Suppose that ${\bf x} \in\mathbb C^n$ is a complex random vector, we know the mean vector and covariance matrix of $\bf x$ are defined as follows:
$${\bf m}\_{\bf x} = \mathbb{E} ({\bf x}) \\
{\bf C}\_{\bf x x} = \mathbb{E} (({\bf x}-{\bf m}\_{\bf x})({\bf x}-{\bf m}\_{\bf x})^H)$$
How is 4th order cumulant of a ${\bf \text{complex random vector}}$ such as ${\bf x} \in\mathbb C^n$ defined?



How is 4th order cumulant of a complex random vector defined?
$\newcommand{\C}{\mathbb C}\newcommand{\ip}[2]{\langle #1,#2\rangle}$First of all, cumulants are defined, rather than derived.


Now, let $X:=\mathbf x$. Suppose $E\|X\|^m<\infty$ for some natural $m$. Let $Y:=X-EX$. Then for the respective characteristic functions $f$ an $g$ of $X$ and $Y$ we have
\begin{equation}
 f(s)=Ee^{i\ip Xs}=e^{i\ip{EX}s}Ee^{i\ip Ys}=e^{i\ip {EX}s}g(s),
\end{equation}
where $s\in\C^n$ and $\ip st:=\sum\_{p=1}^ns\_p\overline{t\_p}$ for $s=(s\_1,\dots,s\_n)$ and $t=(t\_1,\dots,t\_n)$ in $\C^n$.


Then the cumulant of order $m$ of $X$ can be defined as the [$m$-linear form/tensor](https://en.wikipedia.org/wiki/Tensor#As_multilinear_maps) on $\C^m$ that is $i^{-m}$ times the $m$th derivative $\ell^{(m)}(0)$ at $0\in\C^m$ of the function $\ell:=\ln f$. So, for any $s\in\C^n$,
\begin{equation}
 \ell(s)=\ln f(s)=i\ip{EX}s+\ln g(s) 
\end{equation}
and, for any $a,b,c,d$ in $\C^n$,
\begin{equation}
 g'(s)(a)=i\,E\ip Ya e^{i\ip Ys},\quad g'(0)(a)=0,
\end{equation}
\begin{equation}
 g''(s)(a,b)=i^2\,E\ip Ya\ip Yb e^{i\ip Ys},
\end{equation}
\begin{equation}
 g'''(s)(a,b,c)=i^3\,E\ip Ya\ip Yb\ip Yc e^{i\ip Ys},
\end{equation}
\begin{equation}
 g''''(0)(a,b,c,d)=i^4\,E\ip Ya\ip Yb\ip Yc\ip Yd.
\end{equation}
So, for any $s\in\C^n$ and any $a,b,c,d$ in $\C^n$,
\begin{equation}
 \ell'(s)(a)=\ip{EX}a+\frac1{g(s)}\,g'(s)(a),
\end{equation}
\begin{equation}
 \ell''(s)(a,b)=-\frac1{g(s)^2}\,g'(s)(a)g'(s)(b)+\frac1{g(s)}\,g''(s)(a,b),
\end{equation}
\begin{multline\*}
 \ell'''(s)(a,b,c)=\frac2{g(s)^3}\,g'(s)(a)g'(s)(b)g'(s)(c) \\ 
 -\frac1{g(s)^2}\,g''(s)(a,c)g'(s)(b)-\frac1{g(s)^2}\,g''(s)(b,c)g'(s)(a)
 -\frac1{g(s)^2}\,g''(s)(a,b)g'(s)(c) \\
 +\frac1{g(s)}\,g'''(s)(a,b,c),
\end{multline\*}
\begin{multline\*}
 \ell''''(0)(a,b,c,d)= \\ 
 -g''(0)(a,c)g'(s)(b,d)-g''(0)(a,d)g'(s)(b,c)-g''(0)(a,b)g'(s)(c,d) \\
 +g''''(0)(a,b,c,d). 
\end{multline\*}


So, the cumulant of order $4$ of $X$ is the quadrilinear form/tensor on $\C^4$ that is $\ell''''(0)$, and this quadrilinear form is defined by the formula
\begin{multline\*}
 \ell''''(0)(a\_1,\dots,a\_4)
 =E\prod\_{j\in[4]}\ip Y{a\_j} \\ 
 -\frac12\,\sum\_{J\in\binom{[4]}2}E\prod\_{j\in J}\ip Y{a\_j}\;E\prod\_{k\in J^{\mathsf c}}\ip Y{a\_k},
\end{multline\*}
where $a\_1,\dots,a\_4$ are in $\C^n$, $[4]:=\{1,2,3,4\}$, $J^{\mathsf c}:=[4]\setminus J$, and $\binom{[4]}2$ is the set of all subsets of cardinality $2$ of the set $[4]$.


So, the $n^4$ components of the tensor that is the cumulant of order $4$ of $X$ can be obtained by substituting for $a\_1,\dots,a\_4$ the standard basis vectors $e\_1,\dots,e\_n$ of $\C^n$. For instance, letting $(a\_1,\dots,a\_4)=(e\_p,e\_q,e\_q,e\_p)$ for distinct $p$ and $q$ in the set $[n]:=\{1,\dots,n\}$, we see that the $(p,q,q,p)$-component of the cumulant tensor is
\begin{equation\*}
EY\_p^2Y\_q^2-EY\_p^2\,EY\_q^2-2(EY\_pY\_q)^2.
\end{equation\*}


