# Cumulants and moments of Dirac Delta distribution

1. Mar 12, 2007

### amacinho

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

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

Thus, the distribution is found to be the Dirac delta function $$a \delta(x-1)$$.

There are two ways to find the cumulants.

1) First, they are defined as:

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

2) but I also previously showed that:

$$c_1=\langle X \rangle$$
$$c_2=\langle X^2 \rangle - \langle X \rangle^2$$
$$c_3=2\langle X \rangle^3 - 3\langle X \rangle \langle X^2 \rangle + \langle X^3 \rangle$$
$$c_4=-6\langle X \rangle^4 + 12\langle X \rangle^2 \langle X^2 \rangle -3 \langle X^2 \rangle^2 - 4\langle X \rangle \langle X^3 \rangle + \langle X^4 \rangle$$

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

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?