Mixed Cumulants Through Fourth Order

Why Mixed Cumulants Matter

For two random variables, a joint cumulant-generating function records more than the separate cumulants of each variable. Its partial derivatives also describe how the variables occur together. The second-order mixed derivative is familiar: it is covariance. At higher orders, mixed derivatives measure joint departures from lower-order structure, and their formulas involve mixed central moments together with covariance corrections.

The order of a mixed cumulant is the total number of derivatives. Thus \(\kappa_{2,1}\) has total order three, while \(\kappa_{2,2}\) has total order four. Keeping track of total order is useful: it tells us which moments are needed and which lower-order contributions must be removed.

Definition: Let \(X\) and \(Y\) be random variables whose joint moment-generating function \(M(s,t)=E[\exp(sX+tY)]\) is finite on an open neighborhood of \((0,0)\). Their joint cumulant-generating function is \(K(s,t)=\log M(s,t)\). For nonnegative integers \(r,s\) with \(r+s\geq 1\), the mixed cumulant of order \((r,s)\) is \[ \kappa_{r,s} = \left. \frac{\partial^{r+s}}{\partial u^r\partial v^s}K(u,v) \right|_{(u,v)=(0,0)}. \] The total order is \(r+s\). In this definition the variables \(u,v\) are arguments of the function, not random variables.

Finiteness of the joint moment-generating function in a neighborhood of the origin ensures that the derivatives used here exist and can be calculated from its local expansion. The labels \(r\) and \(s\) count derivatives with respect to the first and second arguments, respectively. For example, \(\kappa_{1,2}\) is obtained by differentiating once in the first argument and twice in the second.

Mixed Cumulants Through Order Four

Write \(\mu_X=E[X]\), \(\mu_Y=E[Y]\), and center the variables as \(U=X-\mu_X\) and \(V=Y-\mu_Y\). For \(r+s\geq2\), centering does not change the mixed cumulant: replacing \(X,Y\) by \(U,V\) subtracts only the linear term \(\mu_Xu+\mu_Yv\) from \(K(u,v)\). Define the mixed central moments \(m_{r,s}=E[U^rV^s]\). Then all mixed cumulants through total order four are as follows.

Result: The first-order cumulants are \(\kappa_{1,0}=\mu_X\) and \(\kappa_{0,1}=\mu_Y\). At second order, \[ \kappa_{2,0}=\operatorname{Var}(X),\qquad \kappa_{1,1}=\operatorname{Cov}(X,Y),\qquad \kappa_{0,2}=\operatorname{Var}(Y). \] At third order, \[ \kappa_{3,0}=m_{3,0},\quad \kappa_{2,1}=m_{2,1},\quad \kappa_{1,2}=m_{1,2},\quad \kappa_{0,3}=m_{0,3}. \] At fourth order, \[ \begin{aligned} \kappa_{4,0}&=m_{4,0}-3m_{2,0}^{\,2},\\ \kappa_{3,1}&=m_{3,1}-3m_{2,0}m_{1,1},\\ \kappa_{2,2}&=m_{2,2}-m_{2,0}m_{0,2}-2m_{1,1}^{\,2},\\ \kappa_{1,3}&=m_{1,3}-3m_{0,2}m_{1,1},\\ \kappa_{0,4}&=m_{0,4}-3m_{0,2}^{\,2}. \end{aligned} \]

The pure fourth-order formulas are the familiar fourth cumulants of individual variables. The mixed formulas show the same principle in a joint setting: a fourth central moment is adjusted by the products of covariances associated with ways of pairing its four factors. For \(U^3V\), there are three pairings, each contributing \(\operatorname{Var}(X)\operatorname{Cov}(X,Y)\). For \(U^2V^2\), one pairing contributes \(\operatorname{Var}(X)\operatorname{Var}(Y)\), and two contribute \(\operatorname{Cov}(X,Y)^2\).

Deriving the Formulas

A short expansion of the centered joint generating function proves the formulas and makes the covariance corrections explicit. Let \(M_0(u,v)=E[\exp(uU+vV)]\). Since \(E[U]=E[V]=0\), the expansion of \(M_0-1\) has no terms of degree one. Write its homogeneous terms of degrees two, three, and four as \(H_2,H_3,H_4\). The exponential power series gives

\begin{aligned} H_2&=\frac{1}{2}E[(uU+vV)^2],\\ H_3&=\frac{1}{6}E[(uU+vV)^3],\\ H_4&=\frac{1}{24}E[(uU+vV)^4]. \end{aligned}

Terms of degree at least five do not affect derivatives of total order at most four at the origin. Since \(K_0=\log M_0=\log(1+(M_0-1))\), use \(\log(1+w)=w-w^2/2+w^3/3-\cdots\). As \(M_0-1\) begins at degree two, its square begins at degree four and its cube at degree six. Thus, through degree four,

K_0(u,v)=H_2+H_3+H_4-\frac{1}{2}H_2^2 \quad\text{through total degree four.}

The degree-two and degree-three terms are exactly the corresponding central-moment terms, so their derivatives give the variance, covariance, and third-order formulas. To see the fourth-order subtraction directly, put \(a=m_{2,0}\), \(b=m_{1,1}\), and \(c=m_{0,2}\). Then

H_2=\frac{1}{2}(a u^2+2buv+c v^2).

The degree-four part of \(H_2^2/2\) is therefore

\frac{1}{2}H_2^2 = \frac{1}{8}\left( a^2u^4+4ab u^3v+(2ac+4b^2)u^2v^2+4bcuv^3+c^2v^4 \right).

For any \(r,s\), the coefficient of \(u^rv^s\) in a Taylor expansion is the corresponding derivative divided by \(r!s!\). In \(H_4\), those coefficients are \(m_{r,s}/(r!s!)\); subtracting the displayed coefficients from \(H_2^2/2\) gives precisely the five fourth-order formulas above. This proves the result under the stated moment-generating-function condition. \(\square\)

Worked Calculations

Worked Example: A Bivariate Normal Distribution

Suppose \((X,Y)\) is bivariate normal with means \(\mu_X,\mu_Y\), variances \(\sigma_X^2,\sigma_Y^2\), and covariance \(\rho\sigma_X\sigma_Y\). Its joint cumulant-generating function is

K(u,v)=\mu_Xu+\mu_Yv+ \frac{1}{2}\left(\sigma_X^2u^2+2\rho\sigma_X\sigma_Yuv+\sigma_Y^2v^2\right).

Differentiating once in each argument gives \(\kappa_{1,1}=\rho\sigma_X\sigma_Y\), which agrees with the specified covariance. The pure second derivatives are \(\kappa_{2,0}=\sigma_X^2\) and \(\kappa_{0,2}=\sigma_Y^2\). Every derivative of total order three or greater is zero because \(K\) is a polynomial of degree two. In particular, all five fourth-order mixed cumulants vanish. This does not mean the fourth central moments vanish; rather, their covariance-pairing terms exactly account for them.

Worked Example: A Discrete Pair With Nonzero Fourth-Order Mixed Cumulants

Let \((U,V)\) take the values \((1,1),(-1,-1)\) with probability \(3/8\) each, and \((1,-1),(-1,1)\) with probability \(1/8\) each. Both variables have mean zero and variance one. The probability of matching signs is \(3/4\), so \(E[UV]=3/4-1/4=1/2\). Since \(U^2=V^2=1\), the needed moments are

\begin{aligned} m_{2,0}=m_{0,2}=1,\quad&m_{1,1}=\frac{1}{2},\\ m_{3,1}=m_{1,3}=E[UV]=\frac{1}{2},\quad& m_{2,2}=1. \end{aligned}

All third central moments are zero: for example, \(m_{2,1}=E[V]=0\), \(m_{1,2}=E[U]=0\), and \(m_{3,0}=E[U]=m_{0,3}=E[V]=0\). Applying the fourth-order formulas gives

\begin{aligned} \kappa_{3,1}&=\frac{1}{2}-3(1)\left(\frac{1}{2}\right)=-1,\\ \kappa_{2,2}&=1-(1)(1)-2\left(\frac{1}{2}\right)^2=-\frac{1}{2},\\ \kappa_{1,3}&=\frac{1}{2}-3(1)\left(\frac{1}{2}\right)=-1. \end{aligned}

The pure fourth-order values are \(\kappa_{4,0}=1-3=-2\) and \(\kappa_{0,4}=1-3=-2\). The joint cgf can also be written as \(K(u,v)=\log\{(3/4)\cosh(u+v)+(1/4)\cosh(u-v)\}\); the derivatives just calculated are its mixed derivatives at the origin. The example makes the covariance correction visible: \(m_{2,2}=1\), but \(\kappa_{2,2}=-1/2\).

Worked Example: One Bernoulli Variable Used in Both Coordinates

Let \(Z\sim\operatorname{Bernoulli}(1/3)\), and set \(X=Z\), \(Y=Z\). This joint distribution is supported on \((0,0)\) and \((1,1)\), so its joint cgf is \(K(u,v)=\log(2/3+(1/3)e^{u+v})\). Define \(g(w)=\log(2/3+(1/3)e^w)\). By the chain rule, \(K(u,v)=g(u+v)\), and every mixed derivative of total order \(j\) at zero equals \(g^{(j)}(0)\).

For clarity, put \(q(w)=(1/3)e^w/(2/3+(1/3)e^w)\). Then \(g'(w)=q(w)\) and \(q'(w)=q(w)(1-q(w))\). Consequently \(g'(0)=1/3\), \(g''(0)=2/9\), and \(g'''(0)=(2/9)(1-2/3)=2/27\). Differentiating once more gives \(g''''(0)=q(0)(1-q(0))(1-6q(0)(1-q(0)))=-2/27\). Therefore each cumulant of total order one, two, three, or four, regardless of how its derivatives are split between \(u\) and \(v\), is respectively \(1/3, 2/9, 2/27,\) or \(-2/27\). For example, \(\kappa_{1,1}=2/9=\operatorname{Var}(Z)\), and \(\kappa_{2,2}=-2/27\). The fourth-order central-moment formula confirms the latter: \(E[(Z-1/3)^4]=2/27\), so \(2/27-3(2/9)^2=-2/27\).

Interpretation and a Common Pitfall

A second-order mixed cumulant is covariance, and every third-order cumulant is a third mixed central moment. Beginning at order four, a mixed cumulant is generally not just the matching central moment. It removes contributions formed from pairs of second moments. These corrections leave a quantity that represents the fourth-order part not already accounted for by those pairings.

Common mistake: It is tempting to write \(\kappa_{2,2}=E[(X-\mu_X)^2(Y-\mu_Y)^2]\), treating every cumulant as a central moment. That expression is \(m_{2,2}\), the mixed central moment, not the fourth-order mixed cumulant. The correct formula is \(\kappa_{2,2}=m_{2,2}-\operatorname{Var}(X)\operatorname{Var}(Y)-2\operatorname{Cov}(X,Y)^2\). In the discrete example, the central moment is \(1\), while the cumulant is \(-1/2\); confusing them changes the answer substantially.

When calculating a mixed cumulant, first identify its total order and the required central moments. At second and third order, the formulas are direct. At fourth order, list the covariance pairings carefully: the unequal pattern \(U^3V\) has three identical pairings, whereas \(U^2V^2\) has one variance–variance pairing and two covariance–covariance pairings.

Check Your Understanding

Use the definitions and formulas above to answer these questions.

What is the total order of \(\kappa_{1,3}\), and which partial derivatives define it?
Express \(\kappa_{2,1}\) using centered variables, and state why no covariance correction is subtracted.
Write the formula for \(\kappa_{3,1}\) in terms of mixed central moments, variances, and covariance.
If \(m_{2,0}=4\), \(m_{0,2}=9\), \(m_{1,1}=2\), and \(m_{2,2}=50\), calculate \(\kappa_{2,2}\).
Why do all cumulants of total order three or higher vanish for a bivariate normal distribution?