a ratio of them.
4. (a) The assumption of homoscedasticity is that the variance of the errors is
constant and finite over time. Technically, we write Var (u t )   u2 .
(b) The coefficient estimates would still be the “correct” ones (assuming that
the other assumptions required to demonstrate OLS optimality are satisfied),
but the problem would be that the standard errors could be wrong. Hence if
we were trying to test hypotheses about the true parameter values, we could
end up drawing the wrong conclusions. In fact, for all of the variables except
the constant, the standard errors would typically be too small, so that we
would end up rejecting the null hypothesis too many times.
(c) There are a number of ways to proceed in practice, including
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- Using heteroscedasticity robust standard errors which correct for the
problem by enlarging the standard errors relative to what they would have
been for the situation where the error variance is positively related to one of
the explanatory variables.
- Transforming the data into logs, which has the effect of reducing the effect of
large errors relative to small ones.
5. (a) This is where there is a relationship between the ith and jth residuals.
Recall that one of the assumptions of the CLRM was that such a relationship
did not exist. We want our residuals to be random, and if there is evidence of
autocorrelation in the residuals, then it implies that we could predict the sign
of the next residual and get the right answer more than half the time on
average!
(b) The Durbin Watson test is a test for first order autocorrelation. The test is
calculated as follows. You would run whatever regression you were interested
in, and obtain the residuals. Then calculate the statistic
T
 uˆ
DW  t 2
 uˆt 1 
2
t
T
 uˆ
t 2
2
t
You would then need to look up the two critical values from the Durbin
Watson tables, and these would depend on how many variables and how many
observations and how many regressors (excluding the constant this time) you
had in the model.
The rejection / non-rejection rule would be given by selecting the appropriate
region from the following diagram:
(c) We have 60 observations, and the number of regressors excluding the
constant term is 3. The appropriate lower and upper limits are 1.48 and 1.69
respectively, so the Durbin Watson is lower than the lower limit. It is thus
clear that we reject the null hypothesis of no autocorrelation. So it looks like
the residuals are positively autocorrelated.
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(d) y t   1   2 x 2t   3 x3t   4 x 4t  u t
The problem with a model entirely in first differences, is that once we calculate
the long run solution, all the first difference terms drop out (as in the long run
we assume that the values of all variables have converged on their own long
run values so that yt = yt-1 etc.) Thus when we try to calculate the long run
solution to this model, we cannot do it because there isn’t a long run solution
to this model!
(e) y t   1   2 x 2t   3 x3t   4 x 4t   5 x 2t 1   6 X 3t 1   7 X 4t 1  vt
The answer is yes, there is no reason why we cannot use Durbin Watson in this
case. You may have said no here because there are lagged values of the
regressors (the x variables) variables in the regression. In fact this would be
wrong since there are no lags of the DEPENDENT (y) variable and hence DW
can still be used.
6.
y t   1   2 x 2t   3 x3t   4 y t 1   5 x 2t 1   6 x3t 1   7 x rt  4  u t
The major steps involved in calculating the long run solution are to
- set the disturbance term equal to its expected value of zero
- drop the time subscripts
- remove all difference terms altogether since these will all be zero by the
definition of the long run in this context.
Following these steps, we obtain
0   1   4 y   5 x 2   6 x3   7 x3
We now want to rearrange this to have all the terms in x2 together and so that
y is the subject of the formula:
 4 y    1   5 x 2   6 x3   7 x3
 4 y    1   5 x 2  (  6   7 ) x3

(   4 )

y   1  5 x2  6
x3
4 4
4
The last equation above is the long run solution.
7. Ramsey’s RESET test is a test of whether the functional form of the
regression is appropriate. In other words, we test whether the relationship
between the dependent variable and the independent variables really should
be linear or whether a non-linear form would be more appropriate. The test
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works by adding powers of the fitted values from the regression into a second
regression. If the appropriate model was a linear one, then the powers of the
fitted values would not be significant in this second regression.
If we fail Ramsey’s RESET test, then the easiest “solution” is probably to
transform all of the variables into logarithms. This has the effect of turning a
multiplicative model into an additive one.
If this still fails, then we really have to admit that the relationship between the
dependent variable and the independent variables was probably not linear
after all so that we have to either estimate a non-linear model for the data
(which is beyond the scope of this course) or we have to go back to the drawing
board and run a different regression containing different variables.
8. (a) It is important to note that we did not need to assume normality in order
to derive the sample estimates of  and  or in calculating their standard
errors. We needed the normality assumption at the later stage when we come
to test hypotheses about the regression coefficients, either singly or jointly, so
that the test statistics we calculate would indeed have the distribution (t or F)
that we said they would.
(b) One solution would be to use a technique for estimation and inference
which did not require normality. But these techniques are often highly
complex and also their properties are not so well understood, so we do not
know with such certainty how well the methods will perform in different
circumstances.
One pragmatic approach to failing the normality test is to plot the estimated
residuals of the model, and look for one or more very extreme outliers. These
would be residuals that are much “bigger” (either very big and positive, or very
big and negative) than the rest. It is, fortunately for us, often the case that one
or two very extreme outliers will cause a violation of the normality
assumption. The reason that one or two extreme outliers can cause a violation
of the normality assumption is that they would lead the (absolute value of the)
skewness and / or kurtosis estimates to be very large.
Once we spot a few extreme residuals, we should look at the dates when these
outliers occurred. If we have a good theoretical reason for doing so, we can
add in separate dummy variables for big outliers caused by, for example, wars,
changes of government, stock market crashes, changes in market
microstructure (e.g. the “big bang” of 1986). The effect of the dummy variable
is exactly the same as if we had removed the observation from the sample
altogether and estimated the regression on the remainder. If we only remove
observations in this way, then we make sure that we do not lose any useful
pieces of information represented by sample points.
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9. (a) Parameter structural stability refers to whether the coefficient estimates
for a regression equation are stable over time. If the regression is not
structurally stable, it implies that the coefficient estimates would be different
for some sub-samples of the data compared to others. This is clearly not what
we want to find since when we estimate a regression, we are implicitly
assuming that the regression parameters are constant over the entire sample
period under consideration.
(b)
1981M1-1995M12
rt = 0.0215 + 1.491 rmt
1981M1-1987M10
rt = 0.0163 + 1.308 rmt
1987M11-1995M12
rt = 0.0360 + 1.613 rmt
RSS=0.189 T=180
RSS=0.079 T=82
RSS=0.082 T=98
(c) If we define the coefficient estimates for the first and second halves of the
sample as 1 and 1, and 2 and 2 respectively, then the null and alternative
hypotheses are
H0 : 1 = 2 and 1 = 2
and
H1 : 1  2 or 1  2
(d) The test statistic is calculated as
Test stat. =
RSS  ( RSS 1  RSS 2 ) (T  2k ) 0.189  (0.079  0.082) 180  4
*

*
 15.304
RSS 1  RSS 2
k
0.079  0.082
2
This follows an F distribution with (k,T-2k) degrees of freedom. F(2,176) =
3.05 at the 5% level. Clearly we reject the null hypothesis that the coefficients
are equal in the two sub-periods.
10. The data we have are
1981M1-1995M12
rt = 0.0215 + 1.491 Rmt
1981M1-1994M12
rt = 0.0212 + 1.478 Rmt
1982M1-1995M12
rt = 0.0217 + 1.523 Rmt
RSS=0.189 T=180
RSS=0.148 T=168
RSS=0.182 T=168
First, the forward predictive failure test - i.e. we are trying to see if the model
for 1981M1-1994M12 can predict 1995M1-1995M12.
The test statistic is given by
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RSS  RSS 1 T1  k 0.189  0.148 168  2
*

*
 3.832
RSS 1
T2
0.148
12
Where T1 is the number of observations in the first period (i.e. the period that
we actually estimate the model over), and T2 is the number of observations we
are trying to “predict”. The test statistic follows an F-distribution with (T2, T1k) degrees of freedom. F(12, 166) = 1.81 at the 5% level. So we reject the null
hypothesis that the model can predict the observations for 1995. We would
conclude that our model is no use for predicting this period, and from a
practical point of view, we would have to consider whether this failure is a
result of a-typical behaviour of the series out-of-sample (i.e. during 1995), or
whether it results from a genuine deficiency in the model.
The backward predictive failure test is a little more difficult to understand,
although no more difficult to implement. The test statistic is given by
RSS  RSS 1 T1  k 0.189  0.182 168  2
*

*
 0.532
RSS 1
T2
0.182
12
Now we need to be a little careful in our interpretation of what exactly are the
“first” and “second” sample periods. It would be possible to define T1 as always
being the first sample period. But I think it easier to say that T1 is always the
sample over which we estimate the model (even though it now comes after the
hold-out-sample). Thus T2 is still the sample that we are trying to predict, even
though it comes first. You can use either notation, but you need to be clear and
consistent. If you wanted to choose the other way to the one I suggest, then
you would need to change the subscript 1 everywhere in the formula above so
that it was 2, and change every 2 so that it was a 1.
Either way, we conclude that there is little evidence against the null
hypothesis. Thus our model is able to adequately back-cast the first 12
observations of the sample.
11. By definition, variables having associated parameters that are not
significantly different from zero are not, from a statistical perspective, helping
to explain variations in the dependent variable about its mean value. One
could therefore argue that empirically, they serve no purpose in the fitted
regression model. But leaving such variables in the model will use up valuable
degrees of freedom, implying that the standard errors on all of the other
parameters in the regression model, will be unnecessarily higher as a result. If
the number of degrees of freedom is relatively small, then saving a couple by
deleting two variables with insignificant parameters could be useful. On the
other hand, if the number of degrees of freedom is already very large, the
impact of these additional irrelevant variables on the others is likely to be
inconsequential.
12. An outlier dummy variable will take the value one for one observation in
the sample and zero for all others. The Chow test involves splitting the sample
into two parts. If we then try to run the regression on both the sub-parts but
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the model contains such an outlier dummy, then the observations on that
dummy will be zero everywhere for one of the regressions. For that subsample, the outlier dummy would show perfect multicollinearity with the
intercept and therefore the model could not be estimated.
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