Solved – what is the return value of predict in the fGarch package

armagarchpredictionrtime series

I have a question about a quit sophisticated model for a time series. Suppose $ \{X_t:0\le t\le T\}$ is a time series. The plot of autocorrelation function and partialcorrelation function suggest and ARMA model. However, I also want to model the volatility, hence I use a ARMA(p,q)-GARCH(1,1) model, say. This means

$$ X_t=\mu_t+\sigma_t Z_t$$
$$\mu_t=\mu +\sum_{i=1}^p\phi_i(X_{t-i}-\mu)+\sum_{j=1}^q\theta_j(X_{t_j}-\mu_{t-j}) $$
$$ \sigma_t^2=\omega+\alpha_1(X_{t-1}-\mu_{t-1})^2+\beta_1\sigma_{t-1}^2$$

Then $\mu_t$ models the conditional expectation and $\sigma_tZ_t$ the conditional variance, where $\{Z_t\}$ is a strict white noise.

In R we can use 

model <- garchFit(formula=~arma(p,q)+garch(1,1),cond.dist="std",trace=F)

command where we have to specify $p,q$ and we decided to assume the white noise has a student $t$ distribution. Using the built in functin predict we can get a forecast for the next day:

predict(model,n=1)

However what is the return value of predict? Does it forecast $\sigma_{t+1}^2$, $\mu_{t+1}$ or $X_{t+1}$ and how can I get the other two forecasts within R?

Best Answer

I see that it is an old post, but I came across the same problem recently. I am not sure what this value implies, but, it is not the forecast for sure. One reason is that if you run the following

predict(model,n=1000)

and look at the time series, the values are the same with a greater mean squared error. Nevertheless, the main premise behind the GARCH model is stationarity such that the volatility converges to a fixed value over time.

To demonstrate this, for a stationary time series R, consider the following 

g <- garchFit(formula = ~ garch(1,1),data = R,trace = F)
h.t <- g@h.t
eps <- g@residuals
B <- coef(g)
my.predict <- function(k) {
  h.predict <- B[2] + h.t[length(R)]*B[4] + (eps[length(R)]^2)*B[3]  
  i <- 2
  while(i <= k) {
    h.predict[i] <- B[2] +  h.predict [i-1]*(B[4]+B[3])
    i <- i + 1
    }
  return(h.predict)
  }
plot(my.predict(1000), type = "l")
grid(20)
abline(h = B[2]/(1-B[3]-B[4]), lty = 2)

enter image description here

where the horizontal line is the unconditional variance: $$ \sigma^2 = \frac{\omega}{1-\alpha-\beta} $$

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