Time Series Forecast

Study guide: Exam MAS-II

A time series forecast is the conditional expectation of a future value given the history, $\hat Y_{t+h} = E[Y_{t+h} \mid Y_t, Y_{t-1}, \dots]$ , together with an interval expressing how uncertain it is. Producing and reading one is objective D4.

$\hat{Y}_{t+h} = E\!\left[Y_{t+h} \mid \mathcal{F}_t\right]$

$\hat{Y}_{t+h} \pm z_{1-\alpha/2}\,\sigma\sqrt{\textstyle\sum_{j=0}^{h-1}\psi_j^{2}}$

Build it recursively: replace future shocks with 0, future observations with their forecasts, and known past values with themselves
The $\psi_j$ are the coefficients of the infinite MA representation; for an AR(1), $\psi_j = \phi^{j}$ , so the $h$ -step variance is $\sigma^{2}\dfrac{1-\phi^{2h}}{1-\phi^{2}}$
Where each model converges: a stationary AR decays to the mean with a bounded interval; an MA( $q$ ) hits the mean exactly at $h > q$ ; a Random Walk stays flat at the last value with an interval widening as $\sqrt{h}$
The intervals assume the model and its parameters are correct — they carry process variance only, no parameter or model uncertainty, so they are optimistic in practice
Assess a forecast on a held-out tail of the series, not on the fitted data, using MAE, RMSE or MAPE

Two-Step Forecast Interval for an AR(1) $Y_t = 20 + 0.6Y_{t-1} + \varepsilon_t$ , $\sigma = 4$ , and $Y_{50} = 61$ . Give the two-step forecast and a 95% interval.

Answer $\hat Y_{51} = 20 + 0.6(61) = 56.6 \qquad \hat Y_{52} = 20 + 0.6(56.6) = 53.96$ Two-step error is $\varepsilon_{52} + \phi\varepsilon_{51}$ , with variance $\sigma^2(1 + \phi^{2}) = 16(1.36) = 21.76$ and SD $4.665$ : $53.96 \pm 1.96(4.665) = (44.8,\ 63.1)$ The one-step interval would be the narrower $\pm 1.96(4) = \pm 7.84$ . As $h$ grows the forecast converges to $\mu = 20/0.4 = 50$ and the SD to $4/\sqrt{1-0.36} = 5$ .