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Worked Cumulant Calculations for Mixed Sums

Learn to organize mixed-sum calculations by independent blocks, then apply affine and joint-cumulant identities to obtain checked low-order results.

Advanced 9 min read

What You'll Learn

  • Separate dependent summands into blocks before applying additivity across independent groups.
  • Expand cumulants of a sum into joint cumulants of its components.
  • Transform joint cumulants under affine changes of variables.
  • Calculate first through fourth cumulants for a sum with shared random components.
  • Detect and correct the error of assuming dependent terms have additive marginal cumulants.

Organizing a Mixed-Sum Calculation

A sum may contain both dependent and independent summands. The useful first step is not to apply additivity term by term, but to identify which variables form dependent blocks and which blocks are independent of one another. Within a dependent block, joint cumulants retain the dependence; across independent blocks, cumulants add. Affine transformations then let us express the calculation in terms of simpler underlying variables.

Write \(\kappa(X_1,\ldots,X_r)\) for the joint cumulant of \(X_1,\ldots,X_r\), and write \(\kappa_r(X)=\kappa(X,\ldots,X)\) for the \(r\)th cumulant of \(X\). In particular, \(\kappa_1(X)=E[X]\), \(\kappa_2(X)=\operatorname{Var}(X)\), and \(\kappa_3(X)=E[(X-E[X])^3]\). The fourth cumulant is \(\kappa_4(X)=E[(X-E[X])^4]-3\operatorname{Var}(X)^2\). Assume the moment-generating functions used below are finite in a neighborhood of the origin, so the relevant cumulant-generating functions and derivatives exist.

Calculation principle: Expand a cumulant of a sum using multilinearity, preserve all joint cumulants inside dependent blocks, and use additivity only across independent blocks. Constants affect first cumulants, but not joint cumulants of order at least two.

Two Identities for Mixed Sums

For \(S=\sum_{i=1}^m c_iX_i\), multilinearity gives the order-\(r\) expansion

$$ \kappa_r(S)=\sum_{i_1=1}^m\cdots\sum_{i_r=1}^m c_{i_1}\cdots c_{i_r}\, \kappa(X_{i_1},\ldots,X_{i_r}). $$

The indices range over all ordered choices, including repetitions. This matters: terms such as \(\kappa(X,X,Y)\) are part of a third-order expansion and cannot be dropped merely because two arguments are the same.

Theorem: Suppose the variables \(X_1,\ldots,X_m\) are partitioned into blocks \(B_1,\ldots,B_q\) that are mutually independent as random vectors. For \(S=\sum_i c_iX_i\), and any order \(r\) for which the cumulants exist,
$$ \kappa_r(S)= \sum_{b=1}^q \ \sum_{i_1,\ldots,i_r\in B_b} c_{i_1}\cdots c_{i_r}\, \kappa(X_{i_1},\ldots,X_{i_r}). $$
Thus terms involving variables from more than one independent block contribute zero.

Proof. Let \(K_b(t)\) be the joint cumulant-generating function of the variables in block \(B_b\). Mutual independence implies that the moment-generating function of the full vector factors into the product of the block moment-generating functions near the origin. Taking logarithms gives the sum of the block cumulant-generating functions. For a fixed block, the cumulant-generating function of its weighted sum is \(K_b((c_i s)_{i\in B_b})\). Differentiating \(r\) times with respect to \(s\) at zero, the chain rule gives the sum over ordered indices in that block, with coefficient \(c_{i_1}\cdots c_{i_r}\). Differentiating the sum of block functions adds these within-block results and produces no cross-block terms. This proves the formula. \(\square\)

A second identity handles several affine transformations at once. If \(Y_j=a_j+\sum_i b_{ji}X_i\), then for \(r\geq 2\),

$$ \kappa(Y_{j_1},\ldots,Y_{j_r}) = \sum_{i_1,\ldots,i_r} b_{j_1i_1}\cdots b_{j_ri_r}\, \kappa(X_{i_1},\ldots,X_{i_r}). $$

For order one, the corresponding formula is \(E[Y_j]=a_j+\sum_i b_{ji}E[X_i]\).

Proof. Let \(K_X\) be the joint cumulant-generating function of the vector \(X\). The joint cumulant-generating function of \(Y\) is \(K_Y(t)=\sum_j a_jt_j+K_X((\sum_j b_{ji}t_j)_i)\). For a derivative of order \(r\geq2\), the linear term \(\sum_j a_jt_j\) contributes zero. Each derivative of the composed term selects one index \(i_k\) and contributes the corresponding coefficient \(b_{j_ki_k}\). Summing over all such choices gives the displayed formula. For \(r=1\), differentiating also retains \(a_j\), giving the stated mean. \(\square\)

Worked Calculations

Worked Example: A Dependent Pair and an Independent Summand

Let \(U\sim\operatorname{Bernoulli}(1/3)\), let \(V\sim N(0,1)\), and let \(Z\sim\operatorname{Poisson}(2)\), with \(U,V,Z\) mutually independent. Define \(X=U+V\) and \(Y=2U-V\). Find the first four cumulants of \(S=X+Y+Z\), keeping the dependence between \(X\) and \(Y\) in view.

The pair \((X,Y)\) is a dependent block: both variables contain \(U\) and \(V\). Indeed, \(\operatorname{Cov}(X,Y)=2\operatorname{Var}(U)-\operatorname{Var}(V)=2(2/9)-1=-5/9\), which is nonzero. The variable \(Z\) is an independent block. Within the first block, combine the terms before calculating: \(X+Y=3U\), because \(V-V=0\). Hence \(S=3U+Z\), a sum of independent variables.

For a Bernoulli variable with success probability \(p\), the first four cumulants are \(p\), \(pq\), \(pq(1-2p)\), and \(pq(1-6pq)\), where \(q=1-p\). At \(p=1/3\), these are \(1/3,\ 2/9,\ 2/27,\ -2/27\). A Poisson variable with parameter \(2\) has each of its first four cumulants equal to \(2\). Scaling \(U\) by \(3\) multiplies its order-\(r\) cumulant by \(3^r\); additivity applies to \(3U\) and \(Z\). Therefore

$$ \begin{aligned} \kappa_1(S)&=3(1/3)+2=3,\\ \kappa_2(S)&=3^2(2/9)+2=4,\\ \kappa_3(S)&=3^3(2/27)+2=4,\\ \kappa_4(S)&=3^4(-2/27)+2=-6+2=-4. \end{aligned} $$

The block calculation avoids treating \(X\) and \(Y\) as independent. Their shared components cancel in their sum before the independent-block additivity is used.

Worked Example: A Third-Order Joint Cumulant

Let \(A\sim\operatorname{Bernoulli}(1/3)\) and \(B\sim\operatorname{Poisson}(1)\) be independent. Set \(X=A+B\) and \(Y=2A-B\). Calculate \(\kappa(X,X,Y)\).

Apply multilinearity to the three arguments separately. The contribution using \(A\) in all three positions has coefficient \(1\cdot1\cdot2=2\), so it is \(2\kappa_3(A)\). The all-\(B\) contribution has coefficient \(1\cdot1\cdot(-1)=-1\), so it is \(-\kappa_3(B)\). Every other term mixes \(A\) and \(B\) and is zero: joint cumulants involving arguments from two independent groups vanish when the order is at least two. Here \(\kappa_3(A)=2/27\), while \(\kappa_3(B)=1\). Thus

$$ \kappa(X,X,Y)=2(2/27)-1=-23/27. $$

As a check on the dependence in the pair, \(\operatorname{Cov}(X,Y)=2\operatorname{Var}(A)-\operatorname{Var}(B)=2(2/9)-1=-5/9\). The answer is not obtained by treating \(X\) and \(Y\) as independent; the shared \(A\) component generates their joint cumulant.

Worked Example: An Affine Combination of Dependent Summands

Let \(A\sim\operatorname{Bernoulli}(1/4)\) and \(B\sim\operatorname{Bernoulli}(1/2)\) be independent. Define \(X=A+B\) and \(Y=A-B\), and let \(T=5+2X-3Y\). Find \(E[T]\), \(\operatorname{Var}(T)\), and \(\kappa_3(T)\).

First simplify the affine combination using the definitions of \(X\) and \(Y\): \(T=5+2(A+B)-3(A-B)=5-A+5B\). This expresses \(T\) in terms of independent components, with coefficients \(-1\) and \(5\). For \(A\), the mean, variance, and third cumulant are \(1/4,\ 3/16,\ 3/32\); for \(B\), they are \(1/2,\ 1/4,\ 0\). The zero third cumulant for \(B\) follows from the Bernoulli formula \(pq(1-2p)\) at \(p=1/2\). The affine and independent-sum identities now give

$$ \begin{aligned} E[T]&=5-1/4+5(1/2)=29/4,\\ \operatorname{Var}(T)&=(-1)^2(3/16)+5^2(1/4)=103/16,\\ \kappa_3(T)&=(-1)^3(3/32)+5^3(0)=-3/32. \end{aligned} $$

As a dependence check, \(\operatorname{Cov}(X,Y)=\operatorname{Var}(A)-\operatorname{Var}(B)=3/16-1/4=-1/16\). The simplification works because the dependence is represented by shared components; it does not make \(X\) and \(Y\) independent.

A Reliable Workflow and a Common Pitfall

For a new problem, write each summand in terms of underlying variables, then mark which underlying groups are independent. Combine or transform terms within each dependent block before adding cumulants across blocks. If a joint cumulant is requested, expand each argument separately and retain the coefficient attached to every choice of underlying variable. Finally, use the scaling factor to the cumulant's order: a coefficient \(c\) contributes \(c^r\) to an order-\(r\) cumulant when the same variable fills every argument.

1
Identify dependence.
List the shared random components and partition variables into mutually independent blocks.
2
Rewrite the target.
Collect coefficients within each block, including constants if the first cumulant is needed.
3
Choose the identity.
Use additivity across independent blocks, or expand a joint cumulant by multilinearity when several arguments are involved.
4
Check the order and result.
Raise scale factors to the cumulant order, and verify that dependence has not been discarded.

Common mistake: Since \(S=X+Y\), it is tempting to write \(\kappa_r(S)=\kappa_r(X)+\kappa_r(Y)\) without checking independence. That identity is valid for independent \(X\) and \(Y\), not for arbitrary dependent variables. In general, the expansion includes mixed terms; for example, \(\kappa_2(X+Y)=\operatorname{Var}(X)+\operatorname{Var}(Y)+2\operatorname{Cov}(X,Y)\). The correction is to retain within-block joint cumulants, or first rewrite the sum in terms of independent underlying components as in the examples above.

These calculations are useful whenever a statistic is built from overlapping measurements, shared random effects, or other components that induce dependence. A correct block structure can turn an apparently complicated cumulant into a short calculation, while the joint-cumulant expansion shows exactly which dependence contributions would otherwise be missed.

Check Your Understanding

Try these questions by identifying the independent blocks before calculating.

  1. If \(S=2X-Y\) and \(X,Y\) are dependent, which part of the cumulant expansion must be checked before adding their separate cumulants?
  2. For \(T=a+cX\), how does the \(r\)th cumulant change when \(r\geq2\), and what happens to the first cumulant?
  3. If \(A\) and \(B\) are independent, which terms in the expansion of \(\kappa(A+B,A+B,A)\) vanish, and why?
  4. Suppose \(X=U+V\) and \(Y=2U-V\). Rewrite \(X+Y\) in terms of \(U,V\), and explain why this may simplify its cumulants.
  5. Why does a nonzero covariance between two summands warn against applying independent-sum additivity?

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