These are basic rules of distribution
- the total area beneath a probability density function equal to 1.0
- the probability density function of the standard normal is equal $\displaystyle \frac{\sqrt{2} e^{- \frac{x^{2}}{2}}}{2 \sqrt{\pi}}$
Based on the facts above, is it reasonable to say the integral of pdf of the standard normal is equal to 1? What is the process of derivation?
I am trying to deduct the process. This code
phi = 1/sqrt(2*pi)*exp(-(x**2)/2)
integrate(phi, x)
outputs
$\displaystyle \frac{\operatorname{erf}{\left(\frac{\sqrt{2} x}{2} \right)}}{2}$
How can I get this result and connect it to rule_1?
The Gauss error function is defined as
${\displaystyle {\begin{aligned}\operatorname {erf} (x)&={\frac {1}{\sqrt {\pi }}}\int _{-x}^{x}e^{-t^{2}}\,dt\\[5pt]&={\frac {2}{\sqrt {\pi }}}\int _{0}^{x}e^{-t^{2}}\,dt.\end{aligned}}}$
plugin $\displaystyle \frac{\sqrt{2} x}{2}$
I got
${\displaystyle {\begin{aligned}{\frac {2}{\sqrt {\pi }}}\int _{0}^{\frac{\sqrt{2} x}{2}}e^{-t^{2}}\,dt.\end{aligned}}}$
what is the next?