# How is 4th order cumulant of a complex random vector defined?

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?

• You need more information than the expected value and the variance to find the higher cumulants. Aug 2, 2021 at 2:23
• I see: Given your way of (incorrectly) using the word "derive", I now suspect you meant to ask how the 4th-order cumulant is defined. Aug 2, 2021 at 2:45

$$\newcommand{\C}{\mathbb C}\newcommand{\ip}{\langle #1,#2\rangle}$$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 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 $$\begin{multline*} \ell''''(0)(a_1,\dots,a_4) =E\prod_{j\in}\ip Y{a_j} \\ -\frac12\,\sum_{J\in\binom{}2}E\prod_{j\in J}\ip Y{a_j}\;E\prod_{k\in J^{\mathsf c}}\ip Y{a_k}, \end{multline*}$$ where $$a_1,\dots,a_4$$ are in $$\C^n$$, $$:=\{1,2,3,4\}$$, $$J^{\mathsf c}:=\setminus J$$, and $$\binom{}2$$ is the set of all subsets of cardinality $$2$$ of the set $$$$.
So, the $$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*}$$
• Thanks a lot. In this paper, 4th order cumulant of ${\bf z}_g(t)$ is a $L^2 \times L^2$ matrix (equation (3.6)), where $L$ is the number of sensors. Is it correct to define 4th order cumulant of ${\bf z}_g(t)$ as $$\mathbb{E} \left ( \left({\bf z}_g(t) \otimes {\bf z}^*_g(t)\right ) \left({\bf z}_g(t) \otimes {\bf z}^*_g(t)\right )^H\right )$$ where $\otimes$ is the Kronecker product? Aug 2, 2021 at 11:10
• @MiladA : One can define anything and call it anything. The question is, how reasonable the definition is and on what principle(s) it is based. With your definition, in the case $L=1$, for the "cumulant" we get the fourth absolute moment, which is not what is usually called the fourth "one-dimensional" cumulant. Aug 2, 2021 at 13:05