5

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
  • 505
  • @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 06 '11 at 09:49
  • 5
    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 06 '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 06 '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 06 '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 07 '11 at 12:35
  • @Hans, I think it is closing. The convergence in distribution is slow. Also, doing a proper simulation in this case is pretty hard since we're looking at the maximum modulus. So, algorithmic nonidealities in the random number generation can come into play. For example, though nobody uses the inversion method for normal random variates, if you did, due to asymmetries in IEEE floating point, the smallest possible normal random variate you can generate is about -37.52, while the largest is only 8.21. – cardinal May 07 '11 at 13:08
  • @cardinal - I actually computed the expected values of the maxima as numerical integrals, using Mathematica. But of course there could be other systematic errors in the integration routines. – Hans Engler May 13 '11 at 12:20

2 Answers2

5

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 $\sigma$.

Community
  • 1
Henry
  • 39,459
  • The modulus got dropped in your answer. Including it destroys the linearity, but not the scale equivariance. – cardinal May 06 '11 at 12:23
  • @cardinal: Thanks - it was only half visible in the question, so I read it as a conditional sign. – Henry May 06 '11 at 13:48
2

Intuitively: values of the standard normal distribution (including the absolute maximum) 'tend to be' 1 SD = 1 away from 0.

In non-standard zero-mean normal, the data 'tend to be' 1 SD = sigma away from 0.

You could say that as long as you're doing linear stuff, all distances to zero will blow up by a factor sigma.

Nick Sabbe
  • 12,819
  • 2
  • 37
  • 47
  • so it means that we need to correct the distance estimated from standard normal distribution hypotheses for non-standard data? – K-1 May 06 '11 at 07:32