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If $X_1, \ldots, X_n \sim N(0,1)$, which each of them independent, then a common result from probability theory is that:

$$ E\left(\max_{1 \leq i \leq n}X_i\right) = O\left(\sqrt{\log\ n}\right) $$

The notation on the left is big-Oh notation. Now, I read in a paper that this signifies that while the normal distribution can be good for fitting data, it is hard for the normal distribution to predict extreme events. However, I am not sure why the above result necessarily shows this. Is it because the term $\sqrt{\log\ n}$ increases very slowly and hence if we have a very large $X_i$, we would need a large number of samples before we can predict (via the expectation) largely? Thanks.

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  • $\begingroup$ The maximum, although it naturally tends to grow with the number of samples, grows quite slow compared to, say, a power law. $\endgroup$ – Ian Sep 17 '16 at 3:15
  • $\begingroup$ Are the random variables $X_i$ independent? $\endgroup$ – i707107 Sep 17 '16 at 3:26
  • $\begingroup$ Hi, yes the random variables are independent $\endgroup$ – user321627 Sep 17 '16 at 3:28
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A random variable from $N(0,1)$ has the expectation $0$. But, for large values of $n$, $\max_{1 \leq i \leq n}X_i$ (maximum of normally distributed rv's) goes to infinity (in expectation) as your theory says. This means that the normal distribution is not good at extreme events.

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