Normal Cumulants and Gaussian Characterization

Why the Normal Distribution Has a Special Cumulant Pattern

Cumulants summarize aspects of a distribution through derivatives of its cumulant-generating function. For a normal random variable, this function is especially simple: it is a polynomial of degree at most two. The consequence is an exact pattern—only the first two cumulants can be nonzero. More than a computational shortcut, this pattern has a converse: if a random variable’s cumulant-generating function is quadratic in a neighborhood of zero, its distribution must be normal.

The converse needs a condition. A formal list of moments or cumulants does not, in general, identify a distribution. Here we require the moment-generating function to be finite in an open interval around zero. That local finiteness permits the cumulants to be defined by derivatives and gives the generating function enough information to identify the law.

Definitions and Regularity

Definition: For a random variable \(X\), its moment-generating function is \(M_X(t)=E[e^{tX}]\) at values of \(t\) for which this expectation is finite. If \(M_X(t)\) is finite on an open interval containing zero, the cumulant-generating function is \(K_X(t)=\log M_X(t)\) on that interval. The \(r\)th cumulant is \(\kappa_r(X)=K_X^{(r)}(0)\), for each integer \(r\geq 1\).

Finiteness on an open interval around zero ensures that these derivatives exist at every order. To see the relevant regularity, choose \(b>0\) such that \(M_X(b)\) and \(M_X(-b)\) are finite. Then \(E[e^{b|X|}]<\infty\), since \(e^{b|x|}\leq e^{bx}+e^{-bx}\). For any \(0<a<b\) and fixed integer \(r\geq 0\), \(|x|^r e^{a|x|}\leq C e^{b|x|}\) for some finite constant \(C\). This provides an integrable bound for differentiating \(M_X(t)\) under the expectation when \(|t|<a\). In particular, \(M_X\), and therefore \(K_X\), have derivatives of every order near zero.

The first two cumulants are the mean and variance when these quantities are defined: \(K_X'(0)=E[X]\) and \(K_X''(0)=\operatorname{Var}(X)\). The higher cumulants are derivatives of \(K_X\), not higher central moments themselves.

Worked Example: Deriving the Cumulant-Generating Function of a Normal Variable

Let \(X\sim N(\mu,\sigma^2)\) with \(\sigma^2>0\). Its density is \(f(x)=(\sigma\sqrt{2\pi})^{-1}\exp(-(x-\mu)^2/(2\sigma^2))\). We compute its moment-generating function directly:

\begin{aligned} M_X(t) &=\int_{-\infty}^{\infty} e^{tx} \frac{1}{\sigma\sqrt{2\pi}} \exp\!\left(-\frac{(x-\mu)^2}{2\sigma^2}\right)\,dx\\ &=\exp\!\left(\mu t+\frac{\sigma^2t^2}{2}\right) \int_{-\infty}^{\infty} \frac{1}{\sigma\sqrt{2\pi}} \exp\!\left(-\frac{(x-\mu-\sigma^2t)^2}{2\sigma^2}\right)\,dx\\ &=\exp\!\left(\mu t+\frac{\sigma^2t^2}{2}\right). \end{aligned}

For the second line, the exponents are equal because \(tx-(x-\mu)^2/(2\sigma^2)=\mu t+\sigma^2t^2/2-(x-\mu-\sigma^2t)^2/(2\sigma^2)\). The integral in that line is one: its integrand is a normal density with mean \(\mu+\sigma^2t\) and variance \(\sigma^2\). Thus \(M_X(t)\) is finite for every real \(t\), and taking its logarithm gives \(K_X(t)=\mu t+\sigma^2t^2/2\).

The Normal Cumulants

Theorem: If \(X\sim N(\mu,\sigma^2)\), then \(\kappa_1(X)=\mu\), \(\kappa_2(X)=\sigma^2\), and \(\kappa_r(X)=0\) for every integer \(r\geq 3\). This also holds for the degenerate normal distribution with \(\sigma^2=0\).

For \(\sigma^2>0\), the worked calculation gives \(K_X(t)=\mu t+\sigma^2t^2/2\). Its first derivative at zero is \(\mu\), its second derivative is \(\sigma^2\), and each derivative of order three or higher is identically zero. If \(\sigma^2=0\), then \(X=\mu\) with probability one, \(M_X(t)=e^{\mu t}\), and \(K_X(t)=\mu t\); the same cumulant conclusion follows.

The zero higher cumulants do not mean that higher moments vanish. For example, a standard normal variable has a positive fourth moment, even though its fourth cumulant is zero.

Worked Example: A Zero Fourth Cumulant Is Not a Zero Fourth Moment

Let \(Z\sim N(0,1)\). Its cumulant-generating function is \(K_Z(t)=t^2/2\), so \(\kappa_1(Z)=0\), \(\kappa_2(Z)=1\), and \(\kappa_4(Z)=0\). To calculate the fourth moment, use \(M_Z(t)=e^{t^2/2}\). The terms through degree four in its power series are \(1+t^2/2+t^4/8\). Hence

E[Z^4]=M_Z^{(4)}(0)=4!\left(\frac{1}{8}\right)=3.

Thus \(E[Z^4]=3\) while \(\kappa_4(Z)=0\). Moments and cumulants are related, but they are not interchangeable: cumulants are derivatives of the logarithm of the moment-generating function, whereas moments are derivatives of the moment-generating function itself.

Quadratic Cumulant-Generating Functions Characterize Normality

The converse uses a uniqueness property of moment-generating functions. If two moment-generating functions are finite on an open interval containing zero and agree on some open interval around zero, then the corresponding distributions are equal. One way to see why is that exponential integrability extends each generating function to an analytic function on a complex strip around the imaginary axis. Agreement on a real interval implies agreement throughout that strip by the identity theorem for analytic functions. In particular, the characteristic functions agree; the uniqueness theorem for characteristic functions then gives equality of the distributions.

Theorem: Suppose \(M_X(t)\) is finite on an open interval containing zero. If there are constants \(\mu\in\mathbb R\) and \(\sigma^2\geq 0\) such that \(K_X(t)=\mu t+\sigma^2t^2/2\) throughout an open interval containing zero, then \(X\sim N(\mu,\sigma^2)\), with \(\sigma^2=0\) interpreted as a point mass at \(\mu\).

Proof. Exponentiating the assumed identity gives \(M_X(t)=\exp(\mu t+\sigma^2t^2/2)\) on that interval. For \(\sigma^2>0\), the calculation for the normal distribution shows that this is the moment-generating function of \(N(\mu,\sigma^2)\); that normal MGF is finite for every real \(t\). Since both MGFs are finite near zero and agree there, the MGF uniqueness theorem implies that \(X\) has distribution \(N(\mu,\sigma^2)\).

If \(\sigma^2=0\), the same local identity gives \(M_X(t)=e^{\mu t}\), which is the MGF of the constant random variable equal to \(\mu\). MGF uniqueness again implies that \(X=\mu\) with probability one. This is the degenerate normal case. \(\square\)

In particular, if the cumulant-generating function exists near zero and all cumulants above order two vanish, its Taylor expansion near zero is quadratic. The constant term is \(K_X(0)=\log 1=0\); the linear and quadratic coefficients are \(K_X'(0)=E[X]\) and \(K_X''(0)/2=\operatorname{Var}(X)/2\). The analytic expansion therefore satisfies the theorem’s hypothesis, so \(X\) is normal. The requirement that the generating function exist near zero is essential to this argument.

Worked Example: Identifying a Distribution From a Local MGF

Suppose a random variable \(Y\) has a moment-generating function finite near zero and \(M_Y(t)=\exp(2t+5t^2/2)\) for all \(t\) in some open interval containing zero. Then \(K_Y(t)=2t+5t^2/2\) there, so its first two cumulants are \(2\) and \(5\), and every higher cumulant is zero.

The normal distribution \(N(2,5)\) has exactly this MGF, as follows from the normal MGF formula with \(\mu=2\) and \(\sigma^2=5\). The MGF uniqueness theorem applies because the two MGFs agree on an open interval around zero and are finite there. Therefore \(Y\sim N(2,5)\). The conclusion identifies the whole distribution, not just its mean and variance.

Why Matching Two Cumulants Is Not Enough

The characterization requires the entire cumulant-generating function to be quadratic near zero, not merely that the first two cumulants match those of a normal distribution. A nonnormal variable can have the same mean and variance as a normal variable, while its higher cumulants reveal the difference.

Worked Example: A Symmetric Two-Point Distribution Is Not Normal

Let \(W\) take the values \(-1\) and \(1\), each with probability \(1/2\). Its MGF and cumulant-generating function are \(M_W(t)=(e^{-t}+e^t)/2=\cosh(t)\) and \(K_W(t)=\log(\cosh t)\). Expand near zero:

\begin{aligned} \cosh(t)&=1+\frac{t^2}{2}+\frac{t^4}{24}+O(t^6),\\ \log(\cosh t) &=\frac{t^2}{2} +\left(\frac{1}{24}-\frac{1}{8}\right)t^4+O(t^6)\\ &=\frac{t^2}{2}-\frac{t^4}{12}+O(t^6). \end{aligned}

The coefficient of \(t^2\) gives \(\kappa_2(W)=1\), and the coefficient of \(t^4\) gives \(\kappa_4(W)=4!(-1/12)=-2\). The mean is zero by symmetry. Thus \(W\) has the same mean and variance as \(N(0,1)\), but its fourth cumulant is not zero. Its cumulant-generating function is not quadratic, and \(W\) is not normal.

Common Pitfall and Takeaway

Wrong reasoning: “The first two cumulants determine the mean and variance, and they match a normal distribution, so the variable must be normal.” Matching the first two cumulants establishes only the mean and variance. The two-point example has mean zero and variance one but is not standard normal.

Correction: To use the Gaussian characterization, verify that the cumulant-generating function exists in an open interval around zero and is quadratic throughout such an interval. Equivalently, under that existence condition, all cumulants of order three and higher must vanish. A finite set of matching cumulants does not suffice.

Key takeaway: A normal distribution has cumulant-generating function \(K_X(t)=\mu t+\sigma^2t^2/2\), so only its first two cumulants can be nonzero. Conversely, if a random variable has a cumulant-generating function of that quadratic form near zero, MGF uniqueness forces its distribution to be \(N(\mu,\sigma^2)\).

Check Your Understanding

Try these questions using the definitions and arguments above.

For \(X\sim N(4,9)\), write \(K_X(t)\) and give \(\kappa_1(X)\), \(\kappa_2(X)\), and \(\kappa_5(X)\).
Why does a zero fourth cumulant not imply that the fourth moment is zero? Give the standard normal values of both quantities.
A random variable has a finite MGF near zero and \(K_X(t)=3t+2t^2\) there. Identify its distribution and state its variance.
Why do a mean of zero and variance of one alone not establish that a variable is standard normal?
What regularity condition is needed to apply the local quadratic characterization, and where does it enter the argument?