By Gerhard Bohm, Günter Zech

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**Example text**

3 Moments and Characteristic Functions d2 φ = exp λ(eit − 1) dt2 d2 φ(0) = −(λ2 + λ) . dt2 (λieit )2 − λeit , Thus, the two lowest moments are µ= k =λ, µ2 = k 2 = λ2 + λ and the mean value and the standard deviation are given by k =λ, √ σ= λ. Expanding K(t) = ln φ(t) = λ(eit − 1) = λ[(it) + 1 1 (it)2 + (it)3 + · · ·] , 2! 3! we ﬁnd for the cumulants the simple result κ1 = κ2 = κ3 = · · · = λ . The calculation of the lower central moments is then trivial. For example, skewness and excess are simply given by √ 3/2 γ1 = κ3 /κ2 = 1/ λ , γ2 = κ4 /κ22 = 1/λ .

7 we had calculated the expected value of the energy from the distribution of velocity. Of course, for certain applications it may be necessary to know not only the mean value of the energy but its complete distribution. To derive it, we have to perform a variable transformation. For discrete distributions, this is a trivial exercise: The probability that the event “u has the value u(xk )” occurs, where u is an uniquely invertible function of x, is of course the same as for “x has the value xk ”: P {u = u(xk )} = P {x = xk } .

The result is given by the convolution integral, see Sect. 4, f (z) = g(x)h(z − x) dx = h(y)g(z − y) dy which often is diﬃcult to evaluate analytically. s which obey the simple relation φf (t) = φg (t)φh (t) . 10)we get φf (t) = E(eit(x+y) ) = E(eitx eity ) = E(eitx )E(eity ) = φg (t)φh (t) . The third step requires the independence of the two variates. Applying the inverse Fourier transform to φf (t), we get f (z) = 1 2π e−itz φf (t) dt . The solution of this integral is not always simple. For some functions it can be found in tables of the Fourier transform.