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The normal distribution function $\Phi(z)$ has the definition $\Phi(z) \equiv \frac{1}{\sqrt{2 \pi}} \int_0^z e^{\frac{-x^2}{2}} \, dx$.

However the error function is conventionally defined such that $\frac{1}{2}\operatorname{erf}(\frac{z}{\sqrt{2}})\equiv\Phi(z)$.

I understand the $\frac{1}{2}$ scaling on the error function itself, since the $\Phi$ integral is symmetrical around $0$. But why scale $z$ in $\operatorname{erf}(z)$??

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    $\begingroup$ I suspect that it's because those who adopted that convention were not probabilists. $\qquad$ $\endgroup$ – Michael Hardy Jun 22 '16 at 3:33
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The error function is defined by \begin{equation} \operatorname{erf}(z)=\frac{2}{\sqrt{\pi}}\int_0^z e^{-t^2}\ dt\tag1 \end{equation} and \begin{equation} \Phi (z)=\frac{1}{\sqrt{2\pi}}\int_0^z e^{-x^2/2}\ dx\tag2 \end{equation} Now, setting $x=t\sqrt{2}$ to the equation $(2)$ then the domain $0<x<z$ changes to $0<t<\frac{z}{\sqrt{2}}$ \begin{align} \Phi (z)&=\frac{1}{\sqrt{2\pi}}\int_0^{{z}/{\sqrt{2}}} e^{-t^2}\cdot\sqrt{2}\ dt\\[10pt] &=\frac{1}{\sqrt{\pi}}\int_0^{{z}/{\sqrt{2}}} e^{-t^2}\ dt\\[10pt] &=\frac{1}{2}\operatorname{erf}\left(\!\frac{z}{\sqrt{2}}\!\right) \end{align}

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  • $\begingroup$ I get the conversion. The question is why $erf$ is defined that way. Does this definition provide an elegant solution to some known problem? $\endgroup$ – ThomasMcLeod Jun 22 '16 at 3:55
  • $\begingroup$ @ThomasMcLeod My bad, I didn't read the question perfectly. I don't really know why but maybe this definition is more to the definer's taste. Anyway, your title is misleading so my answer isn't entirely wrong. $\endgroup$ – Sophie Agnesi Jun 22 '16 at 4:01

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