# If $X_1, \ldots, X_n \sim N(0,1)$, then $E(\max_{1 \leq i \leq n}X_i) = O(\sqrt{\log\ n})$. What does this mean for prediction of extreme events?

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.

• The maximum, although it naturally tends to grow with the number of samples, grows quite slow compared to, say, a power law. – Ian Sep 17 '16 at 3:15
• Are the random variables $X_i$ independent? – i707107 Sep 17 '16 at 3:26
• Hi, yes the random variables are independent – user321627 Sep 17 '16 at 3:28

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.