University:
University of Houston
Course:
MATH 3339 | Statistics for the Sciences
Academic year:

2024
Views:

132

Pages:

12
Author:

Edward Ritter

ARCH Estimation: MLE • We compute f.o.c.’s: θ S(𝑦 , θ) = 0 -S(.) = Score vector. - We have a (𝑘+𝑚 x 𝑘+𝑚) system. But, it is a non-linear system. We use numerical optimization. Gauss-Newton or BHHH (approximates H also by the outer product of S(𝑦 , θ)) can be easily implemented. - Given the AR structure, we will need to make assumptions about σ (and ε , ε , ..., ε if we assume an AR(𝑝) process for the mean). - Alternatively, we take σ (& ε , ε , ..., ε ) as parameters to estimate. Note: The 𝛄 0 (𝑘 x1 vector of f.o.c.’s) will give us GLS (or WLS). ∑ 𝛄 ∑ 𝑿′ 𝜀 /𝜎 , 𝑿 =∑ 𝑿 , , , 𝑿′ 𝑦 𝜸 𝑿𝜸 /𝜎 , =0 =0 ARCH Estimation: MLE • MLE = GLS for mean parameters; 𝑦 & 𝑿 weighted by 1/ 𝜎 , . • If the conditional density is well specified and θ0 belongs to Ω, then T S ( y , ) T 1 / 2 (ˆ   0 )  N ( 0 , A01 ), where A01  T 1  t t 0  t 1 • Under the correct specification assumption, A0 = B0, where T B 0  T 1  E [ S t ( y t ,  0 ), S t ( y t ,  0 )' ] t 1 We estimate A0 and B0 by replacing θ0 by its estimated MLE value. • The estimator B0 has a computational advantage over A0.: Only first derivatives are needed. But A0 = B0 only if the distribution is correctly specified. This is very difficult to know in practice. 37 ARCH Estimation: MLE – ARCH(1) Example: ARCH(1) model. Mean equation: 𝑦 Variance equation: 𝜎 𝑿𝜸 𝜀, 𝜔 𝛼 𝜀 𝜀 ~ N(0, 𝜎 ) We write the pdf for the normal distribution, 𝑓 𝜀 |𝛾, 𝜔, 𝛼 exp 𝜸 exp = We form the likelihood L (the joint pdf): L ∏ exp / 2𝜋 ∏ exp We take logs to form the log likelihood, 𝐿 = log L: 𝐿 𝑇 log 2𝜋 2 log 𝑓 1 2 log 𝜎 1 2 𝜀 /𝜎 Then, we maximize 𝐿 with respect to θ = (𝛾, 𝜔, 𝛼 ) the function 𝐿. 75 ARCH Estimation: MLE – ARCH(1) Example (continuation): ARCH(1) model. 𝐿 𝑇 log 2𝜋 2 1 2 log 𝜔 1 2 𝛼 𝜀 𝜔 𝜀 𝛼 𝜀 Taking derivatives with respect to θ = (𝜔, 𝛼 , 𝛄), where 𝛄 is a vector of 𝑘 mean parameters: 𝜕𝐿 𝜕𝜔 𝜕𝐿 𝜕𝛼 𝛄 1 2 1 2 ∑ 𝜔 1 𝛼 𝜀 𝜀 𝜔 𝛼 𝜀 𝑿 𝜀 /𝜎 1 2 1 2 𝜔 𝜀 𝛼 𝜀 𝜀 𝜀 𝜔 𝛼 𝜀 (𝑘x1 vector of derivatives) 38 ARCH Estimation: MLE – Example (in R) • Log likelihood of AR(1)-GARCH(1,1) Model: log_lik_garch11 <- function(theta, data) { mu <- theta[1]; rho1 <- theta[2]; alpha0 <- abs(theta[3]); alpha1 <- abs(theta[4]); beta1 logL_g11 [1] –1745.197 # Log likelihood value > ml_2$par # Extract from ml_2 function parameters [1] -0.0021051742 0.0260003610 0.00012375 0.1900276519 0.7100718082 > I_Var_m2 <- ml_2$hessian # Extract Hessian (matrix of 2nd derivatives) > eigen(I_Var_m2) # Check if Hessian is pd to invert. eigen() decomposition $values # Eigenvalues: if positives => Hessian is pd [1] 1.687400e+08 6.954454e+05 7.200084e+03 5.120984e+02 2.537958e+02 $vectors [,1] [,2] [,3] [,4] [,5] [1,] 4.265907e-05 9.999960e-01 -0.0011397586 0.0018331957 -0.0018541203 [2,] -3.333961e-06 -2.188159e-03 -0.0010048203 0.9769058449 -0.2136566699 [3,] 9.999998e-01 -4.223001e-05 -0.0003544245 0.0001291633 0.0005770707 [4,] -3.599974e-06 -1.702277e-03 -0.8603563865 -0.1097470278 -0.4977344477 [5,] -6.893837e-04 6.416141e-04 -0.5096905472 0.1833226197 0.8405994743 79 ARCH Estimation: MLE – Example (in R) Example 1 (continuation): > sqrt(diag(solve(I_Var_m2))) # Invert Hessian: Parameters Var on diag [1] 1.203690e-03 4.419049e-02 7.749756e-05 5.014454e-02 3.955411e-02 > t_stats <- ml_2$par/sqrt(diag(solve(I_Var_m2))) > t_stats [1] -1.7489333 0.5883701 1.5967743 3.7895984 17.9519078 80 40 ARCH Estimation: MLE – Example (in R) Example 1 (continuation): Summary for CHF/USD changes e , = [log(St) - log(St-1)] = a0 + a1 e 𝜎 𝝎 𝜶𝟏 𝜀 𝜷𝟏 𝜎 , +𝜀 , 𝜀 It-1 ~ N(0, 𝜎 ) • T: 562 (January 1971 - July 2020, monthly). The estimated model for e , is given by: e , = -0.00211 + 0.02600 e , , (.0012) (0.044) 𝜎 = 0.00012 + 0.19003 𝜀 + 0.71007 𝜎 . (0.00096)* (0.050)* (0.040)* 2 Unconditional σ = 0.00012 /(1- 0.19003 - 0.71007) = 0.001201201 Log likelihood: 1745.197 Note: α1 + ß1 = .90 < 1. (Persistent.) 81 ARCH Estimation: MLE – Example (in R) 82 41 ARCH Estimation: MLE – Example (in R) Example 2: Using Robert Shiller’s monthly data set for the S&P 500 (1871:Jan - 2020:Aug, T=1,795), we estimate an AR(1)-GARCH(1,1) model: 𝑟 = [log(Pt) - log(Pt-1)] = a0 + a1 𝑟 𝜎 𝝎 𝜶𝟏 𝜀 𝜷𝟏 𝜎 +𝜀 , 𝜀 It-1 ~ N(0, 𝜎 ) The estimated model for st is given by: 𝑟 = 0.338 + 0.278 𝑟 , (.08)* (0.025)* 𝜎 = 0.756 + 0.126 𝜀 + 0.826 𝜎 . (0.151)* (0.017)* (0.021)* 2 Unconditional σ = 0.756 /(1 - 0.126 - 0.826) = 15.4630 Log likelihood: 4795.08 Note: α1 + ß1 = .952 < 1. (Very persistent.) 83 ARCH Estimation: MLE – Example (in R) Example 2: Below, we plot the time-varying variance. Certain events are clearly different, for example, the 1930 great depression, with a peak variance of 282 (18 times unconditional variance!). The covid-19 volatility similar to the 2008-2009 financial crisis recession: 42 ARCH Estimation: MLE – Regularity Conditions Note: The appeal of MLE is the optimal properties of the resulting estimators under ideal conditions. • Crowder (1976) gives one set of sufficient regularity conditions for the MLE in models with dependent observations to be consistent and asymptotically normally distributed. • Verifying these regularity conditions is very difficult for general ARCH models - proof for special cases like GARCH(1,1) exists. Example: For the GARCH(1,1) model: if E[ln(α1zt2 +β1)] < 0, the model is strictly stationary and ergodic. See Nelson (1990) & Lumsdaine (1996). ARCH Estimation: MLE – Regularity Conditions • Block-diagonality In many applications of ARCH, the parameters can be partitioned into mean parameters, θ1, and variance parameters, θ2. Then, δμt(θ)/δθ2 = 0 and, although, δσt(θ)/δθ1≠0, the Information matrix is block-diagonal (under general symmetric distributions for zt and for particular ARCH specifications). Not a bad result: - Regression can be consistently done with OLS. - Asymptotically efficient estimates for the ARCH parameters can be obtained on the basis of the OLS residuals. 43 ARCH Estimation: MLE – Remarks • But, block diagonality cannot buy everything: - Conventional OLS standard errors could be terrible. - When testing for serial correlation, in the presence of ARCH, the conventional Bartlett s.e. – T-1/2– could seriously underestimate the true standard errors. ARCH Estimation: Non-Normality • The basic GARCH model allows a certain amount of leptokurtosis. It is often insufficient to explain real world data. Solution: Assume a distribution other than the normal which help to allow for the fat tails in the distribution. • t Distribution - Bollerslev (1987) The t distribution has a degrees of freedom parameter which allows greater kurtosis. The t likelihood function is lt  ln((0.5(v  1))(0.5v) 1 (v  2) 1 / 2 (1  zt (v  2) 1 ) ( v1) / 2 )  0.5 ln( 2t ) where Γ is the gamma function and v is the degrees of freedom. As υ→∞, this tends to the normal distribution. • GED Distribution - Nelson (1991) 44 ARCH Estimation: GMM • Suppose we have an ARCH(𝑞). We need moment conditions: (1)  E [ m1 ]  E [ x t ' ( y t  x t γ )]  0 ( 2 )  E [ m 2 ]  E [ε t2 ' ( t2   t2 )]  0 (3)  E [ m 3 ]  E [ t2   /(1   1  ...   q )]  0 Note: (1) refers to the conditional mean, (2) refers to the conditional variance, and (3) to the unconditional mean. • GMM objective function: ^ ^ Q ( X, y; θ )  E [ m ( θ ; X, y )]' W E [ m ( θ ; X, y )] where ^ ^ ^ ^ E [ m ( θ ; X, y )]  [ E [ m 1 ]' E [ m 2 ]' E [ m 3 ]' ]' ARCH Estimation: GMM • γ has 𝑘 free parameters; α has 𝑞 free parameters. Then, we have J = 𝑘 + 𝑞 + 1 parameters. Note that: m(θ; X, 𝒚) has J = 𝑘 + 𝑞 + 1 equations. Dimensions: Q is 1x1; E[m(θ; X, 𝒚)] is Jx1; W is JxJ. • Problem is over-identified: more equations than parameters so cannot solve E[m(θ; X, 𝒚)]=0, exactly. • Choose a weighting matrix W for objective function and minimize using numerical optimization. • Optimal weighting matrix: W = {E[m(θ; X, 𝒚)] E[m(θ; X, 𝒚)]’}-1. Var(θ) = (1/T)[DW-1D’]-1, where D = δE[m(θ; X, 𝒚)]/δθ’ –expressions evaluated at θGMM. 45 ARCH Estimation: Testing • Standard BP test, with auxiliary regression given by: 𝑒 =α +α 𝑒 + .... + α 𝑒 + 𝑣 -Keep R2. H0: α = α = ... = α = 0 (No ARCH). It is not possible to do GARCH test, since we are using the same lagged squared residuals. Then, the LM test is (T - 𝑞)* R2 χ – Engle’s (1982). • In ARCH Models, testing as usual: LR, Wald, and LM tests. Reliable inference from the LM, Wald and LR test statistics generally does require moderately large sample sizes of at least two hundred or more observations. ARCH Estimation: Testing • Issues: - Non-negative constraints must be imposed. θ0 is often on the boundary of Ω. (Two sided tests may be conservative.) - Lack of identification of certain parameters under H0 creates a singularity of the Information matrix under H0. For example, under H0: α = 0 (No ARCH), in the GARCH(1,1), ω and β are not jointly identified. See Davies (1977). • Ignoring ARCH - You suspect 𝑦 has an AR structure: 𝑦 = γ0 + γ1 𝑦 +𝜀 Hamilton (2008) finds that OLS t-test with no correction for ARCH spuriously reject H0: γ1= 0 with arbitrarily high probability for sufficiently large T. White’s (1980) SE help. NW SE help less. 46 ARCH Estimation: Testing Figure. From Hamilton (2008). Fraction of samples in which OLS ttest leads to rejection of H0: γ1= 0 as a function of T for regression with Gaussian errors (solid line) and Student’s t errors (dashed line). Note: H0 is actually true and test has nominal size of 5%. Testing for Heteroscedasticity: ARCH • ARCH Test for the 3 factor F-F model for IBM returns (T=320), with one lag: 𝑟 , – 𝑟 = 𝛼 + β (𝑟 , – 𝑟 ) + β 𝑆𝑀𝐵 + β 𝐻𝑀𝐿 +  , > b <- solve(t(x)%*% x)%*% t(x)%*%y #OLS regression > e <- y - x%*%b > e2 <- e^2 > xx1 <- e2[1:T-1] > fit2 <- lm(e2[2:T]~xx1) > r2_e2 <- summary(fit2)$r.squared > r2_e2 [1] 0.2656472 > lm_t <- (T-1)*r2_e2 > lm_t [1] 84.74147 LM-ARCH Test: 84.74  reject H0 at 5% level (χ2[1],05≈3.84), the usual result for financial time series. 47 GARCH: Forecasting and Persistence (Again) • Consider the forecast in a GARCH(1,1) model: 𝜎 𝜔 𝛼 𝜀 𝛽𝜎 𝜔 𝜎 𝛼 𝑧 𝛽 𝜀 𝜎 𝑧 Taking expectation at time 𝑡 𝐸 𝜎 𝜔 𝜎 𝛼 1 𝛽 Then, by repeated substitutions: 𝜔 ∑ 𝐸 𝜎 • When 𝛼 𝛼 𝛽 𝜎 𝛼 𝛽 𝛽 = 1, today’s volatility affect future forecasts forever: 𝐸 𝜎 𝜎 𝑗𝜔 GARCH: Forecasting and Persistence Example 1: We want to forecast next month (September 2020) variance for CHF/USD changes. Recall we estimated 𝜎 : 𝜎 = 0.00012 + 0.19003 𝜀 + 0.71007 𝜎 . getting 𝜎 = 0.003672220 : We based the 𝜎 Then, 𝛼 𝛽 𝜎 : : 𝜎 𝛼 0.00012 : : = sqrt(0.00367) = 6.1%) 𝛽 𝜎 𝛼 𝛽 = 0.190 + 0.710 = 0.900 We also forecast 𝜎 𝐸 : forecast on: 𝜔 ∗ ∑ 𝐸 𝜎 𝐸 : (=𝜎 0.00367 ∗ 0.9) = 0.003423 : 0.00012 ∗ {1+ (0.9)+ (0.9)2} + 0.00367 ∗ (0.9)3 = 0.00300063 96 48

Lecture 12 Heteroscedasticity 4

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