09.02 26/11/2023, Multiple Choice Questions and Feedback Chapters 8 and 8 to main content inski Chapters 8 and 9 Which of the following features of financial asset return time-series could be captured 1 using a standard GARCH(1,1) model? i) Fat tails in the return distribution ii) Leverage effects iil) Volatility clustering iv) Volatility affecting returns ©) (iand (iv) only © b) Gand (i) only ©.©) (0. (i), and (ii) only C4 (.(id, (id, ana (ivy Correct! (Check your answer If the standard tools for time-series analysis, such as estimation of the acf, pact and 2 spectral analysis, find no evidence of structure in the data, this implies that the data are which of the following? © a) Normally distributed © b) Uncorrelated O ¢) Independent O 4) Fattailed Correct! The standard tools for time-series analysis are all concerned with whether there is a linear relationship (or correlation) between the value of the series at one point in time and its value at other points in time. If such tools do not detect any evidence of these relationships, all we ‘can say is that the observations on the series are (linearly) uncorrelated. These observations could, however, be non-linearly related to one another (as they would be if they followed a GARCH or bilinear process), so we therefore cannot say that the observations are independent of one another, Finally, the observations may or may not be fat-tailed or normally distributed, and neither of these is tested for using the acf, pacf or spectral analysis. (Check your answer Which of the following statements are true concerning a comparison between ARCH(q) and 3 carcH(1,1) models? i) The ARCH(q) model is likely to be the more parsimonious Il) The ARCH(q) model is the more likely to violate non-negativity constraints iil) The ARCH(q) model can allow for an infinite number of previous lags of squared returns to affect the current conditional variance iv) The GARCH(1,1) mode! will usually be sufficient to capture all of the dependence in the conditional variance nitpslwiw.cambridge orgieatureslecanomics/brooksidownloadsimea/ChapterSand9/ch089 him 19(09.02 26/11/2023, Multiple Choice Questions and Feedback Chapters 8 and 8 @ a) (ii) and (iv) only © b) (and (i) only OO. (.(i), and (ip only 4 (9, (i, ana (wv) Correct! The GARCH(1,1) model uses only 3 parameters in the conditional variance equation, and is therefore highly parsimonious. The GARCH(1,1) model can be expressed as an infinite order ARCH model, so that it can allow for an infinite number of previous lags of squared returns. to affect the current conditional variance. Thus the GARCH(1,1) model will usually be sufficient to capture all of the dependence in the conditional variance, implying that bin practice higher order ‘models such as GARCH(2,3) are not necessary. On the other hand, the ARCH(a) model can only allow for the previous q lags of squared returns to affect the current conditional variance, and the value of q required to capture all of the conditional dependence in the variance may be large so that the ARCH(q) model is less parsimonious than the GARCH(1, 1). (Check your answer Which of the following statements are true concerning maximum likelihood (ML) estimation 4 iin the context of GARCH models? 1) Maximum likelihood estimation selects the parameter values that maximise the probability that we would have actually observed the values of the series y that we actually did ii) GARCH models can only be estimated by ML and not by OLS. iit) For estimation of a standard linear model (with no GARCH), the OLS and ML. estimates for the slope and intercept parameters will be identical but the estimator for the variance of the disturbances is slightly different iv) Most computer packages use numerical procedures to estimate GARCH models rather than a set of analytical formulae © a) (il)and (wv) only Od) Wand Gi only 4). Ci), and (i) only © 4) (), i), (ii), and (iv) Correct! All of the possibilities given in the question are correct! (i) is the definition of maximum likelinood, Since GARCH is a non-linear model (for the variance af a series), its parameters cannot be estimated using OLS. For a standard linear model (i. the CLRM with homoscedastic disturbances), the OLS and ML estimates for the slope and intercept parameters will be identical ‘but the estimator for the variance of the disturbances is slightly different. An analytical approach to estimating GARCH models has only been developed for the very simplest examples of the models, and for more complex or extended GARCH models, analytical formulae cannot be used. ‘Thus most computer packages will ust use a numerical search procedure for coefficient estimation, (Check your answer Which of the following statements are true concerning the standardised residuals 5 (residuals divided by their respective conditional standard deviations) from an estimated GARCH model? i) They are assumed to be normally distributed nitpsilwiw.cambridge.orgieaturesleconomics/orooksidownloadsimea/ChapterSand9/ch089 him 29(09.02 26/11/2023, Multiple Choice Questions and Feedback Chapters 8 and 8 ii) Their squares will be related to their lagged squared values if the GARCH model is appropriate iii) In practice, they are likely to have fat tails iv) If the GARCH model is adequate, the standardised residuals and the raw residuals will be identical © a) (il)and (i) only @ b) (and (il) only Oe) (i, Ci), and Gi) only 4) ().Ci), (i), and (iv) Correct! (i) and (i) only are correct. The standardised innovations (disturbances) are assumed to be normally distributed with mean zero and unit variance, although in practice the standardised residuals from a GARCH model estimated on some financial asset return data are likely to be fat- failed, although not as fat-tailed as the raw residuals would be. This arises from the fact that the GARCH model can account for fat tals in the original data y, but such models usually cannot account forall of the observed leptokurticity in the returns, so that some fat-tailedness is left in the standardised residuals! Ifthe GARCH model is an adequate characterisation of the data, the standardised residuals should be independent of one another as well as standard normally distributed. Thus, if the GARCH model proposed is adequate, the squares of the standardised residuals should be independent of their previous squared values. It should also be clear that the residuals and the standardised residuals should be quite different from one another, whether the GARCH model is adequate or not, The residuals will be conditionally heteroscedastic (i.e. will follow a GARCH), while the standardised residuals will have had this conditional heteroscedasticity divided aut of them so that they should be iid (Check your answer Which of the following criticisms of standard ("plain vanilla") GARCH models can be 6 overcome by EGARCH models? i) Estimated coefficient values from GARCH models may be negative i) GARCH models cannot account for leverage effects ill) The responsiveness of future volatility to positive and negative shacks is symmetric under @ GARCH formulation iv) GARCH models cannot allow for a feedback from the volatility to the returns © a) (il)and (iv) only ©») Wand Gi only © ©) (Gi), and (ii) only 4 (,Gi), (ip, and (iv Correct (i), (i) and (ii) only are correct. Ifthe estimated coefficient values from a standard GARCH model are negative, this would not be a sensible model since this could lead to negative estimates for the conditional variance at some points in time which is clearly nonsensical. The EGARCH model gets around this by using a logarithmic formulation for the conditional variance ‘equation such that, even if the estimated coefficients in the EGARCH model aro negative, the conditional variance will always be positive, EGARCH models also allow for the possible asymmetry in the relationship between volatility and innovations of equal sizes but opposite signs, which could be caused by leverage effects. Thus (i) and (i) are two different ways of writing the same thing. However, neither the plain GARCH nor the EGARCH model can allow for feedback from the volatility to the returns - to model such an effect would require a volatility term to be present in the conditional mean equation (as a GARCH-M model would have). nitpsilwiw.cambridge.orgieaturesleconomics/orooksidownloadsimea/ChapterSand9/ch089 him a9(09.02 26/11/2023, Multiple Choice Questions and Feedback Chapters 8 and (Check your answer If there were a leverage effect in practice, what would be the shape of the news impact curve for as model that accounted for that leverage? © @) Itwould rise more quickly for negative disturbances than for positive ones of the same magnitude © b) Itwould be symmetrical about zero © ©) Itwould rise less quickly for negative disturbances than for positive ones of the same ‘magnitude © 4d) Itwould be zero for all positive disturbances Correct! ais correct. The news impact curve plots the value of the lagged innovation, ie. u_(t1), on the x-axis against the value of the conditional variance, h_t, on the y-axis. Again, the leverage: effect implies that negative innovations should lead to a higher level of volatility during the next period than positive innovations of the same magnitude. If there were no asymmetries in the relationship between the sign of the innovation and the next period volatility, the news impact curve would be symmetrical about zero on the x-axis. However, a leverage effect implies that next period volatility would rise more sharply if for negative innovations than positive innovations of the same magnitude. (Check your answer Consider the estimation of a GARCH-M model. If the data employed were a time-series of 8 daily corporate bond percentage returns, which of the following would you expect the value of the GARCH-in-mean parameter estimate to be? © 4) Less than 1 © b) Between -1 ando @ ©) Between 0 and 1 O 4) Bigger than 1 Correct! When a GARCH-M model is estimated, the specification is equivalent to that of a plain vanilla GARCH model except that an additional term (a contemporaneous or lagged value of either the conditional variance or of the conditional standard deviation) is included in the conditional mean equation. The coefficient attached to this variable can be interpreted as the risk- return trade-off observed from the market. Therefore, itis a risk premium of sorts (although to be fully described as a risk premium, the returns would have to be excess retums over a risk-free rate etc.), and hence the value of this coefficient should be positive, Further, we can say that since the dependent variable is measured as daily percentage retums, this daly “risk premium" to bbe very small in magnitude (for example, of the order 0.01), and certainly not bigger than one. Therefore, cis correct. (Check your answer ‘Suppose that we have estimated 2 GARCH model for daily equity returns, and we are interested in producing @ 10-day forecast of the volatility (measured by the standard deviation of returns) for use in a value at risk model. How could such a forecast most validly be calculated? © 2) Produce 1, 2, 3, ... 10 step ahead conditional variance forecasts and add them up nitpsilwiw.cambridge.orgieaturesleconomics/orooksidownloadsimea/ChapterSand9/ch089 him 49(09.02 26/11/2023, Multiple Choice Questions and Feedback Chapters 8 and 8 10 @ b) Produce 1, 2,3, .., 10 step ahead conditional variance forecasts and add them up and take the square root © ©) Produce 1, 2, 3, ... 10 step ahead conditional variance forecasts, take the square roots of each one and add them up © 4) Produce a 1-step ahead conditional variance forecast, take its square root and multiply it by the square root of 10 Correct! b is correct. The first thing to note is that variances are additive over time, so that the variance over a 10-day period is the sum of each of the daily variances over the 10 days. The same is not true of the standard deviations, which scale with the square root of time rather than with time. Therefore, even if the eventually want to have a 10-day standard deviation, we must first work with the daily variances, add them up to get a 10-day variance and then take the square root. Clearly itis important to take the square root at the end since volatity is usually measured as a standard deviation rather than as a variance. Its possible to produce a 1-step ahead conditional variance forecast, take its square root and multiply it by the square root of 10, but this is likely to be sub-optimal. Whilst this approach would use the data in a valid way, this would implicitly be assuming that the optimal daily forecast for the second, third, ..., tanth days was the first day's forecast. Recall that the forecasts from a non-integrated and non-degenerate GARCH_ process will converge upon the long term average variance level as the prediction horizon increases. Therefore, more accurate multi-step ahead forecasts can be produced by iterating the GARCH equation forward rather than assuming that the 1-day forecast is also the forecast for all future days. (Check your answer ‘Suppose that we are interested in testing the null hypothesis that a GARCH(2,2) model can be restricted to a process with a constant conditional variance using the likelihood ratio test approach, Which of the following statements are true? © a) The test statistic will follow a chi-squared distribution with 2 degrees of freedom under the null hypothesis © b) The value ofthe log-likelihood function will almost always be bigger for the restricted model than for the unrestricted mode! © ¢) Ifthe relevant values of the log-likelihood functions are -112.3 and -118.4, the value of tho test statistic is 12.2 © 4) The hkelinood ratio test compares the slopes of the log-Ikelihood function at the maximum and at the restricted parameter value Correct! ¢ is correct. If we restrict a GARCH(2,2) model to have a constant conditional variance, this means that all of the terms in the conditional variance equation except for the intercept must all be set fo zero. Ina GARCH(2,2) this will mean setting the 2 lags of the ‘squared residuals and the 2 lags of the conditional variance (2. 4 parameters) to zero, implying 4 restrictions. Recall that ML maximises the value if the log-likelihood function (LF). ‘Therefore, the LLF cannot be larger when a restriction is imposed, since if there were values of the parameters that lead to a higher value of the LLF then ML would have already found them. ‘As with F-test comparisons of the RSS for restricted and unrestricted models, the values of the LLFs for the restricted and unrestricted models will only be the same when the restriction is already present in the data so that the unrestricted parameter estimat(s) is(are) equal to the value(s) under the null hypothesis. The test statistic is calculated as twice the difference between the LLF for the unrestricted and restricted models -ie, LR = 2x(-112.3 -- 118.4) 12.2. The likelihood ratio test compares the vertical distances (the slopes are compared by the LM test approach) between the likelihood values for the restricted and unrestricted models in a plot of these against the parameter values. (Check your answer nitpsilwiw.cambridge.orgieaturesleconomics/orooksidownloadsimea/ChapterSand9/ch089 him 59(09.02 26/11/2023, Multiple Choice Questions and Feedback Chapters 8 and 8 What is the most appropriate value of a forecast (to 2 decimal places) of the conditional variance for time t+2 If the conditional variance and residual at time t are 0.02 and -0.465 respectively and the model is as follows? 0.003 + us , Ur ( N(O,h,) hy = 0.01 40.1%" 40.6%". Qa) 001 @ >) 0.04 Oe) o.10 O 4 0.05 Correct! The first step is to write the conditional variance equation from the model out for times #1 and t#2 (note that the conditional mean equation is not used to produce the variance forecast): h(t) =0.01+0.1uv2+06ht h_(Q+2) = 0.01 + 0.1 u_(t+t)'2 + 0.6 h_(t+1) Then, we would take the conditional expectation, made at time t, of these 2 equations: E{h_(t+1)] = £10.01 + 0.1 u_'2 + 0.6 ht) Eth_{t+2)] = £10.01 + 0.1 u_{te1)*2 + 0.6 h_(t#1)} ‘Taking the expectations operator through any constants and any values known at time t would give Efh_(t+1)]=0.01 +0.1u¥2+06ht E{h_(t+2)] = 0.01 + 0.1 Efu_(t+1)*2] + 0.6 Efh_(t+t)] It should already be apparent that producing the 1-step ahead forecast is easy since all of the necessary quantities can be directly observed, and the forecast is calculated as E{h_{t+1)] = 0.01 + 0.1 x (-0.465%2) + 0.6 x 0.02, = 0,044 (to 3 decimal places) Now considering the 2-step ahead forecast, it should be obvious that [h_(t+1)] is simply the one step ahead forecast that we have just made for the conditional variance (i.e. 0.044), but what about E[u_(t+1)*2]? It also turns out that this is our best guess of the forecast for the squared residual at time (t+1) as well, so that E{h_(t+2)] = 0.01 + 0.1 x 0.044 + 0.6 x 0.044 = 0.041 (Check your answer ‘Suppose that you were asked to provide a guess at a 20-step ahead forecast for the 1.2 model in question 12, What would the most appropriate guess be (to 2 decimal places)? @) 003 O>) 001 nitpsilwiw.cambridge.orgieaturesleconomics/orooksidownloadsimea/ChapterSand9/ch089 him08.02 26/8/2023 Multiple Choice Questions and Feesback- Chapters 8 and 9 Oe 0.10 4) 0.05 Correct! ais the correct answer. Recall that so long as the GARCH model is stationary in variance (i.e. the sum of the coatfciants on the laggad squared error and the lagged Conditional variance is less than one) and not degenerate (i.e. the intercept coefficient in the variance equation is positive), then the forecasts from a GARCH model will converge upon the Jong term mean variance as the forecast horizon increases. Thus fora forecast as far into the future as 20 steps ahead, a good forecast would be the average variance, which would be given by the intercept coefficient in the variance divided by one minus the sum of the slope Coefficients inthe variance. In this case, this long term mean would be 0.04 / (1 -0.1 - 0.6) = 0.08 (to 2 decimal places). (Check your answer Which one of the following equations is the form that is usually used for the conditional 13 covariance equation in a "diagonal VECH" model (using the standard notation)? ©) Fh on ook on ® Correct! a is correct. Using the standard notation, h_(12t) would be the conditional covariance between the two series 1 and 2 at time t. The question is then, what are the appropriate conditioning variables? In other words, what are the appropriate variables to determine the value of this conditional covariance. The most obvious response would be the lagged conditional covariance between the two series, i.e. h_(12t-1), and not the product of the lagged conditional variances for the 2 series, h_(1t-1) h_(2t-1). The scale of the product of the variances would obviously be different to the scale of the product of the series (the covariances). Thus c and d are both incorrect. Then, consider that we could write the covariance between two random variables u1 and u2 as E[(u1-£(ut))(u2-E[u2))]. Under the assumption that ut and u2 are zero mean, this covariance would simplify to E[u1 u2], Thus, the appropriate lagged information for determining the conditional covariance would be the product, of the disturbances. The appropriate information would not the product of their squares (again, the latter would have the wrong scale, and would measure the covariance between the squares of ut and u2, not the covariance between the levels of u1 and u2). (Check your answer What is the most important disadvantage of the diagonal VECH approach to building 14 muttivariate GARCH models that is overcome by the BEKK formulation? © 2) The diagonal VECH model is hard to interpret intuitively © b) The diagonal VECH model contains too many parameters @ ©) The diagonal VECH model does not ensure a positive-definite variance-covariance matrix O d) The BEKK model reduces the dimensionality problem that arises when a number of series are modelled together. Correct! The VECH model may be subjected to a number of important criticisms, although only the fact that the diagonal VECH model does not ensure a positive-definite variance-covariance matrix is overcome by using a BEKK formulation, The BEKK formulation ensures a positive- definite variance-covariance matrix of the disturbances by using a "quadratic formulation” for the parameter matrices in the variance-covariance equations. The parameter estimates in the nitpslwiw.cambridge orgeaturesleconomics/orooksidownloadsimca/ChapterSand9/ch089 him 719(09.02 26/11/2023, Multiple Choice Questions and Feedback Chapters 8 and 8 diagonal VECH are considerably easier to interpret than those of the BEKK formulation since those of the former model are like those of a univariate GARCH model while the quadratic form complicates matters somewhat. The BEKK model will not contain fewer parameters than the diagonal VECH model, although it will contain fewer than the full unrestricted VECH. Finally, it should be obvious that the BEKK model certainly does not reduce the “dimensionality problem” of multivariate GARCH models, which is that the models get very complex and large very quickly as the number of variables in the system increases. The dimensionality problem is equally serious in the VECH and BEKK formulations, The only way to reduce the dimensionality problem (apart from using fewer variables!) is to use a factor GARCH or orthogonal GARCH model that did not actually use the variables themselves, but just their most, important parts (Check your answer Ifa threshold autoregressive (TAR) model is termed a "SETAK it? ', what must be true about © a) It must follow a Markov process. © b)_ The model must contain only two regimes @ ©} The state-determining variable must be the variable being modelled © d) The number of lagged variables on the RHS of the equations for each regime must be the same Correct! The threshold autoregressive (TAR) model approach to modelling variables whose behaviour switches from one type to another over time is quite different to that of the Markov switching methodology. Under the TAR specification, the variable is not purported to follow a Markov process - in fact quite the opposite since if there were 2 lags or more of the dependent, variable on the RHS, the Markov property would clearly not hold. There can be any number of regimes or states (so long as there are at least 2), and there will be one less thresholds than regimes. By definition, a SETAR is a self-exciting TAR, which is one where the state- determining variable is the one being modelled. There is no restriction that the number of lagged variables on the RHS of the equations for each regime must be the same. (Check your answer Consider the following time series model applied to daily data: 16 where rare the returns, and D1, D2, D3 and D4 are dummy variables. D1 = 1 on Monday and zero otherwise; D2 = 1 on Tuesday and zero otherwise, ..., D4 = 1 on ‘Thursday and zero otherwise. What is the interpretation of the parameter estimate for the intercept? © a) tis the average return on Friday © b) tis the average return on Monday © €) tis the Friday doviation from the moan return for the week © ) tis the Monday deviation from the mean return for the week Correct! For any Friday observation, all of the included dummy variables will be zero (i.e. D1 = D2 = D3 = D4 = 0 on Friday. Therefore the estimated value of beta0 will be the estimate of the average return on Friday. Beta0 + beta‘ will be the average Monday return, beta0 + beta2 will be the average Tuesday return and so on. (Check your answer nitpsilwiw.cambridge.orgieaturesleconomics/orooksidownloadsimea/ChapterSand9/ch089 him a9(09.02 26/11/2023, Muliple Choice Questions and Feedback - Chapters 8 and 9 htips:tinww. cambridge or/features/economicsforooksidounloadsimeq/ChapterSand9/ch089 him