Solutions to the Review Questions at the End of Chapter 4
y t   1   2 x 2t   3 x3t   4 y t 1  u t
1.
y t   1   2 x 2t   3 x3t   4 y t 1  vt .
Note that we have not changed anything substantial between these models in the sense
that the second model is just a re-parameterisation (rearrangement) of the first, where
we have subtracted yt-1 from both sides of the equation.
(i) Remember that the residual sum of squares is the sum of each of the squared
residuals. So lets consider what the residuals will be in each case. For the first model
in the level of y
uˆ t  yt  yˆ t  yt  ˆ1  ˆ 2 x2t  ˆ 3 X 3t  ˆ 4 yt 1
Now for the second model, the dependent variable is now the change in y:
vˆt  y t  yˆ t  y t  ˆ1  ˆ 2 x 2t  ˆ 3 x3t  ˆ 4 y t 1
where y is the fitted value in each case (note that we do not need at this stage to
assume they are the same). Rearranging this second model would give
uˆ t  y t  y t 1  ˆ1  ˆ 2 x 2t  ˆ 3 x3t  ˆ 4 y t 1
 y t  ˆ1  ˆ 2 x 2t  ˆ 3 x 3t  (ˆ 4  1) y t 1
If we compare this formulation with the one we calculated for the first model, we can
see that the residuals are exactly the same for the two models, with ˆ 4  ˆ 4  1 and
ˆ  ˆ (i = 1, 2, 3). Hence if the residuals are the same, the residual sum of squares
i
i
must also be the same. In fact the two models are really identical, since one is just a
rearrangement of the other.
(ii) As for R2, recall how we calculate R2:
RSS
for the first model and
R2  1
 ( yi  y ) 2
R2 1
RSS
in the second case. Therefore since the total sum of squares
 (yi  y ) 2
(the denominator) has changed, then the value of R2 must have also changed as a
consequence of changing the dependent variable.
(iii) By the same logic, since the value of the adjusted R2 is just an algebraic
modification of R2 itself, the value of the adjusted R2 must also change.
2. A researcher estimates the following two econometric models
y t   1   2 x 2t   3 x3t  u t
(1)
y t   1   2 x 2t   3 x3t   4 x 4t  vt
(2)
2
(i) The value of R will almost always be higher for the second model since it has
another variable added to the regression. The value of R2 would only be identical for
the two models in the very, very unlikely event that the estimated coefficient on the x4t
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variable was exactly zero. Otherwise, the R2 must be higher for the second model than
the first.
(ii) The value of the adjusted R2 could fall as we add another variable. The reason for
this is that the adjusted version of R2 has a correction for the loss of degrees of
freedom associated with adding another regressor into a regression. This implies a
penalty term, so that the value of the adjusted R2 will only rise if the increase in this
penalty is more than outweighed by the rise in the value of R2.
3. 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.
4. 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.
5. 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.
ŷt =
0.638 + 0.402 x2t - 0.891 x3t
(0.436) (0.291)
(0.763)
t-ratios 1.46
1.38
1.17
R 2  0.96,R 2  0.89
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
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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 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 (see Section 4.12).
- 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.
6.
(i) The assumption of homoscedasticity is that the variance of the errors is
constant and finite over time. Technically, we write Var (u t )   u2 .
(ii) 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.
(iii) 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.
7.
(i) 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!
(ii) 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:
(iii) 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.
(iv) 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!
(v) 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
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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.
8.
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.
9. 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 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.
10.
(i) 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.
(ii) 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
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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.
11. (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
RSS=0.189 T=180
1981M1-1987M10
rt = 0.0163 + 1.308 rmt
RSS=0.079 T=82
1987M11-1995M12
rt = 0.0360 + 1.613 rmt
RSS=0.082 T=98
(i) 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
(ii) 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.
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12.The data we have are
1981M1-1995M12
rt = 0.0215 + 1.491 Rmt
RSS=0.189 T=180
1981M1-1994M12
rt = 0.0212 + 1.478 Rmt
RSS=0.148 T=168
1982M1-1995M12
rt = 0.0217 + 1.523 Rmt
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
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, T1-k) 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-ofsample (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.
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