Poisson Cumulants and Their Equal-Order Pattern

Why the Poisson Cumulants Have a Simple Pattern

A Poisson random variable records a count, but its distribution has more structure than just a convenient formula for probabilities. The logarithm of its moment-generating function is especially simple: its derivatives at zero are identical at every order. This makes the Poisson law a useful setting for seeing how cumulants encode information that can be harder to spot in raw moments.

We will derive the cumulant-generating function directly, establish the equal-order pattern, and use it to calculate moments and standardized cumulants. The rate parameter is denoted by \(\lambda\), with \(\lambda\geq 0\). When \(\lambda=0\), the distribution is degenerate at zero; standardized cumulants require \(\lambda>0\), since they involve division by a positive power of the variance.

Definition: If \(X\sim\operatorname{Poisson}(\lambda)\), then \(X\) takes values \(k=0,1,2,\ldots\), with \(P(X=k)=e^{-\lambda}\lambda^k/k!\). Its moment-generating function is \(M_X(t)=E[e^{tX}]\). Wherever this function is finite and positive, its cumulant-generating function is \(K_X(t)=\log M_X(t)\), and the \(r\)th cumulant is \(\kappa_r=K_X^{(r)}(0)\), for integers \(r\geq 1\).

Deriving the Poisson CGF

For \(\lambda>0\), sum \(e^{tk}\) against the Poisson probability mass function. The resulting series is finite for every real \(t\), because it is an exponential series with argument \(\lambda e^t\). The case \(\lambda=0\) gives \(M_X(t)=1\) directly.

\begin{aligned} M_X(t) &=\sum_{k=0}^{\infty}e^{tk}e^{-\lambda}\frac{\lambda^k}{k!}\\ &=e^{-\lambda}\sum_{k=0}^{\infty}\frac{(\lambda e^t)^k}{k!}\\ &=e^{-\lambda}e^{\lambda e^t} =\exp\{\lambda(e^t-1)\},\qquad t\in\mathbb{R}. \end{aligned}

Thus the moment-generating function exists on the whole real line, and taking its logarithm gives \(K_X(t)=\lambda(e^t-1)\). Since this function is differentiable of every order, all cumulants are defined. In particular, there is no restriction to a small neighborhood of zero when differentiating this CGF.

Worked Example: Finding the CGF and Cumulants at Rate 3

Let \(X\sim\operatorname{Poisson}(3)\). Substituting \(\lambda=3\) into the derived formula gives

$$ K_X(t)=3(e^t-1). $$

Differentiating gives \(K_X'(t)=3e^t\), \(K_X''(t)=3e^t\), and, more generally, \(K_X^{(r)}(t)=3e^t\) for every integer \(r\geq 1\). Evaluating at zero yields \(\kappa_r=3\) at every order. For example, \(\kappa_1=3\) is the mean and \(\kappa_2=3\) is the variance; the third and fourth cumulants are also 3.

Every Cumulant Equals the Rate

Theorem: If \(X\sim\operatorname{Poisson}(\lambda)\), then \(K_X(t)=\lambda(e^t-1)\) for every real \(t\), and \(\kappa_r=\lambda\) for every integer \(r\geq 1\). This includes \(\lambda=0\).

Proof. The probability generating series calculation above gives \(M_X(t)=\exp\{\lambda(e^t-1)\}\) for \(\lambda>0\); for \(\lambda=0\), \(X=0\) with probability one and the same formula gives \(M_X(t)=1\). Therefore \(K_X(t)=\lambda(e^t-1)\) for all \(\lambda\geq0\). The derivative of \(e^t-1\) is \(e^t\), and every further derivative remains \(e^t\). Hence, for any integer \(r\geq1\), \(K_X^{(r)}(t)=\lambda e^t\), so \(K_X^{(r)}(0)=\lambda\). This is exactly the definition of the \(r\)th cumulant. \(\square\)

The equality is about cumulants, not moments. For instance, the second raw moment is not simply \(\lambda\): it contains both the contribution from the mean and the second cumulant. Recovering moments from cumulants makes this distinction explicit.

Recovering the First Four Moments

The moment-cumulant relations express raw moments in terms of cumulants. Through order four, writing \(m_r=E[X^r]\), they 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. \end{aligned}

For a Poisson variable, substituting \(\kappa_r=\lambda\) gives

\begin{aligned} E[X]&=\lambda,\\ E[X^2]&=\lambda^2+\lambda,\\ E[X^3]&=\lambda^3+3\lambda^2+\lambda,\\ E[X^4]&=\lambda^4+6\lambda^3+7\lambda^2+\lambda. \end{aligned}

The fourth-moment coefficient \(7\) comes from adding the \(6\lambda^3\) term and the contribution \(3\kappa_2^2=3\lambda^2\) alongside \(4\kappa_1\kappa_3=4\lambda^2\): the terms of order \(\lambda^2\) total \(3+4=7\). Thus the result retains contributions from different cumulant combinations, even though all the cumulants themselves are equal.

Worked Example: Raw Moments at Rate 2

Let \(X\sim\operatorname{Poisson}(2)\). Each cumulant is 2. Applying the moment-cumulant formulas gives

\begin{aligned} E[X]&=2,\\ E[X^2]&=2^2+2=6,\\ E[X^3]&=2^3+3(2)(2)+2=8+12+2=22,\\ E[X^4]&=2^4+6(2^3)+7(2^2)+2\\ &=16+48+28+2=94. \end{aligned}

These are raw moments, not central moments. For example, \(E[X^2]=6\), while the variance is \(E[X^2]-(E[X])^2=6-4=2\), agreeing with the second cumulant.

The first cumulant is the mean and the second is the variance. The third cumulant equals the third central moment. The fourth cumulant is the fourth central moment minus three times the squared variance. Consequently, for a Poisson variable,

\begin{aligned} \operatorname{Var}(X)&=\lambda,\\ E[(X-\lambda)^3]&=\lambda,\\ E[(X-\lambda)^4]&=\kappa_4+3\kappa_2^2 =\lambda+3\lambda^2. \end{aligned}

The fourth central moment is not itself the fourth cumulant: it includes the additional \(3\lambda^2\) term. This distinction is important when moving between moments and cumulants.

Worked Example: Checking Central Moments at Rate 2

Continue with \(X\sim\operatorname{Poisson}(2)\). The mean is 2, and the raw moments from the previous calculation are \(m_1=2\), \(m_2=6\), \(m_3=22\), and \(m_4=94\). Expanding the centered powers gives

\begin{aligned} E[(X-2)^2]&=m_2-4m_1+4=6-8+4=2,\\ E[(X-2)^3]&=m_3-6m_2+12m_1-8\\ &=22-36+24-8=2,\\ E[(X-2)^4]&=m_4-8m_3+24m_2-32m_1+16\\ &=94-176+144-64+16=14. \end{aligned}

The general central-moment formulas give variance \(2\), third central moment \(2\), and fourth central moment \(2+3(2^2)=14\), confirming all three direct calculations.

Standardized Cumulants

For a distribution with finite positive variance, the standardized cumulant of order \(r\geq3\) is \(\kappa_r/\kappa_2^{r/2}\). For a Poisson variable with \(\lambda>0\), every cumulant equals \(\lambda\), so its standardized cumulants of orders \(r\geq3\) follow one formula:

\frac{\kappa_r}{\kappa_2^{r/2}} =\frac{\lambda}{\lambda^{r/2}} =\lambda^{1-r/2},\qquad r\geq3.

At order three, this is the skewness, \(\lambda^{-1/2}=1/\sqrt{\lambda}\). At order four, it is the excess kurtosis, \(\lambda^{-1}\). The ordinary kurtosis, defined as the fourth central moment divided by the variance squared, is the excess kurtosis plus 3; it is therefore \(3+1/\lambda\) for the Poisson distribution. The skewness and excess kurtosis decrease as the rate grows, while the unstandardized cumulants remain equal to the rate.

Worked Example: Standardized Cumulants at Rate 9

Let \(X\sim\operatorname{Poisson}(9)\). Its variance is 9, so its standard deviation is 3. The third cumulant is 9, giving skewness

\frac{\kappa_3}{\kappa_2^{3/2}} =\frac{9}{9^{3/2}} =\frac{9}{27} =\frac{1}{3}.

The fourth cumulant is also 9, giving excess kurtosis

\frac{\kappa_4}{\kappa_2^2} =\frac{9}{81} =\frac{1}{9}.

Thus the ordinary kurtosis is \(3+1/9=28/9\). As a check using the central fourth moment, \(E[(X-9)^4]=9+3(9^2)=252\), and \(252/9^2=252/81=28/9\), the same ordinary kurtosis.

Common Mistakes and Why the Pattern Matters

One common mistake is to read “every cumulant equals \(\lambda\)” as “every moment equals \(\lambda\).” That is false: already \(E[X^2]=\lambda^2+\lambda\), not \(\lambda\) except at special parameter values. Cumulants and moments are related but are not interchangeable. A second mistake is to call the fourth central moment the fourth cumulant; for a Poisson variable these are \(\lambda+3\lambda^2\) and \(\lambda\), respectively.

The parameter-zero case also needs care. When \(\lambda=0\), the cumulant statement remains valid: every cumulant is zero. But the standardized cumulants are undefined because the variance is zero. In general, standardization requires a nonzero finite standard deviation; the formula \(\lambda^{1-r/2}\) is stated only for \(\lambda>0\).

The equal-cumulant pattern gives a compact description of the Poisson distribution's moment structure. It provides a quick route from the CGF to the mean and variance, higher raw and central moments, and measures of asymmetry and tail shape. It also supplies a useful check on calculations: if a proposed Poisson cumulant of any positive integer order differs from \(\lambda\), either the derivative or the cumulant calculation needs to be revisited.

Key takeaway: For \(X\sim\operatorname{Poisson}(\lambda)\), the CGF is \(K_X(t)=\lambda(e^t-1)\), so every cumulant is \(\lambda\). For \(\lambda>0\), the skewness is \(1/\sqrt{\lambda}\) and the excess kurtosis is \(1/\lambda\).

Check Your Understanding

Use the CGF and moment-cumulant relationships developed above to answer these questions.

For \(X\sim\operatorname{Poisson}(5)\), write its CGF and state its first four cumulants.
Use cumulants to calculate \(E[X^2]\) and \(E[X^3]\) when \(X\sim\operatorname{Poisson}(4)\).
For \(X\sim\operatorname{Poisson}(16)\), calculate the skewness and excess kurtosis.
Explain why the fourth central moment and the fourth cumulant of a Poisson variable are not generally equal.
What are the cumulants when \(\lambda=0\), and why are standardized cumulants undefined in that case?