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Cumulants and moments of Dirac Delta distribution

  1. Mar 12, 2007 #1
    For my statistical mechanics class I need to find the cumulants of a special distribution of which all moments are constant and equal to a. I followed two different approaches and obtained imcompatible results, something is wrong but I couldn't figure it out.

    Here's what I did:

    Since all the moments are constant I formed the generating function as

    [tex] F(k)&=\sum_{n=0}^{\infty}\frac{(\i k)^n}{n!}a[/tex]
    [tex]=a\sum_{n=0}^{\infty}\frac{(\i k)^n}{n!}[/tex]
    [tex]=a e^{ik}[/tex]

    Thus, the distribution is found to be the Dirac delta function [tex]a \delta(x-1)[/tex].

    There are two ways to find the cumulants.

    1) First, they are defined as:

    [tex]c_n &=\frac{1}{\i^n}\frac{d^n}{dk^n}\ln F(k) \biggr{|}_{k=0}[/tex]

    2) but I also previously showed that:

    [tex]c_1=\langle X \rangle[/tex]
    [tex]c_2=\langle X^2 \rangle - \langle X \rangle^2[/tex]
    [tex]c_3=2\langle X \rangle^3 - 3\langle X \rangle \langle X^2 \rangle + \langle X^3 \rangle[/tex]
    [tex]c_4=-6\langle X \rangle^4 + 12\langle X \rangle^2 \langle X^2 \rangle -3 \langle X^2 \rangle^2 - 4\langle X \rangle \langle X^3 \rangle + \langle X^4 \rangle[/tex]

    The logarithm of the generating function is
    [tex]\ln F(k)= \ln(a e^{\i k})[/tex]

    The first derivative of this function is i, and all the rest are zero. Thus leaving me c_1 = i, and c_n=0 for n>1.

    The results of the first and second methods are different and at least one of them is wrong, but I can't see what's wrong in neither of them. Could someone help me?
     
  2. jcsd
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