# Calculating the distribution of maximal value of $n$ draws from a normal distribution

According to normal probability distribution theory which says that for $n$ independent, identically distributed, standard, normal, random variables $\xi_j$ the expected absolute maximum is

$E(\max|\xi_j|)=\sqrt{2 \ln n}$

Regarding this, why do we need to multiple the above-mentioned estimate by $\sigma$ (Standard Deviation) in order to derive the expected absolute maximum for a normal, random variable with zero mean?

-
@K-1, I assume that you asking about the maximum of absolute values of a similar sequence of random variables $\eta_j$ the difference being that $\xi_j\sim N(0,1)$ and $\eta_j\sim N(0,1)$? If so please correct the question. Now it asks about the absolute maximum of a single normal random variable. The answer to that question is $\infty$. –  mpiktas May 6 '11 at 9:49
I don't think that is an equality. I think that is an asymptotic equivalence. It definitely doesn't hold for $n = 1$. –  cardinal May 6 '11 at 12:12
It seems that something like $E(\max_j |\xi_j|) = \sqrt{2 \ln n} + C_n$ is correct with $C_n \sim -.3...$ as $n \to \infty$. –  Hans Engler May 6 '11 at 13:45
@Hans, Hmm. Can you explain? Leting $(X_n)$ be a sequence of iid standard normals and $b_n = \sqrt{2\log n}$, it is well-known that $b_n(M_n - b_n) \to G$ in distribution where $M_n = \max_{1\leq k \leq n}X_k$ and $G$ is a standard Gumbel random variable. Arguments from extreme-value theory indicate that a dominating random variable exists, so that $\mathbb{E} b_n(M_n-b_n) \to \mathbb{E} G = \gamma \approx 0.57721$. This would seem to contradict your claim. But, perhaps I'm missing something. Maybe the modulus affects things, though this seems doubtful in light of simple symmetry arguments. –  cardinal May 6 '11 at 18:05
@cardinal - Just a numerical observation with $n \le 2000$. In that range it appears that $E(M_n) \sim \sqrt{2 \log n} - .27...$, but of course that gap could be closing. –  Hans Engler May 7 '11 at 12:35

If $\zeta_j = \sigma \xi_j$ for some $\sigma >0$ and some $\mu$ then

$$E[\max|\zeta_j|] = E[\max|\sigma \xi_j |] = E[\sigma \max| \xi_j|]= \sigma E[ \max| \xi_j|]$$

and this tells us how to move from a standard normal with mean $0$ and standard deviation $1$ to a normal distribution with mean $0$ and standard deviation $\sigma$.

-
The modulus got dropped in your answer. Including it destroys the linearity, but not the scale equivariance. –  cardinal May 6 '11 at 12:23
@cardinal: Thanks - it was only half visible in the question, so I read it as a conditional sign. –  Henry May 6 '11 at 13:48
(+1) No problem. –  cardinal May 6 '11 at 18:00