# Is connected correlation/cumulant expansion additive?

Say X is a free field or a Gaussian random variable.

Then I want to analyse the connected correlation, $<(X + a (X^2 - \langle X^2 \rangle))^n>_c$

• I think that for $n \geq 4$ there are no linear terms in "a" in the above.

• I want to know what is the linear term in "a" at $n=3$, the only $n$ where any linear term ever occurs.

• The key thing I want to check is whether for any two variables $Y$ and $Z$ one can say that its additive, as in $\langle Y + Z \rangle _c = \langle Y \rangle _c + \langle Z \rangle_c$ ?

If the above is true then I can binomially expand in my initial expression to linear order in a and see that the terms linear or lower in are, $\langle X^n \rangle _c + na \langle X^{n-1}(X^2 - \langle X^2 \rangle)\rangle_c$ Now one can see that this term is $0$ for any $n \geq 4$ using the fact that for Gaussian fields $<X^m>_c$ is $0$ for any $m \geq 3$.

But this "proof" seems flawed since if I look at a specific example of say $n=5$ then I have the equality,

$\langle (X + a (X^2 - \langle X^2 \rangle))^5 \rangle _c = <(X + a (X^2 - \langle X^2 \rangle))^5> -$ $10<(X + a (X^2 - \langle X^2 \rangle))^3>_c <(X + a (X^2 - \langle X^2 \rangle))^2>_c$

since we had $<(X + a (X^2 - \langle X^2 \rangle))> = 0$

Now on evaluating the RHS I get a linear term of "$90a\langle X^2 \rangle ^3$"!?

But something is getting confusing at the threshold value of $n=3$ when I am trying to find the term linear in $a$.

• If I substitute $n=3$ in $\langle X^n \rangle_c + na \langle X^{n-1}(X^2 - \langle X^2\rangle )\rangle _c$ I am getting $-3a\langle X^2 \rangle^2$ (by again using the fact that higher then 2-point all connected correlators of a free field are $0$.

• But since $\langle X + a(X^2 - \langle X^2 \rangle)\rangle = 0$ it follows that $\langle (X + a(X^2 - \langle X^2 \rangle))^3\rangle = \langle(X + a(X^2 - \langle X^2 \rangle))^3\rangle _c$. Now the term linear in "a" in the LHS (after doing the same binomial expansion) is $3a(\langle X^4 \rangle-\langle X^2 \rangle ^2)$. But for a Gaussian field, $<X^4> = 3<X^2>^2$ and hence the term linear in "a" evaluates to, $6a\langle X^2 \rangle^2$

I am confused as to why the two different ways of thinking gave two different results.

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Can I put my reputation as a bounty on this question? If yes, how? – Anirbit Mar 13 '13 at 21:11
@Anirbit: no, but you will be able to once MO moves to the SE 2.0 network. – Qiaochu Yuan Mar 14 '13 at 17:47
@Qiachu And may be you can yourself help answer this question? :) – Anirbit Mar 14 '13 at 23:28
Please do not deface your question like that. I rolled back your revisions. – Andy Putman Mar 21 '13 at 22:13
@Andy Putman I wanted to remove my question. How can I do that!? – curious Mar 21 '13 at 23:30