Converting Moments to Cumulants With Bell Polynomials

Why Bell Polynomials Help

Cumulants are defined by derivatives of the log moment-generating function, whereas raw moments are derivatives of the moment-generating function itself. Bell polynomials provide a systematic way to connect these two sets of derivatives. Their main practical benefit is organizational: instead of expanding products and collecting terms from scratch at each order, one can use a general coefficient formula and then specialize it.

This tutorial develops the conversion from raw moments to cumulants. Throughout, write $m_j=E[X^j]$ for the $j$th raw moment and $\kappa_j$ for the $j$th cumulant. Assume the moment-generating function $M_X(t)=E[e^{tX}]$ is finite on an open interval containing zero. Then $K_X(t)=\log M_X(t)$ is defined near zero, and its derivatives at zero give the cumulants. The assumption ensures that the moments needed in each finite-order calculation exist.

Partial and Complete Bell Polynomials

The partial exponential Bell polynomial $B_{n,k}$ collects the terms formed by dividing a total order $n$ among exactly $k$ parts. For $1\leq k\leq n$, it is defined by the following finite sum.

Definition: For variables $x_1,x_2,\ldots$, the partial exponential Bell polynomial is $$ B_{n,k}(x_1,\ldots,x_{n-k+1}) = \sum \frac{n!}{j_1!j_2!\cdots j_{n-k+1}!} \prod_{i=1}^{n-k+1}\left(\frac{x_i}{i!}\right)^{j_i}, $$ where the sum is over nonnegative integers $j_1,\ldots,j_{n-k+1}$ satisfying $\sum_{i=1}^{n-k+1}j_i=k$ and $\sum_{i=1}^{n-k+1}i j_i=n$. The complete exponential Bell polynomial is $B_n(x_1,\ldots,x_n)=\sum_{k=1}^n B_{n,k}(x_1,\ldots,x_{n-k+1})$.

For example, $B_{n,1}=x_n$, since one part of size $n$ is the only way to make total order $n$ with one part. At the other extreme, $B_{n,n}=x_1^n$, since $n$ parts must all have size one. The general coefficient records how many ways the different part sizes can occur.

A useful generating identity follows directly from this definition:

\frac{1}{k!}\left(\sum_{j\geq 1}x_j\frac{t^j}{j!}\right)^k = \sum_{n\geq k} B_{n,k}(x_1,\ldots,x_{n-k+1})\frac{t^n}{n!}.

To see why, select $j_i$ copies of $x_i t^i/i!$ from the product on the left. The choices must use $k$ factors in total and contribute power $t^n$, so $\sum_i j_i=k$ and $\sum_i i j_i=n$. The multinomial coefficient for those choices is $k!/(j_1!j_2!\cdots)$; the prefactor $1/k!$ cancels its $k!$. Rewriting the remaining coefficient with denominator $n!$ gives exactly the defining coefficient of $B_{n,k}$. This proves the identity.

The Moment-to-Cumulant Formula

Set $U(t)=M_X(t)-1$. Since $M_X(0)=1$, $U(0)=0$, and the power series for the logarithm gives $\log M_X(t)=\log(1+U(t))$. The coefficient of order $n$ in this logarithm yields the conversion formula.

Theorem: If $M_X(t)$ is finite on an open interval containing zero, then for $n\geq 1$, $$ \kappa_n = \sum_{k=1}^{n}(-1)^{k-1}(k-1)!\, B_{n,k}(m_1,\ldots,m_{n-k+1}). $$ Here $m_j=E[X^j]$. The formula expresses the $n$th cumulant using only raw moments through order $n$.

Proof. Near zero, $M_X(t)=\sum_{j\geq 0}m_j t^j/j!$, with $m_0=1$. Thus $U(t)=\sum_{j\geq 1}m_j t^j/j!$. For a fixed $n$, terms of degree greater than $n$ cannot affect the coefficient of $t^n$, so the logarithm series can be truncated:

\log(1+U(t)) = \sum_{k=1}^{n}(-1)^{k-1}\frac{U(t)^k}{k} + \text{terms of order greater than }n.

By the generating identity, the coefficient of $t^n$ in $U(t)^k$ is $k!B_{n,k}(m_1,\ldots,m_{n-k+1})/n!$. Multiplying the coefficient in the logarithm by $n!$ gives the $n$th derivative at zero, which is $\kappa_n$. The resulting factor is $(-1)^{k-1}k!/k=(-1)^{k-1}(k-1)!$, proving the formula. $\square$

The complete Bell polynomials organize the sum over all possible numbers of parts; the partial polynomials keep the number of parts fixed. In this conversion, the alternating coefficient $(-1)^{k-1}(k-1)!$ weights each partial polynomial. That weight comes from the logarithm series, not from the definition of the Bell polynomial.

Expanding the Conversion Through Order Five

The needed partial Bell polynomials can be listed explicitly. Substituting them into the theorem gives the familiar formulas through order five.

\begin{aligned} B_{1,1}&=m_1,\\ B_{2,1}&=m_2,\qquad B_{2,2}=m_1^2,\\ B_{3,1}&=m_3,\qquad B_{3,2}=3m_1m_2,\qquad B_{3,3}=m_1^3,\\ B_{4,1}&=m_4,\qquad B_{4,2}=4m_1m_3+3m_2^2,\\ B_{4,3}&=6m_1^2m_2,\qquad B_{4,4}=m_1^4,\\ B_{5,1}&=m_5,\qquad B_{5,2}=5m_1m_4+10m_2m_3,\\ B_{5,3}&=10m_1^2m_3+15m_1m_2^2,\qquad B_{5,4}=10m_1^3m_2,\qquad B_{5,5}=m_1^5. \end{aligned}

For instance, $B_{5,2}$ has two possible part-size patterns: one part of size five is not allowed because there must be two parts; the patterns are sizes $1$ and $4$, or sizes $2$ and $3$. Their coefficients are $5$ and $10$, respectively. Applying the theorem to the displayed polynomials gives:

\begin{aligned} \kappa_1&=m_1,\\ \kappa_2&=m_2-m_1^2,\\ \kappa_3&=m_3-3m_1m_2+2m_1^3,\\ \kappa_4&=m_4-4m_1m_3-3m_2^2+12m_1^2m_2-6m_1^4,\\ \kappa_5&=m_5-5m_1m_4-10m_2m_3+20m_1^2m_3+30m_1m_2^2\\ &\qquad-60m_1^3m_2+24m_1^5. \end{aligned}

For order five, the successive weights are $1,-1,2,-6,24$. For example, the term $B_{5,3}$ is multiplied by $2$, producing $20m_1^2m_3+30m_1m_2^2$. Keeping the Bell-polynomial grouping visible is a useful check on both coefficients and signs.

Worked Example: A Two-Point Distribution Through Order Five

Let $X$ take the values $2$ and $4$, each with probability $1/2$. First compute each raw moment by averaging the corresponding powers:

\begin{aligned} m_1&=(2+4)/2=3,\\ m_2&=(2^2+4^2)/2=(4+16)/2=10,\\ m_3&=(2^3+4^3)/2=(8+64)/2=36,\\ m_4&=(2^4+4^4)/2=(16+256)/2=136,\\ m_5&=(2^5+4^5)/2=(32+1024)/2=528. \end{aligned}

Substitution into the formulas gives $\kappa_1=3$ and $\kappa_2=10-3^2=1$. At order three, $\kappa_3=36-3(3)(10)+2(3^3)=36-90+54=0$.

At order four, \[ \kappa_4=136-4(3)(36)-3(10^2)+12(3^2)(10)-6(3^4) =136-432-300+1080-486=-2. \] At order five,

\begin{aligned} \kappa_5 &=528-5(3)(136)-10(10)(36)+20(3^2)(36)\\ &\quad+30(3)(10^2)-60(3^3)(10)+24(3^5)\\ &=528-2040-3600+6480+9000-16200+5832\\ &=0. \end{aligned}

The zero third and fifth cumulants are consistent with the distribution being symmetric about $3$. The calculation also illustrates why each raw moment should be checked before substituting it: errors in a high-order moment propagate through several terms.

Worked Example: Uniform Distribution on the Unit Interval

Let $X$ have the uniform distribution on $[0,1]$. Its density is $1$ on that interval, so $m_j=\int_0^1 x^j\,dx=1/(j+1)$. In particular, $m_1=1/2$, $m_2=1/3$, $m_3=1/4$, and $m_4=1/5$.

The first two conversions give $\kappa_1=1/2$ and $\kappa_2=1/3-1/4=1/12$. The third is \[ \kappa_3=\frac14-3\left(\frac13\right)\left(\frac12\right) +2\left(\frac12\right)^3 =\frac14-\frac12+\frac14=0. \] For the fourth cumulant, the five terms are

\begin{aligned} \kappa_4 &=\frac15-4\left(\frac14\right)\left(\frac12\right) -3\left(\frac13\right)^2 +12\left(\frac12\right)^2\left(\frac13\right) -6\left(\frac12\right)^4\\ &=\frac15-\frac12-\frac13+1-\frac38 =-\frac{1}{120}. \end{aligned}

This example checks the conversion for a continuous law with moments different from those of the two-point distribution. In particular, the zero third cumulant again reflects symmetry about the mean, while the fourth cumulant is not zero.

Worked Example: A Discrete Law With Nonzero Fifth Cumulant

Suppose $P(X=-1)=1/2$, $P(X=0)=1/3$, and $P(X=2)=1/6$. Directly averaging powers gives $m_1=-1/6$, $m_2=7/6$, $m_3=5/6$, $m_4=19/6$, and $m_5=29/6$. The first cumulant is $\kappa_1=-1/6$; the next two are

\begin{aligned} \kappa_2 &=\frac76-\left(-\frac16\right)^2 =\frac{41}{36},\\ \kappa_3 &=\frac56-3\left(\frac76\right)\left(-\frac16\right) +2\left(-\frac16\right)^3 =\frac{38}{27}. \end{aligned}

Using the order-four formula, \[ \kappa_4=\frac{19}{6}+\frac{20}{36} -\frac{147}{36}+\frac{84}{216}-\frac{1}{216} =\frac{5}{216}. \] For order five, each term in the formula contributes as follows:

\begin{aligned} \kappa_5 &=\frac{29}{6}+\frac{95}{36}-\frac{350}{36} +\frac{100}{216}-\frac{1470}{216} +\frac{420}{1296}-\frac{1}{324}\\ &=\frac{6264+3420-12600+600-8820+420-4}{1296}\\ &=-\frac{10720}{1296}=-\frac{670}{81}. \end{aligned}

The denominator $1296$ in the second line is a common denominator for every term in the first line. Writing the common-denominator numerators explicitly makes the final reduction auditable.

Common Pitfalls and a Practical Check

The conversion is algebraic, but several bookkeeping errors are easy to make. The moment $m_j$ is the raw moment $E[X^j]$, not the central moment $E[(X-E[X])^j]$. Also, the alternating signs and factorial weights belong to the outer sum over $k$; they are not additional coefficients inside $B_{n,k}$.

Wrong reasoning: “The fourth cumulant is the fourth raw moment minus the fourth power of the mean.” That omits the $B_{4,2}$ and $B_{4,3}$ contributions.
Correction: Use the complete order-four conversion: include the $m_1m_3$, $m_2^2$, $m_1^2m_2$, and $m_1^4$ terms with their stated coefficients.
Wrong reasoning: “The coefficient of $B_{n,k}$ is always just $(-1)^{k-1}$.”
Correction: The logarithm contributes $(-1)^{k-1}/k$, while the power $U^k$ contributes a factor $k!$ in the Bell-polynomial identity. Together these give $(-1)^{k-1}(k-1)!$.

The formulas are useful whenever cumulants are needed but moments are what have been calculated or supplied. Bell polynomials make the pattern extend to any specified finite order, while the explicit expansions through five provide convenient checks for hand calculations. A reliable workflow is to verify the raw moments, form the relevant partial Bell polynomials, apply the logarithm weights, and then simplify.

Check Your Understanding

Try these questions by using the definitions and conversion formulas above.

What constraints on the indices $j_i$ define $B_{n,k}$, and what does $k$ count?
Use the order-three formula to calculate $\kappa_3$ when $m_1=2$, $m_2=5$, and $m_3=14$.
For $B_{5,2}$, which two part-size patterns are possible, and why are their coefficients different?
Explain why the coefficient multiplying $B_{n,k}$ in the cumulant formula is $(-1)^{k-1}(k-1)!$.
Identify the missing terms in the proposed expression $\kappa_4=m_4-m_1^4$.