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?

  • $\begingroup$ You need more information than the expected value and the variance to find the higher cumulants. $\endgroup$ Aug 2, 2021 at 2:23
  • $\begingroup$ 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. $\endgroup$ Aug 2, 2021 at 2:45

1 Answer 1


$\newcommand{\C}{\mathbb C}\newcommand{\ip}[2]{\langle #1,#2\rangle}$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 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*}

  • $\begingroup$ 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? $\endgroup$
    – Milad A
    Aug 2, 2021 at 11:10
  • $\begingroup$ @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. $\endgroup$ Aug 2, 2021 at 13:05

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .