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Is there an asymptotic or approximate expression for the expectation (maybe even the variance, but my main question is about the mean) of the n-th order statistic, i.e. the maximum, of n independent, identically distributed Gaussian random variables from $\mathcal{N}(\mu,\sigma)$ where n is large?

I have tried plotting the estimate of Blom from here, but for very large quantities ($>2^{20}$) this seems to become overoptimistic.

This question was seemingly answered here, however plotting these formulas they seem to be for a standard normal distribution, so for $\mathcal{N}(0,1)$.

I however have a (very large) sample from $\mathcal{N}(\mu,\sigma)$. How do the formulas of the max-central limit theorem change in this case? For the the mean $\mu$ the mean of the probably just shifts by the same quantity. What happens with the $\sigma$?

Christine Van Vredendaal
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    The mean and standard deviation are location and scale parameters, and have the same effect on the the distributions of the order statistics as they do on the underlying random variable (i.e. multiply by $\sigma$ and then add $\mu$) – Henry Mar 18 '20 at 13:38
  • So what would that mean for the answer of here? That $\mu_n' = \mu_n + \mu$ and $\sigma_n' = \sigma_n*\sigma$? And $m_n$ should then equal $\mu_n' + \gamma\sigma_n'$, as is the average of an extreme value distribution? Or is something more complicated happening? And where does the $\sqrt(2)$ come from in their case? – Christine Van Vredendaal Mar 20 '20 at 13:30
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    I am saying use $\mu+\sigma \sqrt{\log \left(\frac{n^2}{2 \pi \log \left(\frac{n^2}{2\pi} \right)}\right)} \cdot \left(1 + \frac{\gamma}{\log (n)} + \mathcal{o} \left(\frac{1}{\log (n)} \right) \right)$ – Henry Mar 20 '20 at 14:09

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