Cumulants and moments of Dirac Delta distribution

[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

[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?

 

[Jhero]

Dear student,

Thank you for sharing your methods and results with us. It seems that there may be a mistake in your second approach when calculating the cumulants. Let's take a closer look at your calculations:

c_1=\langle X \rangle = a \delta(x-1)

This is correct, as the first cumulant is simply the mean of the distribution, which in this case is a delta function centered at 1.

c_2=\langle X^2 \rangle - \langle X \rangle^2 = a \delta(x-1)^2 - a^2 \delta(x-1)^2 = 0

Here, you have taken the second moment (the mean of the squared distribution) and subtracted the square of the first moment (the mean). However, in this case, both the first and second moments are equal to a delta function centered at 1, so subtracting them will result in 0.

c_3=2\langle X \rangle^3 - 3\langle X \rangle \langle X^2 \rangle + \langle X^3 \rangle = 2a \delta(x-1)^3 - 3a \delta(x-1)^2 \delta(x-1) + a \delta(x-1)^3 = 0

Similarly, taking the third moment and subtracting the first and second moments will also result in 0.

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 = -6a \delta(x-1)^4 + 12a \delta(x-1)^2 \delta(x-1)^2 - 3a^2 \delta(x-1)^4 - 4a \delta(x-1)^3 \delta(x-1) + a \delta(x-1)^4 = 0

Again, all terms in this equation will cancel out and result in 0. This is because the distribution you are working with has all of its moments equal to a constant, so when calculating the cumulants using the formula above, all terms will cancel out and result in 0.

In summary, it seems that your second approach is not incorrect, but rather the