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Lets say I have two time series a and b and I predict the next value and a prediction interval (say lower and upper 80%) of each of these two. I end up with something like that: a: [10, 20, 30] and b: [2, 7, 12] (as [lo80%, prediction, high80%]). As these time series are independent, I see I could just add 20 + 7 to get a prediction of a + b, but what about the prediction interval? I suppose I cant just add them to [12, 27, 42], right?

(Sorry if it is too simple a question)

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You need to make some assumptions about the distribution of each prediction, but since your confidence intervals are symmetric I am writing the Gaussian distribution case.

For a Gaussian RV the 80 percentile corresponds to being $\approx 0.8416 \sigma$ away from the mean, so the standard deviation for each prediction can be calculated as $\sigma = (q_{(0.80)} - \mu) / 0.8416$.

For independent RVs, $\text{Var}(X + Y) = \text{Var}(X) + \text{Var}(Y)$, therefore

$\begin{align} q_{(0.80)} &= 0.8416\cdot\sigma_{X+Y} + \mu_{X+Y} \\ &= 0.8416 \cdot \sqrt{(10 / 0.8416)^2 + (5/.8416)^2} + \mu_{X+Y}\approx 11.18 + \mu_{X+Y} \end{align}$

So the 80% CI and prediction for the sum is [15.82, 27, 38.18].

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    $\begingroup$ +1 Did you notice that the value of $0.8416$ is irrelevant? You can obtain $11.18 = \sqrt{(10^2+5^2)}$. $\endgroup$ – whuber Sep 4 '12 at 12:31
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    $\begingroup$ @whuber I wish I could say the inclusion was solely pedagogical but alas I missed that fact. Of course if one wants something other than the 80th quantile of the sum, this shows how to calculate it from the 80th quantile of the parts. $\endgroup$ – muratoa Sep 4 '12 at 22:17

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