I'd like to solve this equation for $\mu$. Is it possible? If not, why?
$$ 2 P = \operatorname{erf}\left( \frac{\mu – A}{ \sqrt{2 \sigma^2} } \right) – \operatorname{erf}\left( \frac{\mu – B}{\sqrt{2 \sigma^2}} \right) $$
Where $\operatorname{erf}$ is the error function. I'm happy to denote the inverse of $\operatorname{erf}$ as $\operatorname{erf^{-1}}$ and use it as such. I hope to solve for $\sigma$ in terms of $\operatorname{erf^{-1}}$.
Assumptions:
- All variables are real numbers.
- $0 < \sigma$.
- $0 < P < 1$.
- $A < B$
In lieu of a general solution, I'd be happy with an approximation as long as I can calculate and adjust the precision as necessary.
Best Answer
If you want a solution by formula, no. You can simplify the problem a bit by introducing a new variable $x=(\mu-A)/\sqrt{2\sigma^2}$, so it becomes instead $$2P=\operatorname{erf}(x)-\operatorname{erf}(x-2b)$$ for some suitable $b>0$. You may notice that the right hand side goes to zero as $x\to\pm\infty$, and it is increasing on $(-\infty,b)$ and decreasing on $(b,\infty)$. Its maximum value is achieved at $x=b$. If this is greater than $2P$, there will be two solutions. If it is less than $2P$, there will be no solution.