the method more extensively, but let us not be scared, this development should remain relatively basic and accessible, if the reader or student has made it this far! First of all, let us see how we can find the PDF, which may obey some set of assumed constraints. Let us also make the problem the most general possible. Let go(x), gn (x) be the assumed sets of n+1 constraint functions of the variable x and their mean values and (go), (g1),...., (gn) defined by (86)= (6.23) It is implicitly assumed that (go), (g1),.... (gn) are all finite, which is an issue to be discussed later on. The first constraint go(x) = 1 is there to impose the PDF normaliza- tion condition, or (x p(x)dx = 1. Using the entropy definition in Eq. (6.1), the method consists in minimizing the "functional" f as defined as: (6.24) where are the unknown Lagrange multipliers. Minimizing f yields: (6.25) Differential entropy which yields the general, but unique, PDF solution: p(x) exp (6.26) Nicely, the solution is exclusively defined by the discrete set of parameters 20, A1,..., An, which we must now find by using the n + 1 constraints defined in Eq. (6.23). Integrating in Eq. (6.24), and equating the result with f = 0, yields the corresponding expression of maximum entropy: Hmax=1-(20+21(g1)+221 = (g2) +,..., 2n(gn)) (6.27) As previously mentioned, however, an implicit condition is that the means (go), (g1) (gn) be finite. This condition is of consequence in the possible values for the parameters 20, A1,..., An, as we see next. Assume, for instance, that our constraint functions are defined by gk(x) = xk for all k, or (gk) = (xk), which corresponds to the PDF moments of order k. The PDF solution takes then the form: p(x) = (6.28) A = exp(20 - 1). Substituting the PDF solution into the constraints in Eq. (6.23) yields: = + + =(xx) (k = 1,...,n). (6.29) For any of the integrals involved in Eq. (6.29) to converge, a certain number of additional conditions should be met, depending on the integration domain X and the number of constraints n + 1: (a) X i