Compute inverse cdf of normal distribution [closed]

Question

How can I compute inverse CDF of normal distribution using the central limit theorem on uniform distribution (u[0,1])

Answer 1

Let $S_n:=U_1+\cdots+U_n$, where $U_1,U_2,\ldots$ are iid random variables (r.v.'s) each uniformly distributed on $[0,1]$. Then, by the central limit theorem, \begin{equation} Z_n:=\frac{S_n-n/2}{\sqrt{n/12}}\to Z \end{equation} in distribution as $n\to\infty$, where $Z$ is a standard normal r.v. So, for each $p\in(0,1)$, \begin{equation} q_n(p)\to q(p) \end{equation} as $n\to\infty$, where $q_n(p)$ and $q(p)$ are the $p$-quantiles of (the distributions of) $Z_n$ and $Z$, respectively, so that $q$ is the function inverse to the standard normal cdf.

Note that \begin{equation} q_n(p)=\frac{X_n(p)-n/2}{\sqrt{n/12}}, \end{equation} where $X_n(p)$ is the $p$-quantile of $S_n$, so that $X_n$ is the function inverse to the cdf of $S_n$, which let us denote by $F_n$. By the Irwin–Hall formula, $F_n$ is piecewise polynomial; more specifically, \begin{equation} k\in\{1,\dots,n\}\ \&\ k-1\le x<k\implies F_n(x)=\frac1{n!}\sum_{j=0}^{k-1}(-1)^j\binom nj (x-j)^n. \end{equation} So, the quantile/inverse functions $X_n$ and $q_n$ are piecewise algebraic.

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Below is an image of a Mathematica notebook containing graphs of $q_2$ (red), $q_5$ (green), $q_{10}$ (blue), and $q$ (black). We see that $q_2$ already approximates $q$ well except near $0$ and $1$, with $q_5$ and $q_{10}$ significantly closer than $q_2$ to $q$ near $1$ (of course, the graphs of $q_n$ and $q$ are centrally symmetric about the point $(1/2,0)$).