lec5-13
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2 (1 − αB ) ln(σt ) = (1 − α)α0 + g ( t−1 ),
Under normality, E (| t|) = 2/π and the model becomes
2 (1 − αB ) ln(σt )=
α∗ + (θ + γ ) α∗ + (θ − γ )
2 σt
=
2×0.856 −1.001 σt e −1
×
e0.1852 t−1
if
t−1 t−1
≥0 < 0.
e−0.3442 t−1 if
For a standardized shock with magnitude 2, (i.e. two standard deviations), we have
2
t−1 t−1
if if
t−1 t−1
≥ 0, <0
where α∗ = (1 − α)α0 − Tong (1978, 1990). Specifically, we have
2 π γ.
This is a nonlinear fun. similar to that t = 0.743 + 0.048σt + at, σt = 0.812 + 0.123a2 t−1 + .854σt−1 .
Std err of risk premium is 0.141 so that the estimate is not statistically significant at the usual 5% level. R demonstration
2 σt ( t−1 = −2) exp[−0.3442 × (−2)] 0.318 = = e = 1.374. 2( = 2) σt exp(0 . 1852 × 2) t−1
Therefore, the impact of a negative shock of size two-standard deviations is about 37.4% higher than that of a positive shock of the same size. Forecasting: some recursive formula available Another parameterization of EGARCH models |at−1| + γ1at−1 2 + β1 ln(σt −1 ), σt−1 where γ1 denotes the leverage effect.
(θ + γ ) t − γE (| t|) if (θ − γ ) t − γE (| t|) if
t t
≥ 0, < 0.
An EGARCH(m, s) model: at =
2 σt t, ln(σt )
1 + β1B + · · · + βs−1B s−1 = α0 + g( 1 − α1 B − · · · − αm B m
Std. Error t value Pr(>|t|) 0.1540336 4.82229 1.4192e-06 *** 0.1408765 0.34140 0.7327991 0.2858128 2.84168 0.0044877 ** 0.0220596 5.55449 2.7843e-08 *** 0.0219079 38.95885 < 2.22e-16 ***
3
Discussion: Using 2/π ≈ 0.7979 ≈ 0.8, we obtain
2 2 ln(σt ) = −1.0 + 0.856 ln(σt −1 ) +
0.1852
t−1 t−1
if if
t−1 t−1
≥0 < 0.
−0.3442
Taking anti-log transformation, we have
2 ln(σt ) = α0 + α1
Below, I re-analyze the IBM log returns by extending the data to December 2009. The sample size is 1008. The fitted model is rt = 0.012 + at, at = σt
One expects γi to be positive so that prior negative returns have higher impact on the volatility. The Asymmetric Power ARCH (APARCH) Model. This model was introduced by Ding, Engle and Granger (1993) as a general class of volatility models. The basic form is rt = µt + at,
t−1 ).
The leverage parameter θ shows the
effect of the sign of at−1 whereas γ denotes the magnitude effect. See Nelson (1991) for an exmaple of EGARCH model. Another example: Monthly log returns of IBM stock from January 1926 to December 1997 for 864 observations. For textbook, an AR(1)-EGARCH(1,1) is obtained (RATS program): rt = 0.0105 + 0.092rt−1 + at, at = σt t g ( t−1) 2 ln(σt ) = −5.496 + , 1 − .856B g ( t−1) = −.0795 t−1 + .2647[| t−1| − 2/π ], Model checking: ˜t: Q(10) = 6.31(0.71) and Q(20) = 21.4(0.32) For a For a ˜2 t : Q(10) = 4.13(0.90) and Q(20) = 15.93(0.66)
2 σt t σt s m
=
α0 + (αi + γiNt−i)a2 t−i i=1
+
j =1
2 βj σt −j ,
where Nt−i is an indicator variable such that Nt−i =
1 if at−i < 0, 0 otherwise.
> source("garchM.R") > mn=garchM(sp5*100) [1] 2399.822 0: 2380.0229: 0.422452 0.00561297 0.806149 0.121976 0.854361 3: 2379.9525: 0.427596 0.00652832 0.806107 0.121466 0.855756 6: 2378.4838: 0.606807 0.0375055 0.801889 0.126193 0.846833 9: 2377.9587: 0.673997 0.00166937 0.798191 0.125517 0.851915 12: 2377.8474: 0.692209 0.0444464 0.802574 0.122278 0.854141 15: 2377.7922: 0.742796 0.0480959 0.812187 0.122530 0.853507 Maximized log-likehood: 2377.792 Coefficient(s): Estimate mu 0.7427956 gamma 0.0480959 omega 0.8121873 alpha 0.1225297 beta 0.8535072 ---
δ σt = ω+ s i=1
at = σt t,
t
∼ D(0, 1)
m j =1 δ βj σ t −j
αi(|at−i| − γiat−i)δ +
where δ is a non-negative real number. In particular, δ = 2 gives rise to the TGARCH model and δ = 0 corresponds to using log(σt). Theoretically, one can use any power δ to obtain a model. In practice, two things deserve further consideration. First, δ will also affect the
Lecture Note of Bus 41202, Spring 2013: More Volatility Models. Mr. Ruey Tsay The GARCH-M model
2 rt = µ + cσt + at , 2 2 at = σt t, σt = α0 + α1a2 t−1 + β1 σt−1
1
Remarks: This R script is relatively slow. It takes longer time due to its use of recursive loop in evaluating likelihood function. The EGARCH model Asymmetry in responses to past positive and negative returns: g ( t) = θ t + γ [| t| − E (| t|)], with E [g ( t)] = 0. To see asymmetry of g ( t), rewrite it as g( t) =
Under normality, E (| t|) = 2/π and the model becomes
2 (1 − αB ) ln(σt )=
α∗ + (θ + γ ) α∗ + (θ − γ )
2 σt
=
2×0.856 −1.001 σt e −1
×
e0.1852 t−1
if
t−1 t−1
≥0 < 0.
e−0.3442 t−1 if
For a standardized shock with magnitude 2, (i.e. two standard deviations), we have
2
t−1 t−1
if if
t−1 t−1
≥ 0, <0
where α∗ = (1 − α)α0 − Tong (1978, 1990). Specifically, we have
2 π γ.
This is a nonlinear fun. similar to that t = 0.743 + 0.048σt + at, σt = 0.812 + 0.123a2 t−1 + .854σt−1 .
Std err of risk premium is 0.141 so that the estimate is not statistically significant at the usual 5% level. R demonstration
2 σt ( t−1 = −2) exp[−0.3442 × (−2)] 0.318 = = e = 1.374. 2( = 2) σt exp(0 . 1852 × 2) t−1
Therefore, the impact of a negative shock of size two-standard deviations is about 37.4% higher than that of a positive shock of the same size. Forecasting: some recursive formula available Another parameterization of EGARCH models |at−1| + γ1at−1 2 + β1 ln(σt −1 ), σt−1 where γ1 denotes the leverage effect.
(θ + γ ) t − γE (| t|) if (θ − γ ) t − γE (| t|) if
t t
≥ 0, < 0.
An EGARCH(m, s) model: at =
2 σt t, ln(σt )
1 + β1B + · · · + βs−1B s−1 = α0 + g( 1 − α1 B − · · · − αm B m
Std. Error t value Pr(>|t|) 0.1540336 4.82229 1.4192e-06 *** 0.1408765 0.34140 0.7327991 0.2858128 2.84168 0.0044877 ** 0.0220596 5.55449 2.7843e-08 *** 0.0219079 38.95885 < 2.22e-16 ***
3
Discussion: Using 2/π ≈ 0.7979 ≈ 0.8, we obtain
2 2 ln(σt ) = −1.0 + 0.856 ln(σt −1 ) +
0.1852
t−1 t−1
if if
t−1 t−1
≥0 < 0.
−0.3442
Taking anti-log transformation, we have
2 ln(σt ) = α0 + α1
Below, I re-analyze the IBM log returns by extending the data to December 2009. The sample size is 1008. The fitted model is rt = 0.012 + at, at = σt
One expects γi to be positive so that prior negative returns have higher impact on the volatility. The Asymmetric Power ARCH (APARCH) Model. This model was introduced by Ding, Engle and Granger (1993) as a general class of volatility models. The basic form is rt = µt + at,
t−1 ).
The leverage parameter θ shows the
effect of the sign of at−1 whereas γ denotes the magnitude effect. See Nelson (1991) for an exmaple of EGARCH model. Another example: Monthly log returns of IBM stock from January 1926 to December 1997 for 864 observations. For textbook, an AR(1)-EGARCH(1,1) is obtained (RATS program): rt = 0.0105 + 0.092rt−1 + at, at = σt t g ( t−1) 2 ln(σt ) = −5.496 + , 1 − .856B g ( t−1) = −.0795 t−1 + .2647[| t−1| − 2/π ], Model checking: ˜t: Q(10) = 6.31(0.71) and Q(20) = 21.4(0.32) For a For a ˜2 t : Q(10) = 4.13(0.90) and Q(20) = 15.93(0.66)
2 σt t σt s m
=
α0 + (αi + γiNt−i)a2 t−i i=1
+
j =1
2 βj σt −j ,
where Nt−i is an indicator variable such that Nt−i =
1 if at−i < 0, 0 otherwise.
> source("garchM.R") > mn=garchM(sp5*100) [1] 2399.822 0: 2380.0229: 0.422452 0.00561297 0.806149 0.121976 0.854361 3: 2379.9525: 0.427596 0.00652832 0.806107 0.121466 0.855756 6: 2378.4838: 0.606807 0.0375055 0.801889 0.126193 0.846833 9: 2377.9587: 0.673997 0.00166937 0.798191 0.125517 0.851915 12: 2377.8474: 0.692209 0.0444464 0.802574 0.122278 0.854141 15: 2377.7922: 0.742796 0.0480959 0.812187 0.122530 0.853507 Maximized log-likehood: 2377.792 Coefficient(s): Estimate mu 0.7427956 gamma 0.0480959 omega 0.8121873 alpha 0.1225297 beta 0.8535072 ---
δ σt = ω+ s i=1
at = σt t,
t
∼ D(0, 1)
m j =1 δ βj σ t −j
αi(|at−i| − γiat−i)δ +
where δ is a non-negative real number. In particular, δ = 2 gives rise to the TGARCH model and δ = 0 corresponds to using log(σt). Theoretically, one can use any power δ to obtain a model. In practice, two things deserve further consideration. First, δ will also affect the
Lecture Note of Bus 41202, Spring 2013: More Volatility Models. Mr. Ruey Tsay The GARCH-M model
2 rt = µ + cσt + at , 2 2 at = σt t, σt = α0 + α1a2 t−1 + β1 σt−1
1
Remarks: This R script is relatively slow. It takes longer time due to its use of recursive loop in evaluating likelihood function. The EGARCH model Asymmetry in responses to past positive and negative returns: g ( t) = θ t + γ [| t| − E (| t|)], with E [g ( t)] = 0. To see asymmetry of g ( t), rewrite it as g( t) =
