Solved – Computing one-step ahead forecast for AR and ARMA models

arimaautoregressiveforecastingtime series

I'm having trouble computing the one-step ahead forecast for the following time-series models.

For the following models, ${Z_t}$ is a whitenoise process with ${Z_t}$ ~ $WN(0, \sigma^2)$.

I am given the following set of $n=5$ data points:
$$-0.93, -0.89, -0.63, -0.38, 0.76$$

and I am trying to forecast the 6th data point.

The models are:

$$AR(1): X_t = 0.9X_{t-1} + Z_t$$
$$ARMA(1,1): X_t = 0.3X_{t-1} + Z_t – Z_{t-1}$$
$$ARMA(2,1): X_t = 0.1X_{t-1} – 0.5X_{t-2} + Z_t + Z_{t-1}$$

I think I can compute the forecast for the $AR(1)$ process: the best linear predictor is $P_nX_{n+1} = a_n^{'}X_n = \phi X_n$, so the 6th data point $\hat{X}_6 = 0.9*0.76$

For the $ARMA(1,1)$ model, using the recursions of the innovations algorithm, I get $\hat{X}_{n+1} = \phi X_n + \theta_{n1}(X_n – \hat{X}_n)$, but do I have to recursively find the estimates for each data point? Also, how do I find out what the prediction for $\hat{X}_1$ is?

Best Answer

Yes, you do need to work recursively, plugging in previous values. First you plug in $X_t$, and when these are not available any more, you plug in the point predictions.

To get the innovations, you are right that you need the first fitted value. There is no "right" way to get this. Some people (and some software) estimate the initial fits via maximum likelihood, others use heuristics and use, e.g., the first actual observation, or a heuristically smoothed value.

If this is self-study, do whatever your instructor suggested. If you are trying to recreate something your software did, you will need to find out how it fits these initial values. This is often not very well documented.

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

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)

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

Solved – Fit an ARMAX model in R

As far as my research on this topic has taken me, I agree that that the arima/Arima functions from the stats and forecast packages, does fit transfer functions as you mention, but instead a linear model with ARMA errors.

I don't see the possibility to tell the TSA::arimax function that the $\nu(B)$ should be equal to $\phi(B)$. But you can force the order of all the individual polynomial in $B$ to be what you want, including $0$. But that does not really give you your ARMAX model.

The last thing I can suggest is to take a look at the MARIMA package. It should somehow be able to fit an ARMAX model, but I am not 100 % sure about the procedure.

Best Answer

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

Solved – Fit an ARMAX model in R

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