Recovering Standard Deviation from a Normal Forecast Interval

# How to extract standard deviation from normal distribution in R


# How to extract standard deviation from normal distribution in R












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:



- 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.

## Answer by Cettt (score 1, accepted)

https://quant.stackexchange.com/a/51269

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.