Recurrences from the Cumulant-Generating Function

Why a Recurrence Connects Moments and Cumulants

Raw moments and cumulants are two ways to summarize a distribution. Raw moments are derivatives of the moment-generating function, while cumulants are derivatives of its logarithm. The logarithm often makes distributional structure easier to see: for example, cumulants add for independent sums. The recurrence developed here provides the complementary tool: once cumulants are known, it converts them systematically into raw moments.

The recurrence is useful beyond memorizing formulas through fourth order. It gives a single rule for generating moments of any order, and it can be rearranged to recover cumulants from moments. Its binomial coefficients come directly from the product rule, so understanding the derivation also helps prevent errors in higher-order calculations.

Definition: Suppose the moment-generating function \(M(t)=E[e^{tX}]\) is finite on an open interval containing \(0\). Define the cumulant-generating function by \(K(t)=\log M(t)\). The \(r\)th raw moment is \(m_r=E[X^r]=M^{(r)}(0)\), and the \(r\)th cumulant is \(\kappa_r=K^{(r)}(0)\), for integers \(r\geq 1\). Set \(m_0=1\).

The finiteness assumption on an open interval around zero ensures that the derivatives needed here exist. In particular, differentiation under the expectation is justified locally: for any fixed derivative order, choose a smaller interval whose endpoints lie inside the interval of finiteness. On that smaller interval, the absolute value of the differentiated integrand is bounded by a constant times exponential terms at those endpoints, which are integrable. Also, \(M(0)=1\) and \(M(t)>0\) for real \(t\) where it is finite, so \(K(t)=\log M(t)\) is well defined there.

The Moment–Cumulant Recurrence

The identity \(M(t)=e^{K(t)}\) implies, by differentiation, that \(M'(t)=K'(t)M(t)\). Repeated differentiation of this product yields a recurrence involving only values of derivatives at zero.

Theorem: If the moment-generating function of \(X\) is finite on an open interval containing zero, then for every integer \(n\geq 1\),

m_n=\sum_{j=1}^{n}\binom{n-1}{j-1}\kappa_j m_{n-j}.

Equivalently, with the index shifted, \(m_{n+1}=\sum_{j=0}^{n}\binom{n}{j}\kappa_{j+1}m_{n-j}\) for \(n\geq 0\).

Proof. Differentiate \(M'(t)=K'(t)M(t)\) a total of \(n-1\) times. The generalized product rule gives

M^{(n)}(t)=\sum_{j=0}^{n-1}\binom{n-1}{j}K^{(j+1)}(t)M^{(n-1-j)}(t).

All derivatives in this equation exist under the stated moment-generating-function condition. Evaluate at \(t=0\), where \(M^{(n)}(0)=m_n\), \(K^{(j+1)}(0)=\kappa_{j+1}\), and \(M^{(n-1-j)}(0)=m_{n-1-j}\). Relabel \(r=j+1\). Then

m_n=\sum_{r=1}^{n}\binom{n-1}{r-1}\kappa_r m_{n-r},

which is the stated recurrence. \(\square\)

The term with \(j=n\) in the displayed theorem is \(\kappa_n m_0=\kappa_n\), because \(m_0=1\). Thus the highest-order cumulant enters with coefficient one, and all other terms use lower-order cumulants and moments. The first few formulas make the structure visible:

\begin{aligned} m_1 &= \kappa_1,\\ m_2 &= \kappa_1m_1+\kappa_2 = \kappa_1^2+\kappa_2,\\ m_3 &= \kappa_1m_2+2\kappa_2m_1+\kappa_3 = \kappa_1^3+3\kappa_1\kappa_2+\kappa_3,\\ m_4 &= \kappa_1m_3+3\kappa_2m_2+3\kappa_3m_1+\kappa_4\\ &= \kappa_1^4+6\kappa_1^2\kappa_2+3\kappa_2^2 +4\kappa_1\kappa_3+\kappa_4. \end{aligned}

Worked Example: Raw Moments of a Poisson Variable

Let \(X\sim\operatorname{Poisson}(2)\). Its moment-generating function is \(M(t)=\exp\{2(e^t-1)\}\), so \(K(t)=2(e^t-1)\). Differentiating shows that every cumulant is \(\kappa_j=2\), for \(j\geq 1\). Use the recurrence to find the first four raw moments.

For \(n=1\), \(m_1=\kappa_1=2\).

For \(n=2\), the recurrence gives \(m_2=\kappa_1m_1+\kappa_2\), hence \(m_2=2(2)+2=6\).

For \(n=3\), it gives \(m_3=\kappa_1m_2+2\kappa_2m_1+\kappa_3\), hence \(m_3=2(6)+2(2)(2)+2=22\).

For \(n=4\), \(m_4=\kappa_1m_3+3\kappa_2m_2+3\kappa_3m_1+\kappa_4\). Substitution yields \(m_4=2(22)+3(2)(6)+3(2)(2)+2=94\).

As a check, the Poisson factorial moments satisfy \(E[(X)_r]=2^r\), where \((X)_r=X(X-1)\cdots(X-r+1)\). Since \(X^4=(X)_4+6(X)_3+7(X)_2+(X)_1\), its expectation is \(16+6(8)+7(4)+2=94\), agreeing with the recurrence.

Recovering Cumulants From Raw Moments

The same equation can be used in the opposite direction. Isolate its final term, the one with \(j=n\), to obtain a recursive formula for \(\kappa_n\). For example, it says \(\kappa_2=m_2-m_1^2\), the familiar variance identity, and it gives \(\kappa_3=m_3-3m_1m_2+2m_1^3\).

Corollary: Under the same moment-generating-function condition, for \(n\geq 1\),

\kappa_n=m_n-\sum_{j=1}^{n-1}\binom{n-1}{j-1}\kappa_jm_{n-j},

where the sum is empty when \(n=1\).

Derivation. In the moment recurrence, the \(j=n\) summand is \(\binom{n-1}{n-1}\kappa_nm_0=\kappa_n\). The other summands have \(1\leq j\leq n-1\). Subtracting those terms from both sides leaves exactly the displayed formula. It starts with \(\kappa_1=m_1\), and each subsequent cumulant is determined by the moments and cumulants of lower order. \(\square\)

Worked Example: Raw Moments of a Normal Variable

Let \(X\sim N(3,4)\), where the second parameter is the variance. Its cumulant-generating function is \(K(t)=3t+2t^2\), so \(\kappa_1=3\), \(\kappa_2=4\), and \(\kappa_j=0\) for \(j\geq 3\). Compute the first four raw moments.

The first moment is \(m_1=3\). Then \(m_2=\kappa_1m_1+\kappa_2=3(3)+4=13\).

For the third moment, \(m_3=\kappa_1m_2+2\kappa_2m_1+\kappa_3=3(13)+2(4)(3)+0=63\).

For the fourth, \(m_4=\kappa_1m_3+3\kappa_2m_2+3\kappa_3m_1+\kappa_4=3(63)+3(4)(13)+0+0=345\).

Check by writing \(X=3+Y\), where \(E[Y]=0\), \(E[Y^2]=4\), and \(E[Y^4]=3(4^2)=48\). Expanding the fourth power and taking expectations gives \(E[(3+Y)^4]=3^4+6(3^2)(4)+48=81+216+48=345\).

Using the Recurrence for a Nonnegative Continuous Variable

For distributions whose cumulants are not all zero after a low order, the recurrence still gives a practical calculation. The exponential distribution provides a compact example: its cumulants change at every order, but the same recursion generates its raw moments without differentiating its moment-generating function repeatedly.

Worked Example: Moments of an Exponential Variable

Let \(X\) have an exponential distribution with rate \(2\), with density \(f(x)=2e^{-2x}\) for \(x\geq 0\). Its moment-generating function is \(M(t)=2/(2-t)\) for \(t<2\), and therefore \(K(t)=-\log(1-t/2)\). Expanding the logarithm, or differentiating, gives \(\kappa_j=(j-1)!/2^j\).

The first four cumulants are \(\kappa_1=1/2\), \(\kappa_2=1/4\), \(\kappa_3=1/4\), and \(\kappa_4=3/8\). Thus \(m_1=1/2\).

Next, \(m_2=\kappa_1m_1+\kappa_2=(1/2)(1/2)+1/4=1/2\).

Then \(m_3=\kappa_1m_2+2\kappa_2m_1+\kappa_3=(1/2)(1/2)+2(1/4)(1/2)+1/4=3/4\).

Finally, \(m_4=\kappa_1m_3+3\kappa_2m_2+3\kappa_3m_1+\kappa_4\). Each term equals \(3/8\), so \(m_4=3/2\). These values agree with the general exponential raw moment formula \(m_n=n!/2^n\): for \(n=1,2,3,4\) it gives \(1/2,1/2,3/4,3/2\), respectively.

How to Apply the Recurrence Reliably

For a desired order \(n\), work upward from \(m_0=1\) and use cumulants already calculated. The coefficient on \(\kappa_jm_{n-j}\) is \(\binom{n-1}{j-1}\), not \(\binom{n}{j}\); the latter appears in the equivalent recurrence for \(m_{n+1}\). Keeping the two indexing forms distinct is especially helpful at higher orders.

Obtain cumulants.
Find \(K(t)=\log M(t)\), then differentiate at zero to identify the required \(\kappa_j\).

Initialize the moments.
Set \(m_0=1\), and calculate \(m_1=\kappa_1\).

Build successive orders.
For each \(n\), evaluate every term \(\binom{n-1}{j-1}\kappa_jm_{n-j}\) for \(1\leq j\leq n\), then add the terms.

Check interpretation or an independent identity.
Where possible, compare with a direct moment formula or a known relationship such as \(m_2=\operatorname{Var}(X)+E[X]^2\).

A common mistake is to differentiate \(M'=K'M\) repeatedly but omit the binomial coefficients. For example, the third raw moment is not \(\kappa_1m_2+\kappa_2m_1+\kappa_3\): differentiating the product twice produces the middle term twice, so the correct expression is \(m_3=\kappa_1m_2+2\kappa_2m_1+\kappa_3\). The coefficient \(2\) is the product-rule coefficient \(\binom{2}{1}\). A second useful check is that the recurrence at order two gives \(m_2=\kappa_1^2+\kappa_2\), so \(\kappa_2\) is the variance, not the second raw moment.

Key takeaway: Differentiating \(M'=K'M\) gives a complete recursive conversion between raw moments and cumulants. The binomial coefficients record the product rule, and the term \(\kappa_n\) enters the \(n\)th-moment formula with coefficient one.

Check Your Understanding

Use the recurrence and the definitions in this tutorial to answer the following.

Write the moment recurrence for \(m_5\), including every binomial coefficient.
If \(\kappa_1=2\) and \(\kappa_2=3\), what are \(m_1\) and \(m_2\)?
Explain why the \(j=n\) term in the recurrence is exactly \(\kappa_n\).
For a standard normal variable, the cumulants are \(\kappa_1=0\), \(\kappa_2=1\), and \(\kappa_j=0\) for \(j\geq 3\). Use the recurrence to find \(m_3\) and \(m_4\).
What coefficient is missing in the incorrect expression \(m_3=\kappa_1m_2+\kappa_2m_1+\kappa_3\), and where does it come from?