4
$\begingroup$

I am trying to compare a manual computation for the prediction interval for a forecasted value (one step ahead, at 95% confidence) on this data set, to the given prediction interval from R's forecast package.

I have selected an ARIMA(1,1,0)$\times$(0,1,1)$_{12}$ model for the above data set (which has 192 data points).

The 'manual' computation I currently have (as R input) is:

      339.5735+1.96*sqrt(t(res)%*%res/192)

, where:

339.5735 is the Forecast value from

forecast(Arima(dataset, order = c(1,1,0), seasonal = list(order =c(0,1,1), period =12)), h=1 ). 

The "Hi 95" and "Lo 95" entries of this output are also the values I am comparing my manually forecasted values against;

res = as.matrix(Arima(mauna, order = c(1,1,0), seasonal = list(order =c(0,1,1), period =12))$residuals)

, so res is a vector of the residual errors from the selected ARIMA model fitted to our time series;

192 is the number of data points in the time series (wasn't too sure whether to use biased or unbiased estimator of variance, and how many degrees of freedom need to be taken away from the denominator in the case of the unbiased estimator).

My manual computation doesn't match that of R's for the lower bound and upper bound for the prediction interval and I was wondering whether anyone could tell me why? Even a formula for the variance of an ARIMA(p,d,q)(P,D,Q)_s model would be appreciated as I haven't been able to find one that helps my case.

Any assistance with this problem would both help me tremendously and be greatly appreciated.

Thank you.

$\endgroup$
6
$\begingroup$

You must use the standard deviation of the prediction errors, rather than the standard deviation of the residuals. The former varies for each forecast period.

For example, in an AR(1) model

$$y_t=\phi y_{t-1} + \varepsilon_t \,, \quad \varepsilon_t\sim NID(0, \sigma^2) \,, \quad t=1,\dots,T\,,$$

the variance of the prediction errors can be obtained as follows.

The h-steps ahead prediction errors are defined as $e(h) = y_{T+h} - E(y_{T+h})$ (the value determined by the model minus its expectation):

$$ \begin{align} e(1) &= y_{T+1} - E(y_{T+1}) = (\phi y_T+\varepsilon_{T+1}) - \phi y_T = \varepsilon_{T+1} \\ e(2) &= y_{T+2} - E(y_{T+2}) = (\phi \underbrace{y_{T+1}}_{\phi y_T+\varepsilon_{T+1}}+\varepsilon_{T+2}) - E(y_{T+2}) \\ &= \underbrace{(\phi^2y_T+\phi \varepsilon_{T+1}+\varepsilon_{T+2})}_{y_{T+2}} - \phi^2y_T) \\ &= \phi \varepsilon_{T+1} + \varepsilon_{T+2} \\ e(3) &= \underbrace{\phi^3y_T+\phi^2\varepsilon_{T+1}+\phi \varepsilon_{T+1} + \varepsilon_{T+3}}_{y_{T+3}} - E(y_{T+3}) \\ &= \phi^3y_T+\phi^2\varepsilon_{T+1}+\phi \varepsilon_{T+1} + \varepsilon_{T+3} - \phi^3y_T = \phi^2\varepsilon_{T+1}+\phi \varepsilon_{T+2} + \varepsilon_{T+3} \\ \cdots & = \cdots \\ e(h) &= \phi^{h-1}\varepsilon_{T+1}+\dots+\phi \varepsilon_{T+h-1} + \varepsilon_{T+h} \end{align} $$

The variances are straightforward to obtain (just note that $Var(\phi^k \varepsilon_t)=\phi^{2k}\sigma^2$):

$$ \begin{align} Var(e(1)) &= \sigma^2 \\ Var(e(2)) &= \sigma^2(1+\phi^2) \\ Var(e(3)) &= \sigma^2(1+\phi^2+\phi^4) \\ \dots &= \dots \end{align} $$


The expressions of the variances vary with the order of the model. A common way to compute the variances of the prediction errors in a general ARMA(p,q) is by means of the Kalman filter. Here is some code for illustration.

These are your data:

x <- ts(c(319.32, 320.36, 320.82, 322.06, 322.17,321.95,321.20,318.81,317.82, 317.37, 318.93, 319.09, 319.94, 320.98, 321.81, 323.03,323.36, 323.11, 321.65, 319.64, 317.86, 317.25, 319.06, 320.26, 321.65, 321.81,322.36, 323.67, 324.17, 323.39, 321.93, 320.29, 318.58, 318.60,319.98, 321.25, 321.88, 322.47, 323.17, 324.23, 324.88, 324.75, 323.47, 321.34,319.56, 319.45, 320.45, 321.92, 323.40, 324.21, 325.33, 326.31,327.01, 326.24, 325.37, 323.12, 321.85, 321.31, 322.31, 323.72, 324.60, 325.57,326.55, 327.80, 327.80, 327.54, 326.28, 324.63, 323.12, 323.11, 323.99, 325.09, 326.12, 326.61, 327.16, 327.92, 329.14, 328.80, 327.52,325.62, 323.61, 323.80, 325.10, 326.25, 326.93, 327.83, 327.95,329.91, 330.22, 329.25, 328.11, 326.39, 324.97, 325.32, 326.54, 327.71, 328.73, 329.69, 330.47, 331.69, 332.65, 332.24, 331.03, 329.36, 327.60, 327.29, 328.28, 328.79, 329.45, 330.89, 331.63, 332.85, 333.28, 332.47, 331.34, 329.53, 327.57, 327.57, 328.53, 329.69, 330.45, 330.97, 331.64, 332.87, 333.61, 333.55, 331.90, 330.05, 328.58, 328.31, 329.41, 330.63, 331.63, 332.46, 333.36, 334.45, 334.82, 334.32, 333.05, 330.87, 329.24, 328.87, 330.18, 331.50, 332.81, 333.23, 334.55, 335.82, 336.44, 335.99, 334.65, 332.41, 331.32, 330.73, 332.05, 333.53, 334.66, 335.07, 336.33, 337.39, 337.65, 337.57, 336.25, 334.39, 332.44, 332.25, 333.59, 334.76, 335.89, 336.44, 337.63, 338.54, 339.06, 338.95, 337.41, 335.71, 333.68, 333.69, 335.05, 336.53, 337.81, 338.16, 339.88, 340.57, 341.19, 340.87, 339.25, 337.19, 335.49, 336.63, 337.74, 338.36), frequency=12, start=c(1965,1))

These are the results from forecast that you want to reproduce:

require("forecast")
fit <- Arima(x, order = c(1,1,0), seasonal = list(order =c(0,1,1), period =12))
forecast(fit, h=4)
     Point Forecast    Lo 80    Hi 80    Lo 95    Hi 95
# Jan 1981       339.5735 339.1483 339.9987 338.9232 340.2238
# Feb 1981       340.1759 339.6493 340.7025 339.3706 340.9813
# Mar 1981       341.2141 340.5865 341.8417 340.2543 342.1739
# Apr 1981       342.3020 341.5914 343.0125 341.2153 343.3886

The state space representation of the fitted model is available in fit$model, which is the input required to run the Kalman filter.

kf <- KalmanForecast(n.ahead = 4, fit$model)
pred.vars <- kf$var * fit$sigma2
kf$pred - 1.96 * sqrt(pred.vars)
# [1] 338.9232 339.3706 340.2543 341.2153
kf$pred + 1.96 * sqrt(pred.vars)
# [1] 340.2238 340.9813 342.1740 343.3886

The variances of prediction errors returned by the Kalman filter are scaled by the estimated variance of the residuals, $\sigma^2$. Then, the confidence intervals are obtained as you did (you can see that these values match the columns Lo 95 and Hi 95 obtained above).


Edit This may depart from your question, but as mentioned by @IrishStat, if the assumption of Gaussian innovations is not plausible, it is worth considering some alternative methods. For example, the option bootstrap=TRUE computes bootstrapped confidence intervals. After a quick view to the documentation, I think this option resamples the residuals and generates an ensemble of replicates based on the parameters of the fitted model. Although the distribution of the residuals of your fitted model depart from the Gaussian distribution (due to excess of kurtosis as suggested for example by the histogram), graphically there are no major differences in the confidence bands. Here are the numbers:

set.seed(1)
forecast(fit, h=4, bootstrap=TRUE)
#          Point Forecast    Lo 80    Hi 80    Lo 95    Hi 95
# Jan 1981       339.5735 339.2259 339.9744 338.9472 340.2538
# Feb 1981       340.1759 339.7288 340.6966 339.4446 341.0239
# Mar 1981       341.2141 340.6837 341.8561 340.3736 342.2230
# Apr 1981       342.3020 341.6879 343.0394 341.3340 343.4285

Aside note: I think your question is more related to theoretical issues rather than on practical issues and your are using this model and data mainly for illustration. However, as preparing this answer I had a closer look to your data, I would suggest an ARIMA(0,1,1)(0,1,1) model with a temporary change at observation 190 (october 1980). I don't claim it is the best model or that it will perform better for forecasting, but according to the diagnostic of the residuals, it is better than the ARIMA(1,1,0)(0,1,1).

$\endgroup$
  • $\begingroup$ Nice workup ! I have been concerned with the omission of the uncertainty in the estimated parameters and additionally the presumption of symmetry in the confidence limits based upon normality . This has lead me to implementing model errors via re-sampling as an option to obtain limits. $\endgroup$ – IrishStat Jan 16 '17 at 19:43
  • $\begingroup$ @IrishStat Interesting, forecast has an option to obtain bootstrapped confidence intervals, which may be in the vein of your comment. I think the OP is interested in Gaussian confidence intervals, but it is worth mentioning alternative methods that may have been overlooked. $\endgroup$ – javlacalle Jan 16 '17 at 21:15
1
$\begingroup$

Whereas @javlacelle did a great job in developing the forecast variance for a particular case as he stated "The expressions of the variances vary with the order of the model", it is interesting to pursue/present a generalized solution. In general How to obtain confidence limits of predicted values in ARIMA? the variance of a forecast can be computed via the model's psi weights . As a note , psi weights are obtained by expressing the model as pure MA model. See @Hyndaman's comment for more clarification.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.