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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 $\text{erf}(x)$)

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  • $\begingroup$ Write down the definition of $\text{erf}$. $\endgroup$
    – user65203
    Apr 17, 2016 at 16:59
  • $\begingroup$ $\text{erf} = \frac{2}{\sqrt{\pi}} \int_0^x e^{-\phi^2/2}d\phi$ $\endgroup$
    – Naz
    Apr 17, 2016 at 17:13
  • $\begingroup$ the above is for $\text{erf}(x)$ $\endgroup$
    – Naz
    Apr 17, 2016 at 17:20
  • $\begingroup$ 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. $\endgroup$ Apr 17, 2016 at 21:01

1 Answer 1

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The definition of an error function is:

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

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