I'm reading this, and I don't understand how the prediction intervals are calculated for the baseline forecast methods. I agree that we can estimate the SD of the sample from the known residuals $$ \begin{equation} \hat{\sigma} = \sqrt{\frac{1}{T-K-M}\sum_{t=1}^T e_t^2}, \tag{5.1} \end{equation}$$

Where $K$ is number of parameters, and $M$ is number of missing values. And I agree that intuitively they should grow with longer forecast (grow with $h$). But I'm not sure how you actually arrive to these quantities? (Assuming normally distributed and uncorrelated residuals).

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Can anyone show the math behind this?


1 Answer 1


We assume that the residuals from the method are uncorrelated and homoscedastic, with mean 0 and variance $\sigma^2$. We don't need to assume normality. Let $y_1,\dots,y_T$ denote the time series observations, and let $\hat{y}_{T+h|T}$ be the estimated forecast mean (or point forecast). Then we can write $$ y_t = \hat{y}_{t|t-1} + \varepsilon_t $$ where $\varepsilon$ is a white noise process. Let $\hat\sigma_h^2$ be the estimated $h$-step forecast variance.

Random walk/Naive method

For a random walk, the equation above suggests that the appropriate model is $$y_t = y_{t-1} + \varepsilon_t.$$ Therefore \begin{align*} y_{T+1} &= y_T + \varepsilon_T \\ y_{T+2} &= y_{T+1} + \varepsilon_{T+1} = y_T + \varepsilon_T + \varepsilon_{T+1}\\ \vdots \\ y_{T+h} &= y_T + \sum_{i=1}^h \varepsilon_{T+i}, \end{align*} and so \begin{align*} \hat{y}_{T+h|T} &= E(y_{T+h|T}) = y_T\\ \text{and}\qquad \hat{\sigma}^2_{h} &= V(y_{T+h|T}) = h\sigma^2 \end{align*}

Seasonal naive method

Here the model is $$y_t = y_{t-m} + \varepsilon_t,$$ where $m$ is the seasonal period. Thus \begin{align*} y_{T+1} &= y_{T+1-m} + \varepsilon_{T+1} \\ y_{T+2} &= y_{T+2-m} + \varepsilon_{T+2} \\ \vdots \\ y_{T+m} &= y_{T} + \varepsilon_{T+m} \\ y_{T+m+1} &= y_{T+1} + \varepsilon_{T+m+1} = y_{T+1-m} + \varepsilon_{T+1} + \varepsilon_{T+m+1}\\ \vdots\\ y_{T+2m} &= y_{T+m} + \varepsilon_{T+2m} = y_{T} + \varepsilon_{T+m} + \varepsilon_{T+2m}\\ y_{T+2m+1} &= y_{T+m+1} + \varepsilon_{T+2m+1} = y_{T+1-m} + \varepsilon_{T+1} + \varepsilon_{T+m+1} + \varepsilon_{T+2m+1}\\ \vdots\\ y_{T+h} &= y_{T+h-m(k+1)} + \sum_{i=0}^{k} \varepsilon_{T+h+m(i-k)}, \end{align*} where $k$ is the integer part of $(h-1) /m$ (i.e., the number of complete years in the forecast period prior to time $T+h$). Therefore \begin{align*} \hat{y}_{T+h|T} &= E(y_{T+h|T}) = y_{T+h-m(k+1)}\\ \text{and}\qquad \hat{\sigma}^2_{h} &= V(y_{T+h|T}) = (k+1)\sigma^2 \end{align*}

Mean method

The model underpinning the mean method is $$y_t = c + \varepsilon_t$$ for some constant $c$ to be estimated. The least-squares estimate of $c$ is the mean, $\hat{c} = \bar{y} = (y_1+\dots+y_T)/T.$ Thus, \begin{align*} y_{T+1} &= c + \varepsilon_{T+1} \\ y_{T+2} &= c + \varepsilon_{T+1} \\ \vdots \\ y_{T+h} &= c + \varepsilon_{T+h}. \end{align*} Therefore \begin{align*} \hat{y}_{T+h|T} &= E(y_{T+h|T}) = \hat{c} = \bar{y}\\ \text{and}\qquad \hat{\sigma}^2_{h} &= V(y_{T+h|T}) = V(\hat{c}) + \sigma^2 = (T^{-1} + 1)\sigma^2 \end{align*}

Drift method

For a random walk with drift $$y_{t} = c + y_{t-1} + \varepsilon_t.$$ Therefore, \begin{align*} y_{T+1} &= c + y_T + \varepsilon_T \\ y_{T+2} &= 2 + y_{T+1} + \varepsilon_{T+1} = 2c + y_T + \varepsilon_T + \varepsilon_{T+1}\\ \vdots \\ y_{T+h} &= hc + y_T + \sum_{i=1}^h \varepsilon_{T+i}. \end{align*} Now the least squares estimate of $c$ is $\hat{c} = (y_{T}-y_1)/(T-1)$. Therefore \begin{align*} \hat{y}_{T+h|T} &= E(y_{T+h|T}) = h\hat{c} + y_T\\ \text{and}\qquad \hat{\sigma}^2_{h} &= V(y_{T+h|T}) = V(\hat{c}) + h\sigma^2 = \frac{h^2\sigma^2}{T-1} + h\sigma^2. \end{align*} I've also cross-posted this answer on my blog at https://robjhyndman.com/hyndsight/forecasting_variances.html

  • $\begingroup$ Thanks for the answer! I think you need the normality assumption for the 1.96 in the prediction interval. $\endgroup$ Dec 9, 2022 at 8:48
  • $\begingroup$ Yes, that is needed for the PI, but not for the derivations in my answer. $\endgroup$ Dec 10, 2022 at 23:48

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