# Error function relation to the normal cumulative distribution function

A CDF for a normal standard is the following:

$$N(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-\phi^2/2} d\phi$$

I have the following relation in my notes which I am not very sure how they arrived at:

$$N(x) = \frac{1}{2}+\frac{1}{2}\text{erf}(x/\sqrt{2})$$

Not sure how they can have $x$ value in $N(\cdot)$ and $x/\sqrt{2}$ in $\text{erf}(\cdot)$. Here is what I got:

$$N(x) = \frac{1}{\sqrt{2\pi}}\left( \int_{-\infty}^0 e^{-x^2/2} dx + \int_0^x e^{-\phi^2/2} d\phi \right)$$

I reasoned that since $$\int_{-\infty}^{\infty}e^{-x^2/2}dx=\sqrt{2\pi}$$

then

$$\int_{-\infty}^0 e^{-x^2/2} dx = \frac{\sqrt{2\pi}}{2}$$

Therefore:

$$N(x) = \frac{1}{\sqrt{2\pi}}\left( \frac{\sqrt{2\pi}}{2} + \frac{\sqrt{\pi}}{2}\text{erf}(x) \right)$$

Which is:

$$N(x) = \frac{1}{2} + \frac{1}{2\sqrt{2}}\text{erf}(x)$$

And not quite what I have in notes (due to $\text{erf}(x)$)

• Write down the definition of $\text{erf}$.
– user65203
Apr 17, 2016 at 16:59
• $\text{erf} = \frac{2}{\sqrt{\pi}} \int_0^x e^{-\phi^2/2}d\phi$
– nz_
Apr 17, 2016 at 17:13
• the above is for $\text{erf}(x)$
– nz_
Apr 17, 2016 at 17:20
• Your definition of $\text{erf}$ above is not the standard one. The standard one is $\text{erf} = \frac{2}{\sqrt{\pi}} \int_0^x e^{-\phi^2}d\phi$ (without the $1/2$ in the exponent). Sources: en.wikipedia.org/wiki/Error_function and mathworld.wolfram.com/Erf.html ... try to redo your calculations with this definition, and the disagreement should disappear. Apr 17, 2016 at 21:01

$$\text{erf}(x) = \frac{\sqrt{2}}{\pi}\int_{0}^x e^{-s^2}ds$$