I am reading "Introduction to Econophysics" by Stanley and Mantegna and I found the following in Chapter 13.


I can't understand why the 2-Norm of a vector standardized by subtracting the mean and dividing by the SD should be 1. To me, it should be $\sqrt{n}$ where $n$ is the dimension of the vector. Let $v$ be a vector, $\overline{v}$ the average of the vector, $sd(v)$ the (population) standard deviation and $\widetilde{v}$ the standardized vector. The squared 2-norm of the standardized vector is $$ ||\widetilde{v}||^2 = \sum_{i=1}^n (\frac{v_i - \overline{v}}{sd(v)})^2 = \frac{1}{sd(v)^2} \sum_{i=1}^n (v_i - \overline{v})^2 = \frac{1}{sd(v)^2} \frac{\sum_{i=1}^n(v_i - \overline{v})^2}{n}{n} = \frac{1}{sd(v)^2} {sd(v)^2} {n} = n $$

I understand that if we apply the same reasoning to a random variable the conclusion holds. But the norm of a random variable includes the expected value (reference here): $$ ||V||_2 = \sqrt{E[V^2]} $$ $$ sd(V)=\sqrt{E[(V-E[V])^2]} = ||(V-E[V])||_2 $$ The 2-norm of a vector of data, however, does not have the $1/n$ component that is, in a sense, "embedded" in the expected value which leads to saying that the 2-norm of RV minus its mean equals the SD of the RV. The same conclusion is not possible when working with data because (interpret here $\overline{v}$ as a vector of averages of $v$) $$ sd(v) = \frac{||v-\overline{v}||}{\sqrt{n}} $$

So my questions are:

  1. Am I missing something or misinterpreting the text?
  2. How should I deal with data: should I standardize (divide by SD) or normalize (divide by 2-norm)?

Thank you for the help!

  • $\begingroup$ Yes, you are missing something: the division in (13.1) is not by the SD. It lacks the factor of $1/\sqrt n.$ $\endgroup$
    – whuber
    Commented Jun 28, 2023 at 13:18
  • $\begingroup$ @whuber, the angular brackets are time average which implies the $1/n$. Otherwise, why would the text explicitly say "divided by the SD"? $\endgroup$
    – lotak
    Commented Jun 28, 2023 at 13:22
  • $\begingroup$ That, then, comes down to exactly what your text means by the angle brackets and the SD. As far as statistical applications go (addressing your second question), when $n$ doesn't vary, the distinction is irrelevant because it's just a proportional change in values. $\endgroup$
    – whuber
    Commented Jun 28, 2023 at 13:30