Lagrange inversion theorem for the normal cumulative distribution

Question

The Lagrangian inversion theorem gives the Taylor series expansion for the inverse of a function. I was wondering how to apply it to the normal cdf: $$ f(x)=\int_{-\infty}^x g(t)dt, $$ where $g(t)=(2\pi)^{-1/2}e^{-t^2/2}$. Following the notation in https://en.wikipedia.org/wiki/Lagrange_inversion_theorem, how that applies to $f(x)$, what are $a$ and $f(a)$ in this case?

cheers.

Answer 1

You have apparently chosen to confuse yourself by reusing $g$ in a second way. Therefore, we do not use the function $g$ from the wiki page you cite.

Since $$ f'(x) = g(x) > 0 $$ and $g(x)$ is analytic everywhere, you are free to pick any $a \in \Bbb{R}$.

For instances,

• randomly choosing $a = 0$, $f(0) = 1/2$;
• randomly choosing $a = -1$, $f(-1) = \frac{1}{2} (1 - \mathrm{erf}(1/\sqrt{2}))$;
• randomly choosing $a = \pi$, $f(\pi) = \frac{1}{2} (1 + \mathrm{erf}(\pi/\sqrt{2}))$; and
• for any choice of $a$, $f(a) = \frac{1}{2}(1 +\mathrm{erf}(a/\sqrt{2}))$,

Where "erf" is the error function.

Let's apply this with $a = 0$, so $f(a) = 1/2$ and $$ f^{-1}(z) = 0 + \sum_{n-1}^\infty g_n \frac{(z - 1/2)^n}{n!} \text{.} $$ Here we find we are getting a power series centered at $1/2$. Since $f(x) \in (0,1)$, you may center your power series anywhere in that interval. (Since you want an inverse function, you cannot center it anywhere else -- the center is a point of convergence of the series and you want a series that converges on the range of $f$, so the series must be centered somewhere in the range of $f$. I picked $a = 0$ so the center would be at $1/2$, so there was some chance the series would converge over the entire range. The inverse function has singularities at $0$ and $1$, so both endpoints place constraints on the radius of convergence of the series.)

Then $$ g_n = \lim_{w \rightarrow 0} \frac{\mathrm{d}^{n-1}}{\mathrm{d}w^{n-1}}\left( \left( \frac{w - 0}{f(w) - 1/2} \right)^n \right) \text{.} $$ The first few terms of the resulting series are $$ f^{-1}(x) = \sqrt{2\pi}(x - 1/2) + \frac{\sqrt{2}\pi^{3/2}}{3}(x-1/2)^3 + \frac{7 \pi^{5/2}}{15\sqrt{2}}(x-1/2)^5 + \frac{127\pi^{7/2}}{315\sqrt{2}}(x-1/2)^7 + \frac{4369 \pi^{9/2}}{11340 \sqrt{2}}(x-1/2)^{9}+ \cdots $$

Everything in the coefficients except for the sequence $1, 1, 7, 127, 4369, 34807, \dots$ is straightforward to extract into a general term. That sequence of coefficients in the numerator is OEIS A092676. The description of that sequence gives a procedure to produce this sequence of coefficients that are part of the $g_n$ coefficients (in the "%p c:=proc(n) ..." item).