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I have just started reading about the Error function but Distribution theory is not my strong point. So I apologize in advance for asking really simple questions about it.

So the Error function is defined to be $$\displaystyle \mathrm{erf}(x)=\frac{2}{\sqrt{\pi}}\int_{y=0}^{y=x}e^{-y^2}\mathrm{d}y$$

I have learnt about the Normal Distribution function already which is defined to be $$\displaystyle \Phi(z)=\frac{1}{\sqrt{2\pi}}\int_{z=-\infty}^{z=\infty}e^{-\frac{1}{2}z^2}\mathrm{d}z$$ where $z=\cfrac{x-\mu}{\sigma}$ and that it's graph is

Normal Distribution

I understand completely that $\Phi(z=-1.14)\approx 0.1271$ is the area under the graph in the left tail shaded $\color{blue}{\mathrm{blue}}$.

But for the Error function graph

Error function

if $\mathrm{erf}\left(x=\frac{1}{\sqrt{2}}\right)=0.682689492137086$, what does this value even mean? Is this the value of the area under the curve for $-\infty \le x \le \frac{1}{\sqrt{2}}$?

And finally why do we need the Error function in the first place if we have a Normal distribution function?

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    $\begingroup$ When you integrate, the dummy variable goes away. Your definition of $\Phi$ is wrong. It should be $$ \Phi(z) = \frac 1{\sqrt{2\pi}}\int_{-\infty}^z e^{-\frac 12x^2} dx $$ $\endgroup$ – Tunococ Nov 3 '15 at 1:55
  • $\begingroup$ @Tunococ Thanks that's what I meant it to be, I'll change it now. Any ideas on the rest of the questions? $\endgroup$ – BLAZE Nov 3 '15 at 2:13
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$$\Phi(x)=\frac{1}{2}+\frac{1}{2}\mathrm{erf}(x/\sqrt{2})$$

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  • $\begingroup$ Thanks for your answer, it was helpful to see how they were related. Any idea what $\mathrm{erf}\left(\frac{1}{\sqrt{2}}\right)=0.682689492137086$ means? Is it an area? $\endgroup$ – BLAZE Nov 3 '15 at 2:57
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    $\begingroup$ that is the value of the error function evaluated at $x=1/\sqrt{2}$. It can however be interpreted in a different way if you consider the formulate I gave. Namely it is $2\Phi(1)-1$, which is the area of normal density from $-\infty$ to $1$ multiplied by $2$ and the subtracted by $1$ $\endgroup$ – Seyhmus Güngören Nov 3 '15 at 3:07
  • $\begingroup$ That's great, just one more question: Why do we need the error function if we have the Gaussian? $\endgroup$ – BLAZE Nov 3 '15 at 3:29
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    $\begingroup$ @BLAZE they are just some standardized way of calculating the integral of Gaussian functions. Many computer programs give results in terms of error functions, not in terms of Gaussian c.d.f.s. Other usage of error function comes from applications. For example in wireless communication error function is related to the bit error rates. There are also other areas within mathematics where one sees the error function. Wouldnt it be possible to do everything over Gaussian distributions? Yes. But, it seems without knowing Gaussian distributions or without any need for it, error function is in use. $\endgroup$ – Seyhmus Güngören Nov 3 '15 at 3:39
  • $\begingroup$ Thank you very much for your excellent answer. Sorry for not accepting it sooner, I always leave it open for a while to see what other users say. Best regards. $\endgroup$ – BLAZE Nov 5 '15 at 0:45

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