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?
[Math] Cdf of standard normal
normal distribution
normal distribution
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?
Best Answer
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.