Affine Changes to Cumulants

Why Affine Changes Have a Simple Cumulant Rule

A change of location or scale alters the numerical values of a random variable without changing the underlying source of its variation. Cumulants make this distinction precise: adding a constant changes the first cumulant, while scaling changes the $r$th cumulant by the $r$th power of the scale factor. The result follows directly from how an affine transformation changes the cumulant-generating function.

The domain matters as well as the algebra. If $X$ has a moment-generating function that is finite near zero, then the transformed variable $aX+b$ has one on the corresponding transformed domain. In particular, a logarithm arising from an exponential distribution must be evaluated only where its argument is positive.

Definition: The moment-generating function of $X$ is $M_X(t)=E[e^{tX}]$ at values of $t$ for which this expectation is finite. Its domain is $D_X=\{t\in\mathbb{R}:M_X(t)<\infty\}$. 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, and the $r$th cumulant is $\kappa_r(X)=K_X^{(r)}(0)$, for integers $r\geq 1$.

Finiteness on an open interval around zero ensures that the moment-generating function, and hence its logarithm, can be differentiated there to obtain cumulants of every order. This is the setting used below. A distribution can have some finite moments without a moment-generating function near zero; the results here concern cumulants defined through the cumulant-generating function.

The Affine Transformation Rule

Let $Y=aX+b$, where $a,b\in\mathbb{R}$ are constants. The transformation has two distinct effects: scaling replaces the argument $t$ in $K_X$ by $at$, and translation contributes a linear term $bt$.

Theorem: Suppose $M_X(t)$ is finite on an open interval containing zero, and let $Y=aX+b$. If $a\neq 0$, then the domain of $M_Y$ is $D_Y=\{t:at\in D_X\}$, and $$ K_Y(t)=bt+K_X(at), \qquad t\in D_Y. $$ For every integer $r\geq 1$, $$ \kappa_r(Y)=a^r\kappa_r(X)+b\,\mathbf{1}_{\{r=1\}}. $$ If $a=0$, then $Y=b$, $K_Y(t)=bt$ for every real $t$, and the same cumulant formula holds.

Proof. First suppose $a\neq 0$. Since $e^{bt}$ is finite and strictly positive for every real $t$, $$ M_Y(t)=E[e^{t(aX+b)}]=e^{bt}E[e^{(at)X}]=e^{bt}M_X(at) $$ is finite exactly when $M_X(at)$ is finite. Thus $D_Y=\{t:at\in D_X\}$. Taking logarithms on this domain gives $K_Y(t)=bt+K_X(at)$. Differentiating $r$ times at zero, the linear term contributes $b$ when $r=1$ and zero when $r\geq 2$, while repeated use of the chain rule gives $$ K_Y^{(r)}(0)=b\,\mathbf{1}_{\{r=1\}}+a^rK_X^{(r)}(0) =b\,\mathbf{1}_{\{r=1\}}+a^r\kappa_r(X). $$ If $a=0$, then $Y=b$ and $M_Y(t)=e^{bt}$, so $K_Y(t)=bt$. Its first derivative at zero is $b$, and all higher derivatives are zero, as required by the formula. $\square$

The indicator in the theorem makes the location and scale effects easy to separate. For a pure translation, set $a=1$: the first cumulant increases by $b$, and all higher cumulants remain unchanged. For a pure scaling, set $b=0$: every cumulant of order $r$ is multiplied by $a^r$. In particular, a negative scale reverses the sign of odd-order cumulants but not even-order ones.

Transform the CGF.
Write $K_{aX+b}(t)=bt+K_X(at)$, checking that $at$ lies in the domain of $K_X$.

Differentiate by order.
The $r$ derivatives of $K_X(at)$ contribute a factor $a^r$; the term $bt$ contributes only at order one.

Check special cases.
For $a=0$, the variable is constant. For $b=0$, only scaling applies; for $a=1$, only translation applies.

Worked Applications

Worked Example: An Affine Transformation of a Normal Variable

Suppose $X\sim N(2,9)$, and define $Y=4-\frac{1}{3}X$. Find the CGF and cumulants of $Y$.

For $X\sim N(\mu,\sigma^2)$, the CGF is $K_X(t)=\mu t+\frac{\sigma^2t^2}{2}$. Thus here $K_X(t)=2t+\frac{9t^2}{2}$, finite for every real $t$. With $a=-\frac13$ and $b=4$, $$ K_Y(t)=4t+K_X\left(-\frac{t}{3}\right) =4t-\frac{2t}{3}+\frac{9}{2}\frac{t^2}{9} =\frac{10}{3}t+\frac{t^2}{2}. $$

The transformed CGF is finite for all real $t$. Its first derivative at zero is $10/3$, its second derivative is $1$, and all derivatives of order three or higher are zero. Equivalently, the affine rule gives $\kappa_1(Y)=4-\frac13(2)=10/3$, $\kappa_2(Y)=(-1/3)^2(9)=1$, and $\kappa_r(Y)=0$ for $r\geq 3$. These results agree with $Y\sim N(10/3,1)$.

Worked Example: An Exponential Variable and Its Transformed Domain

Let $X$ have an exponential distribution with rate $2$, so its density is $f(x)=2e^{-2x}$ for $x\geq 0$. Set $Y=4+3X$. Find $K_Y$, its domain, and its cumulants.

For $t<2$, direct integration gives $$ M_X(t)=\int_0^\infty e^{tx}2e^{-2x}\,dx =2\int_0^\infty e^{-(2-t)x}\,dx =\frac{2}{2-t}=\frac{1}{1-t/2}. $$ For $t\geq 2$, this integral diverges, so $D_X=(-\infty,2)$ and $K_X(t)=-\log(1-t/2)$ on that domain.

Here $a=3$ and $b=4$. The transformed domain consists of $t$ for which $3t<2$, or $t<2/3$. Therefore $$ K_Y(t)=4t+K_X(3t) =4t-\log(1-3t/2), \qquad t<\frac{2}{3}. $$ The bound follows both from $3t\in D_X$ and from requiring the logarithm's argument $1-3t/2$ to be positive.

For $r\geq 1$, differentiating the series-free expression for $K_X$ gives $$ K_X^{(r)}(0)=\frac{(r-1)!}{2^r}. $$ Indeed, $K_X'(t)=1/(2-t)$, and differentiating $r-1$ more times gives $K_X^{(r)}(t)=(r-1)!/(2-t)^r$. The affine rule now yields $$ \kappa_1(Y)=4+3\left(\frac12\right)=\frac{11}{2}, \qquad \kappa_r(Y)=3^r\frac{(r-1)!}{2^r}=(r-1)!\left(\frac32\right)^r,\quad r\geq 2. $$ For example, $\kappa_2(Y)=9/4$ and $\kappa_3(Y)=27/4$. The CGF formula is not valid at $t=1$: there $1-3t/2<0$, and $E[e^Y]$ is infinite.

Worked Example: A Negative Scale for a Bernoulli Variable

Let $X\sim\operatorname{Bernoulli}(1/4)$, so $P(X=1)=1/4$ and $P(X=0)=3/4$, and put $Y=5-2X$. Compute the first three cumulants of $Y$.

The CGF of $X$ is $K_X(t)=\log(3/4+e^t/4)$, which is finite for every real $t$. Its first three cumulants are $$ \kappa_1(X)=\frac14,\qquad \kappa_2(X)=\frac14\frac34=\frac{3}{16},\qquad \kappa_3(X)=\frac14\frac34\left(1-2\frac14\right)=\frac{3}{32}. $$ These expressions follow by differentiating $\log(1-p+pe^t)$ at zero: its first derivative is $pe^t/(1-p+pe^t)$, and the next two derivatives evaluated at zero are $p(1-p)$ and $p(1-p)(1-2p)$.

For $Y$, the affine constants are $a=-2$ and $b=5$. Hence $$ \kappa_1(Y)=5-2\left(\frac14\right)=\frac92,\qquad \kappa_2(Y)=(-2)^2\frac{3}{16}=\frac34,\qquad \kappa_3(Y)=(-2)^3\frac{3}{32}=-\frac34. $$ A direct check uses $P(Y=5)=3/4$ and $P(Y=3)=1/4$. The mean is $15/4+3/4=9/2$; the variance is $(3/4)(1/2)^2+(1/4)(-3/2)^2=3/4$; and the third central moment is $(3/4)(1/2)^3+(1/4)(-3/2)^3=-3/4$, agreeing with the cumulants calculated above.

Interpretation and Common Pitfalls

Translation changes the origin of measurement, not the shape around that origin. Accordingly, it changes the first cumulant, which is the mean, but leaves every higher-order cumulant unchanged. Scaling changes the units and orientation: a scale factor $a$ multiplies the order-$r$ cumulant by $a^r$. This order dependence is essential; applying the scale factor only once would give incorrect results for every order other than one.

The rule is useful when a variable has already been described by its CGF or cumulants and a later calculation uses a shifted or rescaled version. It avoids recomputing derivatives from scratch. However, the domain transformation must accompany the algebra: a formal expression for $K_X(at)$ does not make sense as a finite CGF where $M_X(at)$ diverges.

Mistake: adding $b$ to every cumulant. The linear term $bt$ has zero derivatives of order two and higher. Only $\kappa_1$ receives the location shift.
Mistake: multiplying every cumulant by $a$. The chain rule contributes one factor of $a$ per derivative, giving $a^r$ at order $r$.
Mistake: overlooking a negative scale. When $a<0$, odd-order cumulants change sign according to $a^r$; even-order cumulants do not.
Mistake: keeping the old domain unchanged. The transformed CGF is finite only when $at\in D_X$. For positive $a$ the boundary rescales, and for negative $a$ the direction of the inequality reverses.

Key takeaway: For $Y=aX+b$, the exact rule is $\kappa_r(Y)=a^r\kappa_r(X)+b\,\mathbf{1}_{\{r=1\}}$, with the CGF domain transformed by requiring $at\in D_X$ when $a\neq 0$.

Check Your Understanding

Use the affine CGF identity and cumulant rule to answer each question.

If $Y=X+7$, which cumulants of $X$ change, and by how much?
If $Y=-3X$, express $\kappa_2(Y)$ and $\kappa_3(Y)$ in terms of the corresponding cumulants of $X$.
If $D_X=(-\infty,5)$ and $Y=2X+1$, what is the domain of $M_Y$?
If $Y=-2X$ and $D_X=(-1,4)$, what is the transformed CGF domain?
For $a=0$, what are the first and all higher cumulants of $Y=aX+b$?

What You'll Learn