MFE5209 ARCH/GARCH
Wu Fuheng
Monday, October 06, 2014
Problem 1
EMAIL: Wu Fuheng(a0095704@nus.edu.sg)
This is a AR(1)/ARCH(1) model, the conditional mean and variance of \(u_t\) will be non-constant.
Since \(\mu=0.4, \phi=0.45, \omega=1\) and \(\alpha_1=0.3\), we can rewrite the equation of \(u_t\) as follows:
\[\begin{align*} u_t &=(1-\phi)\mu + \phi u_{t-1} + h_t\\ &=(1-0.45)0.4+0.45 u_{t-1}+h_t\\ &=0.22+0.45 u_{t-1}+h_t\\ \end{align*}\]
a)Conditional Expectation
The conditional mean of \(u_t\) is
\[\begin{align*} E[u_t|u_{t-1},...] &= 0.22+0.45u_{t-1}+E[h_t|h_{t-1}...]\\ \end{align*}\]
According to P479 of SDAFE, for a ARCH(1) process \(h_t\), the conditional mean \(E[h_t|h_{t-1}...]\)=0. Thus we have
\[\begin{align*} E[u_t|u_{t-1},...] &= 0.22+0.45u_{t-1}\\ \end{align*}\]
So,
\[\begin{align*} E[u_2|u_1=1,u_0=0.2] &= 0.22+0.45u_1\\ &=0.67 \end{align*}\]
b)Conditional Variance
The conditional variance of \(u_t\) is \[\begin{align*} Var[u_t|u_{t-1},...] &= E[(u_t-E[u_t|u_{t-1},...])^2|u_{t-1},...]\\ &= E[(h_t)^2|u_{t-1},...]\\ &= \omega + \alpha_1 h_{t-1}^2\\ \end{align*}\]
Thus, we have \[\begin{align*} Var[u_2|u_1=1,u_0=0.2] &= \omega + \alpha_1 h_{1}^2\\ &= 1 + 0.3 h_{1}^2\\ \end{align*}\]
And we also know \[\begin{align*} u_1-\mu=\phi (u_0 - \mu) + h_1\\ \longrightarrow 1-0.4=0.45(0.2-0.4)+ h_1\\ \longrightarrow h_1=0.6+0.45 \times 0.2=0.69 \end{align*}\]
Therefore, \(Var[u_2|u_1=1,u_0=0.2]=1+0.3 \times 0.69^2=1.14283\)
Problem 2
Assume \(\beta_0=0.06, \beta_1=0.35, \delta=0.22\), we can rewrite the model
\[\begin{align*} Y_t=0.06+0.35X_t+0.22\sigma_t+h_t \end{align*}\]
a)Conditional Expectation
\[\begin{align*} E[Y_t|X_t=0.1,h_{t-1}=0.6] &=0.06+0.35*0.1+0.22\sigma_t\\ &=0.095+0.22\sigma_t\\ \end{align*}\]
b)Conditional Variance
\[\begin{align*} Var[Y_t|X_t=0.1,h_{t-1}=0.6] &=E[(Y_t-E[Y_t|X_t=0.1,h_{t-1}=0.6])^2|X_t=0.1,h_{t-1}=0.6]\\ &=E[h_t^2|X_t=0.1,h_{t-1}=0.6]\\ &=1+0.5h_{t-1}^2\\ &=1+0.5*0.6^2\\ &=1.18 \end{align*}\]
c)Is the conditional distribution of \(Y_t\) given \(X_t\) and \(h_{t-1}\) normal? Why or why not?
Yes, it is normal. Reason: There is only one stochastic item \(\varepsilon\) in the formula which is normal. And we know the linear combination of a normal distribution is normal.
d)Is the marginal distribution of \(Y_t\) normal? Why or why not?
No, it isn’t. Reason: In this case, \(h_t\) is a ARCH(1) process, which is of a non-Gaussian distribution with a heavy tail. So \(Y_t\) is also not Gaussian/normal.
Problem 3
This problem uses monthly observations of the two-month yield, that is, \(Y_T\) with T equal to two months, in the data set Irates in the Ecdat package. The rates are log-transformed to stabilize the variance. To fit a GARCH model to the changes in the log rates, run the following R code.
library(fGarch)
library(Ecdat)
data(Irates)
r = as.numeric(log(Irates[,2]))
n = length(r)
lagr = r[1:(n-1)]
diffr = r[2:n] - lagr
g=garchFit(~arma(1,0)+garch(1,1),data=diffr, cond.dist = "std")a) What model is being fit to the changes in \(r\)? Describe the model in detail.
The model is a AR(1)/GARCH(1,1) model, assuming a t-distributed errors. We are trying to use AR(1) to model the return conditional mean and GARCH(1,1) to model its conditional variance. We can write is mathematically as follows:
\[\begin{align*} \Delta r_t=Y_t=\mu+ar_1 Y_{t-1}+a_t\\ \\ a_t=\varepsilon_t \sqrt{\omega + \alpha_1 a_{t-1}^2+ \beta_1 \sigma_{t-1}^2}, \varepsilon_t \thicksim t(.) \end{align*}\]
In this case, the fitted model is: \[\begin{align} \Delta r_t=Y_t=0.00910873+0.09508060 Y_{t-1}+a_t\\ \\ a_t=\varepsilon_t \sqrt{0.00054864 + 0.32444019 a_{t-1}^2+ 0.74026870 \sigma_{t-1}^2}\\ \\ \varepsilon_t \thicksim t(2.83060758) \end{align}\]
The details of the fitted model are as follows:
summary(g)##
## Title:
## GARCH Modelling
##
## Call:
## garchFit(formula = ~arma(1, 0) + garch(1, 1), data = diffr, cond.dist = "std")
##
## Mean and Variance Equation:
## data ~ arma(1, 0) + garch(1, 1)
## <environment: 0x000000000b8aa668>
## [data = diffr]
##
## Conditional Distribution:
## std
##
## Coefficient(s):
## mu ar1 omega alpha1 beta1 shape
## 0.00910873 0.09508060 0.00054864 0.32444019 0.74026870 2.83060758
##
## Std. Errors:
## based on Hessian
##
## Error Analysis:
## Estimate Std. Error t value Pr(>|t|)
## mu 0.0091087 0.0023895 3.812 0.000138 ***
## ar1 0.0950806 0.0432758 2.197 0.028014 *
## omega 0.0005486 0.0002968 1.848 0.064564 .
## alpha1 0.3244402 0.1368876 2.370 0.017782 *
## beta1 0.7402687 0.0656629 11.274 < 2e-16 ***
## shape 2.8306076 0.4243716 6.670 2.56e-11 ***
## ---
## Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
##
## Log Likelihood:
## 630.9 normalized: 1.19
##
## Description:
## Sat Oct 11 00:48:12 2014 by user: wufuh_000
##
##
## Standardised Residuals Tests:
## Statistic p-Value
## Jarque-Bera Test R Chi^2 1048 0
## Shapiro-Wilk Test R W 0.9114 0
## Ljung-Box Test R Q(10) 7.962 0.6326
## Ljung-Box Test R Q(15) 11.93 0.6846
## Ljung-Box Test R Q(20) 12.76 0.8874
## Ljung-Box Test R^2 Q(10) 13.12 0.2173
## Ljung-Box Test R^2 Q(15) 14.7 0.4736
## Ljung-Box Test R^2 Q(20) 17.79 0.6014
## LM Arch Test R TR^2 14.28 0.2829
##
## Information Criterion Statistics:
## AIC BIC SIC HQIC
## -2.358 -2.310 -2.358 -2.339b) What are the estimates of the parameters of the model?
The estimates of the model parameters are as follows
Coefficient(s):
mu ar1 omega alpha1 beta1 shape
0.00910873 0.09508060 0.00054864 0.32444019 0.74026870 2.83060758
c) What is the estimated ACF of \(\Delta r_t\) ?
acf(diffr)
d) What is the estimated ACF of \(a_t\) ?
\(a_t\) is uncorrelated, aka \(ACF(a_t)=\rho_a(h)=0\) if \(h\neq 0\).
acf(residuals(g))
e) What is the estimated ACF of \({a_t}^2\) ?
acf(residuals(g)^2)