Can $\Phi^{-1}(x)$ be written in terms of $\operatorname{erf}^{-1}(x)$?

Question

Can the inverse CDF of a standard normal variable $\Phi^{-1}(x)$ be written in terms of the inverse error function $\operatorname{erf}^{-1}(x)$, and, if so, how?

This seems like an easy question, but I am struggling with it. I know that $\Phi(x)=\frac{1}{2}\left[1+\operatorname{erf}\left(\frac{x}{\sqrt{2}}\right)\right]$ simply by the change of variables in the definitions of $\Phi(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^xe^{-t^2/2}dt$ and $\operatorname{erf}(x)=\frac{1}{\sqrt{\pi}}\int_{-x}^xe^{-t^2}dt$, however I can't figure out the inverses.

Answer 1

$\Phi(x)=\frac{1}{2}\left[1+\operatorname{erf}\left(\frac{x}{\sqrt{2}}\right)\right]$, which means

$\Phi^{-1}\left(\frac{1}{2}\left[1+\operatorname{erf}\left(\frac{x}{\sqrt{2}}\right)\right]\right)=x$

Now the goal is to make what is inside $\Phi^{-1}(\cdot)$ "equal to" $x$. For this purpose, let us put $\sqrt{2}\mathrm{erf}^{-1}(x)$ instead of $x$:

$\Phi^{-1}\left(\frac{1}{2}\left[1+x\right]\right)=\sqrt{2}\mathrm{erf}^{-1}(x)$

Now put $2x-1$ instead of $x$:

$\Phi^{-1}\left(x\right)=\sqrt{2}\mathrm{erf}^{-1}(2x-1)$.

These can be done in a single line of course.

Answer 2

Write $\Phi(x)=y$ and solve $y=\frac{1}{2}\left[1+\operatorname{erf}\left(\frac{x}{\sqrt{2}}\right)\right]$ for $x$, getting $$x=\sqrt2\operatorname{erf}^{-1}(2y-1).$$ Hence $$\Phi^{-1}(y)=\sqrt2\operatorname{erf}^{-1}(2y-1).$$