Why Location and Scale Matter for Characteristic Functions
A location-scale transformation changes a random variable by moving and stretching its distribution. If \(Z\) is a reference random variable and \(X=\mu+\sigma Z\), the characteristic function of \(X\) can be obtained directly from that of \(Z\). The formula is simple, but its two parts have distinct roles: location contributes a complex phase, while scale changes the argument at which the original characteristic function is evaluated.
This rule is useful whenever a distribution is described through a standardized variable. It also works when moments do not exist: unlike a moment-generating function, a characteristic function is defined for every real-valued random variable.
Here \(\mu\) is a real location parameter, and the conventional scalar scale parameter satisfies \(\sigma>0\). The algebraic identity below also holds for any real \(\sigma\), including \(\sigma=0\); the positive-scale convention is used when describing a nondegenerate location-scale family. No assumption that \(Z\) has a density, finite mean, or finite variance is needed.
The Location-Scale Rule
The proof uses only the definition of the characteristic function and the fact that \(X\) is the stated function of \(Z\). In particular, it does not require differentiating a characteristic function or interchanging a limit with an expectation.
By definition and the almost-sure equality \(X=\mu+\sigma Z\), \(\varphi_X(t)=E[e^{it(\mu+\sigma Z)}]\).
The complex exponential factors as \(e^{it(\mu+\sigma Z)}=e^{it\mu}e^{i(\sigma t)Z}\). Since \(e^{it\mu}\) is deterministic, it can be taken outside the expectation.
The remaining expectation is \(E[e^{i(\sigma t)Z}]=\varphi_Z(\sigma t)\), giving the claimed identity. \(\square\)
The factor \(e^{it\mu}\) has modulus one, so shifting a distribution changes the phase but not the modulus of its characteristic function. Scaling acts differently: it evaluates the original function at \(\sigma t\), expanding or contracting the argument when \(\sigma\) changes.
To verify the corollary, apply the theorem with \(t=u/\sigma\). It gives \(\varphi_X(u/\sigma)=e^{iu\mu/\sigma}\varphi_Z(u)\). Multiplication by \(e^{-iu\mu/\sigma}\) yields the displayed expression. This is an algebraic recovery rule; it does not require \(Z\) to have moments.
Worked Applications
Worked Example: A Normal Location-Scale Transformation
Let \(Z\sim N(0,1)\) and \(X=\mu+\sigma Z\), with \(\sigma>0\). First obtain \(\varphi_Z\), then apply the location-scale rule.
The standard normal density is \(f_Z(z)=(2\pi)^{-1/2}e^{-z^2/2}\), so $$ \varphi_Z(t)=\int_{-\infty}^{\infty}e^{itz}f_Z(z)\,dz. $$
To derive this function, differentiate under the integral. This is justified by dominated convergence: for nonzero \(h\), \(|(e^{ihz}-1)/h|\leq |z|\), and \(|z|f_Z(z)\) is integrable. Thus \(\varphi_Z'(t)=i\int_{-\infty}^{\infty}z e^{itz}f_Z(z)\,dz\). Since \(f_Z'(z)=-zf_Z(z)\), integration by parts, with the boundary term \(e^{itz}f_Z(z)\) tending to zero at both endpoints, gives $$ \int_{-\infty}^{\infty}z e^{itz}f_Z(z)\,dz =-\int_{-\infty}^{\infty}e^{itz}f_Z'(z)\,dz =it\varphi_Z(t). $$ Consequently \(\varphi_Z'(t)=-t\varphi_Z(t)\). Together with \(\varphi_Z(0)=1\), this differential equation has solution \(\varphi_Z(t)=e^{-t^2/2}\).
Now apply the transformation rule: $$ \varphi_X(t)=e^{it\mu}\varphi_Z(\sigma t) =e^{it\mu}e^{-(\sigma t)^2/2} =e^{it\mu-\sigma^2t^2/2}. $$ This is the characteristic function of \(N(\mu,\sigma^2)\). In particular, the scale appears squared in the Gaussian exponential because the standard-normal characteristic function depends on the square of its argument.
Worked Example: A Shifted and Scaled Two-Point Variable
Let \(Z\) take the values \(-1\) and \(1\), each with probability \(1/2\), and set \(X=3+2Z\). The characteristic function of \(Z\) follows directly from its two possible values: $$ \varphi_Z(t)=\frac{1}{2}e^{-it}+\frac{1}{2}e^{it}=\cos(t). $$
The transformed variable takes value \(3+2(-1)=1\) with probability \(1/2\), and value \(3+2(1)=5\) with probability \(1/2\). The rule therefore gives $$ \varphi_X(t)=e^{3it}\varphi_Z(2t)=e^{3it}\cos(2t). $$
Checking by summing over the two values of \(X\) gives $$ \varphi_X(t)=\frac{1}{2}e^{it}+\frac{1}{2}e^{5it} =e^{3it}\frac{e^{-2it}+e^{2it}}{2} =e^{3it}\cos(2t). $$ The direct calculation confirms both parts of the rule: the shift produces \(e^{3it}\), and the scale changes \(t\) to \(2t\).
Worked Example: A Shifted and Scaled Exponential Variable
Let \(Z\) have the exponential distribution with rate \(1\), so its density is \(f_Z(z)=e^{-z}\) for \(z\geq 0\) and zero otherwise. For real \(t\), $$ \varphi_Z(t)=\int_0^\infty e^{itz}e^{-z}\,dz =\int_0^\infty e^{-(1-it)z}\,dz =\frac{1}{1-it}. $$
The integral converges because the real part of \(1-it\) is \(1>0\); evaluating its antiderivative gives \(1/(1-it)\). Now take \(X=4+3Z\). The transformation rule yields $$ \varphi_X(t)=e^{4it}\varphi_Z(3t) =\frac{e^{4it}}{1-3it}. $$
This expression also reflects the support change: since \(Z\geq 0\), \(X\geq 4\). The shift sets the lower endpoint at \(4\), while the positive scale \(3\) stretches distances from that endpoint. The characteristic-function calculation needs neither a new density calculation nor a new integral.
Interpreting and Checking the Formula
The argument \(\sigma t\) is essential. Scaling the random variable by \(\sigma\) changes the frequency at which its characteristic function is evaluated; it does not multiply the function by \(\sigma\). The location term is a separate multiplicative factor, and its exponent has a positive sign because the definition uses \(e^{itZ}\).
The formula also behaves correctly in limiting and special cases. If \(\mu=0\), it reduces to \(\varphi_X(t)=\varphi_Z(\sigma t)\). If \(\sigma=1\), it gives \(\varphi_X(t)=e^{it\mu}\varphi_Z(t)\). If \(\sigma=0\), then \(X=\mu\) almost surely and the formula becomes \(\varphi_X(t)=e^{it\mu}\varphi_Z(0)=e^{it\mu}\), as required for a constant random variable.
A negative real \(\sigma\) also satisfies the identity: it reverses the sign of the argument, exactly as \(Z\) is reflected before shifting. In the conventional location-scale parameterization, however, \(\sigma\) is positive so that it represents scale rather than reflection. If the transformation is written with a minus sign, it is safest to retain the signed coefficient inside \(\varphi_Z(\sigma t)\).
Common Mistakes
- Scaling the characteristic function instead of its argument. The incorrect expression \(e^{it\mu}\sigma\varphi_Z(t)\) does not follow from the expectation. The correct expression is \(e^{it\mu}\varphi_Z(\sigma t)\).
- Putting the location term inside the argument. Location is not represented by replacing \(t\) with \(t+\mu\). It contributes the factor \(e^{it\mu}\).
- Using the wrong sign in the phase. Under the convention \(\varphi_Z(t)=E[e^{itZ}]\), the shift factor is \(e^{it\mu}\), not \(e^{-it\mu}\). The negative sign appears when solving the formula backward.
- Confusing scale with variance. If \(X=\mu+\sigma Z\) and \(Z\sim N(0,1)\), the normal distribution is \(N(\mu,\sigma^2)\), and its characteristic function contains \(-\sigma^2t^2/2\).
Check Your Understanding
Use the definition and the transformation rule to answer these questions.
- If \(\varphi_Z(t)=e^{-t^2/2}\) and \(X=-2+4Z\), what is \(\varphi_X(t)\)?
- A random variable \(Z\) takes values \(0\) and \(2\), each with probability \(1/2\). Find \(\varphi_Z(t)\), then find the characteristic function of \(X=1+Z\).
- For \(X=\mu+\sigma Z\) with \(\sigma\neq 0\), solve the transformation formula for \(\varphi_Z(u)\).
- Why is a characteristic function defined even when a random variable has no finite mean?
- For \(X=\mu+\sigma Z\), identify which part of the formula corresponds to location and which part corresponds to scale.