Prediction Interval

Study guides: Exam SRM, Exam MAS-II

A prediction interval is a range that will contain a future observation, such as next quarter's claim frequency or next month's claim count, with a stated probability. A confidence interval covers a fixed parameter such as a mean. A prediction interval must also allow for the random shock the new observation will carry, so it is always wider.

$\hat{x}_{t+h} \pm z_{1-\alpha/2}\,\mathrm{SE}(e_{t+h})$

$\mathrm{SE}(e_{t+h}) = \sigma\sqrt{\sum_{j=0}^{h-1}\psi_j^{2}}$

$\hat{x}_{t+h}$ is the $h$ -step forecast and $e_{t+h} = x_{t+h} - \hat{x}_{t+h}$ its error. $\sigma$ is the white-noise standard deviation, and the $\psi_j$ are the weights of the model's MA( $\infty$ ) form, with $\psi_0 = 1$ . For $95\%$ , $z = 1.96$ .
How the width grows with $h$ :
- MA( $q$ ): the $\psi_j$ vanish beyond $q$ , so the width stops growing after $q$ steps, at $\sigma\sqrt{1 + \theta_1^2 + \cdots + \theta_q^2}$
- stationary AR(1): $\psi_j = \phi^j$ , so the variance is $\sigma^2(1 - \phi^{2h})/(1 - \phi^2)$ , levelling off at the series' own variance $\sigma^2/(1 - \phi^2)$
- random walk: $\psi_j = 1$ , so the variance is $h\sigma^2$ and the interval widens without limit, the signature of a unit root
Reading software output. In R, predict() on a fitted arima model returns the point forecasts (pred) and their standard errors (se). The interval is pred $\pm\ 1.96 \times$ se. A se column that rises with $h$ is the $\psi$ -sum accumulating.
Regression analogue. For a new observation at $\mathbf{x}_0$ in a Linear Regression, the interval is $\hat{y}_0 \pm t_{n-p-1}\,s\sqrt{1 + \mathbf{x}_0^{\top}(\mathbf{X}^{\top}\mathbf{X})^{-1}\mathbf{x}_0}$ . The " $1 +$ " is the new observation's own noise. It is why a prediction interval does not shrink to zero as $n$ grows, while the confidence interval for the mean does.
Caveats. The usual intervals assume the model is correct, its parameters are known and the errors are normal, so check the residuals with a correlogram and a QQ Plot. They omit parameter and model uncertainty, so real outcomes land outside them more often than the nominal $5\%$ . For a model fitted to logs, build the interval on the log scale and exponentiate the endpoints.

Reading an AR(1) Forecast Table Quarterly claim frequency (claims per $100$ exposures) is fitted as $x_t = 2.4 + 0.6x_{t-1} + \varepsilon_t$ , and the latest value is $x_t = 7.0$ . The software reports:

$h$ pred se

1 $6.600$ $0.500$

2 $6.360$ $0.583$

3 $6.216$ $0.610$

Give $95\%$ prediction intervals for $h = 1$ and $h = 3$ , explain why se grows and where it stops, and recover $\phi$ from the se column.
Answer
 $\begin{align*} \text{PI}_{1} &= 6.600 \pm 1.96(0.500) \\ &= 6.600 \pm 0.980 \\ &= (5.62,\ 7.58) \\ \text{PI}_{3} &= 6.216 \pm 1.96(0.610) \\ &= 6.216 \pm 1.196 \\ &= (5.02,\ 7.41) \end{align*}$ 
Each extra step adds one more unknown shock, so the error variance is $\sigma^2(1 + \phi^2 + \phi^4 + \cdots)$ . The standard error levels off at $\sigma/\sqrt{1 - \phi^2} = 0.5/0.8 = 0.625$ . Far ahead, the interval becomes the long-run mean $2.4/0.4 = 6.0$ plus or minus $1.96(0.625) = 1.225$ .

The ratio of the first two standard errors gives $\phi$ :
 $\begin{align*} 1 + \phi^{2} &= \left(\frac{0.583}{0.500}\right)^{2} \\ &= 1.36 \\ \lvert\phi\rvert &= 0.6 \end{align*}$ 
This matches the fitted coefficient: the se column alone carries the AR structure of the model.

$h$	`pred`	`se`
1	$6.600$	$0.500$
2	$6.360$	$0.583$
3	$6.216$	$0.610$

Back-Transforming a Log-Scale Interval An ARIMA model is fitted to the log of monthly average paid severity. Next month's forecast is $8.70$ on the log scale, with standard error $0.08$ . Give a $95\%$ prediction interval for the severity in dollars.
Answer
 $\begin{align*} \text{log scale} &= 8.70 \pm 1.96(0.08) \\ &= (8.543,\ 8.857) \\ \text{dollars} &= \left(e^{8.543},\ e^{8.857}\right) \\ &= (\$5{,}132,\ \$7{,}022) \end{align*}$ 
The point forecast is $e^{8.70} = \$6{,}003$ . The interval is asymmetric: $\$871$ below and $\$1{,}019$ above, as a lognormal should be. $\$6{,}003$ is the median forecast. The mean is higher, $e^{8.70 + 0.08^2/2} = \$6{,}022$ .