Affine Transformations of Characteristic Functions

Why affine transformations are simple in the characteristic-function domain

A change of location or scale is one of the most common ways to describe a new random variable from an old one. If \(X\) is a measurement, then \(aX+b\) rescales it by \(a\) and shifts it by \(b\). Characteristic functions make these operations especially transparent: scaling changes the frequency at which the original characteristic function is evaluated, while translation contributes a complex phase.

This relationship is useful in calculations and in interpreting formulas. It also makes sign conventions important. A negative scale reflects the variable, and the characteristic function must then be evaluated at a negative argument. The formula remains valid whether or not \(X\) has moments.

Definition: For a real-valued random variable \(X\), its characteristic function is the function \(\varphi_X:\mathbb{R}\to\mathbb{C}\) defined, for every real \(t\), by

\varphi_X(t)=\mathbb{E}\!\left[e^{itX}\right].

For real constants \(a\) and \(b\), the affine transform of \(X\) is the random variable \(Y=aX+b\).

The affine transformation formula

Theorem: If \(X\) is a real-valued random variable and \(a,b\in\mathbb{R}\), then for every \(t\in\mathbb{R}\),

\varphi_{aX+b}(t)=e^{itb}\varphi_X(at).

No moment or integrability assumption on \(X\) is required.

Proof. Fix \(t\in\mathbb{R}\). By the definition of a characteristic function and the identity \(e^{it(aX+b)}=e^{itb}e^{i(at)X}\),

\begin{aligned} \varphi_{aX+b}(t) &=\mathbb{E}\!\left[e^{it(aX+b)}\right]\\ &=\mathbb{E}\!\left[e^{itb}e^{i(at)X}\right]\\ &=e^{itb}\mathbb{E}\!\left[e^{i(at)X}\right]\\ &=e^{itb}\varphi_X(at). \end{aligned}

The expectation exists because the complex exponential in the definition has absolute value \(1\). This proves the identity for every real \(t,a,b\), including \(a=0\). \(\square\)

The two parameters have distinct roles. The factor \(e^{itb}\) records translation by \(b\). The argument \(at\) records scaling by \(a\). The formula is not \(e^{itb}\varphi_X(t)+a\), nor is the phase \(e^{itab}\): the scale acts on the frequency argument, while the shift acts in the phase.

Worked Example: A shift and scale of a two-point variable

Suppose \(X=-2\) with probability \(1/3\) and \(X=1\) with probability \(2/3\). Its characteristic function is

\varphi_X(t)=\frac{1}{3}e^{-2it}+\frac{2}{3}e^{it}.

Let \(Y=2X+3\). The affine formula gives

\begin{aligned} \varphi_Y(t) &=e^{3it}\varphi_X(2t)\\ &=e^{3it}\left(\frac{1}{3}e^{-4it}+\frac{2}{3}e^{2it}\right)\\ &=\frac{1}{3}e^{-it}+\frac{2}{3}e^{5it}. \end{aligned}

This agrees with the transformed values: \(2(-2)+3=-1\) and \(2(1)+3=5\), with the same respective probabilities. Thus direct calculation gives \(\varphi_Y(t)=\frac{1}{3}e^{-it}+\frac{2}{3}e^{5it}\). At \(t=0\), both expressions equal \(1/3+2/3=1\), as a characteristic function must. The shift contributes \(e^{3it}\), and the scale \(2\) changes the argument from \(t\) to \(2t\).

Translation, scaling, and reflection

Several useful cases follow immediately. A pure translation has characteristic function \(e^{itb}\varphi_X(t)\). A pure scaling has characteristic function \(\varphi_X(at)\). Reflection is the special case \(a=-1\), so \(-X\) has characteristic function \(\varphi_X(-t)\). By the conjugate-symmetry property of characteristic functions, \(\varphi_X(-t)=\overline{\varphi_X(t)}\); hence reflection conjugates the characteristic function.

The zero-scale case is worth stating separately. When \(a=0\), the transformed variable is the constant \(b\), regardless of \(X\). Since \(\varphi_X(0)=1\), the formula gives \(\varphi_b(t)=e^{itb}\), exactly the characteristic function of that constant.

Worked Example: Combining reflection, scaling, and translation

Let \(X=-1\) with probability \(1/4\) and \(X=2\) with probability \(3/4\). Then

\varphi_X(t)=\frac{1}{4}e^{-it}+\frac{3}{4}e^{2it}.

Set \(Y=-2X+1\). The scale is negative, so the transformed characteristic function uses \(\varphi_X(-2t)\), not \(\varphi_X(2t)\):

\begin{aligned} \varphi_Y(t) &=e^{it}\varphi_X(-2t)\\ &=e^{it}\left(\frac{1}{4}e^{2it}+\frac{3}{4}e^{-4it}\right)\\ &=\frac{1}{4}e^{3it}+\frac{3}{4}e^{-3it}. \end{aligned}

Directly, \(-2(-1)+1=3\) and \(-2(2)+1=-3\), so \(Y\) takes \(3\) and \(-3\) with probabilities \(1/4\) and \(3/4\). This gives the same expression. At \(t=0\), each formula gives \(1/4+3/4=1\). The negative argument is essential: using \(\varphi_X(2t)\) would correspond to \(2X+1\), a different transformation.

Translation changes only the phase factor, whose absolute value is \(1\). Consequently, it does not change the modulus of the characteristic function. Scaling by a nonzero \(a\) samples the original function at \(at\); if \(a<0\), this also reverses the direction of the frequency axis.

Corollary: For every real \(a,b,t\),

\left|\varphi_{aX+b}(t)\right|=\left|\varphi_X(at)\right|.

Proof. The affine formula and \(|e^{itb}|=1\) give

\left|\varphi_{aX+b}(t)\right| =\left|e^{itb}\right|\left|\varphi_X(at)\right| =\left|\varphi_X(at)\right|.

In particular, translating a variable leaves the modulus unchanged at every frequency. \(\square\)

Standardization

When \(X\) has a finite mean \(\mu=\mathbb{E}[X]\) and finite, strictly positive variance \(\sigma^2=\mathbb{E}[(X-\mu)^2]\), its standard deviation is \(\sigma=\sqrt{\sigma^2}>0\). The standardized variable is \(Z=(X-\mu)/\sigma\). It has mean zero and variance one, and its characteristic function follows by taking \(a=1/\sigma\) and \(b=-\mu/\sigma\).

Standardization formula: If \(X\) has finite mean \(\mu\) and finite positive standard deviation \(\sigma\), then

\varphi_Z(t) =e^{-it\mu/\sigma}\varphi_X\!\left(\frac{t}{\sigma}\right), \qquad Z=\frac{X-\mu}{\sigma}.

The condition \(\sigma>0\) is necessary: if the variance is zero, this standardization divides by zero.

Worked Example: Standardizing a two-point variable

Let \(X\) take the values \(0\) and \(2\), each with probability \(1/2\). Its characteristic function is

\varphi_X(t)=\frac{1}{2}+\frac{1}{2}e^{2it}=e^{it}\cos t.

The mean is \(\mu=(0+2)/2=1\), and the variance is \(\sigma^2=((0-1)^2+(2-1)^2)/2=1\), so \(\sigma=1\). Therefore \(Z=(X-1)/1\), and the standardization formula gives

\begin{aligned} \varphi_Z(t) &=e^{-it}\varphi_X(t)\\ &=e^{-it}\left(e^{it}\cos t\right)\\ &=\cos t. \end{aligned}

Indeed, \(Z\) takes \(-1\) and \(1\) with equal probability, so direct evaluation gives \(\varphi_Z(t)=\frac{1}{2}e^{-it}+\frac{1}{2}e^{it}=\cos t\). At \(t=0\), both the formula and direct expression equal \(1\); at \(t=\pi\), both equal \(-1\). The centering factor removes the phase \(e^{it}\) associated with the original location, while the unit scale leaves the frequency argument unchanged.

Composition and a useful check

Affine transformations compose according to ordinary algebra. If \(Y=aX+b\) and \(W=cY+d\), then \(W=(ca)X+(cb+d)\). Applying the formula first to \(Y\) and then to \(W\) yields

\begin{aligned} \varphi_W(t) &=e^{itd}\varphi_Y(ct)\\ &=e^{itd}e^{itcb}\varphi_X(cat)\\ &=e^{it(cb+d)}\varphi_X((ca)t), \end{aligned}

which is exactly the formula for the single affine transformation \(W=(ca)X+(cb+d)\). This is a useful consistency check when several shifts and rescalings are performed in sequence.

Another immediate check is \(t=0\). Every characteristic function satisfies \(\varphi_X(0)=1\), and the affine formula gives \(e^0\varphi_X(0)=1\). If a calculation of a transformed characteristic function does not equal \(1\) at zero, an algebraic error has occurred.

Why the formula matters

Affine transformations occur whenever data are shifted to a new origin, converted to different units, reflected, or standardized. The formula lets these changes be carried through a characteristic-function calculation without reconstructing the transformed distribution from scratch. It also separates two effects that are easy to confuse: location produces a phase, and scale changes the frequency argument.

A common pitfall is to overlook the sign of the scale. For \(a<0\), the correct argument is \(at\), a negative frequency when \(t>0\). Another is to assume that moments are needed to use the transformation rule. They are not: the defining exponential is bounded, so the expectation exists for every real-valued \(X\). Finite mean and positive finite variance are required for the particular operation of standardization, not for the general affine identity.

Check Your Understanding

Answer each question using the affine transformation formula.

If \(Y=X+b\), what factor multiplies \(\varphi_X(t)\), and what happens to its modulus?
For \(Y=-3X\), at what argument of \(\varphi_X\) is the transformed characteristic function evaluated?
Why does the formula remain valid when \(a=0\), and what random variable does \(aX+b\) then represent?
If \(X\) has finite mean \(\mu\) and standard deviation \(\sigma>0\), write the characteristic function of \((X-\mu)/\sigma\).
For \(X\) equal to \(-1\) with probability \(1/4\) and \(2\) with probability \(3/4\), identify the values and probabilities of \(-2X+1\), and state how they appear in its characteristic function.