1 ECMT2130 – Replacement final exam solutions
1. (10 points) Andy’s differencing
Andy obtains 3000 observations, x1, x2, …, x3000 from the stochastic process: xt = 0.99xt−1 + εt
whereεt ∼ N 0, σ2 are independently and identically distributed for all t.
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He tests for a unit root and mistakenly decides that the stochastic process is integrated of order 1. He
then differences the data to obtain yt = ∆xt or all t = 2,…,3000.
(a) (2 points) Write out the equation describing the stochastic process for the differenced data, yt.
(b) (2 points) If you were to fit an ARMA model to the differenced data, yt, what should the order of the AR and MA components be?
(c) (2 points) Is the ARMA model for yt weakly stationary? Explain your answer.
(d) (2 points) Is the ARMA model for yt invertible? Explain your answer.
(e) (1 point) What pattern would be exhibited by the ACF for yt?
(f) (1 point) What pattern would be exhibited by the PACF for yt?
(a) The differenced model would be:
yt = 0.99yt−1 + εt − εt−1
(b) The model for the differenced data should be an ARMA(1,1) because the maximum lag of the
dependent variable is 1 and the maximum lag of the shock is 1.
(c) The ARMA model for yt is weakly stationary because the model can be expressed as:
(1 − 0.99L)yt = (1 − L)εt
The model for yt is weakly stationary if the root of 1 − 0.99L = 0 does not lie on the unit circle in the complex plane. The root is L = 1/0.99 > 1 so it does lie outside the unit circle. This is sufficient to establish that the process is weakly stationary. In other words, its mean, variance and auto-covariances are invariant to a time shift.
(d) The model for yt is invertible if the root of 1 − L = 0 lies outside the unit circle in the complex plane. The root is L = 1 so it lies on the unit circle. Thus the process is not invertible.
(e) The ACF would taper off very slowly because the model has an AR component with a very high degree of persistence.
(f) The PACF would also taper off slowly because the model has an MA component.
2. (10 points) Clare’s ARMA-GARCH
(a) (4 points) GARCH(1,1) models have tended to replace the use of ARCH models in empirical liter- ature. Explain the difficulties in applying ARCH models that have motivated this trend.
Clare estimates an ARMA(1,1)-GARCH(1,1) model for a univariate time series of daily returns on a broad market index. She uses a Maximum Likelihood Estimator.
The ARMA(1,1)-GARCH(1,1) model can be written as:
rt =μ+θrt−1 + ut +φut−1
• ut = εtσt εt ∼ N (0, 1) are independently and identically distributed through time
• σ2 = α + α u2
• α1 + β1 < 1
She uses a likelihood ratio test to test joint exclusion restrictions on the lagged daily return and the lagged shock. The null hypothesis is:
and the alternative hypothesis is:
H1: θ̸=0and/orφ̸=0
(b) (2 points) What are the economic implications if Clare rejects the null hypothesis?
(c) (4 points) Clare’s likelihood ratio test statistic is 52.54. It has a Chi-squared distribution under the null hypothesis. Set out her hypothesis test at the 1% significance level, indicating the number of degrees of freedom of the Chi-squared distribution, the decision rule, and the conclusion of the test.
(a) A GARCH model is more parsimonious than an ARCH model that allows for the same amount of volatility clustering so it avoids overfitting problems, Being more parsimonious, a GARCH model is less likely to breach the non-negativity constraints on the parameters. By reducing the probability of those problems with constraints, the models are less likely to permit negative variances.
(b) The null hypothesis H0 : ψ1 = θ2 = 0 against the alternative H1 : not H0.
If she rejects the null hypothesis, that implies that daily returns have a degree of predictability
that constitutes a violation the weak form of the efficient markets hypothesis.
(c) The test statistic is LR = −2(LLFr − LLFu). With sufficient data, we know that, under the null hypothesis, the statistic has an asymptotic Chi-squared distribution with 2 degrees of freedom (because we are jointly testing 2 restrictions).
We perform the test at the 1% level of significance.
The test is an upper-tail test and so, from the Chi-squared distribution with 2 degrees of freedom the critical value is 9.2103. The rejection region is (9.2103, ∞].
The decision rule is “reject H0 at the 5% significance level if LR > 9.2103. Otherwise fail to
reject the null hypothesis.”.
The conclusion is: reject H0 at the 5% significance level because LR = 52.54 > 9.2103”. There
is sufficient evidence, at the 1% level, to reject the null hypothesis. There is sufficient evidence, at the 1% level, to reject the null hypothesis. Thus, we can conclude, at the 5% level of significance, that the daily financial returns have a degree of predictability using a ARMA(1,1) model.
3. (10 points) Frank’s EWMA
(a) (2 points) What advantage does an EWMA model have over ARMA models when forecasting? [2 points]
In the 1960s, Frank was analysing the growth patterns in the airline industry to inform his view about prospective earnings if he invested heavily in airlines. He had access to the data show in the following time series plot of monthly data on airline passengers in the US (in thousands of people).
He fitted this data (yt for t = 1..T ) with the following Holt-Winters model: Level: lt =α yt +(1−α)(lt−1 +bt−1)
Slope: bt = β(lt − lt−1) + (1 − β)bt−1
Seasonal: st = γ yt + (1 − γ)st−p lt−1 +bt−1
The coefficients have been estimated to provide the best possible fit. They are α = 0.25, β = 0.05, and γ = 0.8.
(b) (2 points) What value would p have in this model? Explain why.
(c) (2 points) Explain why he chose to introduce seasonality multiplicatively into the model?
(d) (2 points) Given that γ = 0.8, what does this say about the stability of the seasonal pattern?
(e) (2 points) As a function of the level, slope and seasonal components in period T, what would you
report if you were asked to provide a de-seasonalised estimate of the value of yT+4?
(a) An EWMA-based Holt-Winters model is an ad-hoc modelling approach that does not impose stationarity requirements on the data being modelled. Thus, it adapts more easily to features like structural breaks. For example, breaks in the trend, mean shifts, and seasonality can all be handled using an EWMA-based Holt-Winters model where an ARMA model would fail.
(b) The data is monthly. It shows a strong seasonal pattern that repeats each year. Thus, p, the seasonal period, should be equal to 12.
(c) The amplitude of the seasonal pattern increases as the airline passenger data trends upwards. This strongly suggests that the seasonal component multiplies the level of the series rather than adds to the series.
(d) γ = 0.8, suggesting that the seasonal pattern evolves very quickly. New information has a strong impact on the seasonal multiplier in each period.
(e) The deseasonalised estimate is lT+4. Now:
= lT +3 + bT +3
= lT +2 + bT +2 + bT +3
= lT +2 + bT +2 + bT +3
= lT +1 + bT +1 + bT +2 + bT +3 =lT +bT +bT+1 +bT+2 +bT+3 =lT +bT +bT +bT +bT
4. (10 points) Hamish’s ARIMA model
Hamish has a data sample produced by the stochastic process:
(1−0.2L)∆xt =(1+0.5L)εt
where εt ∼ N 0, σ2 is independently and identically distributed over time.
(a) (2 points) As an ARIMA(P,I,Q) model, what would the values of P and Q be?
(b) (2 points) Explain why the process xt is not weakly stationary.
(c) (4 points) The researcher performs an Augmented (ADF) test with no drift or trend using 60 observations. The ADF test statistic is equal to -1.94. Perform the ADF test at the 1% level of significance, stating the null hypothesis and the alternative hypothesis, the critical region for the test, and the conclusion of the test. The critical value for the test is -2.61 at the 1% level of significance.
(d) (2 points) If the researcher estimates a regression explaining the behaviour of an I(1) data series, zt,using an intercept and xt as a regressor, would a slope coefficient estimate with a high t-ratio indicate evidence of an economically meaningful relationship? If not, why not.
(a) P =1andQ=1.
(b) The process is not weakly stationary because the autoregressive part of the model has a unit
root. The relevant equation is:
(1−0.2L)∆ = (1−0.2L)(1−L) = 0
The roots of this equation are L = 5 and L = 1. The first root lies outside the unit circle but the second lies on the unit circle and that implies that the process, xt is not weakly stationary.
(c) The regression used for the ADF test is:
∆yt =πyt−1 +γi∆yt−i +ut
The null hypothesis is H0 : π = 0 implying at least one unit root in xt. The alternative hypothesis is H0 : π < 0 implying that there are no unit roots in xt and instead, xt is causal.
Under the null hypothesis of at least one unit root, the test statistic has a distribution with lower tail critical values described in the distribution tables.
The decision rule, at the 1% level of significance is, reject the null hypothesis if the test statistic, τ is less than the critical value, -2.61 and otherwise fail to reject the null hypothesis because there is insufficient evidence to do so.
The test statistic estimate is -1.94. This does not lie in the rejection region so we do not have enough evidence at the 1% level of significance to reject the null hypothesis that xt has one or more unit roots.
(d) The described regression would be an example of a spurious regression. The t-ratio on xt would be high but the spurious relationship would just reflect the fact that the dependent variable and the regressor both have stochastic trends. The OLS estimate of the slope coefficient would just maximise the extent to which the sample mean shift in xt could be used to capture the sample mean shift in the dependent variable, zt.
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