# Use of escort distribution in nonextensive stat. mech

In some of the articles which I read recently, I happen to see the following statement.

In Nonextensive statistical physics, it is inappropriate to use the original distribution $P=(p_i)$ instead of the escort distribution $P'=(p_i')$ where $p_i'=\frac{p_i^q}{\sum_j p_j^q}$ in calculating the expectation values of physical quantities.

Can someone clarify this point?

-
If you use the original distribution, you just won't get nonextensive stat phys. It's just a recipe to generate nonextensiveness. –  Raskolnikov Sep 21 '11 at 11:27
Can you point to the articles? I am curious about this now. –  Ron Maimon Sep 22 '11 at 6:21
For example sciencedirect.com/science/article/pii/S0378437198004373 and pre.aps.org/abstract/PRE/v79/i4/e041116. Chapter 9 of the book Thermodynamics of Chaotic Systems: An Introduction by C.Beck and F.Schlogl is on the escort distributions. –  Ashok Sep 22 '11 at 10:29

I don't know if this helps, but the escort distribution has a simple interpretation: you have q replicas of the system, and you are selecting to view only those instances when all q replicas have exactly the same state. This situation occurs with probability $p_i^q$, so the formula follows.