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I found here that

The cdf for the max is the cdf for the normal raised to the power of the sample size. [say, $n$]

Since the cdf for the normal is the error function,

$$F_X(x)=\frac{1}{\sigma\sqrt{ 2\pi}} \int_{-\infty}^x \exp{\left(\frac{-(t -\mu)^2}{2\sigma^2} \right)}\mathrm dt=\frac{1}{2}\Bigg[1+ \mathrm{erf} \left( \frac{x-\mu}{\sigma \sqrt 2}\right)\Bigg],$$

and the pdf of the standard normal is

$$f(x) = \frac{e^{-x^{2}/2}} {\sqrt{2\pi}}$$

I presume there is no other workaround to

$$f_{\max(\mathrm X)}(x_i)=n \quad \frac{e^{-x^{2}/2}} {\sqrt{2\pi}}\quad \frac{1}{2}\Bigg[1+ \mathrm{erf} \left( \frac{x-\mu}{\sigma \sqrt 2}\right)\Bigg]^{n-1}$$

Would this be correct? It doesn't get any simpler?

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  • $\begingroup$ @GordonSmyth Corrected now. $\endgroup$ – Antoni Parellada Feb 4 '18 at 21:49
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Yes it is. Some related ideas/references/bounds are disused in

https://math.stackexchange.com/questions/89030/expectation-of-the-maximum-of-gaussian-random-variables

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