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Characteristic Functions · Tutorial 548 of 994

Affine Transformations of a Random Variable

Learn how shifting, scaling, reflecting, or standardizing a random variable changes its characteristic function, and why the same formula works for discrete and continuous laws.

Advanced 9 min read

What You'll Learn

  • Derive the characteristic function of \(aX+b\) directly from its definition.
  • Identify the distinct effects of translation and rescaling on the argument and phase.
  • Handle negative scale factors as reflections and the zero scale factor as a constant.
  • Calculate transformed characteristic functions for Bernoulli, uniform, and exponential variables.
  • Standardize a finite-variance random variable and verify its mean and variance.
  • Recognize and correct common sign and scaling errors.

Why Affine Transformations Matter

A change of origin or measurement scale changes how a random variable is described, but it does not require starting a characteristic-function calculation from scratch. If \(X\) has characteristic function \(\phi_X\), then the characteristic function of \(aX+b\) can be obtained from \(\phi_X\) by changing its argument and multiplying by a phase factor. This covers translation, rescaling, reflection, and standardization in one formula.

The rule is useful because it applies whether \(X\) is discrete, continuous, or has a more general distribution. It relies only on the definition of a characteristic function, not on a probability mass function or density. We use the convention \(\phi_X(t)=E[e^{itX}]\), for real \(t\).

Definition: Let \(X\) be a real-valued random variable and let \(a,b\in\mathbb{R}\). The affine transformation of \(X\) with scale \(a\) and shift \(b\) is \(Y=aX+b\). Its characteristic function is \(\phi_Y(t)=E[e^{itY}]\), for \(t\in\mathbb{R}\).

The Affine Transformation Formula

Theorem: For every real-valued random variable \(X\) and all real \(a,b,t\), \[ \phi_{aX+b}(t)=e^{itb}\phi_X(at). \] No moment or density assumptions are needed.

The two parts of the formula have different roles. The shift \(b\) contributes the factor \(e^{itb}\); it does not change the argument at which \(\phi_X\) is evaluated. The scale \(a\) changes that argument from \(t\) to \(at\). In particular, a negative \(a\) reverses the sign of the argument, while \(a=0\) removes all dependence on \(X\).

1
Insert the transformed variable.
Write the characteristic function as \(E[e^{it(aX+b)}]\).
2
Separate the constant factor.
Use \(e^{it(aX+b)}=e^{itb}e^{i(at)X}\), and pull the nonrandom factor \(e^{itb}\) outside the expectation.
3
Recognize the original characteristic function.
The remaining expectation is \(E[e^{i(at)X}]=\phi_X(at)\).

In full, the proof is

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

The expectation is well-defined because the complex exponential has modulus one. Thus the proof does not require \(E[|X|]\) to be finite. \(\square\)

Some special cases are worth recording. A translation \(X+b\) has characteristic function \(e^{itb}\phi_X(t)\). A rescaling \(aX\) has characteristic function \(\phi_X(at)\). A reflection followed by a shift, \(-X+b\), has characteristic function \(e^{itb}\phi_X(-t)\). If \(a=0\), then \(aX+b=b\) is constant and the formula gives \(\phi_b(t)=e^{itb}\phi_X(0)=e^{itb}\), since \(\phi_X(0)=1\).

Derived property: If \(Y=aX+b\) and \(Z=cY+d\), then \(Z=(ca)X+(cb+d)\). Applying the affine formula in either stage gives \[ \phi_Z(t)=e^{it(cb+d)}\phi_X(cat). \] This confirms that successive changes of scale and origin combine according to ordinary affine algebra.

Indeed, applying the formula first to \(Z=cY+d\) gives \(\phi_Z(t)=e^{itd}\phi_Y(ct)\). Applying it to \(Y=aX+b\) at argument \(ct\) gives \(\phi_Y(ct)=e^{ictb}\phi_X(act)\). Multiplication yields \(e^{itd}e^{ictb}\phi_X(act)=e^{it(cb+d)}\phi_X(cat)\), as claimed.

Worked Applications

Worked Example: Shifting and Rescaling a Bernoulli Variable

Let \(X\sim\operatorname{Bernoulli}(1/3)\), so \(P(X=1)=1/3\) and \(P(X=0)=2/3\). Find the characteristic function of \(Y=4X-2\).

First, \[ \phi_X(t)=E[e^{itX}] =\frac23 e^{it\cdot 0}+\frac13 e^{it\cdot 1} =\frac23+\frac13 e^{it}. \] Here \(a=4\) and \(b=-2\). Therefore,

$$ \phi_Y(t)=e^{-2it}\phi_X(4t) =e^{-2it}\left(\frac23+\frac13 e^{i4t}\right) =\frac23e^{-2it}+\frac13e^{2it}. $$

This also agrees with the values of \(Y\): when \(X=0\), \(Y=-2\), and when \(X=1\), \(Y=2\). Thus \(E[e^{itY}]=\frac23e^{-2it}+\frac13e^{2it}\). At \(t=0\), the result is \(2/3+1/3=1\), as every characteristic function must be.

Worked Example: Reflection and Translation of a Uniform Variable

Suppose \(X\sim\operatorname{Uniform}(-1,1)\), and let \(Y=3-2X\). The characteristic function of \(X\), obtained by integrating its density \(1/2\) over \((-1,1)\), is

$$ \phi_X(t)=\frac12\int_{-1}^{1}e^{itx}\,dx =\frac12\left[\frac{e^{itx}}{it}\right]_{-1}^{1} =\frac{\sin t}{t},\qquad t\ne 0, $$

with \(\phi_X(0)=1\), the continuous value at zero. Since \(Y=(-2)X+3\), the affine formula gives

$$ \phi_Y(t)=e^{3it}\phi_X(-2t) =e^{3it}\frac{\sin(-2t)}{-2t} =e^{3it}\frac{\sin(2t)}{2t},\qquad t\ne0. $$

At zero the value is again \(1\). The negative scale reflects \(X\) before multiplying its values by \(2\), and the shift then moves the result by \(3\). The transformed support is \([1,5]\), consistent with \(Y=3-2X\) as \(X\) ranges from \(1\) down to \(-1\).

Worked Example: Standardizing an Exponential Variable

Let \(X\) have an exponential distribution with rate \(2\), with density \(f_X(x)=2e^{-2x}\) for \(x\geq0\). Its characteristic function is

$$ \phi_X(t)=\int_0^\infty 2e^{-2x}e^{itx}\,dx =2\int_0^\infty e^{-(2-it)x}\,dx =\frac{2}{2-it}. $$

The exponential mean and variance are \(E[X]=1/2\) and \(\operatorname{Var}(X)=1/4\), so its standard deviation is \(1/2\). Its standardized version is therefore \(Z=(X-1/2)/(1/2)=2X-1\). Substituting \(a=2\), \(b=-1\) yields

$$ \phi_Z(t)=e^{-it}\phi_X(2t) =e^{-it}\frac{2}{2-2it} =\frac{e^{-it}}{1-it}. $$

For a check, \(Z=2X-1\) has mean \(2(1/2)-1=0\) and variance \(2^2(1/4)=1\). The transformation formula preserves the full characteristic function, while this calculation also verifies the intended centering and scaling.

Standardization and the Moment Check

Standardization is an affine transformation, but it is defined only when the variance is finite and positive. If \(E[X]=\mu\) and \(0<\operatorname{Var}(X)=\sigma^2<\infty\), then \(\sigma\) denotes the positive square root of the variance and \(Z=(X-\mu)/\sigma\). The formula immediately gives

$$ \phi_Z(t)=e^{-it\mu/\sigma}\phi_X(t/\sigma). $$

The moment properties of this standardized variable follow directly. Since \(E[X]\) exists,

$$ E[Z]=E\!\left[\frac{X-\mu}{\sigma}\right] =\frac{E[X]-\mu}{\sigma}=0. $$

And, using the definition of variance,

$$ \operatorname{Var}(Z) =E[(Z-E[Z])^2] =E\!\left[\left(\frac{X-\mu}{\sigma}\right)^2\right] =\frac{\operatorname{Var}(X)}{\sigma^2}=1. $$

More generally, when \(E[X^2]<\infty\), any \(Y=aX+b\) has \(E[Y]=aE[X]+b\) and \(\operatorname{Var}(Y)=a^2\operatorname{Var}(X)\). For the variance identity, subtract the mean to obtain \(Y-E[Y]=a(X-E[X])\), square, and take expectations. These moment checks are useful for confirming the direction and magnitude of a standardization, but the characteristic-function transformation itself remains valid even when moments do not exist.

Common Mistakes and Interpretation

A frequent error is to attach the shift to the argument and the scale to the phase, writing \(e^{iatb}\phi_X(t)\) or \(e^{itb}\phi_X(t/a)\). Neither expression follows from \(e^{it(aX+b)}=e^{itb}e^{i(at)X}\). The shift contributes \(e^{itb}\); the multiplier of \(X\) changes the argument to \(at\).

  • Forgetting the sign under reflection. For \(Y=-X\), the correct expression is \(\phi_Y(t)=\phi_X(-t)\), not generally \(\phi_X(t)\). If \(X\) is symmetric about zero, these agree, but symmetry is an additional property, not a general rule.
  • Dividing by the scale in the argument. If \(Y=aX\), the argument is \(at\), not \(t/a\). A quick check with \(a=0\) gives \(\phi_Y(t)=\phi_X(0)=1\), as required for the constant zero.
  • Using standardization when the variance is zero or infinite. The expression \((X-\mu)/\sigma\) requires a finite mean and a finite, strictly positive standard deviation. The affine characteristic-function formula itself needs no such assumptions.

Translations multiply the characteristic function by a complex number of modulus one, while rescaling changes the frequency at which the original function is evaluated. These different effects explain why location and scale transformations can be tracked separately. In particular, since \(|e^{itb}|=1\), translating a random variable does not change the modulus of its characteristic function; the affine formula gives \(|\phi_{aX+b}(t)|=|\phi_X(at)|\).

Key takeaway: For \(Y=aX+b\), always start from \(\phi_Y(t)=E[e^{it(aX+b)}]\). Separating the constant shift from the random term gives \(\phi_Y(t)=e^{itb}\phi_X(at)\), including shifts, rescalings, reflections, and standardization as special cases.

Check Your Understanding

Use the affine formula and show the key substitution in each answer.

  1. If \(Y=X+5\), express \(\phi_Y(t)\) in terms of \(\phi_X(t)\). Which part of the formula reflects the shift?
  2. If \(Y=-3X\), what argument of \(\phi_X\) appears in \(\phi_Y(t)\)? Explain why the sign matters.
  3. A random variable \(X\) has mean \(4\) and variance \(9\). Write the standardization \(Z\) and its characteristic function in terms of \(\phi_X\).
  4. For \(Y=0X+b\), simplify the affine formula and explain why the result cannot depend on the distribution of \(X\).
  5. If \(E[X^2]<\infty\), derive the variance of \(aX+b\). Why does the shift \(b\) not affect it?

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