Gaussian and Cauchy Characteristic Functions

Why These Two Characteristic Functions Matter

The normal and Cauchy distributions are both location-scale families, but their characteristic functions behave differently as frequency increases. For the normal law, the modulus decreases at a rate determined by the variance; for the Cauchy law, it decreases at a rate determined by the scale through an absolute value. Deriving these formulas makes the distinction precise and shows how a Fourier integral reflects the shape of a distribution.

Use the characteristic-function convention \(\varphi_X(t)=E[e^{itX}]\), for real \(t\). For a continuous random variable with density \(f\), this is the Fourier integral \(\int_{-\infty}^{\infty} e^{itx}f(x)\,dx\). The calculations below derive the two formulas directly from their densities.

The Normal Characteristic Function

Begin with the standard normal density \(\phi_0(x)=(2\pi)^{-1/2}e^{-x^2/2}\), and let \(Z\sim N(0,1)\). Its characteristic function is \(\varphi_Z(t)=\int_{-\infty}^{\infty}e^{itx}\phi_0(x)\,dx\). Rather than attempting to complete a square with a complex exponent, differentiate this integral with respect to \(t\), then use integration by parts.

Proof. First consider \(Z\sim N(0,1)\). Differentiation under the integral is valid: the difference quotients of \(e^{itx}\) are bounded in absolute value by \(|x|\), and \(\int |x|\phi_0(x)\,dx<\infty\). Dominated convergence therefore gives \(\varphi_Z'(t)=i\int_{-\infty}^{\infty}x e^{itx}\phi_0(x)\,dx\). Since \(\phi_0'(x)=-x\phi_0(x)\), integration by parts yields \[ \int_{-\infty}^{\infty}x e^{itx}\phi_0(x)\,dx =-\int_{-\infty}^{\infty} e^{itx}\phi_0'(x)\,dx =it\int_{-\infty}^{\infty} e^{itx}\phi_0(x)\,dx. \] The boundary term vanishes because \(\phi_0(x)\to0\) as \(x\to\pm\infty\). Thus \(\varphi_Z'(t)=-t\varphi_Z(t)\). Also \(\varphi_Z(0)=1\), since \(\phi_0\) integrates to one. Multiplying the differential equation by \(e^{t^2/2}\) gives \[ \frac{d}{dt}\left(e^{t^2/2}\varphi_Z(t)\right)=0. \] The expression in parentheses is therefore constant, and its value at zero is one. Hence \(\varphi_Z(t)=e^{-t^2/2}\).

For \(X\sim N(\mu,\sigma^2)\), the substitution \(x=\mu+\sigma z\) in the density integral gives \[ \varphi_X(t)=\int_{-\infty}^{\infty}e^{it(\mu+\sigma z)}\phi_0(z)\,dz =e^{i\mu t}\varphi_Z(\sigma t) =\exp(i\mu t-\sigma^2t^2/2). \] This proves the formula. \(\square\)

The Cauchy Characteristic Function

For the Cauchy law, the Fourier integral is governed by the poles of the rational density when extended to a complex variable. First evaluate the centered integral \[ I_\gamma(t)=\int_{-\infty}^{\infty}\frac{e^{itx}}{x^2+\gamma^2}\,dx, \qquad \gamma>0. \] The residue theorem gives a direct calculation, with the choice of contour depending on the sign of \(t\).

Proof. First take \(t\geq0\) and integrate \(e^{itz}/(z^2+\gamma^2)\) around the upper semicircle of radius \(R>\gamma\), together with the real line segment from \(-R\) to \(R\). The only enclosed pole is \(z=i\gamma\), whose residue is \(e^{-\gamma t}/(2i\gamma)\). The residue theorem makes the contour integral \(2\pi i\) times this residue, or \(\pi e^{-\gamma t}/\gamma\).

The curved part of the contour vanishes as \(R\to\infty\). On that arc, \(|e^{itz}|=e^{-t\,\operatorname{Im}(z)}\leq1\), and \(|z^2+\gamma^2|\geq R^2-\gamma^2\). Thus the arc integral has absolute value at most \(\pi R/(R^2-\gamma^2)\), which tends to zero. Consequently \(I_\gamma(t)=\pi e^{-\gamma t}/\gamma\) for \(t\geq0\), including \(t=0\).

For \(t<0\), close the contour in the lower half-plane. There \(|e^{itz}|=e^{-t\,\operatorname{Im}(z)}\leq1\), and the same arc bound applies. The contour is clockwise and encloses the pole \(z=-i\gamma\), with residue \(e^{\gamma t}/(-2i\gamma)\). The contour integral is therefore \(-2\pi i\) times this residue, giving \(I_\gamma(t)=\pi e^{\gamma t}/\gamma\). Combining the cases, \[ I_\gamma(t)=\frac{\pi}{\gamma}e^{-\gamma|t|}. \] For the centered Cauchy density, multiplying the integrand by \(\gamma/\pi\) gives the characteristic function \(e^{-\gamma|t|}\). For location \(\mu\), substitute \(x=\mu+y\) in the density integral. The factor \(e^{itx}\) becomes \(e^{i\mu t}e^{ity}\), so the characteristic function is \(e^{i\mu t-\gamma|t|}\). \(\square\)

Reading Location and Scale From the Formulas

The formulas share a location factor \(e^{i\mu t}\), but their scale factors differ: \[ \varphi_{N(\mu,\sigma^2)}(t)=e^{i\mu t}e^{-\sigma^2t^2/2}, \qquad \varphi_{\mathrm{Cauchy}(\mu,\gamma)}(t)=e^{i\mu t}e^{-\gamma|t|}. \] For either law, the location factor has modulus one, so changing location rotates the characteristic function without changing its modulus. The scale determines the rate at which the modulus decreases away from \(t=0\): quadratically in the normal exponent and linearly in \(|t|\) in the Cauchy exponent.

These formulas are also consistent with affine changes of a random variable. If a centered normal variable is multiplied by a positive constant \(a\), its standard deviation is multiplied by \(a\), and the quadratic exponent acquires the factor \(a^2\). If a centered Cauchy variable is multiplied by \(a\), its scale becomes \(a\gamma\), and the exponent acquires the factor \(a\). This difference is important: \(\sigma\) is the normal standard deviation, whereas \(\gamma\) is a Cauchy scale parameter, not a standard deviation.

Common Pitfalls and Why the Formulas Matter

• Using the variance as the normal standard deviation. In \(N(\mu,\sigma^2)\), the characteristic function has \(-\sigma^2t^2/2\) in its exponent. If the variance is \(4\), this term is \(-2t^2\).
• Dropping the absolute value in the Cauchy formula. The scale contribution is \(-\gamma|t|\) for positive and negative frequencies. It must be nonpositive on both sides of zero.
• Calling the Cauchy location a mean. The parameter \(\mu\) shifts the density and appears in the phase factor, but the Cauchy distribution has no finite expectation.
• Interpreting Cauchy scale as standard deviation. A Cauchy distribution has no finite variance, so \(\gamma\) is a scale parameter rather than a standard deviation.

The contrast between the two formulas is useful whenever frequency behavior is used to study distributions. The normal characteristic function is smooth at zero and has Gaussian decay in frequency; the Cauchy characteristic function has a cusp at zero and exponential decay in absolute frequency. In both cases, the location-scale structure is visible directly in the Fourier integral.