Solutions to the Review Questions at the End of Chapter 5
1. ARMA models are of particular use for financial series due to their flexibility. They
are fairly simple to estimate, can often produce reasonable forecasts, and most
importantly, they require no knowledge of any structural variables that might be
required for more “traditional” econometric analysis. When the data are available at
high frequencies, we can still use ARMA models while exogenous “explanatory”
variables (e.g. macroeconomic variables, accounting ratios) may be unobservable at
any more than monthly intervals at best.
2.
yt = yt-1 + ut
yt = 0.5 yt-1 + ut
yt = 0.8 ut-1 + ut
(1)
(2)
(3)
(i) The first two models are roughly speaking AR(1) models, while the last is an
MA(1). Strictly, since the first model is a random walk, it should be called an
ARIMA(0,1,0) model, but it could still be viewed as a special case of an
autoregressive model.
(ii) We know that the theoretical acf of an MA(q) process will be zero after q lags, so
the acf of the MA(1) will be zero at all lags after one. For an autoregressive process,
the acf dies away gradually. It will die away fairly quickly for case (2), with each
successive autocorrelation coefficient taking on a value equal to half that of the
previous lag. For the first case, however, the acf will never die away, and in theory
will always take on a value of one, whatever the lag.
Turning now to the pacf, the pacf for the first two models would have a large positive
spike at lag 1, and no statistically significant pacf’s at other lags. Again, the unit root
process of (1) would have a pacf the same as that of a stationary AR process. The pacf
for (3), the MA(1), will decline geometrically.
(iii) Clearly the first equation (the random walk) is more likely to represent stock
prices in practice. The discounted dividend model of share prices states that the
current value of a share will be simply the discounted sum of all expected future
dividends. If we assume that investors form their expectations about dividend
payments rationally, then the current share price should embody all information that is
known about the future of dividend payments, and hence today’s price should only
differ from yesterday’s by the amount of unexpected news which influences dividend
payments.
Thus stock prices should follow a random walk. Note that we could apply a similar
rational expectations and random walk model to many other kinds of financial series.
If the stock market really followed the process described by equations (2) or (3), then
we could potentially make useful forecasts of the series using our model. In the latter
case of the MA(1), we could only make one-step ahead forecasts since the “memory”
of the model is only that length. In the case of equation (2), we could potentially make
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a lot of money by forming multiple step ahead forecasts and trading on the basis of
these.
Hence after a period, it is likely that other investors would spot this potential
opportunity and hence the model would no longer be a useful description of the data.
(iv) See the book for the algebra. This part of the question is really an extension of the
others. Analysing the simplest case first, the MA(1), the “memory” of the process
will only be one period, and therefore a given shock or “innovation”, ut, will only
persist in the series (i.e. be reflected in yt) for one period. After that, the effect of a
given shock would have completely worked through.
For the case of the AR(1) given in equation (2), a given shock, ut, will persist
indefinitely and will therefore influence the properties of yt for ever, but its effect
upon yt will diminish exponentially as time goes on.
In the first case, the series yt could be written as an infinite sum of past shocks, and
therefore the effect of a given shock will persist indefinitely, and its effect will not
diminish over time.
3.
(i) Box and Jenkins were the first to consider ARMA modelling in this logical
and coherent fashion. Their methodology consists of 3 steps:
Identification - determining the appropriate order of the model using graphical
procedures (e.g. plots of autocorrelation functions).
Estimation - of the parameters of the model of size given in the first stage. This can be
done using least squares or maximum likelihood, depending on the model.
Diagnostic checking - this step is to ensure that the model actually estimated is
“adequate”. B & J suggest two methods for achieving this:
- Overfitting, which involves deliberately fitting a model larger than that
suggested in step 1 and testing the hypothesis that all the additional coefficients can
jointly be set to zero.
- Residual diagnostics. If the model estimated is a good description of the data,
there should be no further linear dependence in the residuals of the estimated model.
Therefore, we could calculate the residuals from the estimated model, and use the
Ljung-Box test on them, or calculate their acf. If either of these reveal evidence of
additional structure, then we assume that the estimated model is not an adequate
description of the data.
If the model appears to be adequate, then it can be used for policy analysis and for
constructing forecasts. If it is not adequate, then we must go back to stage 1 and start
again!
(ii) The main problem with the B & J methodology is the inexactness of the
identification stage. Autocorrelation functions and partial autocorrelations for actual
data are very difficult to interpret accurately, rendering the whole procedure often
little more than educated guesswork. A further problem concerns the diagnostic
checking stage, which will only indicate when the proposed model is “too small” and
would not inform on when the model proposed is “too large”.
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(iii) We could use Akaike’s or Schwarz’s Bayesian information criteria. Our objective
would then be to fit the model order that minimises these.
We can calculate the value of Akaike’s (AIC) and Schwarz’s (SBIC) Bayesian
information criteria using the following respective formulae
AIC = ln (  2 ) + 2k/T
SBIC = ln (  2 ) + k ln(T)/T
The information criteria trade off an increase in the number of parameters and
therefore an increase in the penalty term against a fall in the RSS, implying a closer fit
of the model to the data.
4. The best way to check for stationarity is to express the model as a lag polynomial
in yt.
y t  0.803 y t 1  0.682 y t 2  ut
Rewrite this as
yt (1  0.803 L  0.682 L2 )  ut
We want to find the roots of the lag polynomial (1  0.803 L  0.682 L2 )  0 and
determine whether they are greater than one in absolute value. It is easier (in my
opinion) to rewrite this formula (by multiplying through by -1/0.682, using z for the
characteristic equation and rearranging) as
z2 + 1.177 z - 1.466 = 0
Using the standard formula for obtaining the roots of a quadratic equation,
 1177
.
 1177
. 2  4 * 1 * 1466
.
= 0.758 or 1.934
z
2
Since ALL the roots must be greater than one for the model to be stationary, we
conclude that the estimated model is not stationary in this case.
5. Using the formulae above, we end up with the following values for each criterion
and for each model order (with an asterisk denoting the smallest value of the
information criterion in each case).
ARMA (p,q) model order
(0,0)
(1,0)
(0,1)
(1,1)
(2,1)
(1,2)
(2,2)
(3,2)
(2,3)
(3,3)
log (  2 )
0.932
0.864
0.902
0.836
0.801
0.821
0.789
0.773
0.782
0.764
AIC
0.942
0.884
0.922
0.866
0.841
0.861
0.839
0.833*
0.842
0.834
SBIC
0.944
0.887
0.925
0.870
0.847
0.867
0.846
0.842*
0.851
0.844
The result is pretty clear: both SBIC and AIC say that the appropriate model is an
ARMA(3,2).
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6. We could still perform the Ljung-Box test on the residuals of the estimated models
to see if there was any linear dependence left unaccounted for by our postulated
models.
Another test of the models’ adequacy that we could use is to leave out some of the
observations at the identification and estimation stage, and attempt to construct out of
sample forecasts for these. For example, if we have 2000 observations, we may use
only 1800 of them to identify and estimate the models, and leave the remaining 200
for construction of forecasts. We would then prefer the model that gave the most
accurate forecasts.
7. This is not true in general. Yes, we do want to form a model which “fits” the data
as well as possible. But in most financial series, there is a substantial amount of
“noise”. This can be interpreted as a number of random events that are unlikely to be
repeated in any forecastable way. We want to fit a model to the data which will be
able to “generalise”. In other words, we want a model which fits to features of the data
which will be replicated in future; we do not want to fit to sample-specific noise.
This is why we need the concept of “parsimony” - fitting the smallest possible model
to the data. Otherwise we may get a great fit to the data in sample, but any use of the
model for forecasts could yield terrible results.
Another important point is that the larger the number of estimated parameters (i.e. the
more variables we have), then the smaller will be the number of degrees of freedom,
and this will imply that coefficient standard errors will be larger than they would
otherwise have been. This could lead to a loss of power in hypothesis tests, and
variables that would otherwise have been significant are now insignificant.
8. (a) We class an autocorrelation coefficient or partial autocorrelation coefficient as
1
significant if it exceeds  1.96
=  0.196. Under this rule, the sample
T
autocorrelation functions (sacfs) at lag 1 and 4 are significant, and the spacfs at lag 1,
2, 3, 4 and 5 are all significant.
This clearly looks like the data are consistent with a first order moving average
process since all but the first acfs are not significant (the significant lag 4 acf is a
typical wrinkle that one might expect with real data and should probably be ignored),
and the pacf has a slowly declining structure.
(b) The formula for the Ljung-Box Q* test is given by
m
Q*  T (T  2)
k 1
 k2
T k

m2
using the standard notation.
In this case, T=100, and m=3. The null hypothesis is H0: 1 = 0 and 2 = 0 and
3 = 0. The test statistic is calculated as
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 0.420 2 0.104 2 0.032 2 
Q*  100  102  


  19.41.
100  1 100  2 100  3 
The 5% and 1% critical values for a 2 distribution with 3 degrees of freedom are 7.81
and 11.3 respectively. Clearly, then, we would reject the null hypothesis that the first
three autocorrelation coefficients are jointly not significantly different from zero.
9. (a) To solve this, we need the concept of a conditional expectation,
i.e. Et 1 ( yt yt 2 , yt 3 ,...)
For example, in the context of an AR(1) model such as , yt  a0  a1 yt 1  ut
If we are now at time t-1, and dropping the t-1 subscript on the expectations operator
E ( yt )  a0  a1 yt 1
E ( yt 1 )  a0  a1 E ( yt )
= a0  a1 yt 1 (a0  a1 yt 1 )
= a0  a0a1  a12 yt 1
E ( yt 2 )  a0  a1 E ( yt 1 )
= a0  a1 (a0  a1 E ( yt ))
=
=
=
=
a0  a0a1  a12 E ( yt )
a0  a0a1  a12 E ( yt )
a0  a0a1  a12 (a0  a1 yt 1 )
a0  a0a1  a12 a0  a13 yt 1
etc.
f t 1,1  a 0  a1 yt 1
f t 1,2  a 0  a1 f t 1,1
f t 1,3  a0  a1 f t 1,2
To forecast an MA model, consider, e.g.
yt  ut  b1ut 1
E ( yt yt 1 , yt 2 ,...)
So
=
E (u t  b1u t 1 )
b1u t 1
=
ft-1,1
=
=
E (u t 1  b1u t )
=
0
b1u t 1
But
E ( yt 1 yt 1 , yt 2 ,...)
Going back to the example above,
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yt  0.036  0.69 yt 1  0.42u t 1  ut
Suppose that we know t-1, t-2,... and we are trying to forecast yt.
Our forecast for t is given by
E ( yt yt 1 , yt 2 ,...) = f t 1,1  0.036  0.69 y t 1  0.42u t 1  u t
ft-1,2 = E( yt 1
= 0.036 +0.693.4+0.42(-1.3)
= 1.836
yt 1 , yt 2 ,...)  0.036  0.69 yt  0.42ut  ut 1
But we do not know yt or ut at time t-1.
Replace yt with our forecast of yt which is ft-1,1.
ft-1,2
= 0.036 +0.69 ft-1,1
= 0.036 + 0.69*1.836
= 1.302
ft-1,3
= 0.036 +0.69 ft-1,2
= 0.036 + 0.69*1.302
= 0.935
etc.
(b) Given the forecasts and the actual value, it is very easy to calculate the MSE by
plugging the numbers in to the relevant formula, which in this case is
1 N
MSE   ( x t 1 n  f t 1, n ) 2
N n 1
if we are making N forecasts which are numbered 1,2,3.
Then the MSE is given by
1
MSE   (1836
.
 0.032) 2  (1302
.
 0.961) 2  (0.935  0.203) 2   3.489  0116
.
 0.536  4.141
3
Notice also that 84% of the total MSE is coming from the error in the first forecast.
Thus error measures can be driven by one or two times when the model fits very
badly. For example, if the forecast period includes a stock market crash, this can lead
the mean squared error to be 100 times bigger than it would have been if the crash
observations were not included. This point needs to be considered whenever
forecasting models are evaluated. An idea of whether this is a problem in a given
situation can be gained by plotting the forecast errors over time.
(c) This question is much simpler to answer than it looks! In fact, the inclusion of the
smoothing coefficient is a “red herring” - i.e. a piece of misleading and useless
information. The correct approach is to say that if we believe that the exponential
smoothing model is appropriate, then all useful information will have already been
used in the calculation of the current smoothed value (which will of course have used
the smoothing coefficient in its calculation). Thus the three forecasts are all 0.0305.
(d) The solution is to work out the mean squared error for the exponential smoothing
model. The calculation is
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MSE 


1
(0.0305  0.032) 2  (0.0305  0.961) 2  (0.0305  0.203) 2
3

1
0.0039  0.8658  0.0298  0.2998
3
Therefore, we conclude that since the mean squared error is smaller for the
exponential smoothing model than the Box Jenkins model, the former produces the
more accurate forecasts. We should, however, bear in mind that the question of
accuracy was determined using only 3 forecasts, which would be insufficient in a real
application.
10. (a) The shapes of the acf and pacf are perhaps best summarised in a table:
Process
acf
pacf
White noise
No significant coefficients
No significant coefficients
AR(2)
Geometrically declining or damped
First 2 pacf coefficients significant,
sinusoid acf
all others insignificant
MA(1)
First acf coefficient significant, all
Geometrically declining or damped
others insignificant
sinusoid pacf
ARMA(2,1)
Geometrically declining or damped
Geometrically declining or damped
sinusoid acf
sinusoid pacf
A couple of further points are worth noting. First, it is not possible to tell what the
signs of the coefficients for the acf or pacf would be for the last three processes, since
that would depend on the signs of the coefficients of the processes. Second, for mixed
processes, the AR part dominates from the point of view of acf calculation, while the
MA part dominates for pacf calculation.
(b) The important point here is to focus on the MA part of the model and to ignore the
AR dynamics. The characteristic equation would be
(1+0.42z) = 0
The root of this equation is -1/0.42 = -2.38, which lies outside the unit circle, and
therefore the MA part of the model is invertible.
(c) Since no values for the series y or the lagged residuals are given, the answers
should be stated in terms of y and of u. Assuming that information is available up to
and including time t, the 1-step ahead forecast would be for time t+1, the 2-step ahead
for time t+2 and so on. A useful first step would be to write the model out for y at
times t+1, t+2, t+3, t+4:
y t 1  0.036  0.69 y t  0.42u t  u t 1
y t  2  0.036  0.69 y t 1  0.42u t 1  u t  2
y t 3  0.036  0.69 y t  2  0.42u t  2  u t 3
y t  4  0.036  0.69 y t 3  0.42u t 3  u t  4
The 1-step ahead forecast would simply be the conditional expectation of y for time
t+1 made at time t. Denoting the 1-step ahead forecast made at time t as ft,1, the 2-step
ahead forecast made at time t as ft,2 and so on:
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E( yt 1 yt , yt 1 ,...)  f t ,1  Et [ yt 1 ]  Et [0.036  0.69 yt  0.42ut  ut 1 ]  0.036  0.69 yt  0.42ut
since Et[ut+1]=0. The 2-step ahead forecast would be given by
E( yt  2 yt , yt 1,...)  ft , 2  Et [ yt  2 ]  Et [0.036  0.69 yt 1  0.42ut 1  ut  2 ]  0.036  0.69 f t ,1
since Et[ut+1]=0 and Et[ut+2]=0. Thus, beyond 1-step ahead, the MA(1) part of the
model disappears from the forecast and only the autoregressive part remains.
Although we do not know yt+1, its expected value is the 1-step ahead forecast that was
made at the first stage, ft,1.
The 3-step ahead forecast would be given by
E( yt 3 yt , yt 1,...)  ft ,3  Et [ yt 3 ]  Et [0.036  0.69 yt  2  0.42ut  2  ut 3 ]  0.036  0.69 f t , 2
and the 4-step ahead by
E( yt  4 yt , yt 1,...)  ft , 4  Et [ yt  4 ]  Et [0.036  0.69 yt 3  0.42ut 3  ut  4 ]  0.036  0.69 f t ,3
(d) A number of methods for aggregating the forecast errors to produce a single
forecast evaluation measure were suggested in the paper by Makridakis and Hibon
(1995) and some discussion is presented in Sections 5.12.8 and 5.12.9. Any of the
methods suggested there could be discussed. A good answer would present an
expression for the evaluation measures, with any notation introduced being carefully
defined, together with a discussion of why the measure takes the form that it does and
what the advantages and disadvantages of its use are compared with other methods.
(e) Moving average and ARMA models cannot be estimated using OLS – they are
usually estimated by maximum likelihood. Autoregressive models can be estimated
using OLS or maximum likelihood. Pure autoregressive models contain only lagged
values of observed quantities on the RHS, and therefore, the lags of the dependent
variable can be used just like any other regressors. However, in the context of MA and
mixed models, the lagged values of the error term that occur on the RHS are not
known a priori. Hence, these quantities are replaced by the residuals, which are not
available until after the model has been estimated. But equally, these residuals are
required in order to be able to estimate the model parameters. Maximum likelihood
essentially works around this by calculating the values of the coefficients and the
residuals at the same time. Maximum likelihood involves selecting the most likely
values of the parameters given the actual data sample, and given an assumed statistical
distribution for the errors. This technique will be discussed in greater detail in the
section on volatility modelling in Chapter 8.
11. (a) Some of the stylised differences between the typical characteristics of
macroeconomic and financial data were presented in Section 1.2 of Chapter 1. In
particular, one important difference is the frequency with which financial asset return
time series and other quantities in finance can be recorded. This is of particular
relevance for the models discussed in Chapter 5, since it is usually a requirement that
all of the time-series data series used in estimating a given model must be of the same
frequency. Thus, if, for example, we wanted to build a model for forecasting hourly
changes in exchange rates, it would be difficult to set up a structural model containing
macroeconomic explanatory variables since the macroeconomic variables are likely to
be measured on a quarterly or at best monthly basis. This gives a motivation for using
pure time-series approaches (e.g. ARMA models), rather than structural formulations
with separate explanatory variables.
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It is also often of particular interest to produce forecasts of financial variables in real
time. Producing forecasts from pure time-series models is usually simply an exercise
in iterating with conditional expectations. But producing forecasts from structural
models is considerably more difficult, and would usually require the production of
forecasts for the structural variables as well.
(b) A simple “rule of thumb” for determining whether autocorrelation coefficients and
partial autocorrelation coefficients are statistically significant is to classify them as
1
significant at the 5% level if they lie outside of  1.96 *
, where T is the sample
T
size. In this case, T = 500, so a particular coefficient would be deemed significant if it
is larger than 0.088 or smaller than –0.088. On this basis, the autocorrelation
coefficients at lags 1 and 5 and the partial autocorrelation coefficients at lags 1, 2, and
3 would be classed as significant. The formulae for the Box-Pierce and the Ljung-Box
test statistics are respectively
m
Q  T   k2
k 1
m
and
Q*  T (T  2)
k 1
 k2
T k
.
In this instance, the statistics would be calculated respectively as
Q  500  [0.307 2  (0.013 2 )  0.086 2  0.0312  (0.197 2 )]  70.79
and
 0.307 2 (0.013 2 ) 0.086 2
0.0312 (0.197 2 ) 
Q*  500  502  




  71.39
500  2
500  3 500  4
500  5 
 500  1
The test statistics will both follow a 2 distribution with 5 degrees of freedom (the
number of autocorrelation coefficients being used in the test). The critical values are
11.07 and 15.09 at 5% and 1% respectively. Clearly, the null hypothesis that the first 5
autocorrelation coefficients are jointly zero is resoundingly rejected.
(c) Setting aside the lag 5 autocorrelation coefficient, the pattern in the table is for the
autocorrelation coefficient to only be significant at lag 1 and then to fall rapidly to
values close to zero, while the partial autocorrelation coefficients appear to fall much
more slowly as the lag length increases. These characteristics would lead us to think
that an appropriate model for this series is an MA(1). Of course, the autocorrelation
coefficient at lag 5 is an anomaly that does not fit in with the pattern of the rest of the
coefficients. But such a result would be typical of a real data series (as opposed to a
simulated data series that would have a much cleaner structure). This serves to
illustrate that when econometrics is used for the analysis of real data, the data
generating process was almost certainly not any of the models in the ARMA family.
So all we are trying to do is to find a model that best describes the features of the data
to hand. As one econometrician put it, all models are wrong, but some are useful!
(d) Forecasts from this ARMA model would be produced in the usual way. Using the
same notation as above, and letting fz,1 denote the forecast for time z+1 made for x at
time z, etc:
Model A: MA(1)
f z ,1  0.38  0.10u t 1
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f z , 2  0.38  0.10  0.02  0.378
f z , 2  f z ,3  0.38
Note that the MA(1) model only has a memory of one period, so all forecasts further
than one step ahead will be equal to the intercept.
Model B: AR(2)
xˆ t  0.63  0.17 xt 1  0.09 xt 2
f z ,1  0.63  0.17  0.31  0.09  0.02  0.681
f z , 2  0.63  0.17  0.681  0.09  0.31  0.718
f z ,3  0.63  0.17  0.718  0.09  0.681  0.690
f z , 4  0.63  0.17  0.690  0.09  0.716  0.683
(e) The methods are overfitting and residual diagnostics. Overfitting involves
selecting a deliberately larger model than the proposed one, and examining the
statistical significances of the additional parameters. If the additional parameters are
statistically insignificant, then the originally postulated model is deemed acceptable.
The larger model would usually involve the addition of one extra MA term and one
extra AR term. Thus it would be sensible to try an ARMA(1,2) in the context of
Model A, and an ARMA(3,1) in the context of Model B. Residual diagnostics would
involve examining the acf and pacf of the residuals from the estimated model. If the
residuals showed any “action”, that is, if any of the acf or pacf coefficients showed
statistical significance, this would suggest that the original model was inadequate.
“Residual diagnostics” in the Box-Jenkins sense of the term involved only examining
the acf and pacf, rather than the array of diagnostics considered in Chapter 4.
It is worth noting that these two model evaluation procedures would only indicate a
model that was too small. If the model were too large, i.e. it had superfluous terms,
these procedures would deem the model adequate.
(f) There are obviously several forecast accuracy measures that could be employed,
including MSE, MAE, and the percentage of correct sign predictions. Assuming that
MSE is used, the MSE for each model is
1
MSE ( Model A)  (0.378  0.62) 2  (0.38  0.19) 2  (0.38  0.32) 2  (0.38  0.72) 2   0.175
4
MSE ( Model B) 


1
(0.681  0.62) 2  (0.718  0.19) 2  (0.690  0.32) 2  (0.683  0.72) 2  0.326
4
Therefore, since the mean squared error for Model A is smaller, it would be concluded
that the moving average model is the more accurate of the two in this case.
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“Introductory Econometrics for Finance” © Chris Brooks 2002