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If I have some point forecast and an 80% confidence interval, with the forecast assumed to be normally distributed with a constant variance, how do I extract the actual variance?

Let us work with the following data:

  • Point Forecast: 6511

  • Lower 80%: 6476

  • Upper 80%: 6547

By definition, we thus know that:

\begin{equation} \int_{6476}^{6547} \frac{1}{\sqrt{2 \pi \sigma^2}}exp \bigg(\frac{x-6511}{2 \sigma^2} \bigg) dx = 0.8 \end{equation}

Can you provide some R-Code to find $\sigma$?

Thank you very much.

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    $\begingroup$ For an 80% confidence interval you need 10% in the lower tail, 80% in the confidence region and 10% in the upper tail. Using the R function qnorm(0.9,0,1) we get that the upper limit (or half width) is at 1.28 standard deviations. The half width is also 6547-6511= 36. So if $1.28 \sigma = 36$ then $\sigma = 28.125$ $\endgroup$ – noob2 Feb 18 at 15:55
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    $\begingroup$ So the R code (which I have not tested) is sigma:=((upper-lower)/2.0)/qnorm(1.0-(1.0-ci)/2.0,0,1) where in this case ci is equal to 0.80, upper is 6547 and lower is 6476. $\endgroup$ – noob2 Feb 18 at 16:35
  • $\begingroup$ I understand. I have tried it myself and apparently: \begin{equation} \frac{90th quantile-10th quantile}{\sigma}=constant \end{equation} for any $\sigma$, for the normal distribution. But could you prove this rigorously? $\endgroup$ – Mathias Barreto Feb 19 at 12:00
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this I just want to add some details to noob2 comments and also answer your second question inside the comments.

So first lets consider a standard normal distribution $X_0$. This means that $X_0$ has mean zero ($\mu = 0$) and standard deviation 1 ($\sigma = 1$). Let's write $ \Phi^{-1}(\alpha)$ for the $\alpha$ quantile of $X_0$. In R you can compute $ \Phi^ {-1}(\alpha)$ using qnorm, i.g. $\Phi^{-1}(0.1)$ = qnorm(0.1).

Now consider $X \sim \mathcal{N}(\mu, \sigma)$. Then we have that $$ X \stackrel{\text{d}}{=} \mu + \sigma \cdot X_0. $$ As a consequence the quantiles of $X$ can be written as $\Phi^{-1}(\alpha) \cdot \sigma + \mu$. You can easily verify this in R:

mu <- 3
sigma <- 1.2
alpha <- 0.45 

qnorm(0.45) * sigma + mu
[1] 2.849206
qnorm(0.45, mean = mu, sd = sigma)
[1] 2.849206

Therefore

$$ \frac{90th quantilie - 10th quantile}{\sigma} = \frac{\Phi^{-1}(0.9) \cdot \sigma + \mu - \Bigl(\Phi^{-1}(0.1) \cdot \sigma + \mu \Bigr)}{\sigma} =\\ \Phi^{-1}(0.9) - \Phi^{-1}(0.1). $$ I hope this answers your question from the comments.

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