Cdf of standard normal

Question

Evaluate,

$$\dfrac{1}{\sqrt{2\,\pi}}\int_{-\infty}^{0.5365}e^{-x^2/2}\,dx$$

In other words, how do I find $N (0.5365)$, where $N(x)$ denotes the $cdf$ of the standard normal random variable?

Answer 1

You can make a series expansion for $e^{-x^2/2}$ at $x=0$:

$$e^{-x^2/2}=1-\frac{x^2}{2}+\frac{x^4}{8}-\frac{x^6}{48}+\frac{x^6}{484}-\ldots$$

We want the integral

$$\int_0^x e^{-t^2/2} \,dt=\frac{x}{\color{blue}1\cdot 1}-\frac{x^3}{\color{blue}2\cdot 3}+\frac{x^5}{\color{blue}8\cdot 5}-\frac{x^7}{\color{blue}{48}\cdot 7}+\frac{x^9}{\color{blue}{384}\cdot 9}-\ldots$$

Without the blue factors the series is

$$\sum_{k=0}^{\infty} (-1)^k\frac{x^{2k+1}}{2k+1}$$

A formula for the sequence $1,2,8,48,484,..$ can be found using OEIS. The result is $2^k\cdot k!$. Therefore $$\int_0^x e^{-t^2/2} \,dt=\sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{(2k+1)\cdot 2^k\cdot k!}$$

And finally

$$\frac1{\sqrt{2\pi}}\int_{-\infty}^x e^{-t^2/2} \,dt=\frac12+\frac1{\sqrt{2\pi}}\cdot \sum_{k=0}^{\infty}(-1)^k\frac{x^{2k+1}}{(2k+1)\cdot 2^k\cdot k!}$$

If $x=0.5365$, already for $k=3$ one gets a good approximation since the approximated value is 0.704193, which is identical to the result of the online calculator.

Answer 2

If you look at the definition of the error function, you will easily understand that $$I(a)=\dfrac{1}{\sqrt{2\,\pi}}\int_{-\infty}^{a}e^{-x^2/2}\,dx=\frac{1}{2} \left(1+\text{erf}\left(\frac{a}{\sqrt{2}}\right)\right)$$ The Wikipedia page also gives the Taylor expansion of it $$\operatorname{erf}(z)= \frac{2}{\sqrt{\pi}}\sum_{n=0}^\infty\frac{(-1)^n z^{2n+1}}{n! (2n+1)} =\frac{2}{\sqrt{\pi}} \left(z-\frac{z^3}{3}+\frac{z^5}{10}-\frac{z^7}{42}+\frac{z^9}{216}-\ \cdots\right)$$ Using it, you then have $$I(a)=\frac{1}{2}+\frac{a}{\sqrt{2 \pi }}-\frac{a^3}{6 \sqrt{2 \pi }}+\frac{a^5}{40 \sqrt{2 \pi }}-\frac{a^7}{336 \sqrt{2 \pi }}+\frac{a^9}{3456 \sqrt{2 \pi }}+O\left(a^{11}\right)$$ Using it for different values of $n$ and $a=0.5365$, you would get $$\left( \begin{array}{cc} n & I(0.5365) \\ 0 & 0.7140325334 \\ 1 & 0.7037649558 \\ 2 & 0.7042082568 \\ 3 & 0.7041930668 \\ 4 & 0.7041934919 \\ 5 & 0.7041934818 \\ 6 & 0.7041934820 \end{array} \right)$$ If you prefer not to use series expansion, look at this post where are proposed two rather simple approximations $$\mathrm{erf}\!\left(X\right)\approx \sqrt{1-\exp\Big(-\frac {4X^2} {\pi} \Big)}\tag 1$$ $$\mathrm{erf}\!\left(X\right)\approx\sqrt{1-\exp\Big(-\frac 4 {\pi}\,\frac{1+\alpha\, X^2}{1+\beta\, X^2}\,X^2 \Big)}\tag 2$$ where $$\alpha=\frac{10-\pi ^2}{5 (\pi -3) \pi }\qquad \text{and}\qquad \beta=\frac{120-60 \pi +7 \pi ^2}{15 (\pi -3) \pi }$$ Approximation $(1)$ would give $0.704592$ and approximation $(2)$ would give $0.704193$.

Answer 3

I use Simpson's Method to compute such function. There is no analytic exact solution to this integral.