Edited out incorrect formula
Can someone please solve this equation for x? I have no idea what to do with the $\mathrm{erf}$ (error function).

Edit: Hm, it did not work correctly... here is the function I meant to solve for x in symbolic form:

$$f(x) = a*\left(0.5*\mathrm{erf}\left(\frac{x-b}{c\sqrt{2}}+.5\right)\right)+d$$

Coefficients (with 99% confidence
       a =       1.412  (1.411, 1.412)
       b =       1.259  (1.259, 1.259)
       c =       1.003  (1.002, 1.003)
       d =      0.3016  (0.3014, 0.3017)

When I solve for f(x) with x=1 I get ans = 0.5460

I want to plug 0.5460 into a formula and get 1 back.
Chris, this is the Finv function you spoke of in my other question.

  • $\begingroup$ Your original had a missing parenthesis. Do check if my corrections are accurate. $\endgroup$ – J. M. is a poor mathematician May 11 '11 at 18:49
  • $\begingroup$ In any event, what computing environment are you using? If there is no implementation of the inverse error function, then you will have to use Newton-Raphson... $\endgroup$ – J. M. is a poor mathematician May 11 '11 at 18:50
  • $\begingroup$ @J.M Yeah your correction is correct. I am using MatLab, and there is an inverse error function (erfinv). I realize that erfinv(erf(x)) = x, but... I really suck at math :( $\endgroup$ – Mike Furlender May 11 '11 at 19:05
  • $\begingroup$ I am not MatLab user, but I suspect that it has a solve function for this kind of trivial manipulations. $\endgroup$ – Phira May 11 '11 at 21:03
  • $\begingroup$ @user9325 I don't think so; it said that it can't compute it. $\endgroup$ – Mike Furlender May 11 '11 at 21:44

Let's start with the "general form" you say you had:


and go over the inversion slo-o-owly...




Now, we can employ the inverse error function:





and that's the inverse you need.

Now, if it were


instead (which I think is more likely, since you're starting from the normal distribution CDF), things are a bit different. What you should end up with is


which is probably the expression you actually need for those confidence intervals...

  • $\begingroup$ M Thanks for the extremely thorough explanation! The data is actually (apparently) closest to a half-normal distribution, but flattened at its max. $\endgroup$ – Mike Furlender May 12 '11 at 0:40

$$x = 1.003\sqrt{2}\left(\mathrm{erf}^{-1}\left(\frac{Y/1.42 + 0.4042}{0.5}\right) - 0.05\right) + 0.5501$$

  • $\begingroup$ Awesome! Thanks so much. You have helped me immeasurably with this project. $\endgroup$ – Mike Furlender May 11 '11 at 19:34
  • $\begingroup$ @ Chris Hmm, it didn't work. Perhaps I wrote it incorrectly. Here it is in symbolic form: a*(0.5*erf((1-b)/(c*sqrt(2))+.5))+d $\endgroup$ – Mike Furlender May 11 '11 at 20:01

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