Answers Review Questions Econometrics

Solutions to the end of Chapter Exercises

Chapter 2 1. (a) The use of vertical rather than horizontal distances relates to the idea that the explanatory variable, x, is fixed in repeated samples, so what the model tries to do is to fit the most appropriate value of y using the model for a given value of x. Taking horizontal distances would have suggested that we had fixed the value of y and tried to find the appropriate values of x. (b) When we calculate the deviations of the points, yt, from the fitted values, yˆ t , some points will lie above the line (yt > yˆ t ) and some will lie below the line (yt < yˆ t ). When we calculate the residuals ( uˆ t = yt – yˆ t ), those corresponding to points above the line will be positive and those below the line negative, so adding them would mean that they would largely cancel out. In fact, we could fit an infinite number of lines with a zero average residual. By squaring the residuals before summing them, we ensure that they all contribute to the measure of loss and that they do not cancel. It is then possible to define unique (ordinary least squares) estimates of the intercept and slope. (c) Taking the absolute values of the residuals and minimising their sum would certainly also get around the problem of positive and negative residuals cancelling. However, the absolute value function is much harder to work with than a square. Squared terms are easy to differentiate, so it is simple to find analytical formulae for the mean and the variance. 2. The population regression function (PRF) is a description of the model that is thought to be generating the actual data and it represents the true relationship between the variables. The population regression function is also known as the data generating process (DGP). The PRF embodies the true values of  and , and for the bivariate model, could be expressed as y t    xt  u t

Note that there is a disturbance term in this equation. In some textbooks, a distinction is drawn between the PRF (the underlying true relationship between y and x) and the DGP (the process describing the way that the actual observations on y come about).

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The sample regression function, SRF, is the relationship that has been estimated using the sample observations, and is often written as

yˆ t  ˆ  ˆxt Notice that there is no error or residual term in the equation for the SRF: all this equation states is that given a particular value of x, multiplying it by  and adding  will give the model fitted or expected value for y, denoted yˆ . It is also possible to write

y t  ˆ  ˆxt  uˆ t This equation splits the observed value of y into value from the model, and a residual term. The values of the PRF. That is the estimates  and sample data.

two components: the fitted SRF is used to infer likely  are constructed, for the

3. An estimator is simply a formula that is used to calculate the estimates, i.e. the parameters that describe the relationship between two or more explanatory variables. There are an infinite number of possible estimators; OLS is one choice that many people would consider a good one. We can say that the OLS estimator is “best” – i.e. that it has the lowest variance among the class of linear unbiased estimators. So it is optimal in the sense that no other linear, unbiased estimator would have a smaller sampling variance. We could define an estimator with a lower sampling variance than the OLS estimator, but it would either be non-linear or biased or both! So there is a trade-off between bias and variance in the choice of the estimator.

4. A list of the assumptions of the classical linear regression model’s disturbance terms is given in Box 2.3 on p.44 of the book. We need to make the first four assumptions in order to prove that the ordinary least squares estimators of  and  are “best”, that is to prove that they have minimum variance among the class of linear unbiased estimators. The theorem that proves that OLS estimators are BLUE (provided the assumptions are fulfilled) is known as the Gauss-Markov theorem. If these assumptions are violated (which is dealt with in Chapter 4), then it may be that OLS estimators are no longer unbiased or “efficient”. That is, they may be inaccurate or subject to fluctuations between samples. We needed to make the fifth assumption, that the disturbances are normally distributed, in order to make statistical inferences about the population parameters from the sample data, i.e. to test hypotheses about the coefficients. Making this assumption implies that test statistics will follow a t-distribution (provided that the other assumptions also hold).

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5. If the models are linear in the parameters, we can use OLS. (2.57) Yes, can use OLS since the model is the usual linear model we have been dealing with. (2.58) Yes. The model can be linearised by taking logarithms of both sides and by rearranging. Although this is a very specific case, it has sound theoretical foundations (e.g. the Cobb-Douglas production function in economics), and it is the case that many relationships can be “approximately” linearised by taking logs of the variables. The effect of taking logs is to reduce the effect of extreme values on the regression function, and it may be possible to turn multiplicative models into additive ones which we can easily estimate. (2.59) Yes. We can estimate this model using OLS, but we would not be able to obtain the values of both  and, but we would obtain the value of these two coefficients multiplied together. (2.60) Yes, we can use OLS, since this model is linear in the logarithms. For those who have done some economics, models of this kind which are linear in the logarithms have the interesting property that the coefficients (  and) can be interpreted as elasticities. (2.61). Yes, in fact we can still use OLS since it is linear in the parameters. If we make a substitution, say qt = xtzt, then we can run the regression yt =  +qt + ut as usual. So, in fact, we can estimate a fairly wide range of model types using these simple tools. 6. The null hypothesis is that the true (but unknown) value of beta is equal to one, against a one sided alternative that it is greater than one: H0 :  = 1 H1 :  > 1 The test statistic is given by

test stat 

ˆ   * 1.147  1   2.682 0.0548 SE ( ˆ )

We want to compare this with a value from the t-table with T-2 degrees of freedom, where T is the sample size, and here T-2 =60. We want a value with 5% all in one tail since we are doing a 1-sided test. The critical t-value from the t-table is 1.671:

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f(x)

5% rejection region

+1.671

The value of the test statistic is in the rejection region and hence we can reject the null hypothesis. We have statistically significant evidence that this security has a beta greater than one, i.e. it is significantly more risky than the market as a whole. 7. We want to use a two-sided test to test the null hypothesis that shares in Chris Mining are completely unrelated to movements in the market as a whole. In other words, the value of beta in the regression model would be zero so that whatever happens to the value of the market proxy, Chris Mining would be completely unaffected by it. The null and alternative hypotheses are therefore: H0 :  = 0 H1 :   0 The test statistic has the same format as before, and is given by: test stat 

ˆ   * 0.214  0   1.150 SE (  ) 0.186

We want to find a value from the t-tables for a variable with 38-2=36 degrees of freedom, and we want to look up the value that puts 2.5% of the distribution in each tail since we are doing a two-sided test and we want to have a 5% size of test over all:

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2.5% rejection region

-2.03

2.5% rejection region

+2.03

The critical t-value is therefore 2.03. Since the test statistic is not within the rejection region, we do not reject the null hypothesis. We therefore conclude that we have no statistically significant evidence that Chris Mining has any systematic risk. In other words, we have no evidence that changes in the company’s value are driven by movements in the market. 8. A confidence interval for beta is given by the formula:

( ˆ  SE ( ˆ )  t crit , ˆ  SE ( ˆ )  t crit ) Confidence intervals are almost invariably 2-sided, unless we are told otherwise (which we are not here), so we want to look up the values which put 2.5% in the upper tail and 0.5% in the upper tail for the 95% and 99% confidence intervals respectively. The 0.5% critical values are given as follows for a t-distribution with T-2=38-2=36 degrees of freedom:

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0.5% rejection region

-2.72

0.5% rejection region

+2.72

The confidence interval in each case is thus given by (0.2140.186*2.03) for a 95% confidence interval, which solves to (-0.164, 0.592) and (0.2140.186*2.72) for a 99% confidence interval, which solves to (0.292,0.720) There are a couple of points worth noting. First, one intuitive interpretation of an X% confidence interval is that we are X% sure that the true value of the population parameter lies within the interval. So we are 95% sure that the true value of beta lies within the interval (-0.164, 0.592) and we are 99% sure that the true population value of beta lies within (-0.292, 0.720). Thus in order to be more sure that we have the true vale of beta contained in the interval, i.e. as we move from 95% to 99% confidence, the interval must become wider. The second point to note is that we can test an infinite number of hypotheses about beta once we have formed the interval. For example, we would not reject the null hypothesis contained in the last question (i.e. that beta = 0), since that value of beta lies within the 95% and 99% confidence intervals. Would we reject or not reject a null hypothesis that the true value of beta was 0.6? At the 5% level, we should have enough evidence against the null hypothesis to reject it, since 0.6 is not contained within the 95% confidence interval. But at the 1% level, we would no longer have sufficient evidence to reject the null hypothesis, since 0.6 is now contained within the interval. Therefore we should always if possible conduct some sort of sensitivity analysis to see if our conclusions are altered by (sensible) changes in the level of significance used. 9. We test hypotheses about the actual coefficients, not the estimated values. We want to make inferences about the likely values of the population parameters (i.e. to test hypotheses about them). We do not need to test hypotheses about the estimated values since we know exactly what our estimates are because we calculated them!

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Chapter 3 1. It can be proved that a t-distribution is just a special case of the more general F-distribution. The square of a t-distribution with T-k degrees of freedom will be identical to an F-distribution with (1,T-k) degrees of freedom. But remember that if we use a 5% size of test, we will look up a 5% value for the F-distribution because the test is 2-sided even though we only look in one tail of the distribution. We look up a 2.5% value for the t-distribution since the test is 2-tailed. Examples at the 5% level from tables T-k 20 40 60 120

F critical value 4.35 4.08 4.00 3.92

t critical value 2.09 2.02 2.00 1.98

2. (a) H0 : 3 = 2 We could use an F- or a t- test for this one since it is a single hypothesis involving only one coefficient. We would probably in practice use a t-test since it is computationally simpler and we only have to estimate one regression. There is one restriction. (b) H0 : 3 + 4 = 1 Since this involves more than one coefficient, we should use an F-test. There is one restriction. (c) H0 : 3 + 4 = 1 and 5 = 1 Since we are testing more than one hypothesis simultaneously, we would use an F-test. There are 2 restrictions. (d) H0 : 2 =0 and 3 = 0 and 4 = 0 and 5 = 0 As for (c), we are testing multiple hypotheses so we cannot use a t-test. We have 4 restrictions. (e) H0 : 23 = 1 Although there is only one restriction, it is a multiplicative restriction. We therefore cannot use a t-test or an F-test to test it. In fact we cannot test it at all using the methodology that has been examined in this chapter. 3. THE regression F-statistic would be given by the test statistic associated with hypothesis iv) above. We are always interested in testing this hypothesis since it tests whether all of the coefficients in the regression (except the constant) are jointly insignificant. If they are then we have a completely useless regression, where none of the variables that we have said influence y actually do. So we would need to go back to the drawing board!

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The alternative hypothesis is: H1 : 2  0 or 3  0 or 4  0 or 5  0 Note the form of the alternative hypothesis: “or” indicates that only one of the components of the null hypothesis would have to be rejected for us to reject the null hypothesis as a whole. 4. The restricted residual sum of squares will always be at least as big as the unrestricted residual sum of squares i.e. RRSS  URSS To see this, think about what we were doing when we determined what the regression parameters should be: we chose the values that minimised the residual sum of squares. We said that OLS would provide the “best” parameter values given the actual sample data. Now when we impose some restrictions on the model, so that they cannot all be freely determined, then the model should not fit as well as it did before. Hence the residual sum of squares must be higher once we have imposed the restrictions; otherwise, the parameter values that OLS chose originally without the restrictions could not be the best. In the extreme case (very unlikely in practice), the two sets of residual sum of squares could be identical if the restrictions were already present in the data, so that imposing them on the model would yield no penalty in terms of loss of fit. 5. The null hypothesis is: H0 : 3 + 4 = 1 and 5 = 1 The first step is to impose this on the regression model: yt = 1 + 2x2t + 3x3t + 4x4t + 5x5t + ut subject to 3 + 4 = 1 and 5 = 1. We can rewrite the first part of the restriction as 4 = 1 - 3 Then rewrite the regression with the restriction imposed yt = 1 + 2x2t + 3x3t + (1-3)x4t + x5t + ut which can be re-written yt = 1 + 2x2t + 3x3t + x4t - 3x4t + x5t + ut and rearranging (yt – x4t – x5t ) = 1 + 2x2t + 3x3t - 3x4t + ut (yt – x4t – x5t) = 1 + 2x2t + 3(x3t –x4t)+ ut

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Now create two new variables, call them Pt and Qt: pt = (yt - x3t - x4t) qt = (x2t -x3t) We can then run the linear regression: pt = 1 + 2x2t + 3qt+ ut , which constitutes the restricted regression model. The test statistic is calculated as ((RRSS-URSS)/URSS)*(T-k)/m In this case, m=2, T=96, k=5 so the test statistic = 5.704. Compare this to an F-distribution with (2,91) degrees of freedom, which is approximately 3.10. Hence we reject the null hypothesis that the restrictions are valid. We cannot impose these restrictions on the data without a substantial increase in the residual sum of squares. 6.

ri = 0.080 + 0.801Si + 0.321MBi + 0.164PEi - 0.084BETAi (0.064) (0.147) (0.136) (0.420) (0.120) 1.25 5.45 2.36 0.390 -0.700

The t-ratios are given in the final row above, and are in italics. They are calculated by dividing the coefficient estimate by its standard error. The relevant value from the t-tables is for a 2-sided test with 5% rejection overall. T-k = 195; tcrit = 1.97. The null hypothesis is rejected at the 5% level if the absolute value of the test statistic is greater than the critical value. We would conclude based on this evidence that only firm size and market to book value have a significant effect on stock returns. If a stock’s beta increases from 1 to 1.2, then we would expect the return on the stock to FALL by (1.2-1)*0.084 = 0.0168 = 1.68% This is not the sign we would have expected on beta, since beta would be expected to be positively related to return, since investors would require higher returns as compensation for bearing higher market risk. 7. We would thus consider deleting the price/earnings and beta variables from the regression since these are not significant in the regression - i.e. they are not helping much to explain variations in y. We would not delete the constant term from the regression even though it is insignificant since there are good statistical reasons for its inclusion. yt  1   2 x2t   3 x3t   4 yt 1  u t yt   1   2 x2t   3 x3t   4 yt 1  vt .

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(a) 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 x 2t  ˆ 3 X 3t  ˆ 4 yt 1 Now for the second model, the dependent variable is now the change in y: vˆt  yt  yˆ t  yt  ˆ1  ˆ 2 x 2t  ˆ 3 x3t  ˆ 4 yt 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 x3t  (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 i  ˆ i (i = 1, 2, 3). Hence if the residuals are the same, the residual sum of squares must also be the same. In fact the two models are really identical, since one is just a rearrangement of the other. (b) As for R2, recall how we calculate R2:

R2  1

RSS for the first model and  ( yi  y ) 2

R2 1

RSS in the second case. Therefore since the total sum of  (yi  y ) 2

squares (the denominator) has changed, then the value of R2 must have also changed as a consequence of changing the dependent variable. (c) 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. 8. A researcher estimates the following two econometric models y t  1   2 x 2t   3 x3t  u t yt  1   2 x2t   3 x3t   4 x4t  vt

(a) The value of R2 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

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coefficient on the x4t variable was exactly zero. Otherwise, the R2 must be higher for the second model than the first. (b) 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. 11. R2 may be defined in various ways, but the most common is ESS R2  TSS Since both ESS and TSS will have units of the square of the dependent variable, the units will cancel out and hence R2 will be unit-free!

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.

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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. 0.638 + 0.402 x2t - 0.891 x3t R2  0.96,R 2  0.89 (0.436) (0.291) (0.763) t-ratios 1.46 1.38 -1.17

yˆ t =

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 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.

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-

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 - 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

t

2  uˆt 1 

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

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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. (d) yt  1   2 x2t   3 x3t   4 x4t  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) yt  1   2 x2t   3 x3t   4 x4t   5 x2t 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.

yt  1   2 x 2t   3 x3t   4 yt 1   5 x2t 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

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- 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 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.

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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.

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

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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 RSS  RSS1 T1  k 0.189  0.148 168  2 *  *  3.832 RSS1 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 17/59

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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 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.

Chapter 5 1. Autoregressive models specify the current value of a series yt as a function of its previous p values and the current value an error term, ut, while moving average models specify the current value of a series yt as a function of the current and previous q values of an error term, ut. AR and MA models have different characteristics in terms of the length of their “memories”, which has implications for the time it takes shocks to yt to die away, and for the shapes of their autocorrelation and partial autocorrelation functions. 2. 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.

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3.

yt = yt-1 + ut yt = 0.5 yt-1 + ut yt = 0.8 ut-1 + ut

(1) (2) (3)

(a) 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. (b) 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. (c) 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 yesterdays 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 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. (d) 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 19/59

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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. 4. (a) 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! (b) 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”. (c) We could use Akaike’s or Schwarz’s Bayesian information criteria. Our objective would then be to fit the model order that minimises these.

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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. 5. The best way to check for stationarity is to express the model as a lag polynomial in yt. yt  0803 . yt 1  0.682 yt 2  ut

Rewrite this as yt (1  0.803L  0.682 L2 )  ut

We want to find the roots of the lag polynomial (1  0.803L  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,

z

 1177 .  1177 . 2  4 * 1 * 1466 . = 0.758 or 1.934 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. 6. 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)

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log (  2 ) 0.932 0.864 0.902 0.836 0.801 0.821 0.789

AIC 0.942 0.884 0.922 0.866 0.841 0.861 0.839

SBIC 0.944 0.887 0.925 0.870 0.847 0.867 0.846

“Introductory Econometrics for Finance” © Chris Brooks 2008

(3,2) 0.842* (2,3) (3,3)

0.773

0.833*

0.782 0.764

0.842 0.834

0.851 0.844

The result is pretty clear: both SBIC and AIC say that the appropriate model is an ARMA(3,2). 7. 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. 8. 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. 9. (a) We class an autocorrelation coefficient or partial autocorrelation 1 coefficient as significant if it exceeds  1.96 =  0.196. Under this rule, the T sample 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.

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(b) The formula for the Ljung-Box Q* test is given by m

Q*  T (T  2)

 k2

k 1 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

 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. 10. (a) To solve this, we need the concept of a conditional expectation, i.e. Et 1 ( y t y t 2 , y t 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 ) 2 = a0  a0a1  a1 yt 1 E ( yt 2 )  a0  a1 E ( yt 1 ) = a0  a1 (a0  a1 E ( yt )) 2 = a0  a0a1  a1 E ( yt ) 2 = a0  a0a1  a1 E ( yt ) 2 = a0  a0a1  a1 (a0  a1 yt 1 ) 2 3 = a0  a0a1  a1 a0  a1 yt 1 etc. f t 1,1  a0  a1 yt 1 f t 1,2  a0  a1 f t 1,1 f t 1,3  a0  a1 f t 1,2

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To forecast an MA model, consider, e.g. yt  ut  b1ut 1

E ( yt yt 1 , yt 2 ,...)

So

ft-1,1

=

=

E (ut  b1ut 1 ) b1u t 1

= =

E (ut 1  b1ut ) 0

b1u t 1

But E ( yt 1 yt 1 , yt 2 ,...)

Going back to the example above, yt  0.036  0.69 yt 1  0.42ut 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 yt 1  0.42ut 1  ut = 0.036 +0.693.4+0.42(-1.3) = 1.836 ft-1,2 = E ( yt 1 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 MSE 

1 N

N

 n 1

( xt 1 n  f t 1, n ) 2

if we are making N forecasts which are numbered 1,2,3. Then the MSE is given by

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1  (1.836  0.032) 2  (1.302  0.961) 2  (0.935  0.203) 2  3 1  (3.489  0.116  0.536)  1.380 3

MSE 

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 1 MSE  (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. 11. (a) The shapes of the acf and pacf are perhaps best summarised in a table: Process White noise AR(2) MA(1) ARMA(2 ,1)

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acf No significant coefficients

pacf No significant coefficients

Geometrically declining or damped sinusoid acf

First 2 pacf coefficients significant, all others insignificant Geometrically declining or damped sinusoid pacf Geometrically declining or damped sinusoid pacf

First acf coefficient significant, all others insignificant Geometrically declining or damped sinusoid acf

“Introductory Econometrics for Finance” © Chris Brooks 2008

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: yt 1  0.036  0.69 yt  0.42ut  u t 1 yt  2  0.036  0.69 yt 1  0.42u t 1  u t  2 yt 3  0.036  0.69 yt  2  0.42u t  2  u t 3 yt  4  0.036  0.69 yt 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:

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

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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 the book. 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. 12. (a) Some of the stylised differences between the typical characteristics of macroeconomic and financial data were presented in 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 timeseries approaches (e.g. ARMA models), rather than structural formulations with separate explanatory variables. 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

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is to classify them as significant at the 5% level if they lie outside of 1 , where T is the sample size. In this case, T = 500, so a particular  1.96 * T 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

and

 k2

m

Q*  T (T  2) k 1

T k

.

In this instance, the statistics would be calculated respectively as Q  500  [0.307 2  (0.0132 )  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:

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Model A: MA(1) f z ,1  0.38  0.10u t 1 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) xˆt  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 MSE ( Model A)  MSE ( Model B) 





1 (0.378  0.62) 2  (0.38  0.19) 2  (0.38  0.32) 2  (0.38  0.72) 2  0.175 4





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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Chapter 6 1. (a) This is simple to accomplish in theory, but difficult in practice as a result of the algebra. The original equations are (renumbering them (1), (2) and (3) for simplicity) y1t   0  1 y 2 t   2 y 3t   3 X 1t   4 X 2 t  u1t

(1)

y 2 t  0  1 y 3t  2 X 1t  3 X 3t  u2 t

(2)

y 3t   0   1 y1t   2 X 2 t   3 X 3t  u3t

( 3)

The easiest place to start (I think) is to take equation (1), and substitute in for y3t, to get y1t  0  1 y2t  2 ( 0   1 y1t   2 X 2t   3 X 3t  u3t )  3 X 1t  4 X 2t  u1t

Working out the products that arise when removing the brackets, y1t  0  1 y2t  2 0  2 1 y1t  2 2 X 2t  2 3 X 3t  2 u3t  3 X 1t  4 X 2t  u1t

Gathering terms in y1t on the LHS: y1t  2 1 y1t  0  1 y2t  2 0  2 2 X 2t  2 3 X 3t  2 u3t  3 X 1t  4 X 2t  u1t y1t (1  2 1 )  0  1 y2t  2 0  2 2 X 2t  2 3 X 3t  2 u3t  3 X 1t  4 X 2t  u1t (4) Now substitute into (2) for y3t from (3). y2t  0  1 ( 0   1 y1t   2 X 2t   3 X 3t  u3t )  2 X 1t  3 X 3t  u2t

Removing the brackets y2t  0  1 0  1 1 y1t  1 2 X 2t  1 3 X 3t  1u3t  2 X 1t  3 X 3t  u2t

(5)

Substituting into (4) for y2t from (5), y1t (1  2 1 )  0  1 ( 0  1 0  1 1 y1t    2 X 2 t  1 3 X 3t  1u3t  2 X 1t 

3 X 3t  u2 t )  2 0  2 2 X 2 t  2 3 X 3t  2 u3t  3 X 1t  4 X 2 t  u1t Taking the y1t terms to the LHS: y1t (1  2 1  1 1 1 )  0  1 0  1 1 0  1   2 X 2 t  1 1 3 X 3t  1 1u3t  1 2 X 1t 

13 X 3t  1u2 t  2 0  2 2 X 2 t  2 3 X 3t  2 u3t  3 X 1t  4 X 2 t  u1t

Gathering like-terms in the other variables together:

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y1t (1   2  1  1 1 1 )   0  1 0  1 1 0   2  0  X 1t (1 2   3 )  X 2 t (1 1 2   2  2   4 )  X 3t (1 1 3  1 3   2  3 )  u3t (1 1   2 )  1 u2 t  u1t

(6) Multiplying all through equation (3) by (1  2 1  1 1 1 ) : y3t (1  2 1  11 1 )   0 (1  2 1  11 1 )   1 y1t (1  2 1  11 1 ) 

 2 X 2 t (1  2 1  11 1 )   3 X 3t (1  2 1  11 1 )  u3t (1  2 1  11 1 ) Replacing y1t (1  2 1  11 1 )

(7) in (7) with the RHS of (6),

 0  1 0  1 1 0   2  0  X 1t (1 2   3 )    y 3t (1   2  1  11 1 )   0 (1   2  1  11 1 )   1  X 2 t (1 1 2   2  2   4 )  X 3t (1 1 3  1 3    2  3 )  u3t (1 1   2 )  1u2 t  u1t    2 X 2 t (1   2  1  11 1 )   3 X 3t (1   2  1  11 1 )  u3t (1   2  1  11 1 ) (8) Expanding the brackets in equation (8) and cancelling the relevant terms y3t (1  2 1  11 1 )   0   10   11 0  X 1t (1 2 1   1 3 )  X 2 t ( 2   14 )  X 3t ( 11 3   3 )  u3t   11u2 t   1u1t

(9) Multiplying all through equation (2) by (1  2 1  1 1 1 ) : y2 t (1   1  1 1 12 )  0 (1   1  1 1 12 )  1 y3t (1   1  1 1 12 ) 

2 X 1t (1   1  11 12 )  3 X 3t (1   1  1 1 12 )  u2 t (1   1  1 1 12 ) (10) Replacing y3t (1  2 1  11 1 )

in (10) with the RHS of (9),

 0   1 0   11 0  X 1t (1 2  1   1 3 )     y 2 t (1   1  1 1 1 2 )  0 (1   1  1 1 12 )  1  X 2 t ( 2   1 4 )  X 3t ( 3   11 3 )  u3t    11u2 t   1u1t   2 X 1t (1   1  1 1 12 )  3 X 3t (1   1  1 1 12 )  u2 t (1   1  1 1 12 ) (11) Expanding the brackets in (11) and cancelling the relevant terms y2t (1   1  1 ( 1 12 )  0  02 1   1 0 1 10  X1t 1 1 3  2  22 1 )  X 2 t (  1 2   1 14 )  X 3t ( 1 3  3  32 1 )  1u3t  u2 t (1  2 1 )   1 1u1t

(12)

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Although it might not look like it (!), equations (6), (12), and (9) respectively will give the reduced form equations corresponding to (1), (2), and (3), by doing the necessary division to make y1t, y2t, or y3t the subject of the formula. From (6),

 0  1 0  1 1 0   2  0 (1 2   3 ) (     2  2   4 )  X 1t  1 1 2 X 2t  (1   2  1  1 1 1 ) (1   2  1  1 1 1 ) (1   2  1  1 1 1 ) (1 1 3  1 3   2  3 ) u (    2 )  1u2 t  u1t X 3t  3t 1 1 (1   2  1  1 1 1 ) (1   2  1  1 1 1 ) (13) From (12), y1t 

y2 t 

0  02 1  1 01 10 ( 1 1 3  2  22 1 ) ( 1 2  1 14 )  X1t  X  (1   1  11 12 ) (1   1  11 12 ) (1   1  11 12 ) 2 t

( 1 3  3  32 1 )  u  u (1  2 1 )  1 1u1t X 3t  1 3t 2 t (1   1  11 12 ) (1   1  11 12 ) (14) From (9),

y 3t 

 0   10   11 0 (1 2  1   1 3 ) ( 2   1 4 )  X 1t  X  (1   2  1  11 1 ) (1   2  1  11 1 ) (1   2  1  11 1 ) 2 t

( 11 3   3 ) u   11u2 t   1u1t X 3t  3t (1   2  1  11 1 ) (1   2  1  11 1 )

(15)

Notice that all of the reduced form equations (13)-(15) in this case depend on all of the exogenous variables, which is not always the case, and that the equations contain only exogenous variables on the RHS, which must be the case for these to be reduced forms. (b) The term “identification” refers to whether or not it is in fact possible to obtain the structural form coefficients (the , , and ’s in equations (1)-(3)) from the reduced form coefficients (the ’s) by substitution. An equation can be over-identified, just-identified, or under-identified, and the equations in a system can have differing orders of identification. If an equation is underidentified (or not identified), then we cannot obtain the structural form coefficients from the reduced forms using any technique. If it is just identified, we can obtain unique structural form estimates by back-substitution, while if it is over-identified, we cannot obtain unique structural form estimates by substituting from the reduced forms. There are two rules for determining the degree of identification of an equation: the rank condition, and the order condition. The rank condition is a necessary and sufficient condition for identification, so if the rule is satisfied, it guarantees that the equation is indeed identified. The rule centres around a restriction on the rank of a sub-matrix containing the reduced form

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coefficients, and is rather complex and not particularly illuminating, and was therefore not covered in this course. The order condition, can be expressed in a number of ways, one of which is the following. Let G denote the number of structural equations (equal to the number of endogenous variables). An equation is just identified if G-1 variables are absent. If more than G-1 are absent, then the equation is overidentified, while if fewer are absent, then it is not identified. Applying this rule to equations (1)-(3), G=3, so for an equation to be identified, we require 2 to be absent. The variables in the system are y1, y2, y3, X1, X2, X3. Is this the case? Equation (1): X3t only is missing, so the equation is not identified. Equation (2): y1t and X2t are missing, so the equation is just identified. Equation (3): y2t and X1t are missing, so the equation is just identified. However, the order condition is only a necessary (and not a sufficient) condition for identification, so there will exist cases where a given equation satisfies the order condition, but we still cannot obtain the structural form coefficients. Fortunately, for small systems this is rarely the case. Also, in practice, most systems are designed to contain equations that are overidentified. (c). It was stated in Chapter 4 that omitting a relevant variable from a regression equation would lead to an “omitted variable bias” (in fact an inconsistency as well), while including an irrelevant variable would lead to unbiased but inefficient coefficient estimates. There is a direct analogy with the simultaneous variable case. Treating a variable as exogenous when it really should be endogenous because there is some feedback, will result in biased and inconsistent parameter estimates. On the other hand, treating a variable as endogenous when it really should be exogenous (that is, having an equation for the variable and then substituting the fitted value from the reduced form if 2SLS is used, rather than just using the actual value of the variable) would result in unbiased but inefficient coefficient estimates. If we take the view that consistency and unbiasedness are more important that efficiency (which is the view that I think most econometricians would take), this implies that treating an endogenous variable as exogenous represents the more severe mis-specification. So if in doubt, include an equation for it! (Although, of course, we can test for exogeneity using a Hausman-type test). (d). A tempting response to the question might be to describe indirect least squares (ILS), that is estimating the reduced form equations by OLS and then substituting back to get the structural forms; however, this response would be WRONG, since the question tells us that the system is over-identified. A correct answer would be to describe either two stage least squares (2SLS) or instrumental variables (IV). Either would be acceptable, although IV requires the user to determine an appropriate set of instruments and hence 2SLS is simpler in practice. 2SLS involves estimating the reduced form equations, and obtaining the fitted values in the first stage. In the second stage, the structural 33/59

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form equations are estimated, but replacing the endogenous variables on the RHS with their stage one fitted values. Application of this technique will yield unique and unbiased structural form coefficients. 2. (a) A glance at equations (6.97) and (6.98) reveals that the dependent variable in (6.97) appears as an explanatory variable in (6.98) and that the dependent variable in (6.98) appears as an explanatory variable in (6.97). The result is that it would be possible to show that the explanatory variable y2t in (6.97) will be correlated with the error term in that equation, u1t, and that the explanatory variable y1t in (6.98) will be correlated with the error term in that equation, u2t. Thus, there is causality from y1t to y2t and from y2t to y1t, so that this is a simultaneous equations system. If OLS were applied separately to each of equations (6.97) and (6.98), the result would be biased and inconsistent parameter estimates. That is, even with an infinitely large number of observations, OLS could not be relied upon to deliver the appropriate parameter estimates. (b) If the variable y1t had not appeared on the RHS of equation (6.98), this would no longer be a simultaneous system, but would instead be an example of a triangular system (see question 3). Thus it would be valid to apply OLS separately to each of the equations (6.97) and (6.98). (c) The order condition for determining whether an equation from a simultaneous system is identified was described in question 1, part (b). There are 2 equations in the system of (6.97) and (6.98), so that only 1 variable would have to be missing from an equation to make it just identified. If no variables are absent, the equation would not be identified, while if more than one were missing, the equation would be over-identified. Considering equation (6.97), no variables are missing so that this equation is not identified, while equation (6.98) excludes only variable X2t, so that it is just identified. (d) Since equation (6.97) is not identified, no method could be used to obtain estimates of the parameters of this equation, while either ILS or 2SLS could be used to obtain estimates of the parameters of (6.98), since it is just identified. ILS operates by obtaining and estimating the reduced form equations and then obtaining the structural parameters of (6.98) by algebraic backsubstitution. 2SLS involves again obtaining and estimating the reduced form equations, and then estimating the structural equations but replacing the endogenous variables on the RHS of (6.97) and (6.98) with their reduced form fitted values. Comparing between ILS and 2SLS, the former method only requires one set of estimations rather than two, but this is about its only advantage, and conducting a second stage OLS estimation is usually a computationally trivial exercise. The primary disadvantage of ILS is that it is only applicable to just identified equations, whereas many sets of equations that we may wish to estimate are over-identified. Second, obtaining the structural form coefficients via algebraic substitution can be a very tedious exercise in the context of large systems (as the solution to question 1, part (a) shows!).

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(e) The Hausman procedure works by first obtaining and estimating the reduced form equations, and then estimating the structural form equations separately using OLS, but also adding the fitted values from the reduced form estimations as additional explanatory variables in the equations where those variables appear as endogenous RHS variables. Thus, if the reduced form fitted values corresponding to equations (6.97) and (6.98) are given by y1t and y2t respectively, the Hausmann test equations would be y1t   0   1 y 2t   2 X 1t   3 X 2t   4 y 2t 'u1t y 2t   0   1 y1t   2 X 1t   3 y1t ' u1t

.

Separate tests of the significance of the y1t and y2t terms would then be performed. If it were concluded that they were both significant, this would imply that additional explanatory power can be obtained by treating the variables as endogenous. 3. An example of a triangular system was given in Section 6.7. Consider a scenario where there are only two “endogenous” variables. The key distinction between this and a fully simultaneous system is that in the case of a triangular system, causality runs only in one direction, whereas for a simultaneous equation, it would run in both directions. Thus, to give an example, for the system to be triangular, y1 could appear in the equation for y2 and not vice versa. For the simultaneous system, y1 would appear in the equation for y2, and y2 would appear in the equation for y1. 4. (a) p=2 and k=3 implies that there are two variables in the system, and that both equations have three lags of the two variables. The VAR can be written in long-hand form as: y1t   10   111 y1t 1   211 y 2t 1   112 y1t  2   212 y 2t  2   113 y1t 3   213 y 2t 3  u1t y 2t   20   121 y1t 1   221 y 2t 1   122 y1t  2   222 y 2t  2   123 y1t 3   223 y 2t 3  u 2t  10   y1t   u1t  where 0   , yt   , ut    , and the  coefficients on the lags of yt 20   y2 t  u2 t 

are defined as follows: ijk refers to the kth lag of the ith variable in the jth equation. This seems like a natural notation to use, although of course any sensible alternative would also be correct. (b) This is basically a “what are the advantages of VARs compared with structural models?” type question, to which a simple and effective response would be to list and explain the points made in the book. The most important point is that structural models require the researcher to specify some variables as being exogenous (if all variables were endogenous, then none of the equations would be identified, and therefore estimation of the structural equations would be impossible). This can be viewed as a

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restriction (a restriction that the exogenous variables do not have any simultaneous equations feedback), often called an “identifying restriction”. Determining what are the identifying restrictions is supposed to be based on economic or financial theory, but Sims, who first proposed the VAR methodology, argued that such restrictions were “incredible”. He thought that they were too loosely based on theory, and were often specified by researchers on the basis of giving the restrictions that the models required to make the equations identified. Under a VAR, all the variables have equations, and so in a sense, every variable is endogenous, which takes the ability to cheat (either deliberately or inadvertently) or to mis-specify the model in this way, out of the hands of the researcher. Another possible reason why VARs are popular in the academic literature is that standard form VARs can be estimated using OLS since all of the lags on the RHS are counted as pre-determined variables. Further, a glance at the academic literature which has sought to compare the forecasting accuracies of structural models with VARs, reveals that VARs seem to be rather better at forecasting (perhaps because the identifying restrictions are not valid). Thus, from a purely pragmatic point of view, researchers may prefer VARs if the purpose of the modelling exercise is to produce precise point forecasts. (c) VARs have, of course, also been subject to criticisms. The most important of these criticisms is that VARs are atheoretical. In other words, they use very little information form economic or financial theory to guide the model specification process. The result is that the models often have little or no theoretical interpretation, so that they are of limited use for testing and evaluating theories. Second, VARs can often contain a lot of parameters. The resulting loss in degrees of freedom if the VAR is unrestricted and contains a lot of lags, could lead to a loss of efficiency and the inclusion of lots of irrelevant or marginally relevant terms. Third, it is not clear how the VAR lag lengths should be chosen. Different methods are available (see part (d) of this question), but they could lead to widely differing answers. Finally, the very tools that have been proposed to help to obtain useful information from VARs, i.e. impulse responses and variance decompositions, are themselves difficult to interpret! – See Runkle (1987). (d) The two methods that we have examined are model restrictions and information criteria. Details on how these work are contained in Sections 6.12.4 and 6.12.5. But briefly, the model restrictions approach involves starting with the larger of the two models and testing whether it can be restricted down to the smaller one using the likelihood ratio test based on the determinants of the variance-covariance matrices of residuals in each case. The alternative approach would be to examine the value of various information criteria and to select the model that minimises the criteria. Since there are only two models to compare, either technique could be used. The restriction approach assumes normality for the VAR error terms, while use of 36/59

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the information criteria does not. On the other hand, the information criteria can lead to quite different answers depending on which criterion is used and the severity of its penalty term. A completely different approach would be to put the VARs in the situation that they were intended for (e.g. forecasting, making trading profits, determining a hedge ratio etc.), and see which one does best in practice!

Chapter 7 1. (a) Many series in finance and economics in their levels (or log-levels) forms are non-stationary and exhibit stochastic trends. They have a tendency not to revert to a mean level, but they “wander” for prolonged periods in one direction or the other. Examples would be most kinds of asset or goods prices, GDP, unemployment, money supply, etc. Such variables can usually be made stationary by transforming them into their differences or by constructing percentage changes of them. (b) Non-stationarity can be an important determinant of the properties of a series. Also, if two series are non-stationary, we may experience the problem of “spurious” regression. This occurs when we regress one non-stationary variable on a completely unrelated non-stationary variable, but yield a reasonably high value of R2, apparently indicating that the model fits well. Most importantly therefore, we are not able to perform any hypothesis tests in models which inappropriately use non-stationary data since the test statistics will no longer follow the distributions which we assumed they would (e.g. a t or F), so any inferences we make are likely to be invalid. (c) A weakly stationary process was defined in Chapter 5, and has the following characteristics: 1. E(yt) =  2. E ( yt   )( yt   )   2   3. E ( yt1   )( yt 2   )   t 2 t1  t1 , t2 That is, a stationary process has a constant mean, a constant variance, and a constant covariance structure. A strictly stationary process could be defined by an equation such as Fx t1 , xt 2 ,..., xtT ( x1 ,..., xT )  Fx t1  k , xt 2  k ,..., xtT  k ( x1 ,..., xT )

for any t1 , t2 , ..., tT  Z, any k Z and T = 1, 2, ...., and where F denotes the joint distribution function of the set of random variables. It should be evident from the definitions of weak and strict stationarity that the latter is a stronger definition and is a special case of the former. In the former case, only the first two moments of the distribution has to be constant (i.e. the mean and variances (and covariances)), whilst in the latter case, all moments of the distribution (i.e. the whole of the probability distribution) has to be constant. 37/59

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Both weakly stationary and strictly stationary processes will cross their mean value frequently and will not wander a long way from that mean value. An example of a deterministic trend process was given in Figure 7.5. Such a process will have random variations about a linear (usually upward) trend. An expression for a deterministic trend process yt could be yt =  + t + ut where t = 1, 2,…, is the trend and ut is a zero mean white noise disturbance term. This is called deterministic non-stationarity because the source of the non-stationarity is a deterministic straight line process. A variable containing a stochastic trend will also not cross its mean value frequently and will wander a long way from its mean value. A stochastically non-stationary process could be a unit root or explosive autoregressive process such as yt = yt-1 + ut where   1. 2. (a)The null hypothesis is of a unit root against a one sided stationary alternative, i.e. we have H0 : yt  I(1) H1 : yt  I(0) which is also equivalent to H0 :  = 0 H1 :  < 0 (b) The test statistic is given by  / SE ( ) which equals -0.02 / 0.31 = -0.06 Since this is not more negative than the appropriate critical value, we do not reject the null hypothesis. (c) We therefore conclude that there is at least one unit root in the series (there could be 1, 2, 3 or more). What we would do now is to regress 2yt on yt-1 and test if there is a further unit root. The null and alternative hypotheses would now be H0 : yt  I(1) i.e. yt  I(2) H1 : yt  I(0) i.e. yt  I(1) If we rejected the null hypothesis, we would therefore conclude that the first differences are stationary, and hence the original series was I(1). If we did not reject at this stage, we would conclude that yt must be at least I(2), and we would have to test again until we rejected.

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(d) We cannot compare the test statistic with that from a t-distribution since we have non-stationarity under the null hypothesis and hence the test statistic will no longer follow a t-distribution. 3. Using the same regression as above, but on a different set of data, the researcher now obtains the estimate =-0.52 with standard error = 0.16. (a) The test statistic is calculated as above. The value of the test statistic = 0.52 /0.16 = -3.25. We therefore reject the null hypothesis since the test statistic is smaller (more negative) than the critical value. (b) We conclude that the series is stationary since we reject the unit root null hypothesis. We need do no further tests since we have already rejected. (c) The researcher is correct. One possible source of non-whiteness is when the errors are autocorrelated. This will occur if there is autocorrelation in the original dependent variable in the regression (yt). In practice, we can easily get around this by “augmenting” the test with lags of the dependent variable to “soak up” the autocorrelation. The appropriate number of lags can be determined using the information criteria. 4. (a) If two or more series are cointegrated, in intuitive terms this implies that they have a long run equilibrium relationship that they may deviate from in the short run, but which will always be returned to in the long run. In the context of spot and futures prices, the fact that these are essentially prices of the same asset but with different delivery and payment dates, means that financial theory would suggest that they should be cointegrated. If they were not cointegrated, this would imply that the series did not contain a common stochastic trend and that they could therefore wander apart without bound even in the long run. If the spot and futures prices for a given asset did separate from one another, market forces would work to bring them back to follow their long run relationship given by the cost of carry formula. The Engle-Granger approach to cointegration involves first ensuring that the variables are individually unit root processes (note that the test is often conducted on the logs of the spot and of the futures prices rather than on the price series themselves). Then a regression would be conducted of one of the series on the other (i.e. regressing spot on futures prices or futures on spot prices) would be conducted and the residuals from that regression collected. These residuals would then be subjected to a Dickey-Fuller or augmented Dickey-Fuller test. If the null hypothesis of a unit root in the DF test regression residuals is not rejected, it would be concluded that a stationary combination of the non-stationary variables has not been found and thus that there is no cointegration. On the other hand, if the null is rejected, it would be concluded that a stationary combination of the non-stationary variables has been found and thus that the variables are cointegrated. Forming an error correction model (ECM) following the Engle-Granger approach is a 2-stage process. The first stage is (assuming that the original series are non-stationary) to determine whether the variables are cointegrated.

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If they are not, obviously there would be no sense in forming an ECM, and the appropriate response would be to form a model in first differences only. If the variables are cointegrated, the second stage of the process involves forming the error correction model which, in the context of spot and futures prices, could be of the form given in equation (7.57) on page 345. (b) There are many other examples that one could draw from financial or economic theory of situations where cointegration would be expected to be present and where its absence could imply a permanent disequilibrium. It is usually the presence of market forces and investors continually looking for arbitrage opportunities that would lead us to expect cointegration to exist. Good illustrations include equity prices and dividends, or price levels in a set of countries and the exchange rates between them. The latter is embodied in the purchasing power parity (PPP) theory, which suggests that a representative basket of goods and services should, when converted into a common currency, cost the same wherever in the world it is purchased. In the context of PPP, one may expect cointegration since again, its absence would imply that relative prices and the exchange rate could wander apart without bound in the long run. This would imply that the general price of goods and services in one country could get permanently out of line with those, when converted into a common currency, of other countries. This would not be expected to happen since people would spot a profitable opportunity to buy the goods in one country where they were cheaper and to sell them in the country where they were more expensive until the prices were forced back into line. There is some evidence against PPP, however, and one explanation is that transactions costs including transportation costs, currency conversion costs, differential tax rates and restrictions on imports, stop full adjustment from taking place. Services are also much less portable than goods and everybody knows that everything costs twice as much in the UK as anywhere else in the world. 5. (a) The Johansen test is computed in the following way. Suppose we have p variables that we think might be cointegrated. First, ensure that all the variables are of the same order of non-stationary, and in fact are I(1), since it is very unlikely that variables will be of a higher order of integration. Stack the variables that are to be tested for cointegration into a p-dimensional vector, called, say, yt. Then construct a p1 vector of first differences, yt, and form and estimate the following VAR yt =  yt-k + 1 yt-1 + 2 yt-2 + ... + k-1 yt-(k-1) + ut Then test the rank of the matrix . If  is of zero rank (i.e. all the eigenvalues are not significantly different from zero), there is no cointegration, otherwise, the rank will give the number of cointegrating vectors. (You could also go into a bit more detail on how the eigenvalues are used to obtain the rank.) (b) Repeating the table given in the question, but adding the null and alternative hypotheses in each case, and letting r denote the number of cointegrating vectors:

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Null Hypothesis r=0 r=1 r=2 r=3 r=4

Alternative Hypothesis r=1 r=2 r=3 r=4 r=5

max 38.962 29.148 16.304 8.861 1.994

95% value 33.178 27.169 20.278 14.036 3.962

Critical

Considering each row in the table in turn, and looking at the first one first, the test statistic is greater than the critical value, so we reject the null hypothesis that there are no cointegrating vectors. The same is true of the second row (that is, we reject the null hypothesis of one cointegrating vector in favour of the alternative that there are two). Looking now at the third row, we cannot reject (at the 5% level) the null hypothesis that there are two cointegrating vectors, and this is our conclusion. There are two independent linear combinations of the variables that will be stationary. (c) Johansen’s method allows the testing of hypotheses by considering them effectively as restrictions on the cointegrating vector. The first thing to note is that all linear combinations of the cointegrating vectors are also cointegrating vectors. Therefore, if there are many cointegrating vectors in the unrestricted case and if the restrictions are relatively simple, it may be possible to satisfy the restrictions without causing the eigenvalues of the estimated coefficient matrix to change at all. However, as the restrictions become more complex, “renormalisation” will no longer be sufficient to satisfy them, so that imposing them will cause the eigenvalues of the restricted coefficient matrix to be different to those of the unrestricted coefficient matrix. If the restriction(s) implied by the hypothesis is (are) nearly already present in the data, then the eigenvectors will not change significantly when the restriction is imposed. If, on the other hand, the restriction on the data is severe, then the eigenvalues will change significantly compared with the case when no restrictions were imposed. The test statistic for testing the validity of these restrictions is given by p

T

[ln(1   )  ln(1   )]  2(p-r)

i  r 1

* i

i

where

i* are the characteristic roots (eigenvalues) of the restricted model

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i are the characteristic roots (eigenvalues) of the unrestricted model r is the number of non-zero (eigenvalues) characteristic roots in the unrestricted model p is the number of variables in the system. If the restrictions are supported by the data, the eigenvalues will not change much when the restrictions are imposed and so the test statistic will be small. (d) There are many applications that could be considered, and tests for PPP, for cointegration between international bond markets, and tests of the expectations hypothesis were presented in Sections 7.9, 7.10, and 7.11 respectively. These are not repeated here. (e) Both Johansen statistics can be thought of as being based on an examination of the eigenvalues of the long run coefficient or  matrix. In both cases, the g eigenvalues (for a system containing g variables) are placed ascending order: 1  2  ...  g. The maximal eigenvalue (i.e. the max) statistic is based on an examination of each eigenvalue separately, while the trace statistic is based on a joint examination of the g-r largest eigenvalues. If the test statistic is greater than the critical value from Johansen’s tables, reject the null hypothesis that there are r cointegrating vectors in favour of the alternative that there are r+1 (for max) or more than r (for trace). The testing is conducted in a sequence and under the null, r = 0, 1, ..., g-1 so that the hypotheses for trace and max are as follows

Null hypothesis for both tests Max alternative H0: H0: H0: H0:

r=0 r=1 r=2 ... r = p-1

H1: 0 < r  g H1: 1 < r  g H1: 2 < r  g ... H1: r = g

Trace alternative H1: r = 1 H1: r = 2 H1: r = 3 ... H1: r = g

Thus the trace test starts by examining all eigenvalues together to test H0: r = 0, and if this is not rejected, this is the end and the conclusion would be that there is no cointegration. If this hypothesis is not rejected, the largest eigenvalue would be dropped and a joint test conducted using all of the eigenvalues except the largest to test H0: r = 1. If this hypothesis is not rejected, the conclusion would be that there is one cointegrating vector, while if this is rejected, the second largest eigenvalue would be dropped and the test statistic recomputed using the remaining g-2 eigenvalues and so on. The testing sequence would stop when the null hypothesis is not rejected.

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The maximal eigenvalue test follows exactly the same testing sequence with the same null hypothesis as for the trace test, but the max test only considers one eigenvalue at a time. The null hypothesis that r = 0 is tested using the largest eigenvalue. If this null is rejected, the null that r = 1 is examined using the second largest eigenvalue and so on. 6. (a) The operation of the Johansen test has been described in the book, and also in question 5, part (a) above. If the rank of the  matrix is zero, this implies that there is no cointegration or no common stochastic trends between the series. A finding that the rank of  is one or two would imply that there were one or two linearly independent cointegrating vectors or combinations of the series that would be stationary respectively. A finding that the rank of  is 3 would imply that the matrix is of full rank. Since the maximum number of cointegrating vectors is g-1, where g is the number of variables in the system, this does not imply that there 3 cointegrating vectors. In fact, the implication of a rank of 3 would be that the original series were stationary, and provided that unit root tests had been conducted on each series, this would have effectively been ruled out. (b) The first test of H0: r = 0 is conducted using the first row of the table. Clearly, the test statistic is greater than the critical value so the null hypothesis is rejected. Considering the second row, the same is true, so that the null of r = 1 is also rejected. Considering now H0: r = 2, the test statistic is smaller than the critical value so that the null is not rejected. So we conclude that there are 2 cointegrating vectors, or in other words 2 linearly independent combinations of the non-stationary variables that are stationary. 7. The fundamental difference between the Engle-Granger and the Johansen approaches is that the former is a single-equation methodology whereas Johansen is a systems technique involving the estimation of more than one equation. The two approaches have been described in detail in Chapter 7 and in the answers to the questions above, and will therefore not be covered again. The main (arguably only) advantage of the Engle-Granger approach is its simplicity and its intuitive interpretability. However, it has a number of disadvantages that have been described in detail in Chapter 7, including its inability to detect more than one cointegrating relationship and the impossibility of validly testing hypotheses about the cointegrating vector.

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Chapter 8 1. (a). A number of stylised features of financial data have been suggested at the start of Chapter 8 and in other places throughout the book: - Frequency: Stock market prices are measured every time there is a trade or somebody posts a new quote, so often the frequency of the data is very high - Non-stationarity: Financial data (asset prices) are covariance nonstationary; but if we assume that we are talking about returns from here on, then we can validly consider them to be stationary. - Linear Independence: They typically have little evidence of linear (autoregressive) dependence, especially at low frequency. - Non-normality: They are not normally distributed – they are fat-tailed. - Volatility pooling and asymmetries in volatility: The returns exhibit volatility clustering and leverage effects. Of these, we can allow for the non-stationarity within the linear (ARIMA) framework, and we can use whatever frequency of data we like to form the models, but we cannot hope to capture the other features using a linear model with Gaussian disturbances. (b) GARCH models are designed to capture the volatility clustering effects in the returns (GARCH(1,1) can model the dependence in the squared returns, or squared residuals), and they can also capture some of the unconditional leptokurtosis, so that even if the residuals of a linear model of the form given by the first part of the equation in part (e), the uˆ t ’s, are leptokurtic, the standardised residuals from the GARCH estimation are likely to be less leptokurtic. Standard GARCH models cannot, however, account for leverage effects. (c) This is essentially a “which disadvantages of ARCH are overcome by GARCH” question. The disadvantages of ARCH(q) are: - How do we decide on q? - The required value of q might be very large -

Non-negativity constraints might be violated.

When we estimate an ARCH model, we require i >0  i=1,2,...,q (since variance cannot be negative) GARCH(1,1) goes some way to get around these. The GARCH(1,1) model has only three parameters in the conditional variance equation, compared to q+1 for the ARCH(q) model, so it is more parsimonious. Since there are less 44/59

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parameters than a typical qth order ARCH model, it is less likely that the estimated values of one or more of these 3 parameters would be negative than all q+1 parameters. Also, the GARCH(1,1) model can usually still capture all of the significant dependence in the squared returns since it is possible to write the GARCH(1,1) model as an ARCH(), so lags of the squared residuals back into the infinite past help to explain the current value of the conditional variance, ht. (d) There are a number that you could choose from, and the relevant ones that were discussed in Chapter 8, inlcuding EGARCH, GJR or GARCH-M. The first two of these are designed to capture leverage effects. These are asymmetries in the response of volatility to positive or negative returns. The standard GARCH model cannot capture these, since we are squaring the lagged error term, and we are therefore losing its sign. The conditional variance equations for the EGARCH and GJR models are respectively  u t 1 u 2 log(  t2 )     log(  t21 )   t 1         t 1   t 1 And

t2 = 0 + 1 ut21 +t-12+ut-12It-1 where It-1

= 1 if ut-1  0 = 0 otherwise

For a leverage effect, we would see  > 0 in both models. The EGARCH model also has the added benefit that the model is expressed in terms of the log of ht, so that even if the parameters are negative, the conditional variance will always be positive. We do not therefore have to artificially impose non-negativity constraints. One form of the GARCH-M model can be written yt =  +other terms + t-1+ ut , ut  N(0,ht) t2 = 0 + 1 ut21 +t-12 so that the model allows the lagged value of the conditional variance to affect the return. In other words, our best current estimate of the total risk of the asset influences the return, so that we expect a positive coefficient for . Note that some authors use t (i.e. a contemporaneous term). (e). Since yt are returns, we would expect their mean value (which will be given by ) to be positive and small. We are not told the frequency of the data,

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but suppose that we had a year of daily returns data, then  would be the average daily percentage return over the year, which might be, say 0.05 (percent). We would expect the value of 0 again to be small, say 0.0001, or something of that order. The unconditional variance of the disturbances would be given by 0/(1-(1 +2)). Typical values for 1 and 2 are 0.8 and 0.15 respectively. The important thing is that all three alphas must be positive, and the sum of 1 and 2 would be expected to be less than, but close to, unity, with 2 > 1. (f) Since the model was estimated using maximum likelihood, it does not seem natural to test this restriction using the F-test via comparisons of residual sums of squares (and a t-test cannot be used since it is a test involving more than one coefficient). Thus we should use one of the approaches to hypothesis testing based on the principles of maximum likelihood (Wald, Lagrange Multiplier, Likelihood Ratio). The easiest one to use would be the likelihood ratio test, which would be computed as follows: 1. Estimate the unrestricted model and obtain the maximised value of the log-likelihood function. 2. Impose the restriction by rearranging the model, and estimate the restricted model, again obtaining the value of the likelihood at the new optimum. Note that this value of the LLF will be likely to be lower than the unconstrained maximum. 3. Then form the likelihood ratio test statistic given by LR = -2(Lr - Lu)  2(m) where Lr and Lu are the values of the LLF for the restricted and unrestricted models respectively, and m denotes the number of restrictions, which in this case is one. 4. If the value of the test statistic is greater than the critical value, reject the null hypothesis that the restrictions are valid. (g) In fact, it is possible to produce volatility (conditional variance) forecasts in exactly the same way as forecasts are generated from an ARMA model by iterating through the equations with the conditional expectations operator. We know all information including that available up to time T. The answer to this question will use the convention from the GARCH modelling literature to denote the conditional variance by ht rather than t2. What we want to generate are forecasts of hT+1 T, hT+2 T, ..., hT+s T where T denotes all information available up to and including observation T. Adding 1 then 2 then 3 to each of the time subscripts, we have the conditional variance equations for times T+1, T+2, and T+3: hT+1 = 0 + 1 u T2 + hT (1)

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hT+2 = 0 + 1 u T2 1 + hT+1 (2) hT+3 = 0 + 1 u T2  2 +hT+2

(3)

Let h1,fT be the one step ahead forecast for h made at time T. This is easy to calculate since, at time T, we know the values of all the terms on the RHS. Given h1,fT , how do we calculate h2,f T , that is the 2-step ahead forecast for h made at time T? From (2), we can write

h2,f T = 0 + 1 ET( uT2 1 )+ h1,fT

(4)

where ET( uT2 1 ) is the expectation, made at time T, of uT2 1 , which is the squared disturbance term. The model assumes that the series t has zero mean, so we can now write Var(ut) = E[(ut -E(ut))2]= E[(ut)2]. The conditional variance of ut is ht, so ht  t = E[(ut)2] Turning this argument around, and applying it to the problem that we have, ET[(uT+1)2] = hT+1 but we do not know hT+1 , so we replace it with h1,fT , so that (4) becomes

h2,f T = 0 + 1 h1,fT + h1f,T = 0 + (1+) h1,fT What about the 3-step ahead forecast? By similar arguments,

h3,f T = ET(0 + 1 uT2  2 + hT+2) = 0 + (1+) h2,f T = 0 + (1+)[ 0 + (1+) h1,fT ] And so on. This is the method we could use to forecast the conditional variance of yt. If yt were, say, daily returns on the FTSE, we could use these volatility forecasts as an input in the Black Scholes equation to help determine the appropriate price of FTSE index options. (h) An s-step ahead forecast for the conditional variance could be written

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s 1

f s ,T

h

  0  (1   ) i 1  (1   ) s 1 h1f,T

(x)

i 1

For the new value of , the persistence of shocks to the conditional variance, given by (1+) is 0.1251+ 0.98 = 1.1051, which is bigger than 1. It is obvious from equation (x), that any value for (1+) bigger than one will lead the forecasts to explode. The forecasts will keep on increasing and will tend to infinity as the forecast horizon increases (i.e. as s increases). This is obviously an undesirable property of a forecasting model! This is called “nonstationarity in variance”. For (1+) 0.1, and zero otherwise. Note that the question does not specify what is “volatility”, so it is assumed in this answer that it is equated with “conditional variance”. Again, this dummy variable would only allow the intercept in the conditional variance (i.e. the unconditional variance) to vary according to the previous day’s volatility. A similar dummy variable could be applied to the lagged squared error or lagged conditional variance terms to allow them to vary with the size of the previous day’s volatility.

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“Introductory Econometrics for Finance” © Chris Brooks 2008

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“Introductory Econometrics for Finance” © Chris Brooks 2008