tweak: add explanation for magic number

This commit is contained in:
NunoSempere 2022-04-14 16:22:21 -04:00
parent e5655dc2d1
commit bd3f2c99d1
2 changed files with 36 additions and 2 deletions

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@ -1,5 +1,8 @@
open SymbolicDistTypes
let normal95confidencePoint = 1.6448536269514722
// explained in website/docs/internal/ProcessingConfidenceIntervals
module Normal = {
type t = normal
let make = (mean: float, stdev: float): result<symbolicDist, string> =>
@ -11,7 +14,7 @@ module Normal = {
let from90PercentCI = (low, high) => {
let mean = E.A.Floats.mean([low, high])
let stdev = (high -. low) /. (2. *. 1.6448536269514722)
let stdev = (high -. low) /. (2. *. normal95confidencePoint)
#Normal({mean: mean, stdev: stdev})
}
let inv = (p, t: t) => Jstat.Normal.inv(p, t.mean, t.stdev)
@ -120,7 +123,7 @@ module Lognormal = {
let logLow = Js.Math.log(low)
let logHigh = Js.Math.log(high)
let mu = E.A.Floats.mean([logLow, logHigh])
let sigma = (logHigh -. logLow) /. (2.0 *. 1.6448536269514722)
let sigma = (logHigh -. logLow) /. (2.0 *. normal95confidencePoint)
#Lognormal({mu: mu, sigma: sigma})
}
let fromMeanAndStdev = (mean, stdev) => {

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# Processing confidence intervals
This page explains what we are doing when we take a 95% confidence interval, and we get a mean and a standard deviation from it
## For normals
```js
module Normal = {
//...
let from90PercentCI = (low, high) => {
let mean = E.A.Floats.mean([low, high])
let stdev = (high -. low) /. (2. *. 1.6448536269514722)
#Normal({mean: mean, stdev: stdev})
}
//...
}
```
We know that for a normal with mean $\mu$ and standard deviation $\sigma$,
$$
a \cdot Normal(\mu, \sigma) = Normal(a\cdot \mu, |a|\cdot \sigma)
$$
We can now look at the inverse cdf of a $Normal(0,1)$. We find that the 95% point is reached at $1.6448536269514722$. ([source](https://stackoverflow.com/questions/20626994/how-to-calculate-the-inverse-of-the-normal-cumulative-distribution-function-in-p)) This means that the 90% confidence interval is $[-1.6448536269514722, 1.6448536269514722]$, which has a width of $2 \cdot 1.6448536269514722$.
So then, if we take a $Normal(0,1)$ and we multiply it by $\frac{(high -. low)}{(2. *. 1.6448536269514722)}$, it's 90% confidence interval will be multiplied by the same amount. Then we just have to shift it by the mean to get our target normal.
## For lognormals