# Why is the domain of the error function scaled by $\sqrt{2}$

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 $\operatorname{erf}(z)$??

• I suspect that it's because those who adopted that convention were not probabilists. $\qquad$ – Michael Hardy Jun 22 '16 at 3:33

The error function is defined by $$\operatorname{erf}(z)=\frac{2}{\sqrt{\pi}}\int_0^z e^{-t^2}\ dt\tag1$$ and $$\Phi (z)=\frac{1}{\sqrt{2\pi}}\int_0^z e^{-x^2/2}\ dx\tag2$$ 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}
• I get the conversion. The question is why $erf$ is defined that way. Does this definition provide an elegant solution to some known problem? – ThomasMcLeod Jun 22 '16 at 3:55