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How to prove

A)$$F(\mu+n \sigma)-F(\mu-n \sigma)=\Phi(n)-\Phi(-n)=\operatorname{erf}\left(\frac{n}{\sqrt{2}}\right)$$

B) $$F(x)=\Phi\left(\frac{x-\mu}{\sigma}\right)=\frac{1}{2}\left[1+\operatorname{erf}\left(\frac{x-\mu}{\sigma \sqrt{2}}\right)\right]$$

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:

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

The following neglects the term "$\dfrac{x}{\sqrt{2}}$" so it proves nothing:

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 $\text{erf}(x)$) instead of $erf(x/\sqrt(2))$

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1 Answer 1

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The definition of the error function is a slightly different integral from the CDF of a standard normal: $$\operatorname{erf}(x) = \frac2{\sqrt\pi} \int_{t=0}^x e^{-t^2}\,\mathrm dt.$$ Inside this integral, we can make the substitution $u = t \sqrt 2$, so that in the exponent we get $t^2 = (u/\sqrt 2)^2 = u^2/2$. But $\mathrm du = \sqrt2 \,\mathrm dt$, so when we make this substitution, we get $$ \operatorname{erf}(x) = \frac{2}{\sqrt{2\pi}} \int_{u=0}^{x \sqrt 2} e^{-u^2/2}\,\mathrm du. $$ To get from here to the standard normal CDF, we replace $x$ by $x/\sqrt2$ (to make the upper bound on the integral correct), and divide by $2$: $$ \frac12 \operatorname{erf}(x/\sqrt 2) = \frac1{\sqrt{2\pi}} \int_{u=0}^x e^{-u^2/2}\,\mathrm du = N(x) - N(0). $$ Since $N(0) = \frac12$ by symmetry, we get $$ N(x) = N(0) + \frac12 \operatorname{erf}(x/\sqrt2) = \frac12 + \frac12 \operatorname{erf}(x/\sqrt2). $$

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