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**Extra resources for Direct interaction approximation, the statistically stationary problem**

**Example text**

In particular, the moment of order 0 always exists (that, due to the normalization condition, is α0 = 1) and those of even order, if exist, are non-negative. A specially important moment is that order 1: the mean (mean value) μ = E[X ] that has two important properties: n n • It is a linear operator since X = c0 + i=1 ci X i −→ E[X ] = c0 + i=1 ci E[X i ] n n • If X = i=1 ci X i with {X i }i=1 independent random quantities, then E[X ] = n i=1 ci E[X i ]. We can define as well the moments (βn ) with respect to any point c ∈ R as: de f.

Tn ) = ei(t1 x1 + · · · + tn xn ) P(X 1 = x1 , . . , X n = xn ) ... 5 Integral Transforms 47 and for the continuous case: (t1 , . . , tn ) = +∞ −∞ d x1 . . +∞ −∞ d xn ei(t1 x1 + · · · + tn xn ) p(x1 , . . , xn ) The n-dimensional Characteristic Function is such that: (1) (0, . . , 0) = 1 (2) | (t1 , . . , tn )| ≤ 1 (3) (−t1 , . . , −tn ) = (t1 , . . , tn ) Laplace Transform: For a function f (x) : R+ →C defined as f (x) = 0 for x < 0, we may consider also the Laplace Transform defined as ∞ L(s) = e−sx f (x) d x 0 with s ∈ C provided it exists.

The uniqueness of the result holds with respect to this strip so, in fact, the Mellin Transform consists on the pair M(s) together with the band a, b . 23 It is clear that to determine the function f (x) from the transform F(s) we have to specify the strip of analyticity for, otherwise, we do not know which poles should be included. Let’s see as an example f 1 (x) = e−x . We have that ∞ M1 (z) = e−x x z−1 d x = (z) 0 holomorphic in the band 0, ∞ so, for the inverse transform, we shall include the poles z = 0, −1, −2, .