Die Gaußsche Fehlerfunktion

\text{Als Gaußsche Fehlerfunktion bezeichnet man das folgende Integral:}\

\text{     }\scalebox{1.1}\bold{erf(x) = \frac{2}{\sqrt{\pi}}\int_{0}^{x}{e^{-\tau}^2 }d\tau}  \text{      diese Funktion hat keine explizite Darstellung!}\
\text{Wie man damit die Verteilungsfunktion } F_{(\mu,\sigma)}(x) \text{ für eine beliebige N(\mu,\sigma^2)-Normalverteilung berechnet, }\
\text{soll hier hergeleitet werden:}\  \;  \
\text{Eine N(\mu,\sigma)-Normalverteilung hat folgende Dichtefunktion:}\
\scalebox{1.1}\bold{   φ_{(\mu,\sigma)}(x)\; = \frac{1}{\sqrt{2\pi\sigma^2}} \cdot e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}}\
\int_{a}^{x}{φ_{(\mu,\sigma)}(\tau)d\tau =  \int_{a}^{x}{\frac{1}{\sqrt{2\pi\sigma^2}} \cdot e^{-\frac{1}{2}\left(\frac{\tau-\mu}{\sigma}\right)^2}}d\tau  }
=\frac{1}{\sqrt{\pi}}  \int_{a}^{x}{\frac{1}{\sigma \sqrt{2 }} \cdot e^{-  \left(\frac{\tau-\mu}{\sigma\sqrt{2 }}\right)^2}}d\tau  
=\frac{1}{2}\left(   \frac{2}{\sqrt{\pi}}\int_{a}^{x}{e^{-\left(\frac{\tau-\mu}{\sigma\sqrt{2 }}\right)^2}}\cdot\frac{1}{\sigma \sqrt{2 }} \; d\tau   \right)\
\text{              Substitution:    u(\tau)=\frac{\tau-\pi}{\sqrt{2}·\sigma}    =>u'(\tau)=\frac{1}{\sqrt{2}·\sigma}    ⇒\
\text{  =}\frac{1}{2}\left(   \frac{2}{\sqrt{\pi}}\int_{a}^{x}{e^{-  u(\tau)^2}}· u'(\tau) \;  d\tau   \right) 
=\frac{1}{2}\left(   \frac{2}{\sqrt{\pi}}\int_{u(a)}^{u(x)}{e^{-  u^2}} \;  du  \right) 
=\frac{1}{2}\left(   \frac{2}{\sqrt{\pi}}\int_{u(a)}^{0}{e^{-  u^2}}  \;  du + \frac{2}{\sqrt{\pi}} \int_{0}^{u(x)}{e^{-  u^2}}  \;  du  \right)  \
\text{  =}\frac{1}{2}\left( + \frac{2}{\sqrt{\pi}} \int_{0}^{u(x)}{e^{-  u^2}}  \;  du -  \frac{2}{\sqrt{\pi}}\int_{0}^{u(a)}{e^{-  u^2}}  \;  du   \right) 
=\frac{1}{2}\left(   erf(u(x)) -   erf(u(a)) \right) 
=\frac{1}{2}\left(  erf\left(    \frac{x-\pi}{\sqrt{2}·\sigma}\right)-erf\left(    \frac{a-\pi}{\sqrt{2}·\sigma}\right) \right)} \
F_{(\mu,\sigma)}(x)= \int_{-∞}^{x}{φ_{(\mu,\sigma)}(\tau)d\tau = \lim_{ a \to \infty }\int_{a}^{x}{φ_{(\mu,\sigma)}(\tau)d\tau =  \lim_{ a \to \infty }\int_{a}^{x}{\frac{1}{\sqrt{2\pi\sigma^2}} \cdot e^{-\frac{1}{2}\left(\frac{\tau-\mu}{\sigma}\right)^2}}d\tau  }
= \lim_{ a \to \infty }\frac{1}{2}\left(  erf\left(    \frac{x-\pi}{\sqrt{2}·\sigma}\right)-erf\left(    \frac{a-\pi}{\sqrt{2}·\sigma}\right) \right)}\
 =\frac{1}{2}\left(   erf\left(    \frac{x-\pi}{\sqrt{2}·\sigma}\right)- (-1) \right) =\frac{1}{2}\left(   erf\left(    \frac{x-\pi}{\sqrt{2}·\sigma}\right)+1\right)\text{ speziell für die N(0,1) Standard-Normalverteilung folgt:}\
\phi_{(1,0}(x)= \int_{-∞}^{x}{φ_{(0,1)}(\tau)d\tau =\frac{1}{2}\left(   erf  \left(    \frac{x}{\sqrt{2}}\right)+1\right)

$\bold\scalebox{1.2}{erf(x)}$

$\bold\scalebox{1.2}{F_{(\mu,\sigma)}}$

Werner Knobloch, 2015

$\bold \scalebox{1.2}{μ}$

f(x) = (2 / sqrt(π) * ℯ^((-x^(2))))
g(x) = erf(x)
Herleitung = false
μ = 4 (slider from -5 to 5, step 0.1)
σ = 0.6 (slider from 0.1 to 2, step 0.1)
φ_{(μ,σ)}(x) = (1 / ((σ * sqrt((2 * π)))) * ℯ^(((-(x - μ)^(2))) / ((2 * σ^(2)))))
F_{(\mu,\sigma)}(x) = (1 / 2 * (erf((x - μ) / ((sqrt(2) * σ))) + 1))
d = Segment((μ - (2 * σ), 0), (μ + (2 * σ), 0))
A = Point(d)
B = Point(d)
a: x(A)
b: x(B)
c = Integral(φ_{(μ,σ)}, a, b)
F_a: F_{(\mu,\sigma)}(a)
F_b: F_{(\mu,\sigma)}(b)
Text6: "" + (LaTeX[LaTeX[c, true, false]]) + " =    " + (LaTeX[LaTeX[c, false, false]]) + "=\int_{a}^{b}{" + (LaTeX[LaTeX[φ_{(μ,σ)}]]) + " }=F(b)-F(a)=" + (LaTeX[F_b]) + "-" + (LaTeX[F_a]) + "=" + (LaTeX[F_b - F_a]) + ""
Text5: "\textcolor{blue}{" + (LaTeX[LaTeX[f, true, true]]) + "}"
Text4: "\textcolor{red}{" + (LaTeX[LaTeX[g, true, true]]) + "= \frac{2}{\sqrt{\pi}}\int_{0}^{x}{e^{-\tau}^2 }d\tau }"
Text3: "\textcolor{#7f00ff}{" + (LaTeX[LaTeX[φ_{(μ,σ)}, true, true]]) + "}"
Text2: " \textcolor{#009900}{" + (LaTeX[LaTeX[F_{(\mu,\sigma)}, true, true]]) + "}"
Text7 = TableText({{Text5, Text4, Text3, Text2}}, "v")
e = Segment((μ, (-0.08)), (μ, φ_{(μ,σ)}(μ) + 0.08))