Cumulants of Scaled Sums and Sample Means

Why Scaling Changes Cumulants by Order

A sum and an average contain the same observations but have different scales. More generally, a weighted sum can amplify some observations, shrink others, or even reverse their signs. Cumulants keep track of these changes in a particularly orderly way: multiplying a random variable by a constant multiplies its cumulant of order $r$ by the $r$th power of that constant, and cumulants add across independent variables. Combining these facts gives a simple formula for weighted sums and, as a special case, sample means.

The first four cumulants are especially interpretable. They describe location, spread, asymmetry, and excess kurtosis. Their scaling shows precisely how these features of a sample mean depend on sample size.

Definitions and Assumptions

Definition: If the moment-generating function $M_X(t)=E[e^{tX}]$ is finite on an open interval containing zero, the cumulant-generating function is $K_X(t)=\log M_X(t)$. The cumulant of order $r$ is $\kappa_r(X)=K_X^{(r)}(0)$, for $r=1,2,\ldots$. In particular, $\kappa_1(X)=E[X]$, $\kappa_2(X)=\operatorname{Var}(X)$, and, when the fourth moment exists, $\kappa_3(X)=E[(X-E[X])^3]$ and $\kappa_4(X)=E[(X-E[X])^4]-3\operatorname{Var}(X)^2$.

The cumulant-generating function condition is convenient for deriving identities at every order. For the first four cumulants, the same formulas can also be defined and used when the moment-generating function does not exist: it is enough that the corresponding moments exist. In particular, the mean, variance, skewness, and excess-kurtosis formulas below require finite fourth moments, not an existing moment-generating function.

Definition: If $X$ has finite fourth moment and $\operatorname{Var}(X)>0$, its skewness is $\gamma_1(X)=\kappa_3(X)/\kappa_2(X)^{3/2}$, and its excess kurtosis is $\gamma_2(X)=\kappa_4(X)/\kappa_2(X)^2$. Excess kurtosis is the ordinary standardized fourth central moment minus $3$.

Cumulants of a Weighted Sum

Let $X_1,\ldots,X_n$ be independent and identically distributed, and let $a_1,\ldots,a_n$ be fixed real numbers. For $S=\sum_{i=1}^n a_iX_i$, the cumulant of order $r$ depends on the weights through their $r$th powers. If the moment-generating function of $X_1$ is finite near zero, the formula holds at every order for which the cumulants are considered.

Theorem: If $X_1,\ldots,X_n$ are independent copies of $X$, and $K_X$ is finite on an open interval containing zero, then $$ \kappa_r\!\left(\sum_{i=1}^n a_iX_i\right) = \kappa_r(X)\sum_{i=1}^n a_i^r. $$

This follows from the cumulant-generating function, using independence to factor the moment-generating function. For $t$ sufficiently close to zero, each $a_it$ lies in the domain of $K_X$, and

K_S(t) =\log E\!\left[\exp\!\left(t\sum_{i=1}^n a_iX_i\right)\right] =\sum_{i=1}^n K_X(a_it).

Taking $r$ derivatives at zero gives $K_S^{(r)}(0)=\sum_i a_i^rK_X^{(r)}(0)$, since differentiating $K_X(a_it)$ $r$ times contributes the factor $a_i^r$. This proves the formula. $\square$

For the first four cumulants, the result also holds under the weaker assumption that $X$ has finite fourth moment. To see the fourth-order point directly, write $Y_i=X_i-E[X]$, $\mu_2=E[Y_i^2]$, and $\mu_4=E[Y_i^4]$. Independence and $E[Y_i]=0$ give

E\!\left[\left(\sum_i a_iY_i\right)^2\right]=\mu_2\sum_i a_i^2, \qquad E\!\left[\left(\sum_i a_iY_i\right)^4\right] =\mu_4\sum_i a_i^4+6\mu_2^2\sum_{i<j}a_i^2a_j^2.

Subtracting three times the squared variance from the fourth central moment cancels the terms involving pairs of distinct indices. The result is $(\mu_4-3\mu_2^2)\sum_i a_i^4=\kappa_4(X)\sum_i a_i^4$. The analogous first-, second-, and third-order calculations give the same weighted-sum formula. Thus finite fourth moments suffice for every formula through order four used here.

Worked Example: A Weighted Sum of Poisson Variables

Let $X_1$ and $X_2$ be independent Poisson random variables with mean $2$, and set $S=X_1+2X_2$. For a Poisson variable with mean $2$, every cumulant is $2$. The weights are $1$ and $2$, so

\kappa_r(S)=2(1^r+2^r).

For $r=1,2,3,4$, this gives $\kappa_1(S)=6$, $\kappa_2(S)=10$, $\kappa_3(S)=18$, and $\kappa_4(S)=34$. Hence $E[S]=6$ and $\operatorname{Var}(S)=10$. The standardized measures are

\gamma_1(S)=\frac{18}{10^{3/2}}\approx 0.569, \qquad \gamma_2(S)=\frac{34}{10^2}=0.34.

The coefficient $2$ contributes $2^r$, not $2$, to the cumulant of order $r$. In particular, its contributions to the second, third, and fourth cumulants are $4$, $8$, and $16$, respectively.

Sample Mean Scaling

The sample mean is the weighted sum with every weight equal to $1/n$. Substituting those weights into the theorem gives a power of $n$ that depends on cumulant order.

Theorem: Let $X_1,\ldots,X_n$ be independent and identically distributed with cumulant $\kappa_r(X_1)$, and define $\bar X=n^{-1}\sum_{i=1}^n X_i$. Whenever the cumulant-generating functions exist near zero, $$ \kappa_r(\bar X)=\frac{\kappa_r(X_1)}{n^{r-1}}. $$ For orders one through four, this also holds if $X_1$ has finite fourth moment.

Indeed, the weighted-sum formula gives $\kappa_r(\bar X)=\kappa_r(X_1)\sum_{i=1}^n(1/n)^r$. There are $n$ equal terms, so this is $n\kappa_r(X_1)/n^r=\kappa_r(X_1)/n^{r-1}$. $\square$

If $E[X_1]=\mu$, $\operatorname{Var}(X_1)=\sigma^2>0$, and the fourth moment is finite, the first four cumulants of the sample mean are therefore

E[\bar X]=\mu,\qquad \operatorname{Var}(\bar X)=\frac{\sigma^2}{n},\qquad \kappa_3(\bar X)=\frac{\kappa_3(X_1)}{n^2},\qquad \kappa_4(\bar X)=\frac{\kappa_4(X_1)}{n^3}.

Standardizing divides the third cumulant by the variance to the power $3/2$, and the fourth cumulant by the variance squared. Thus

\gamma_1(\bar X)=\frac{\gamma_1(X_1)}{\sqrt n}, \qquad \gamma_2(\bar X)=\frac{\gamma_2(X_1)}{n}.

For example, the skewness shrinks at the rate $n^{-1/2}$, while excess kurtosis shrinks at the rate $n^{-1}$. These statements concern standardized shape; they do not say that the third or fourth cumulant itself shrinks at those rates.

Worked Example: The Mean of Bernoulli Observations

Suppose $X_1,\ldots,X_{25}$ are independent Bernoulli variables with success probability $p=0.2$. For a Bernoulli variable, $\kappa_1=p$, $\kappa_2=p(1-p)$, $\kappa_3=p(1-p)(1-2p)$, and $\kappa_4=p(1-p)[1-6p(1-p)]$. Here $p(1-p)=0.2(0.8)=0.16$, so the individual cumulants are

\kappa_1=0.2,\quad \kappa_2=0.16,\quad \kappa_3=0.16(0.6)=0.096,\quad \kappa_4=0.16(1-0.96)=0.0064.

For $\bar X$, the first four cumulants are

\kappa_1(\bar X)=0.2,\quad \kappa_2(\bar X)=\frac{0.16}{25}=0.0064,\quad \kappa_3(\bar X)=\frac{0.096}{25^2}=0.0001536,\quad \kappa_4(\bar X)=\frac{0.0064}{25^3}=0.0000004096.

The mean is $0.2$ and the variance is $0.0064$. The skewness and excess kurtosis are

\gamma_1(\bar X)=\frac{0.096}{0.16^{3/2}\sqrt{25}}=\frac{1.5}{5}=0.3, \qquad \gamma_2(\bar X)=\frac{1-6(0.16)}{25(0.16)}=\frac{0.04}{4}=0.01.

As a check, the individual Bernoulli variable has skewness $0.096/0.16^{3/2}=1.5$ and excess kurtosis $0.0064/0.16^2=0.25$. Dividing these by $\sqrt{25}=5$ and $25$, respectively, gives $0.3$ and $0.01$, as required.

Worked Example: The Mean of Exponential Observations

Let $X_1,\ldots,X_4$ be independent exponential variables with mean $2$. Their cumulants are $\kappa_r(X_i)=(r-1)!2^r$, so the first four are $2,4,16,96$. For the sample mean, the corresponding divisors are $1,4,16,64$:

\kappa_1(\bar X)=2,\qquad \kappa_2(\bar X)=\frac{4}{4}=1,\qquad \kappa_3(\bar X)=\frac{16}{16}=1,\qquad \kappa_4(\bar X)=\frac{96}{64}=1.5.

Thus $\bar X$ has mean $2$, variance $1$, skewness $1/1^{3/2}=1$, and excess kurtosis $1.5/1^2=1.5$. The original exponential variable has skewness $2$ and excess kurtosis $6$; the sample-mean formulas give $2/\sqrt{4}=1$ and $6/4=1.5$, confirming the calculation.

Reading the Formula for General Weights

For a weighted sum $S=\sum_i a_iX_i$ of independent copies with variance $\sigma^2>0$, its variance is $\sigma^2\sum_i a_i^2$. Provided the fourth moment is finite and $\sum_i a_i^2>0$, its standardized skewness and excess kurtosis are

\gamma_1(S)=\gamma_1(X)\frac{\sum_i a_i^3}{(\sum_i a_i^2)^{3/2}}, \qquad \gamma_2(S)=\gamma_2(X)\frac{\sum_i a_i^4}{(\sum_i a_i^2)^2}.

The third-power sum can be negative or zero, so a weighted sum may reverse or cancel skewness. The fourth-power sum is nonnegative, and the fourth-order formula retains the sign of the original excess kurtosis. If all weights are equal to $1/n$, these expressions reduce to the sample-mean rates above.

Common Mistakes and Why the Scaling Matters

Using the same divisor for every cumulant. The cumulant of order $r$ for a sample mean is divided by $n^{r-1}$, not by $n$ at every order. The third cumulant is divided by $n^2$, and the fourth by $n^3$.
Confusing cumulants with standardized measures. The sample mean’s fourth cumulant is divided by $n^3$, but its excess kurtosis is divided by $n$, because standardization also divides by the square of its variance.
Forgetting the power on a weight. A weight $a_i$ contributes $a_i^r$ to the order-$r$ cumulant. Replacing this with $a_i$ is correct only for the first cumulant.
Assuming scaling changes the mean. Averaging preserves the common mean: $\kappa_1(\bar X)=\mu$. Its variance and higher cumulants, rather than its mean, depend on sample size.

These relationships help distinguish changes in location and spread from changes in standardized shape. They also provide quick checks on algebra in sampling calculations: the mean remains fixed, the variance decreases inversely with sample size, and the standardized skewness and excess kurtosis decrease at their respective rates when the observations are independent and identically distributed with finite fourth moments.

Check Your Understanding

Use the weighted-sum formula and the first-four-cumulant relationships to answer the following.

If $S=3X_1-X_2$ for independent identically distributed variables, what factor multiplies $\kappa_3(X_1)$ to give $\kappa_3(S)$?
A distribution has variance $9$ and third cumulant $27$. What is its skewness, and what is the skewness of the mean of four independent copies?
If a variable has excess kurtosis $2$, what is the excess kurtosis of the mean of ten independent copies?
Why does the fourth cumulant of a sample mean scale by $n^{-3}$, while its excess kurtosis scales by $n^{-1}$?
What condition on the weights makes the variance of a weighted sum positive when the common variance is positive?