ARCH and GARCH Processes

Transcription

1 Chapter 8 ARCH and GARCH Processes 8.1 Introduction When modelling time series, there are broadly speaking two approaches, the fundamentalist and the data analyst. The first of these aims to construct a model based on the fundamental principles of the situation and then use data to estimate the parameters so that the model may be used for forecasting and prediction. The second of these makes no attempt to search for underlying principles and simply tries to find a model that fits the available data. The ARCH and GARCH processes, motivated by observed volatility of financial series, fall firmly into the second category. The ARCH and GARCH processes are designed for situations where, after fitting a stationary time series model with residuals that satisfy {X t } WN(0,σ 2 ), the residuals do not correspond to IID(0,σ 2 ), because the volatility seems variable, with bursts of high volatility. More information on the nature of the residuals can lead to more accurate prediction. The ARCH / GARCH processes are one attempt to study the situation of variable volatility. Let {X t : t Z} be a stationary, mean zero time series. That is, the mean and possible trend have already been removed. Suppose that the {X t } appear uncorrelated, but where the WN(0,σ 2 ) does not appear to be justified by a time series plot; the noise is not constant. One approach is to model this by processes of the form: { Xt = σ t Z t {Z t } IIDN(0,1) σ t F (Z) t 1 (8.1) where F (Z) t 1 denotes the collection {Z s : s t 1} and σ t F (Z) t 1 denotes that σ t is a function of these random variables. 139

2 8.2 The ARCH Process A prominent model of this class is the so-called ARCH process, introduced by R.F. Engle in [4] (1982). The abbreviation ARCH stands for autoregressive conditional heteroscedasticity. Definition 8.1. The process {X t : t Z} is said to be an ARCH(p) process if it is stationary and { X t = σ t Z t {Z t } IIDN(0,1) σt 2 = α 0 +α 1 Xt α pxt p 2 (8.2) α 0 > 0, α j 0 j = 1,...,p The requirement that α 0 > 0 and α j 0 for j = 1,...,p ensure that σ t > 0. The coefficients α 0,α 1,...,α p have to be chosen such that the process {X t : t Z} defined in this way is stationary. The following lemma, which holds for all processes of the form given by Equation (8.1), helps to establish when {Xt} 2 may be represented as a stationary process with WN(0,σ 2 ) innovations, for some well defined σ 2 < +. Lemma 8.2. Let {X t } be a stationary process satisfying Equation (8.1). Let Z t = Xt 2 σt 2 = σt(z 2 t 2 1). (8.3) Assume that {σ t : t Z} is a stationary process satisfying E[σ 4 t] < +, then { Z t : t Z} is WN(0,σ 2 ) where σ 2 = 2E[σ 4 t]. Proof By construction, E[ Z t ] = E[σ 2 t(z 2 t 1)] = E[σ 2 t](e[z 2 t] 1) = 0 and E[ Z 2 t] = E[σ 4 t]e[z 4 t 2Z 2 t +1] = 2E[σ 4 t] = σ 2. and the proof is complete. Lemma 8.3. Let {X t : t Z} be an ARCH(p) process defined by Equation (8.2) satisfying E[Xt] 4 < +. Let Z be defined by Equation (8.3). Then Xt 2 α j Xt j 2 = α 0 + Z t. j=1 If {X t } is defined by Equation (8.2), it follows that {X 2 t} is an AR(p) process with mean if and only if E[X 2 t] = µ = α 0 1 p j=1 α j 1. p j=1 α j < 1 and 2. E[X 4 t] <

3 Proof giving X 2 t σ 2 t = Z t = X 2 t α 0 X 2 t α j Xt j 2 j=1 α j Xt j 2 = α 0 + Z t. This has the form of an AR(p) process. Taking expectations gives µ 1 = α 0 from which it follows that j=1 µ = j=1 α j α 0 1 p j=1 α. j It remains to show that E[ Z 2 t] < + if and only if E[X 4 t] < +. This follows directly from: [ ) ] 2 E ( Zt +α 0 = 1 +2 α j (1 j=1 = I +II +III. j=1 α j 2 E[X 4 t] α k )E[Xt(X 2 t 2 Xt j)]+ 2 k=1 j=1 k=1 α j α k E[(Xt 2 Xt j)(x 2 t 2 Xt k 2 )] Firstly, E[ Z 2 t] < + E[( Z +α 0 ) 2 ] < +. Clearly each of these terms is non-negative; I = c 2 E[X 4 t] c = 1 α j > 0. j=1 Since E[X 2 tx 2 t j ] E[X4 t] by Hölder together with stationarity, it follows that 0 II 2 1 α j α j j=1 j=1 E[X 4 t] = 2c(1 c)e[x 4 t] Finally, let γ(j,k) = E[(Xt 2 Xt j 2 )(X2 t X t k ) 2 )] then γ is a covariance matrix and hence non negative definite. It follows that α t γα > 0, so that III > 0. It follows that 0 III 4(1 c) 2 E[Xt]. 4 The if and only if condition now follows directly. Although it is the process {X t } that is of interest, the process {Xt} 2 and its representation as an AR(p) process, is very useful for parameter estimation. Most statistical tests rely on a central limit theorem effect, which requires the innovations to have a well defined finite variance. 141

4 For general models, there may be difficulties establishing conditions on the coefficients so that the innovations for the AR(p) process corresponding to {Xt} 2 are WN(0,σ 2 ) for a well defined σ 2 < +. The following example shows the parameter range for the ARCH(1) process. Example 8.1 (ARCH(1) process). Let {X t } satisfy { X t = σ t Z t σt 2 = α 0 +α 1 Xt 1 2. {Z t } IIDN(0,1) Suppose that E[X 2 t] = µ < +. By stationarity, it follows from the first equation that From the second, it therefore follows that in line with the general result. E[X 2 t] = E[σ 2 t] = µ > 0. µ = α 0 +α 1 µ µ = E[X 2 t] = α 0 1 α 1, Now consider the time series {X 2 t : t Z}. With Z t = Z 2 t σ 2 t, so that X 2 t σ 2 t = Z t = X 2 t α 0 α 1 X 2 t 1 X 2 t α 1 X 2 t 1 = α 0 + Z t. It follows that, provided E[ Z 2 t] < + and α 1 < 1, the process {X 2 t} is a causal AR(1) process with mean µ = α 0 1 α 1. For the AR(1) process, precise conditions on the coefficients can be established to ensure that {X 2 t} is an AR(1) process; the following computation shows that this is the case if and only if α 2 1 < 1 3. Using X t = σ t Z t X 4 t = σ 4 tz 4 t = (α 0 +α 1 X 2 t 1) 2 Z 4 t, it follows, using E[Z 4 t] = 3, that E[Xt] 4 = 3E[α0 2 +2α 0 α 1 Xt 2 +α1x 2 t] 4 ( = 3 α α2 0 α ) 1 +3α 1 α 1E[X 2 t] 4 = 3α2 0 (1+α 1) +3α 1 1 α 1E[X 2 t] 4 1 Clearly, E[X 4 t] = + for 3α For 3α2 1 < 1, E[X 4 t] = 3α 2 0 (1+α 1) (1 α 1 )(1 3α 2 1 ). 142

5 It follows that {X 2 t} is a causal AR(1) process if and only if α 1 < 1 3. Since E[X t ] = 0, it follows that the kurtosis of X t is: E[Xt] 4 V(X t ) 2 3 = 3 α0 2(1+α 1) (1 α 1 )(1 3α1 2) (1 α 1) 2 α0 2 3 = 3 1 α α α1 2 = 6α α 2 1 The kurtosis of an ARCH(p) process is always greater than 0, indicating thicker tails than a normal distribution. > 0. Generating Polynomials for the ARCH(p) Process Let α(z) = α 1 z +...+α p z p. Note that the α 0 has not been included in the polynomial. Then the equation for σt 2 may be written in the form: σ 2 t = α 0 +α(b)x 2 t. In this notation, p j=1 α j = α(1) and hence the formula for E[X 2 t] may be written as: ARCH - t Process E[X 2 t] = α 0 +α(1)e[x 2 t] E[X 2 t] = α 0 1 α(1). Sometimes the underlying noise {Z t } is not IID N(0,1); it is heavier tailed. Instead, IID t may be used; the degrees of freedom of the t distribution may be altered to fit the empirical distribution of the residuals. 8.3 Testing for the ARCH Effect Let (z t ) T t=1 denote the estimates of the innovations of the time series. There are two prominent tests available to determine whether a data set exhibits an ARCH effect. The first is simply the Ljung - Box test. The Ljung-Box test determines whether or not the acf is significantly different from the acf of white noise. This is applied to both the series (z t ) T t=1 and (z2 t) T t=1. If the first series is white noise, but the second is not, then this is an indicator of ARCH effects Ljung - Box test The Ljung Box test (named for Greta M. Ljung and George E. P. Box) is a statistical test of whether a of a group of autocorrelations of a time series are different from zero. The Ljung Box test is defined as follows. Let 143

6 The test statistic is: { H 0 : The data are an observed random sample from WN(0,σ 2 ) H 1 The data are not an observed random sample from WN(0,σ 2 ). Q = T (T +2) h k=1 ρ 2 k T k where T is the sample size, ρ k is the sample autocorrelation at lag k, and h is the number of lags being tested. For significance level α, the critical region for rejection of the null hypothesis H 0 is Q > χ 2 h,α where χ 2 h.α is the 1 α quantile of the chi-squared distribution with h degrees of freedom. Note that the test is applied to the residuals of the fitted model, not the original series. This means, for example, that if a causal invertible ARMA(p,q) is fitted to the original series {X t } t Z : then {Z t } t Z is recovered by: φ(b)x t = θ(b)z t Z t = θ(b) 1 φ(b)x t and the Ljung-Box test applied to {Z t } t Z. In such applications the hypothesis being tested is that the residuals have no autocorrelation. When testing the residuals of an estimated model, the degrees of freedom need to be adjusted to reflect the number of parameters estimated. In a model with p + q estimated parameters (for example ARMA(p,q)), the degrees of freedom should be set to h p q. The distribution of the test statistic is derived under the assumption that the residuals are normally distributed; when the data is not normal, it is approximate and relies on a central limit theorem effect Engle s test The second test available is the Lagrange multiplier test of Engle [4] (1982). This test is equivalent to the usual F statistic for testing α j = 0 for all j = 1,...,p in the linear regression z 2 t = α 0 +α 1 z 2 t α p z 2 t p +ǫ t t = p+1,...,t. (8.4) where ǫ t denotes the error term, p is a pre-specified positive integer and T is the sample size. The null hypothesis is: H 0 : α 1 =... = α m = 0. LetQ 0 = T t=p+1 (z2 t ω) 2, where ω = 1 T T t=1 z2 t is the sample mean of (zt) 2 T t=1 and letq 1 = T t=p+1ê2 t where ê t is the least squares residual estimate for the regression problem of Equation (8.4). Then 144

7 F = (SSR 0 SSR 1 )/p SSR 1 /(T 2p 1) T + χ 2 p. The decision rule is to reject H 0 if F > χ 2 p,α. Again, note that the statistic is based on a central limit theorem effect Example Consider the monthly log stock returns of Intel Corporation from , found in the file m-intc7308.txt. Figures 8.1, 8.2, 8.3, 8.4 and 8.5 indicate that there is conditional heteroscedasticity. Firstly, the time series plot indicates clusters of high volatility. Secondly, while the acf of the log stock returns shows no significant serial correlations with only minor exceptions at 7 and 14, the acf and pacf of the squared log returns indicates show serial correlations, indicating that the monthly returns are not independent. intc Index Figure 8.1: Intel log stock returns time series plot These observations are confirmed by the Ljung - Box test: the Q(m) statistics of the log return series give Q(12) = with a p value of 0.11 confirming no serial correlations in the data (significance level 5%). On the other hand, the Ljung - Box test on the squared log return series gives a value of Q(12) = with a p value close to 0, thus indicating that strong ARCH effects are present. > m.intc7308 <- read.table("~/download/m-intc7308.txt", header=t, quote="\"") > View(m.intc7308) > da = m.intc7308 > intc = log(da[,2]+1) > plot(intc, type = "l") > acf(intc) > pacf(intc) > acf(intc^2) 145

8 Series intc ACF Lag Figure 8.2: Intel log stock returns acf Series intc Partial ACF Lag Figure 8.3: Intel log stock returns pacf > pacf(intc^2) > Box.test(intc,lag=12,type= Ljung ) Box-Ljung test data: intc X-squared = , df = 12, p-value = > at=intc-mean(intc) > Box.test(at^2,lag=12,type= Ljung ) Box-Ljung test 146

9 Series intc^2 ACF Lag Figure 8.4: Intel log stock returns squared acf Series intc^2 Partial ACF Lag Figure 8.5: Intel log stock returns squared pacf data: at^2 X-squared = , df = 12, p-value = 5.274e Building an ARCH model Using the fact that Xt 2 is an AR(p) process, it follows that the pacf of {Xt} 2 is 0 for lags greater than p. The sample pacfs may therefore be used to select an ARCH model. Having selected the order, there are several methods for estimating the parameters. One way is to consider {Xt 2 : t Z}, subtract the mean, and apply the techniques of fitting an AR(p) process. 147

10 Likelihood Method Under the distributional assumption that {Z t } IIDN(0,1), the likelihood function of an ARCH(p) model is: f(z 1,...,z T α) = f(z T F T 1 )f(z T 1 F T 2 )...f(z p+1 F p )f(z 1,...,z p α) T } = 1 exp{ z 2 t f(z 2πσ 2 t 2σt 2 1,...,z p α), t=p+1 where α = (α 0,...,α p ) t. It follows that the log likelihood function L is: where L(z p+1,...z T α) = const T t=p+1 { 1 2 ln(σ2 t)+ 1 2 σ 2 t = α 0 +α 1 z 2 t α p z 2 t p. zt 2 σt 2 }, The conditional likelihood is usually used since f(z 1,...,z p α) is usually rather complicated. In heavy tailed applications, the normal density may be replaced by a standardised Student-t distribution with degrees of freedom chosen to reflect the desired tail. Other distributions may be chosen to reflect the fact that the innovations are often skewed. 8.5 Forecasting Forecasts from the ARCH model can be obtained recursively in exactly the same way as those for an AR model. Consider an ARCH(p) model; at the forecast origin h, the 1-step ahead forecast of σ 2 h+1 is: σ 2 h (1) = α 0 +α 1 X 2 h +...+α px 2 h+1 p. Since σ 2 h+1 is F h measurable, it follows that σ 2 h (1) = σ2 h+1. The two step ahead forecast is: σ 2 h (2) = α 0 +α 1 σ 2 h (1)+α 2X 2 h +...+α px 2 h+2 p which follows because E[X 2 h+1 F h] = σ 2 h+1 = σ2 h (1). Similarly, the l step ahead forecast for σ 2 h+l is: where σ 2 h (l) = α 0 + = α 0 + i=1 α i E [ Xh+l 1 2 F ] h α i σh 2 (l i) i=1 σ 2 h (l i) = X2 h+l i l i

11 Example The following output uses the fgarch package with the command garchfit. Again, the data considered is the log stock returns of Intel Corporation from The ARCH(1) model returned is: { R t = X t σt 2 = Xt 1 2 All the parameter estimates are highly significant. The acf of the residuals {x t } gives Q(10) = with a p value of 0.25 and those of {x 2 t} give Q(10) = with a p value of 0.10, neither significant at the 5% level, indicating that the model fits. > m.intc7308 <- read.table("~/data/m-intc7308.txt", header=t, quote="\"") > View(m.intc7308) > install.packages("fgarch") > library(fgarch) > da = m.intc7308 > intc=log(da[,2]+1) > m1 = garchfit(intc~garch(1,0),data=intc,trace=f) > summary(m1) Title: GARCH Modelling Call: garchfit(formula = intc ~ garch(1, 0), data = intc, trace = F) Mean and Variance Equation: data ~ garch(1, 0) <environment: 0x5c7c490> [data = intc] Conditional Distribution: norm Coefficient(s): mu omega alpha Std. Errors: based on Hessian Error Analysis: Estimate Std. Error t value Pr(> t ) 149

12 mu * omega < 2e-16 *** alpha ** --- Signif. codes: 0 *** ** 0.01 * Log Likelihood: normalized: Description: Thu May 23 16:38: by user: john Standardised Residuals Tests: Statistic p-value Jarque-Bera Test R Chi^ Shapiro-Wilk Test R W e-08 Ljung-Box Test R Q(10) Ljung-Box Test R Q(15) Ljung-Box Test R Q(20) Ljung-Box Test R^2 Q(10) Ljung-Box Test R^2 Q(15) Ljung-Box Test R^2 Q(20) LM Arch Test R TR^ Information Criterion Statistics: AIC BIC SIC HQIC > predict(m1,5) #obtain 1 to 5 step predictions meanforecast meanerror standarddeviation Now try fitting a GARCH model (this will be considered later); the tests indicate that this is a substantially better fit. > m2 = garchfit(intc~garch(1,1),data=intc,trace=f) > summary(m2) 150

13 Title: GARCH Modelling Call: garchfit(formula = intc ~ garch(1, 1), data = intc, trace = F) Mean and Variance Equation: data ~ garch(1, 1) <environment: 0x5f20c70> [data = intc] Conditional Distribution: norm Coefficient(s): mu omega alpha1 beta Std. Errors: based on Hessian Error Analysis: Estimate Std. Error t value Pr(> t ) mu omega * alpha ** beta <2e-16 *** --- Signif. codes: 0 *** ** 0.01 * Log Likelihood: normalized: Description: Thu May 23 16:40: by user: john Standardised Residuals Tests: Statistic p-value 151

14 Jarque-Bera Test R Chi^ Shapiro-Wilk Test R W e-07 Ljung-Box Test R Q(10) Ljung-Box Test R Q(15) Ljung-Box Test R Q(20) Ljung-Box Test R^2 Q(10) Ljung-Box Test R^2 Q(15) Ljung-Box Test R^2 Q(20) LM Arch Test R TR^ Information Criterion Statistics: AIC BIC SIC HQIC Now try fitting an ARCH(1) model with Student t distribution. The statistics illustrate that this model does not give an adequate fit for the volatility. > m3 = garchfit(intc~garch(1,0),data=intc,trace=f,cond.dist= std ) > summary(m3) Title: GARCH Modelling Call: garchfit(formula = intc ~ garch(1, 0), data = intc, cond.dist = "std", trace = F) Mean and Variance Equation: data ~ garch(1, 0) <environment: 0x61f48f0> [data = intc] Conditional Distribution: std Coefficient(s): mu omega alpha1 shape Std. Errors: based on Hessian 152

15 Error Analysis: Estimate Std. Error t value Pr(> t ) mu ** omega e-14 *** alpha ** shape *** --- Signif. codes: 0 *** ** 0.01 * Log Likelihood: normalized: Description: Thu May 23 16:44: by user: john Standardised Residuals Tests: Statistic p-value Jarque-Bera Test R Chi^ Shapiro-Wilk Test R W e-08 Ljung-Box Test R Q(10) Ljung-Box Test R Q(15) Ljung-Box Test R Q(20) Ljung-Box Test R^2 Q(10) Ljung-Box Test R^2 Q(15) Ljung-Box Test R^2 Q(20) LM Arch Test R TR^ Information Criterion Statistics: AIC BIC SIC HQIC The GARCH Process The GARCH process is defined as: X t = σ t Z t {Z t } IIDN(0,1) σ 2 t = α 0 +α 1 X 2 t α px 2 t p +β 1 σ 2 t β qσ 2 t q α 0 > 0, α j 0 j = 1,...,p β k 0 k = 1,...,q The equation for σ 2 may be written, in terms of generating polynomials, as 153

16 where σ 2 t = α 0 +α(b)x 2 t +β(b)σ 2 t, { α(z) = α 1 z +...+α p z p β(z) = β 1 z +...+β q z q Using E[X 2 t] = E[σ 2 t] together with stationarity gives: E[X 2 t] = α 0 +(α(1)+β(1))e[x 2 t] E[X 2 t] = α 0 1 α(1) β(1). Again using Z t = X 2 t σ 2 t = σ 2 t(z 2 t 1), so that Z t WN(0,σ 2 ) provided that E[σ 4 t] < +, it follows that This may be written as X 2 t Z t = α 0 +α(b)x 2 t +β(b)(x 2 t Z t ). X 2 t (α(b)+β(b))x 2 t = α 0 +(1 β(b)) Z t. Provided E[ Z 2 t] < +, this is an ARMA(max(p,q),q) process with mean µ = E[X 2 t] given by: µ (α(1)+β(1))µ = α 0 µ = 8.7 An Illustrative Example α 0 1 (α(1)+β(1)). Specifying the order of a GARCH process is not so easy. Only lower order GARCH processes are used in most applications. Consider the monthly excess returns of S&P 500 index starting from 1926 for 792 observations. The ACF is shown in Figure 8.7 while the partial autocorrelations for hte squared excess returns is shown in Figure 8.8. The series of returns has some serial correlations at lags 1 and 3, but the key feature of the pacf of x 2 t is that it shows strong linear dependence and hence an MA(3) model may seem appropriate. An AR(3) model has the same number of parameters; it is considered here. The fitted model is: { x t 0.088x t x t x t = a t σ 2 a = A joint estimation of the AR(3) - GARCH(1,1) model gives: { x t = x t x t x t 3 +a t σ 2 t = a 2 t σ2 t 1 From the volatility equation, the implied (unconditional) variance of a t is: =

17 which is in line with the variance estimate for the AR(3) model. In the AR(3) - GARCH(1,1) model, however, the parameters for the three AR coefficients are insignificant at the 5% level. It therefore seems appropriate to redefine the model by dropping all AR parameters. The redefined model is: { x t = a t σt 2 = a 2 t σ2 t 1 The Ljung-Box statistics suggest that the model is adequate. The analysis is as follows: > sp500 <- read.table("~/data/sp500.dat", quote="\"") > View(sp500) > m1 = garchfit(~arma(3,0)+garch(1,1),data=sp500,trace=f) > summary(m1) Title: GARCH Modelling Call: garchfit(formula = ~arma(3, 0) + garch(1, 1), data = sp500, trace = F) Mean and Variance Equation: data ~ arma(3, 0) + garch(1, 1) <environment: 0xac52190> [data = sp500] Conditional Distribution: norm Coefficient(s): mu ar1 ar2 ar3 omega alpha1 beta e e e e e e e-01 Std. Errors: based on Hessian Error Analysis: Estimate Std. Error t value Pr(> t ) mu 7.708e e e-06 *** ar e e ar e e ar e e

18 omega 7.975e e ** alpha e e e-08 *** beta e e < 2e-16 *** --- Signif. codes: 0 *** ** 0.01 * Log Likelihood: normalized: Description: Thu May 23 17:09: by user: john Standardised Residuals Tests: Statistic p-value Jarque-Bera Test R Chi^ e-16 Shapiro-Wilk Test R W e-07 Ljung-Box Test R Q(10) Ljung-Box Test R Q(15) Ljung-Box Test R Q(20) Ljung-Box Test R^2 Q(10) Ljung-Box Test R^2 Q(15) Ljung-Box Test R^2 Q(20) LM Arch Test R TR^ Information Criterion Statistics: AIC BIC SIC HQIC Now fit a GARCH(1,1) model with Student t distribution. > m2 = garchfit(~garch(1,1),data=sp5,trace=f,cond.dist="std") > summary(m2) Title: GARCH Modelling Call: garchfit(formula = ~garch(1, 1), data = sp5, cond.dist = "std", trace = F) 156

19 Mean and Variance Equation: data ~ garch(1, 1) <environment: 0xb13e2d0> [data = sp5] Conditional Distribution: std Coefficient(s): mu omega alpha1 beta1 shape Std. Errors: based on Hessian Error Analysis: Estimate Std. Error t value Pr(> t ) mu 8.455e e e-08 *** omega 1.248e e ** alpha e e e-05 *** beta e e < 2e-16 *** shape 7.003e e e-05 *** --- Signif. codes: 0 *** ** 0.01 * Log Likelihood: normalized: Description: Thu May 23 17:11: by user: john Standardised Residuals Tests: Statistic p-value Jarque-Bera Test R Chi^ Shapiro-Wilk Test R W e-08 Ljung-Box Test R Q(10) Ljung-Box Test R Q(15) Ljung-Box Test R Q(20) Ljung-Box Test R^2 Q(10)

20 Ljung-Box Test R^2 Q(15) Ljung-Box Test R^2 Q(20) LM Arch Test R TR^ Information Criterion Statistics: AIC BIC SIC HQIC > #obtain standardised residuals > stresi = residuals(m2,standardize=t) > plot(stresi,type="l") > Box.test(stresi,10,type= Ljung ) Box-Ljung test data: stresi X-squared = , df = 10, p-value = > predict(m2,5) meanforecast meanerror standarddeviation stresi Index Figure 8.6: Standard and Poor standardised residuals 158

21 Series sp500 ACF Lag Figure 8.7: Standard and Poor acf for (excess) return series Series sp500^2 Partial ACF Lag Figure 8.8: Standard and Poor partial autocorrelation for squared excess return series 8.8 Further Extensions of ARCH / GARCH There are numerous associated models: Non-linear ARCH (NARCH) σ γ t = α 0 +α(b) X t γ +β(b)σ γ t. The choice γ = 2 yields the GARCH model; γ = 1 is sometimes used. The standard approach for constructing variations on ARCH / GARCH is as follows: some financial data sets are considered. For each set, the hypothesis H 0 : γ = 2 (GARCH) is tested against the alternative H 1 : γ 2 (or rather: the null hypothesis is that the data may be modelled by one of the current models; the alternative is that the wider class of models is more appropriate). The null 159

22 hypothesis is generally rejected. The creator of the generalisation is then happy and some papers are written. The typical situation is that we have some further parameters to play with in H 0 H 1. Since most (in fact all) models are simplifications of reality, a null hypothesis H 0 which is much smaller than H 0 H 1 is (almost) always rejected if the data set is large enough. Asymmetric ARCH Processes One drawback with ARCH, GARCH and NARCH is that positive and negative values have a summetric effect on the volatility. Many financial time series are strongly asymmetric; negative returns are followed by larger increase in the volatility than positive returns. Typical examples of this may be prices of gasoline or the interest on a loan. Let X t + = max(x t,0) Xt = min(x t,0) There are several examples which address this problem; for example, the exponential GARCH (EGARCH) and quadratic GARCH (QGARCH). The EGARCH is: lnσt 2 = α 0 +βlnσt 1 2 +λz t 1 +φ( Z t 1 E[ Z t 1 ]). Other examples may be found in the literature. 160

23 Tutorial 7 1. Consider the monthly simple returns of Intel stock from January 1973 to December 2008 m-intc7308.txt. Transform the returns into log returns. Build a GARCH model for the transformed series and compute 1-step to 5-step ahead volatility forecasts at the forecast origin December The file m-mrk4608.txt contains monthly simple returns of Merck stock from June 1946 to December The file has two columns denoting date and simple return. Transform the simple returns into log returns. (a) Is there any evidence of serial correlations in the log returns? Use autocorrelations and 5% significance level to answer the question. If yes, remove the serial correlations. This is done by fitting an ARMA model. One is then interested in ARCH-GARCH effects in the residuals. (b) Is there any evidence of ARCH effects in the log returns? Use the residual series if there are serial correlations in part (a). Use Ljung-Box statistics for the squared returns (or residuals) with 6 and 12 lags of autocorrelations and 5% significance level to answer the question. (c) Identify an ARCH model for the data and fit the identified model. Write down the fitted model. 3. The file m-3m4608.txt contains two columns. They are date and the monthly simple return for 3M stock. Transform the returns to log returns. (a) Is there any evidence of ARCH effects in the log returns? Use Ljung-Box statistics with 6 and 12 lags of autocorrelations and 5% significance level to answer the question. (b) use the PACF of the squared returns to identify an ARCH model. What is the fitted model? (c) There are 755 data points. Refit the model using the first 750 observations and use the fitted model to predict the folatilities for times 751 to 755 (the forecast origin is 750). (d) Build an EGARCH model for the log return series of 3M stocks using the first 750 observations. Use the fitted model to compute the 1-step to 5-step-ahead volatility forecasts at the forecast origin h = 750. For EGARCH, you may find the package betategarch useful. Look at the documentation. 4. The file m-gmsp5008.txt contains the dates and monthy simple returns of General Motors stock and the S&P index from (a) Build a GARCH model with Gaussian innovations for the log returns of GM stock. Check the model and write down the fitted model. (b) Build a GARCH model with Student-t distribution for the log returns of GM stock including estimation of the degrees of freedom. Write down the fitted model. Let ν be the degrees of freedom. Test the hypothesis H 0 : ν = 6 versus H 1 : ν 6 at a significance level of 5%. (c) Build an EGARCH model for the log returns of GM stock. What is the fitted model? 161

24 (d) Obtain 1-step to 6-step-ahead volatility forecasts for all models obtained. Compare the forecasts. 5. Consider again the data in m-gmsp5008.txt. (a) Build a Gaussian GARCH model for the monthly log returns of the S&P 500 index. (b) Is there are summer effect on the volatility of the index return? This requires writing an R script. Make a variable Try fitting the model: u t = { 1 month is: June, July, August 0 other months. σ 2 (t) = α 0 +α 1 X 2 (t)+β 1 σ 2 (t 1)+u t ( α00 +α 01 X 2 (t)+β 01 σ 2 (t 1) ). Then the coefficients are (α 0,α 1,β 1 ) for September to May and (α 0 +α 00,α 1 +α 01,β 1 +β 01 ) for June, July and August. Are α 00,α 01,β 01, which represent the differences, significant? (c) Are the lagged returns of GM stock useful for modelling the index volatility? Use your GARCH model as a baseline for comparison. 6. The purpose of this exercise is to simulate a GARCH(1,1) process u(t) = v(t) h(t) h(t) = a 0 +a 1 u(t 1) 2 +b 1 h(t 1) a 0 = 0.1 a 1 = 0.4 b 1 = 0.2 > v<-rnorm(1000) > u<-array(0,1000) > h<-array(0,1000) > for(t in 2:1000){} > for(t in 2:1000){} > a0 = 0.1; a1 = 0.4; b1 = 0.2 > for(t in 2:1000){ + h[t] <- a0 + a1*(u[t-1]^2)+b1*h[t-1] + u[t]<-v[t]*sqrt(h[t]) + } > plot(u,type="l") > acf(u) > acf(u^2) > library("fgarch") > u.garch <- garchfit(~garch(1,1),u,trace=f) > u.garch@fit$matcoef How do the estimated coefficients compare with the true coefficients? 162

25 Exercises (These exercises should be considered at home) 1. Derive multistep-ahead forecasts of the volatility ] for a GARCH(1,2) model at the forecast origin h. Do this by noting thatσh [σ 2(j) = E h+j 2 F h, usinge[e[x Y]] = E[X] and thate[xt F 2 t 1 ] = σt 2. Obtain a recursive formula by computing σh 2(1), σ2 h (2) and σ2 h (j) for j > 2. Compute σh 2 ( ), assuming it exists and state conditions such that the answer is well defined. 2. Derive multistep-ahead forecasts of the volatility for a GARCH(2,1) model at the origin h. Comute σh 2 ( ) assuming it exists and state conditions such that the answer is well defined. 3. Suppose that r 1,...,r n are observations of a return series that follows the AR(1)-GARCH(1,1) model: X t = σ t Z t {Z t } IIDN(0,1) σ 2 t = α 0 +α 1 X 2 t 1 +β 1σ 2 t 1 R t = µ+φ 1 R t 1 +X t Derive the conditional log-likelihood function of the data. Base this on r 3,...,r n and use as an initialisation r 1 = r 0 = σ0 2 = Consider the AR(1)-GARCH(1,1) model of the previous exercise, with the difference that {Z t } is IID t ν. Derive the conditional log-likelihood function of the data. 163

26 Answers 1. For the GARCH(1,2), { X t = σ t Z t {Z t } IID N(0,1) It follows that σ 2 t = α 0 +α 1 X 2 t 1 +β 1σ 2 t 1 +β 2σ 2 t 2 E[X 2 t F t 1 ] = σ 2 t so that σ 2 h (1) = α 0 +α 1 X 2 h +β 1σ 2 h +β 2σ 2 h 1 σ 2 h (2) = α 0 +α 1 σ 2 h (1)+β 1σ 2 h (1)+β 2σ 2 h = α 0 +(α 1 +β 1 )σ 2 h (1)+β 2σ 2 h = α 0 +(α 1 +β 1 )(α 0 +α 1 X 2 h +β 2σ 2 h 1 )+(β 1(α 1 +β 1 )+β 2 )σ 2 h For j 3, σh 2 (j) = α 0 +α 1 E[Xh+j 1 2 F h]+β 1 σh 2 (j 1)+β 2σh 2 (j 2) = α 0 +(α 1 +β 1 )σ 2 h (j 1)+β 2σ 2 (j 2). σ 2 h ( )(1 α 1 β 1 β 2 ) = α 0 σ 2 h ( ) = α 0 1 α 1 β 1 β σ 2 h (1) = α 0 +α 1 X 2 h +α 2X 2 h 1 +β 1σ 2 h σ 2 h (2) = α 0 +α 1 σ 2 h (1)+α 2X 2 h +β 1σ 2 h (1) α h (j) = α 0 +(α 1 +β 1 )σh 2(j 1)+α 2σh 2 (j 2) (if and only if α 1 +α 2 +β 1 < 1) σ 2 h ( ) = α 0 1 α 1 α 2 β 1 3. Let θ = (α 0,α 1,β 1,µ,φ 1 ) denote the vector of parameters and let f(r 1,...,r n ) be the joint density, then f(r 1,...,r n θ) = f(r 1,r 2 θ) Take L(θ) = f(r 3,...,r n θ,r 1,r 2 ), then L(θ) = n 2 2 ln(2π) n f(r j θ,r 1,...,r j 1 ) j=3 n lnσj j=3 Use an initialisation of σ 2 0 = 0, r 0 = r 1 = 0. n (r j µ φ 1 r j 1 ) 2 j=3 σ 2 t = β 1 σ 2 t 1 +α 0 +α 1 (r t 1 µ φ 1 r t 2 ) σ 2 j

27 4. The student t on ν degrees of freedom has density proportional to: f(x) ) (ν+1)/2 (1+ x2 ν and the model is: so that X t := (R t ν φ 1 R t 1 ) t ν σ t σ t L(θ) = const ν +1 2 Use an initialisation of σ 2 0 = 0, r 0 = r 1 = 0. n ln (1+ (r j ν φ 1 r j 1 ) 2 ) j=3 νσ 2 t σ 2 t = β 1 σ 2 t 1 +α 0 +α 1 (r t 1 µ φ 1 r t 2 )