Recovering Moments from Cumulants

From Cumulants Back to Moments

Cumulants and moments encode the same local information about a distribution when the relevant derivatives exist, but they organize it differently. Cumulants are especially useful in algebraic calculations; moments are often the quantities needed to describe scale, skewness-related features, or expected powers of a random variable. Recovering moments from cumulants is therefore a useful conversion, not a new source of information.

The conversion is governed by set partitions. Each partition groups the indices $1,\ldots,n$ into blocks; a block of size $r$ contributes the $r$th cumulant. Summing over all such groupings gives the $n$th raw moment. Complete Bell polynomials collect these terms compactly.

Definition: Let $X$ have a moment-generating function $M(t)=E[e^{tX}]$ that is finite on an open interval containing $0$. Its cumulant-generating function is $K(t)=\log M(t)$, and its $r$th cumulant is $\kappa_r=K^{(r)}(0)$. Its $r$th raw moment is $m_r=E[X^r]$. In particular, $m_0=1$.

The neighborhood condition on $M$ ensures that these derivatives exist and that its Taylor expansion near zero can be used to identify moments. The moment $m_r$ is raw: it is the expectation of $X^r$, not of a power of $X-E[X]$.

The Set-Partition Formula

A set partition of $\{1,\ldots,n\}$ is a collection of nonempty, disjoint subsets, called blocks, whose union is the whole set. Write $\Pi_n$ for the set of all such partitions. For a block $B$, let $|B|$ denote its number of elements.

Theorem: If $M(t)$ is finite on an open interval containing zero, then for every integer $n\geq 1$, $$ m_n=\sum_{\pi\in\Pi_n}\ \prod_{B\in\pi}\kappa_{|B|}. $$ Equivalently, $m_n=B_n(\kappa_1,\ldots,\kappa_n)$, where $B_n$ is the $n$th complete Bell polynomial.

Here is why the formula holds. The Taylor expansions at zero are $M(t)=\sum_{n\geq0}m_n t^n/n!$ and $K(t)=\sum_{r\geq1}\kappa_r t^r/r!$. Since $M(t)=e^{K(t)}$, expanding the exponential groups terms according to how many times each cumulant occurs.

Expand the exponential.
$e^{K(t)}=\sum_{q\geq0}K(t)^q/q!$. The $q$th power selects an ordered list of $q$ cumulant terms.

Collect the coefficient of $t^n$.
A selection of orders $r_1,\ldots,r_q$ contributes only when $r_1+\cdots+r_q=n$, with product $\kappa_{r_1}\cdots\kappa_{r_q}/(r_1!\cdots r_q!q!)$.

Interpret the terms as partitions.
For a partition with block sizes $r_1,\ldots,r_q$, the factorials count the ways to assign $n$ labeled indices to those blocks. The factor $q!$ removes the ordering of the blocks.

More formally, the number of partitions with exactly $j_r$ blocks of size $r$, for each $r=1,\ldots,n$, is $\displaystyle \frac{n!}{\prod_{r=1}^n (r!)^{j_r}j_r!}$, provided $\sum_{r=1}^n rj_r=n$. The factors $r!$ account for the orderings within blocks, and $j_r!$ account for the orderings among blocks of equal size. Each such partition contributes $\prod_{r=1}^n\kappa_r^{j_r}$. Grouping the partition sum by its block counts therefore proves the coefficient formula

B_n(\kappa_1,\ldots,\kappa_n) = n!\!\sum_{\substack{j_1,\ldots,j_n\geq0\\ \sum_{r=1}^n rj_r=n}} \prod_{r=1}^n \frac{1}{j_r!}\left(\frac{\kappa_r}{r!}\right)^{j_r}.

These coefficient arguments can be read as Taylor-series calculations near zero: the assumed moment-generating function is finite in a neighborhood of zero, so its derivatives there identify the moments and cumulants. The exponential identity then equates the coefficients at each finite order. This proves the set-partition formula and its Bell-polynomial form. $\square$

The First Five Moments

For low orders, listing the partitions makes the formula concrete. At order four, for example, the partitions have block sizes $1+1+1+1$, $2+1+1$, $2+2$, $3+1$, or $4$. Their respective contributions are $\kappa_1^4$, $6\kappa_1^2\kappa_2$, $3\kappa_2^2$, $4\kappa_1\kappa_3$, and $\kappa_4$.

Result: The first five raw moments in terms of cumulants are $$ \begin{aligned} m_1&=\kappa_1,\\ m_2&=\kappa_1^2+\kappa_2,\\ m_3&=\kappa_1^3+3\kappa_1\kappa_2+\kappa_3,\\ m_4&=\kappa_1^4+6\kappa_1^2\kappa_2+3\kappa_2^2 +4\kappa_1\kappa_3+\kappa_4,\\ m_5&=\kappa_1^5+10\kappa_1^3\kappa_2+15\kappa_1\kappa_2^2 +10\kappa_1^2\kappa_3+10\kappa_2\kappa_3 +5\kappa_1\kappa_4+\kappa_5. \end{aligned} $$

The coefficients count partitions, not arbitrary algebraic choices. For instance, at order five there are ten partitions with one block of size three and one block of size two, giving $10\kappa_3\kappa_2$. The coefficient $5$ on $\kappa_1\kappa_4$ counts the five choices of the singleton block.

Worked Example: Recovering Moments of a Symmetric Two-Point Variable

Let $X$ equal $-1$ or $1$, each with probability $1/2$. Its moment-generating function is $M(t)=\cosh(t)$, so $K(t)=\log\cosh(t)$. Through order five, its Taylor expansion is $K(t)=t^2/2-t^4/12+O(t^6)$. Since the coefficient of $t^r$ in $K(t)$ is $\kappa_r/r!$, this gives $\kappa_1=0$, $\kappa_2=1$, $\kappa_3=0$, $\kappa_4=-2$, and $\kappa_5=0$.

Substituting these values gives $m_1=0$, $m_2=1$, and $m_3=0$. At order four, $m_4=3\kappa_2^2+\kappa_4=3-2=1$. At order five, every term is zero, so $m_5=0$. Directly, the odd powers of $X$ have expectation zero by symmetry, while $X^2=X^4=1$; all five results agree.

Worked Example: A Shifted Two-Point Distribution

Let $Y$ equal $0$ or $2$, each with probability $1/2$. Then $M_Y(t)=(1+e^{2t})/2=e^t\cosh(t)$, and $K_Y(t)=t+\log\cosh(t)$. The expansion above gives cumulants $\kappa_1=1$, $\kappa_2=1$, $\kappa_3=0$, $\kappa_4=-2$, and $\kappa_5=0$.

The formulas yield $m_1=1$, $m_2=1+1=2$, and $m_3=1+3=4$. Next, $m_4=1+6+3-2=8$. Finally, $m_5=1+10+15-10=16$; the other terms vanish because $\kappa_3=\kappa_5=0$. Direct evaluation gives $E[Y^r]=(0^r+2^r)/2=2^{r-1}$ for $r\geq1$, which is $1,2,4,8,16$ for the five orders.

Worked Example: A Three-Point Distribution

Let $Z$ take values $-1,0,1$ with probabilities $1/4,1/2,1/4$, respectively. Its moment-generating function is $M_Z(t)=1/2+\cosh(t)/2=1+t^2/4+t^4/48+O(t^6)$. Using $\log(1+u)=u-u^2/2+O(u^3)$ with $u=t^2/4+t^4/48+O(t^6)$, we obtain $K_Z(t)=t^2/4-t^4/96+O(t^6)$. Thus $\kappa_1=0$, $\kappa_2=1/2$, $\kappa_3=0$, $\kappa_4=-1/4$, and $\kappa_5=0$.

The first and third moments are zero, and the second is $m_2=\kappa_2=1/2$. The fourth moment is $m_4=3\kappa_2^2+\kappa_4=3(1/2)^2-1/4=3/4-1/4=1/2$. The fifth moment is zero because all the terms in its formula contain a zero cumulant. Directly, $Z^2=Z^4$ equals one with probability $1/2$, and the distribution is symmetric, confirming the results.

Interpreting the Conversion Carefully

The first two cumulants have familiar interpretations: $\kappa_1=E[X]$ and $\kappa_2=\operatorname{Var}(X)$. Consequently, $m_2=\kappa_1^2+\kappa_2$ is the identity $E[X^2]=(E[X])^2+\operatorname{Var}(X)$. At higher orders, cumulants should not be mistaken for raw or central moments. For example, $\kappa_4$ is not generally $E[X^4]$; the fourth raw moment also includes contributions from lower-order cumulants.

A common mistake is to use the fourth-moment expression for a central moment, or to omit terms because a cumulant looks like a moment of the same order. The correction is to identify the target first: the formulas here recover $E[X^r]$. If a central moment is needed, it must be computed from the raw moments or derived using the centered variable. Another useful check is the order-two identity: if a proposed formula gives $m_2=\kappa_2$ when $\kappa_1\ne0$, it has lost the contribution from the nonzero mean.

The partition form also explains why Bell polynomials are useful. They organize all possible products of cumulants at a given order and keep the integer coefficients attached to the number of ways each block pattern can occur. This makes higher-order calculations systematic: enumerate block sizes, or use the complete Bell polynomial, rather than guessing terms.

Key takeaway: For a random variable with a moment-generating function finite near zero, the $n$th raw moment is the complete Bell polynomial $B_n(\kappa_1,\ldots,\kappa_n)$. Each term corresponds to a set partition, and its coefficient counts how many partitions have that pattern of block sizes.

Check Your Understanding

Use the set-partition interpretation and the first five formulas to check your reasoning.

Write the set-partition formula for $m_3$ and identify the partition that contributes $\kappa_1\kappa_2$.
If $\kappa_1=0$ and $\kappa_2=4$, what are $m_2$ and $m_4$ in terms of $\kappa_4$?
Why does the term $10\kappa_2\kappa_3$ appear in the formula for $m_5$?
For the distribution on $-1,0,1$ in the third example, explain why the fifth raw moment is zero using both its symmetry and the cumulant formula.
Explain why $\kappa_4$ alone is not generally the fourth raw moment.

Recovering Moments from Cumulants

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