Statistical exam

1 ECMT2130 - Practice final exam Author: Geoff Shuetrim 1. (10 points) EWMA for drift processes The Exponentially Weighted Moving Average (EWMA) model: lt = λxt + (1 − λ)lt−1 is used to make forecasts of the stochastic process, xt. Using λ = 0.1, and given xt = 6 and lt−1 = 5: (a) (3 points) Showing your working, what is the estimate of lt? (b) (3 points) Showing your working, what is the estimate of lt+3? (c) (4 points) How should this model be changed if xt is actually defined by the stochastic process: xt = µ + xt−1 + t Explain your reasoning and write out your suggested version of the Holt-Winters model in full. Solution: (a) lt = 0.1 × 6 + 0.9 × 5 = 5.1 (b) lt+3 = 1 × lt+2 = 1 × lt+1 = 1 × lt = 5.1 (c) If the stochastic process is a random walk with drift, then the model would not be appropriate. It would need a slope EWMA to enable the model to capture the drift component. By just having a level component in the model, all forecasts are going to be unchanged from the final level estimate, so there is no ability to capture the drift. To address this the model augmented with a slope component would be: The level EWMA would be: lt = λxt + (1 − λ)(lt−1 + bt−1) The slope EWMA would be: bt = β(lt − lt−1) + (1 − β)bt−1 2. (10 points) EWMA with seasonals (a) (2 points) When might it be preferable to use a Holt-Winters EWMA model instead of an ARMA model? (b) (2 points) What period would you use for the seasonal component of a Holt-Winters EWMA model applied to quarterly data? (c) (2 points) How would you decide whether to use an additive seasonal or a multiplicative seasonal in a quarterly Holt-Winters EWMA model? Modelling the data (yt for t = 1 . . . T) with the following Holt-Winters model: Level: lt = α yt st−p + (1 − α)(lt−1 + bt−1) Slope: bt = β(lt − lt−1) + (1 − β)bt−1 Seasonal: st = γ yt lt−1+bt−1 + (1 − γ)st−p (d) (2 points) The smoothing parameter for the seasonal component is 0.03 while the smoothing parameter for the slope is 0.2. What does this say about the speed of adjustment of the seasonal pattern, compared to the speed of adjustment of the slope? (e) (2 points) What is the de-seasonalised estimate of the value of yT +3 as a function of the level, slope and seasonal components? Solution: (a) A Holt-Winters EWMA model can be used for forecasting time-series that are non-stationary in various ways, for example, having time-varying rates of drift or additive or multipicative seasonal elements. ARMA models are not suitable for such time-series. Also, a Holt-Winters EWMA model can adapt to a structural break more effectively than an ARMA model. For example, if the rate of drift changes, that cannot be accommodated by a standard ARMA model but the slope component of the Holt-Winters EWMA model can adapt over a few observations, depending on the value of the slope smoothing parameter. (b) If the data is quarterly, I would use a seasonal period of 4 to ensure that the seasonal component is associated the same season or quarter in each year. (c) I would use a multiplicative seasonal if the magnitude of the seasonal variation was an increasing function of the level of the series. I would use an additive seasonal if the size of the seasonal variation was unrelated to the level of the series. Put another way, if a series is trending upwards and the seasonal amplitude remained constant, I would consider an additive seasonal. If the series trends upward and the seasonal amplitude scales up in line with the level of the series, then I would use a multiplicative seasonal. (d) The seasonal component has a lower smoothing parameter so it adjusts more slowly to new information, compared to the adjustment rate for the slope. (e) The deseasonalised estimate of yT +3 is the level in period T + 3. lT +3 = lT +2 + bT +2 = lT +1 + bT +1 + bT +2 = lT + bT + bT +1 + bT +2 = lT + bT + bT + bT = lT + 3bT Page 2 3. (10 points) GARCH parameter constraints (a) (3 points) What stylised facts characterise daily data on financial rates of return? (b) (3 points) What advantages does a GARCH model have over an ARCH model? Consider the Normal GARCH(1,1) model: rt − µ = t = utσt σ 2 t = a0 + a1 2 t−1 + b1σ 2 t−1 where: • ut ∼ N (0, 1) is independently and identically distributed over time; • a0 and a1 are both strictly greater than 0; • b1 is greater than or equal to 0; and • a1 + b1 < 1 and b1 < 1. (c) (2 points) What is the problem if a1 + b1 > 1? (d) (2 points) What is the problem if a0 = 0? Solution: (a) Daily financial returns are characterised by: • Clustered sets of observations with high volatility (volatility clustering) • very little, if any, serial correlation (supporting the efficient markets hypothesis) • Leptokurtosis (fat-tails in the distribution) suggesting deviation from normality • Negative skew, with more negative extreme values than positive extreme values • Leverage effects, whereby a large negative shock tends to cause volatility to rise more than a large positive shock. (b) A GARCH model is more parsimonious than an ARCH model so it avoids overfitting problems, Being more parsimonious, a GARCH model is less likely to breach the non-negativity constraints on the parameters. (c) If a1 + b1 > 1, then the variance of the shocks will continue to grow at an increasing rate over time. This is an unrealistic characterisation of daily financial return volatility. (d) If a0 = 0, then the unconditional variance of the returns is zero because: V ar(rt) = a0 1 − a1 − b1 = 0 Again this is an unrealistic characterisation of daily financial return volatility, Page 3 4. (10 points) Gwen’s ARMA model identification Gwendoline is wanting to model a univariate time-series using an autoregressive moving average model. (a) (3 points) Show that the following model is over-parameterised and write out the appropriately simplified model. xt = 0.6xt−1 + t − 0.7t−1 + 0.06t−2 (b) (2 points) Given a large enough set of data on xt, what features would you expect to observe in the autocorrelation function? (c) (2 points) Given a large enough set of data on xt, what features would you expect to observe in the partial autocorrelation function? (d) (1 point) Is the model invertible? (e) (1 point) Is the model weakly stationary? (f) (1 point) Is the model causal? Solution: (a) The model can be written as: xt − 0.6xt−1 = t − 0.7t−1 + 0.06t−2 Using the lag operator: (1 − 0.6L)xt = (1 − 0.7L + 0.06L 2 )t Factoring the right hand side: (1 − 0.6L)xt = (1 − 0.6L)(1 − 0.1L)t Cancelling the common factor(1 − 0.6L) xt = (1 − 0.1L)t Without the lag operator we have xt = t − 0.1t−1 The model is an MA(1). (b) The ACF would cut off abruptly after the autocorrelation at lag 1. (c) The PACF would taper off to insignificant PACF estimates as the lag length increased. There would be no abrupt cut-off. (d) The model is invertible because the root of 1 −0.1L = 0 isL = 1/0.1 = 10 which is greater than one so it lies outside the unit circle. (e) The model has no AR component so it is weakly stationary. (f) The model has no AR component so it is causal. Page 4 5. (10 points) Max’s ARIMA model Max has a lot of data from stochastic process: xt = 0.98xt−1 + t − 0.1t−1 where t ∼ N 0, σ2  are Gaussian white noise. 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. (a) (2 points) Write out the ARMA model for yt. (b) (2 points) If you were to fit an ARMA model to yt, what would the order of the AR and MA components be? (c) (1 point) Is xt causal? (d) (1 point) Is the ARMA model for yt stationary? (e) (2 points) Is the ARMA model for yt invertible? (f) (1 point) Knowing the ARMA model for yt = ∆xt, would you expect the ACF for yt to cut off abruptly after 2 lags? (g) (1 point) Knowing the ARMA model for yt = ∆xt, would you expect the PACF for yt to cut off abruptly after 1 lag? Solution: (a) The ARMA model for yt = ∆xt is: yt = 0.98yt−1 + ∆t − 0.1∆t−1 This can be simplified to: yt = 0.98yt−1 + t − 1.1t−1 + 0.1t−2 It is an ARMA(1,2) process. (b) The root of 1 − 0.98L = 0 is L = 1/0.98 > 1 so it lies outside the unit circle. The process is causal. (c) The root of 1 − 0.98L = 0 is L = 1/0.98 > 1 so it lies outside the unit circle. The process is stationary. (d) The roots of (1 − L)(1 − 0.1L) = 0 are L = 1 and L = 10. One is on the unit circle and one is outside the unit circle. Because one root is on the unit circle, the process is not invertible. Invertibility requires all roots for the MA component of the model to lie outside the unit circle. (e) The ACF for yt would taper off slowly because the model has an AR order P = 1. (f) The PACF for yt would taper off slowly because the model has an MA order Q = 2. Page 6. (10 points) Real national output unit root test (a) (2 points) Would you expect that real national output is weakly stationary? Explain your answer in terms of the definition of weakly stationary time series and the characteristics of how real output grows over time. (b) (2 points) An analyst performs an Augmented Dickey Fuller test using quarterly data, allowing for a drift and trend and including 4 lagged differences of the quarterly logged data as regressors. Write out the regression equation that you would use to conduct the test, defining all the variables. [2 points] (c) (4 points) The test uses data from January 2000 through to the December 2019. The test statistic is -2.54. The critical value defining the rejection region for the Augmented Dickey-Fuller test with the number of observations that have been used is reported as -3.495 at the 5% level of significance. Perform the Augmented Dickey-Fuller test at the 5% level of significance, stating both the null hypothesis and the alternative hypothesis, the decision rule, and the conclusion of the test. (d) (2 points) Given the test result, would you use the level or the first difference of logged real national output as an explanatory variable in a model of quarterly excess financial returns (over the risk-free rate of return) for a broad stock market index? Explain your reasoning. Solution: (a) Real national output grows over time because of increases in productivity and increases in the availability of factors of production, like labour and capital.The first moment E(yt) is a function of time, t, and so is not time-shift invariant. Thus, the mean would change over time so it would not be weakly stationary. (b) There are 20 years of 4 observations per year so there are 80 observations. This is enough data to rely on the asymptotic properties of the ADF test. let yt be the log of real national output in quarter t. The regression used for the ADF test is: ∆yt = α + βt + πyt−1 + X 4 i=1 γi∆yt−i + ut (c) The null hypothesis is H0 : π = 0 implying at least one unit root in the log of quarterly real national output. The alternative hypothesis is H0 : π < 0 implying that there are no unit roots in the log of quarterly real national output. It is weakly stationary around a deterministic trend. Under the null hypothesis of at least one unit root, the test statistic has a distribution with lower tail critical values described in the Dickey Fuller distribution tables. The decision rule, at the 5% level of significance is, reject the null hypothesis if the test statistic, τ is less than the critical value, -3.495 and otherwise fail to reject the null hypothesis because there is insufficient evidence to do so. The test statistic estimate is -2.54. This does not lie in the rejection region so we do not have enough evidence at the 5% level of significance to reject the null hypothesis that the log of quarterly real national output has one or more unit roots. (d) I would use the first difference of logged real national output as an anomaly regressor in a CAPM model of quarterly excess financial returns (over the risk-free rate of return) for an index fund that invests in mining because excess returns do not trend upwards. The regression will be miss-specified if we try to explain variation in a variable with a non-trending mean with a stochastically trending variable. Page 6 7. (10 points) Betty’s LR test Betty wants to test the efficient markets hypothesis while also allowing for volatility clustering. the following AR(2)-GARCH(1,1) model for daily financial returns using with 3,000 observations. rt = µ + φ1rt−1 + φ2rt−2 + t t = utσt σ 2 t = a0 + a1 2 t−1 + b1σ 2 t−1 where: • ut ∼ N (0, 1) is independently and identically distributed over time; • a0 and a1 are both strictly greater than 0; • b1 is greater than or equal to 0; and • a1 + b1 < 1 and b1 < 1. (a) (1 point) Does volatility clustering violate the weak form of the efficient markets hypothesis? (b) (1 point) Betty tests the null hypothesis H0 : φ1 = φ2 = 0 against an alternative hypothesis. State the appropriate alternative hypothesis. [1 point] (c) (1 point) How many restrictions is Betty testing jointly? (d) (2 points) Explain why the null hypothesis in B implies the weak-form of the efficient markets hypothesis. (e) (1 point) Betty tests her proposed null hypothesis by estimating two models, the unrestricted AR(2)-GARCH(1,1) model, and the restricted GARCH(1,1) model. She obtains two maximised Log Likelihood Function (LLF) values: one for the unrestricted regression Lu = 6199.555, and one for the restricted regression, Lr = 6197.794. Showing your working, compute the likelihood ratio test statistic that Betty should use in testing her hypothesis. (f) (1 point) State the distribution of the test statistic under the null hypothesis. (g) (1 point) Specify the decision rule that Betty should use for her hypothesis if the test is being conducted at the 5% level of significance. (h) (1 point) Draw a distribution diagram showing the rejection region for the test. (i) (1 point) State the hypothesis test conclusion formally and state what conclusion should Betty draw from her test. Solution: (a) The weak-form of the efficient markets hypothesis states that today’s prices reflect all the information in past prices and that no form of technical analysis of past prices can be effectively utilized to aid investors in making trading decisions. Typically this is interpreted as stating that you cannot improve your forecast of the mean of the future price (or the mean return) using data on past prices. It is not typically taken to mean that return volatility changes are unpredictable over the short-term. Thus volatility clustering is not typically regarded as a violation of the efficient markets hypothesis. (b) H1 : φ1 6= 0 and/or φ2 6= 0. (c) Betty is jointly testing 2 exclusion restrictions, one on φ1, and one on φ2. Page 7 (d) Under the alternative, expected returns are predictable given information on past prices so it implies a violation of the weak form of the efficient markets hypothesis. Under the null hypothesis, deviations of the expected return for the next period, from its unconditional mean, µ, are not predictable using past prices so the null hypothesis is consistent with the weak form of the efficient markets hypothesis. (e) The likelihood ratio test statistic is LR = −2(Lr − Lu) = −2(6197.794 − 6199.555) = 3.52. (f) Under the null hypothesis, given that we have a large sample, the test statistic will have a Chi-squared distribution with 2 degrees of freedom, because 2 restrictions are being tested jointly. (g) The test is upper-tailed. At the 5% level of significance, the critical value is χ 2 (0.95,2) = 5.9915 (Obtain in R using the function “qchisq(0.95,2)” or look this up in the Chi-squared distribution table). Thus, the decision rule is: “Reject the null hypothesis if the test statistic is greater than 5.9915. Otherwise, fail to reject the null hypothesis.” (h) A suitable diagram is shown below. The rejection region corresponds to the part of the density function that has been shaded with yellow. (i) . The test statistic, 3.52, does not lie in the rejection region. Thus, at the 5% level of significance, there is insufficient evidence to reject the null hypothesis. Betty should conclude that there is insufficient evidence to support a view that the weak form of the efficient markets hypothesis is violated. Page 8

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