Why is the maximum of i.i.d. Gaussians asymptotically $\sqrt{2 \log n}$? Assuming that $\xi$ is bounded (as a function of $x$?), the claim is that given the equation:
$$\xi \frac{\sqrt{2\pi}}{n} = \frac{1}{x} e^{-\frac{x^2}{2}} \left( 1 + O\left(\frac{1}{x^2} \right) \right)  $$
one can solve ("after some calculation") for $x$ to get:
$$x = \sqrt{2 \log n} - \frac{\log \log n + \log 4 \pi}{2 \sqrt{2 \log n}} - \frac{\log \xi}{\sqrt{2 \log n}} + O\left( \frac{1}{\log n} \right) \,. $$

Question: Would it be possible to get some hints about how to solve for $x$ in this situation? 

There are several issues about this which I don't understand:


*

*How is the assumption that $\xi$ is bounded used or otherwise relevant?

*Why is the "imprecise knowledge of $x$ transferred to $n$" when solving for $x$, and not "transferred" to some other variable? If we require an assumption on $\xi$, then why isn't the "imprecise knowledge transferred" to $\xi$? (Why don't we get an $O(f(\xi))$ term for some $f$?

*Do we have to use/calculate some inverse function to $g(x) := \frac{1}{x} e^{-\frac{x^2}{2}}$? This function isn't even defined, strictly speaking, at $x=0$, but perhaps it has a continuous extension over the entire real line? If it does have a continuous extension over $\mathbb{R}$, is that continuous extension even an invertible function, such that talking about $g^{-1}(x)$ even makes sense?

*If we do calculate such an inverse function, do we then basically proceed by applying that function to both sides of the equation and ignoring the $O(x^{-2})$ term, with the understanding that the "uncertainty" contained within it now needs to be transferred somewhere else? If so, this leads back to the above question about where the $O((\log n)^{-1})$ term could come from.


Context: I don't think the context is actually relevant to solving this problem, but for the record this comes up on p. 374 of Cramer's 1946 Mathematical Methods of Statistics where an asymptotic form for the maximum of i.i.d. Gaussian random variables is sought.
 A: You are correct, essentially this is an asymptotic expansion for the inverse of the survival function of the standard normal distribution. The first equation holds for $x \to \infty$, we are not concerned with the behavior of $n(x)$ around $x = 0$. We're looking for an expansion of $x(n)$ when $n \to \infty$, which can be obtained by applying the method of dominant balance. Taking logarithms gives
$$- \ln n + \ln(\xi \sqrt {2 \pi}) \approx
-\frac {x^2} 2 - \ln x.$$
Since $\xi$ is bounded, the dominant terms are $-\ln n$ and $-x^2/2$, from which we get the first approximation $x = \sqrt {2 \ln n}$. Then we look for the next approximation of the form $x = \sqrt{2 \ln n} \,(1 + \alpha), \,\alpha = o(1)$:
$$-\ln n + \underbrace {\ln(\xi \sqrt{2 \pi})}_{O(1)} \approx
-\underbrace {(1 + \alpha)^2 \ln n}_{(1 + 2 \alpha) \ln n + o(\alpha \ln n)} -
 \underbrace {\ln \sqrt {2 \ln n}}_{\ln (\ln n)/2 + O(1)} -
 \underbrace {\ln(1 + \alpha)}_{o(1)}, \\
0 \approx - 2 \alpha \ln n - \frac {\ln \ln n} 2,$$
which gives $\alpha = -\ln(\ln n)/(4 \ln n).$ In the same way, the next step is
$$x = \sqrt {2 \ln n} - \frac {\ln \ln n} {2 \sqrt {2 \ln n}} (1 + \alpha), \\
x^2 = 2 \ln n - (1 + \alpha) \ln \ln n + o(1),
\quad \ln x = \ln \sqrt {2 \ln n} + o(1), \\
\ln( \xi \sqrt {2 \pi}) \approx
\frac {(1 + \alpha) \ln \ln n} 2 - \ln \sqrt {2 \ln n}, \\
\alpha = \frac {\ln 4 \pi \xi^2} {\ln \ln n}.$$
See also the reference in this answer.
