# Solving equation involving error function and normal distribution

Let $\phi(x)$ be a normal distribution, that is $\phi=\frac{1}{\sigma}\phi(x|\mu,\sigma^2)$ and $\Phi(x)$ be the CDF of $\phi$, i.e., $\Phi(x|\mu,\sigma^2)=\frac{1}{2}[1+erf(\frac{x-\mu}{\sqrt{2}\sigma})]$, where erf is the error function). Given some fixed constant $a$, how can I get a good approximation (ideally solve) for x that satisfies the equation
$x=[\phi(x)/\Phi(x)]-a$?

• Given $\mu,\sigma,a$ ? – Claude Leibovici Mar 16 '17 at 12:57
• Yes given $\mu,\sigma,a$. – Doron Mar 18 '17 at 6:51

So, you want to solve for $x$ the equation $$x=\frac{\sqrt{\frac{2}{\pi }} e^{-\frac{(x-\mu )^2}{2 \sigma ^2}}}{\sigma \left(1+\text{erf}\left(\frac{x-\mu }{\sigma }\right)\right)}-a$$ For simplicity, let $x=\mu +\sigma y$ which makes $$\sigma (a+\mu) +\sigma^2 y=\sqrt{\frac{2}{\pi }}\frac{ e^{-\frac{y^2}{2}}}{1+\text{erf}(y)}$$ which means that you search for the intersection of a straight line (given by the lhs) with a nice monotonic curve (given by the rhs).
We can approximate the rhs using Pade approximants; for example $$\frac{ e^{-\frac{y^2}{2}}}{1+\text{erf}(y)}\approx\frac{1+\frac{4 \sqrt{\pi } }{3 (\pi -8)}y+\frac{(8-3 \pi ) }{6 (\pi -8)}y^2}{1+\frac{2 (5 \pi -24) }{3 (\pi -8) \sqrt{\pi }}y}$$ would let you with a quadratic equation in $y$. But, gain, this is an approximation.