Multilinearity and Symmetry of Joint Cumulants

Why Symmetry and Multilinearity Matter

A joint cumulant records how several random variables contribute together at a specified order. Its algebra becomes especially useful when the variables of interest are sums or linear combinations of other variables. Two properties make those calculations manageable: the order of the arguments does not matter, and each argument can be expanded linearly. These properties hold whether or not the variables are independent.

The distinction from independence is important. Multilinearity is an algebraic property of cumulants; it does not say that mixed cumulants vanish. Independence can make certain mixed cumulants zero, but that is a separate fact. Here the goal is to establish the algebra that lets us identify and calculate every term before making any independence simplification.

Definition Through Moments

Let $X_1,\ldots,X_r$ be random variables, where $r\geq 1$. Assume, for example, that each $X_i$ has a finite absolute moment of order $r$. This condition ensures that every product moment appearing below is finite, by Hölder's inequality. A partition of $\{1,\ldots,r\}$ is a collection of nonempty, disjoint blocks whose union is the whole set.

Definition: The joint cumulant of $X_1,\ldots,X_r$ is $$ \kappa(X_1,\ldots,X_r) = \sum_{\pi\in\mathcal P_r} (|\pi|-1)!(-1)^{|\pi|-1} \prod_{B\in\pi} E\!\left[\prod_{i\in B}X_i\right], $$ where $\mathcal P_r$ is the set of partitions of $\{1,\ldots,r\}$, and $|\pi|$ is the number of blocks in partition $\pi$. Each argument has its own index, even when two or more arguments are the same random variable.

For example, the first three orders are

\begin{aligned} \kappa(X)&=E[X],\\ \kappa(X,Y)&=E[XY]-E[X]E[Y]=\operatorname{Cov}(X,Y),\\ \kappa(X,Y,Z) &=E[XYZ]-E[XY]E[Z]-E[XZ]E[Y]-E[YZ]E[X]\\ &\quad+2E[X]E[Y]E[Z]. \end{aligned}

The third-order expression also equals $E[(X-E[X])(Y-E[Y])(Z-E[Z])]$, as follows by expanding the product. In particular, it is unchanged if any argument is centered before the cumulant is taken.

Why the Partition Formula Matches the Cumulant-Generating Function

When a moment-generating function is finite in a neighborhood of the origin, joint cumulants can be obtained by differentiating the logarithm of that function. The partition definition also works under finite-moment assumptions alone, even when such a moment-generating function does not exist near the origin. The following formal calculation explains the connection without assuming convergence of an infinite series.

Introduce formal variables $t_1,\ldots,t_r$ and impose $t_i^2=0$ for every $i$. In this finite-dimensional algebra, set $H=E[\exp(\sum_{i=1}^r t_iX_i)]-1$. The coefficient of $\prod_{i\in A}t_i$, for a subset $A$, in $H$ is $E[\prod_{i\in A}X_i]$. Since every term of $H$ has positive degree and all degrees above $r$ vanish, $\log(1+H)=\sum_{k=1}^r(-1)^{k-1}H^k/k$ is a finite algebraic calculation. To obtain the coefficient of $t_1\cdots t_r$ in $H^k$, assign the indices $1,\ldots,r$ to $k$ ordered, nonempty blocks. Each partition into $k$ blocks appears in $k!$ orders. Multiplication by $(-1)^{k-1}/k$ therefore gives the partition weight $(-1)^{k-1}(k-1)!$. Thus the coefficient of $t_1\cdots t_r$ in this formal logarithm is exactly the joint cumulant defined above.

If the joint moment-generating function is finite on an open neighborhood of the origin, its derivatives there are justified by exponential integrability, and the formal coefficient agrees with the corresponding mixed derivative of the log moment-generating function at zero. The partition definition, however, needs only the stated product-moment integrability.

Symmetry and Multilinearity

Theorem: If the product moments in the partition definition are finite, then $\kappa(X_1,\ldots,X_r)$ is symmetric in its arguments and linear in each argument separately.

Proof of symmetry. Permute the indices $1,\ldots,r$. This permutation gives a one-to-one correspondence between partitions before and after the permutation: each block is carried to the block containing its permuted indices. The number of blocks is unchanged, so the factor $(|\pi|-1)!(-1)^{|\pi|-1}$ is unchanged. The product moments are merely relabelled. Consequently, the sum over partitions has the same value after any permutation of the arguments. This proves symmetry. $\square$

Proof of linearity in one argument. Fix an index $j$, and replace $X_j$ by $aU+bV$, where $a,b$ are scalars and all moments needed in the formula are finite. For each partition, index $j$ belongs to exactly one block, say $B$. In the moment for that block,

E\!\left[\left(\prod_{i\in B,\ i\ne j}X_i\right)(aU+bV)\right] = aE\!\left[U\prod_{i\in B,\ i\ne j}X_i\right] +bE\!\left[V\prod_{i\in B,\ i\ne j}X_i\right].

Every other block moment is unaffected. Thus the contribution from this partition is $a$ times its contribution with $U$ in slot $j$, plus $b$ times its contribution with $V$ there. Summing over the partitions gives

\kappa(X_1,\ldots,X_{j-1},aU+bV,X_{j+1},\ldots,X_r) = a\kappa(X_1,\ldots,X_{j-1},U,X_{j+1},\ldots,X_r) +b\kappa(X_1,\ldots,X_{j-1},V,X_{j+1},\ldots,X_r).

This holds for each slot $j$, so the joint cumulant is multilinear. $\square$

Transforming Cumulants of Linear Combinations

Repeatedly applying multilinearity expands each argument into its component random variables. If $Y_j=\sum_{a=1}^{m_j}c_{ja}X_a$, for $j=1,\ldots,r$, then

\kappa(Y_1,\ldots,Y_r) = \sum_{a_1=1}^{m_1}\cdots\sum_{a_r=1}^{m_r} c_{1a_1}\cdots c_{ra_r}\, \kappa(X_{a_1},\ldots,X_{a_r}).

The formula requires the relevant moments to be finite, but it does not require independence. Symmetry can then combine terms that have the same arguments in a different order. At order two, this is the familiar covariance transformation; at higher orders, the expansion retains mixed cumulants as well as the cumulants involving only one component.

Worked Example: Permuting the Arguments of a Third-Order Cumulant

Let $U\sim\operatorname{Bernoulli}(1/4)$. Calculate $\kappa(U,U,1-U)$, and check that changing the order of the arguments leaves its value unchanged.

For three arguments the cumulant is the product of the three centered variables in expectation. The mean of $U$ is $1/4$, and the mean of $1-U$ is $3/4$. Thus

\kappa(U,U,1-U) = E\!\left[(U-\tfrac14)^2\big((1-U)-\tfrac34\big)\right].

If $U=1$, the centered product is $(3/4)^2(-3/4)=-27/64$, and this outcome has probability $1/4$. If $U=0$, it is $(-1/4)^2(1/4)=1/64$, and this outcome has probability $3/4$. Therefore

\kappa(U,U,1-U) = \tfrac14(-\tfrac{27}{64})+\tfrac34(\tfrac{1}{64}) = -\tfrac{27}{256}+\tfrac{3}{256} = -\tfrac{3}{32}.

The centered-product formula is symmetric in its three factors, so $\kappa(1-U,U,U)$, for example, has the same value. This agrees with the general partition argument, which does not depend on this particular distribution.

Worked Example: Covariance of Two Linear Combinations

Let $U$ and $V$ be independent $\operatorname{Bernoulli}(1/2)$ variables, and define $X=U$ and $Y=U+V$. Find $\operatorname{Cov}(2X-Y,X+Y)$ using multilinearity, then verify it directly.

First, $E[X]=1/2$, $E[Y]=1$, $\operatorname{Var}(X)=1/4$, and $\operatorname{Var}(Y)=1/2$. Also, $\operatorname{Cov}(X,Y)=\operatorname{Cov}(U,U+V)=1/4$, since $U$ and $V$ are independent. By symmetry, $\operatorname{Cov}(Y,X)=1/4$. Expand the covariance, which is the second-order joint cumulant:

\begin{aligned} \operatorname{Cov}(2X-Y,X+Y) &=2\operatorname{Cov}(X,X)+2\operatorname{Cov}(X,Y)\\ &\quad-\operatorname{Cov}(Y,X)-\operatorname{Cov}(Y,Y)\\ &=2(\tfrac14)+2(\tfrac14)-\tfrac14-\tfrac12 =\tfrac14. \end{aligned}

For a direct check, the four equally likely outcomes $(U,V)=(0,0),(0,1),(1,0),(1,1)$ give the pairs $(2X-Y,X+Y)=(0,0),(-1,1),(1,2),(0,3)$. The means of these two coordinates are $0$ and $3/2$. Their mean product is $(0-1+2+0)/4=1/4$, so their covariance is $1/4-0(3/2)=1/4$, as found by expansion.

Worked Example: A Third Cumulant of a Linear Combination

Let $U$ take the values $-1,0,2$, each with probability $1/3$, and set $X=U$, $Y=U^2$, and $W=X+Y$. Expand $\kappa(W,W,W)$ and verify the result from the distribution of $W$.

The needed raw moments are $E[U]=1/3$, $E[U^2]=5/3$, $E[U^3]=7/3$, $E[U^4]=17/3$, $E[U^5]=31/3$, and $E[U^6]=65/3$, obtained by averaging the corresponding powers of $-1,0,2$. For three variables, the joint cumulant is the expectation of the product of the centered variables. Substitution gives

\begin{aligned} \kappa(X,X,X) &=E[(U-\tfrac13)^3]=\tfrac{20}{27},\\ \kappa(X,X,Y) &=E[(U-\tfrac13)^2(U^2-\tfrac53)]=\tfrac{46}{27},\\ \kappa(X,Y,Y) &=E[(U-\tfrac13)(U^2-\tfrac53)^2]=\tfrac{68}{27},\\ \kappa(Y,Y,Y) &=E[(U^2-\tfrac53)^3]=\tfrac{70}{27}. \end{aligned}

For clarity, the second line, for example, follows by expanding its product:

\begin{aligned} \kappa(X,X,Y) &=E[U^4]-\tfrac23E[U^3]+\tfrac19E[U^2] -\tfrac53\left(E[U^2]-\tfrac19\right)\\ &=\tfrac{17}{3}-\tfrac{14}{9}+\tfrac{5}{27}-\tfrac{70}{27} =\tfrac{46}{27}. \end{aligned}

By multilinearity there are three equal terms of each mixed type, because the cumulant is symmetric:

\begin{aligned} \kappa(W,W,W) &=\kappa(X+Y,X+Y,X+Y)\\ &=\kappa(X,X,X)+3\kappa(X,X,Y)+3\kappa(X,Y,Y)+\kappa(Y,Y,Y)\\ &=\frac{20+3(46)+3(68)+70}{27} =\frac{432}{27}=16. \end{aligned}

Directly, $W=U+U^2$ takes the values $0,0,6$ with equal probabilities. Its mean is $2$, so its third cumulant is its third centered moment:

E[(W-2)^3] = \frac{(-2)^3+(-2)^3+4^3}{3} = \frac{-8-8+64}{3} =16.

The direct calculation confirms both the expansion and the coefficients: the factor $3$ counts the three positions in which the single $Y$ can occur, and likewise for the term with two $Y$'s.

Common Pitfall: Multilinearity Is Not Independence

A common mistake is to assume that a mixed cumulant disappears simply because it contains different random variables. Multilinearity says how to expand a cumulant when an argument is a sum; symmetry says that reordered arguments have the same value. Neither property alone makes a mixed term zero. For instance, in the covariance example, $X$ and $Y=U+V$ are dependent, and $\operatorname{Cov}(X,Y)=1/4$, not zero.

The correction is to expand first and retain every term. A term may then be set to zero only when a separate condition justifies it, such as an applicable independence result. This distinction is useful in dependent-sum calculations: multilinearity supplies the full expansion, while dependence determines which mixed cumulants remain.

Key takeaway: Joint cumulants are symmetric and linear in each slot. Therefore, cumulants of linear combinations expand into weighted sums of joint cumulants of the component variables; independence is not part of that algebraic rule.

Check Your Understanding

Use the partition definition and the transformation rule to answer the following.

Write the partition formula for $\kappa(X,Y)$ and identify it with a familiar second-order quantity.
Why does permuting the indices in a partition not change the value of a joint cumulant?
Expand $\kappa(aX+bY,Z)$ using multilinearity.
For $W=X+Y$, write the full expansion of $\kappa(W,W,W)$, combining terms using symmetry.
Does multilinearity alone imply $\kappa(X,Y)=0$ when $X$ and $Y$ are different random variables? Explain.