Modulus Bounds and Conjugate Symmetry

Three Identities Every Characteristic Function Must Satisfy

A characteristic function is a complex-valued summary of a probability distribution. Its values need not be real, but they are not arbitrary complex numbers: the definition forces a normalization, a bound on their size, and a precise relation between values at opposite arguments. These facts are useful both for understanding the structure of a characteristic function and for ruling out an incorrect proposed formula.

Let \(X\) be a real-valued random variable. Using the convention \(e^{itX}\), its characteristic function is the function \(\varphi_X:\mathbb{R}\to\mathbb{C}\) defined by the expectation below. The expectation exists for every real \(t\) because the integrand has absolute value one.

Definition: The characteristic function of a real-valued random variable \(X\) is \(\varphi_X(t)=E[e^{itX}]\), for \(t\in\mathbb{R}\). Here \(i^2=-1\), and the expectation of a complex random variable is defined by taking expectations of its real and imaginary parts.

The key identities apply to every probability distribution on the real line, without any moment or continuity assumptions.

Theorem: For every real-valued random variable \(X\) and every \(t\in\mathbb{R}\), \(\varphi_X(0)=1\), \(|\varphi_X(t)|\leq 1\), and \(\varphi_X(-t)=\overline{\varphi_X(t)}\), where the bar denotes complex conjugation.

Proof of the Basic Identities

At zero, the exponential is identically one, so normalization follows immediately. For the modulus bound, we use the inequality \(|E[Z]|\leq E[|Z|]\) for an integrable complex random variable \(Z\). To see why this inequality holds, let \(c=E[Z]\). If \(c=0\), the inequality is immediate. If \(c\ne 0\), multiply \(Z\) by the unit complex number \(\overline{c}/|c|\). Then \[ |c|=\operatorname{Re}\left(E\left[\frac{\overline{c}}{|c|}Z\right]\right) \leq E\left[\left|\frac{\overline{c}}{|c|}Z\right|\right] =E[|Z|]. \] The inequality uses \(\operatorname{Re}(w)\leq |w|\) for every complex number \(w\), and the unit multiplier has absolute value one.

Apply this inequality to \(Z=e^{itX}\). Since \(|e^{itX}|=1\) for every outcome, the resulting upper bound is one. Finally, complex conjugation commutes with the expectation of an integrable complex random variable: this follows by writing it as \(U+iV\), with real integrable \(U,V\), and using \(\overline{E[U+iV]}=E[U-iV]\). Because \(\overline{e^{itX}}=e^{-itX}\), conjugate symmetry follows as well.

\varphi_X(0)=E[1]=1,\qquad |\varphi_X(t)|=|E[e^{itX}]|\leq E[|e^{itX}|]=1,\qquad \varphi_X(-t)=E[e^{-itX}]=\overline{E[e^{itX}]}=\overline{\varphi_X(t)}.

This proves all three identities. The modulus inequality can be an equality, but it need not be; it is an upper bound, not a claim that every value of the characteristic function lies on the unit circle. \(\square\)

Real and Imaginary Parts

Euler's formula, \(e^{itX}=\cos(tX)+i\sin(tX)\), gives a useful decomposition. Both sine and cosine are bounded, so their expectations exist whether or not \(X\) has a finite mean or variance.

Theorem: Write \(\varphi_X(t)=a(t)+ib(t)\), where \(a(t)=\operatorname{Re}\varphi_X(t)\) and \(b(t)=\operatorname{Im}\varphi_X(t)\). Then \(a(t)=E[\cos(tX)]\), \(b(t)=E[\sin(tX)]\), \(a(-t)=a(t)\), and \(b(-t)=-b(t)\).

To prove the expectation formulas, take real and imaginary parts of \(E[\cos(tX)+i\sin(tX)]\). For the parity identities, cosine is even and sine is odd, so

a(-t)=E[\cos(-tX)]=E[\cos(tX)]=a(t), \qquad b(-t)=E[\sin(-tX)]=-E[\sin(tX)]=-b(t).

The same conclusion follows from conjugate symmetry: if \(\varphi_X(t)=a(t)+ib(t)\), then its conjugate is \(a(t)-ib(t)\), which must equal \(\varphi_X(-t)\). In particular, the real part is even and the imaginary part is odd. Also, since \(|\varphi_X(t)|^2=a(t)^2+b(t)^2\leq 1\), each part has absolute value at most one. \(\square\)

These are properties of the two component functions for every real-valued \(X\). They do not say that the imaginary part must vanish: that would be an additional restriction, not a consequence of the identities proved here.

Using the Identities to Check Formulas

A proposed characteristic-function formula must have value one at zero, stay within the closed unit disk in the complex plane, and take conjugate values at opposite arguments. Any failure rules the formula out. Passing these checks is useful, but does not by itself establish that a function is a characteristic function.

Worked Example: A Bernoulli Random Variable

Suppose \(X\) takes the value \(1\) with probability \(p\) and \(0\) with probability \(1-p\), where \(0\leq p\leq 1\). Calculate its characteristic function and verify its normalization, modulus bound, and conjugate symmetry.

By summing over the two possible values, \(\varphi_X(t)=(1-p)e^{it\cdot 0}+pe^{it\cdot 1}=1-p+pe^{it}\). At \(t=0\), this is \(1-p+p=1\).

For the modulus, multiply by the complex conjugate: \[ |\varphi_X(t)|^2=(1-p+pe^{it})(1-p+pe^{-it}) =(1-p)^2+p^2+2p(1-p)\cos t. \] Since \((1-p)^2+p^2+2p(1-p)=1\), this becomes \[ |\varphi_X(t)|^2=1-2p(1-p)(1-\cos t). \] Here \(p(1-p)\geq 0\) and \(1-\cos t\geq 0\), so \(|\varphi_X(t)|^2\leq 1\), and consequently \(|\varphi_X(t)|\leq 1\). The squared modulus is nonnegative because it is the squared absolute value of a complex number.

Finally, \(\varphi_X(-t)=1-p+pe^{-it}=\overline{1-p+pe^{it}}=\overline{\varphi_X(t)}\). The real and imaginary parts are \(1-p+p\cos t\) and \(p\sin t\); the first is even and the second is odd, as required.

Worked Example: A Two-Point Distribution With Unequal Weights

Let \(P(X=-1)=1/3\) and \(P(X=2)=2/3\). Find the real and imaginary parts of the characteristic function, then check the identities, including the modulus bound.

The expectation over the two support points gives \(\varphi_X(t)=\frac13e^{-it}+\frac23e^{2it}\). Using Euler's formula, \[ a(t)=\frac13\cos t+\frac23\cos(2t), \qquad b(t)=-\frac13\sin t+\frac23\sin(2t). \] Cosine is even, so \(a(-t)=a(t)\); sine is odd, so \(b(-t)=-b(t)\). Therefore \(\varphi_X(-t)=a(t)-ib(t)=\overline{\varphi_X(t)}\).

At zero, the two probabilities sum to one, giving \(\varphi_X(0)=1/3+2/3=1\). For the bound, expand the squared modulus: \[ |\varphi_X(t)|^2 =\left(\frac13\right)^2+\left(\frac23\right)^2 +2\left(\frac13\right)\left(\frac23\right)\cos(2t-(-t)) =\frac59+\frac49\cos(3t). \] Because \(\cos(3t)\leq 1\), this is at most \(\frac59+\frac49=1\). Thus \(|\varphi_X(t)|\leq 1\). This calculation illustrates that the real and imaginary parts can both be nonzero even though their parity is constrained.

Worked Example: The Uniform Distribution on an Interval

Let \(X\) have the uniform distribution on \([0,1]\). Evaluate its characteristic function and verify the three basic identities using its real and imaginary parts.

For \(t\ne 0\), integrate the density, which is one on \([0,1]\): \[ \varphi_X(t)=\int_0^1e^{itx}\,dx=\frac{e^{it}-1}{it}. \] At \(t=0\), the defining expectation is \(1\). Separating real and imaginary parts in the integral gives, for \(t\ne 0\), \[ a(t)=\int_0^1\cos(tx)\,dx=\frac{\sin t}{t}, \qquad b(t)=\int_0^1\sin(tx)\,dx=\frac{1-\cos t}{t}. \] The first expression is even, since both \(\sin t\) and \(t\) change sign when \(t\) is replaced by \(-t\). The second is odd, since its numerator is unchanged and its denominator changes sign. At zero the values are \(a(0)=1\) and \(b(0)=0\), consistent with these parity relations.

For \(t\ne0\), the squared modulus is \[ |\varphi_X(t)|^2 =\frac{\sin^2t+(1-\cos t)^2}{t^2} =\frac{2-2\cos t}{t^2} =\frac{4\sin^2(t/2)}{t^2}. \] The elementary inequality \(|\sin u|\leq |u|\), applied with \(u=t/2\), gives \(|\varphi_X(t)|\leq 1\). At \(t=0\), the same bound holds because \(\varphi_X(0)=1\). The evenness of \(a\) and oddness of \(b\) also give conjugate symmetry.

Common Mistakes and Practical Checks

The three identities are quick diagnostics, but using them correctly requires attention to what each one does and does not assert.

Assuming every characteristic function is real. Conjugate symmetry says \(\varphi_X(-t)=\overline{\varphi_X(t)}\), not that \(\varphi_X(t)=\overline{\varphi_X(t)}\) at each \(t\). The imaginary part may be nonzero.
Replacing the modulus bound with equality. The argument gives \(|E[e^{itX}]|\leq E[|e^{itX}|]=1\). The expectation can have modulus strictly less than one because complex contributions can cancel.
Checking only the value at zero. The candidate \(q(t)=1+it\) satisfies \(q(0)=1\) and \(q(-t)=\overline{q(t)}\), but \(|q(t)|=\sqrt{1+t^2}>1\) for every \(t\ne0\). It cannot be a characteristic function.
Treating the identities as a complete test. They are necessary conditions: every characteristic function satisfies them. Their validity alone is not a proof that a proposed function comes from a probability distribution.

When checking a formula, first substitute \(t=0\), then examine its modulus, and finally compare its values at \(t\) and \(-t\). If it passes, also separate its real and imaginary parts and confirm that they are even and odd, respectively. These checks are simple, but they often expose algebraic or sign errors before more involved arguments begin.

Key takeaway: Every characteristic function is normalized at zero, bounded in modulus by one, and conjugate-symmetric. Equivalently, its real part is even and its imaginary part is odd; these necessary identities provide a compact consistency check for any proposed formula.

Check Your Understanding

Use the definitions and identities above to answer the following questions.

Why does \(E[e^{itX}]\) exist for every real \(t\), even when \(X\) has no finite expectation?
If \(\varphi_X(t)=a(t)+ib(t)\), state the parity of \(a\) and \(b\), and explain how it follows from conjugate symmetry.
For the two-point distribution with probabilities \(1/3\) at \(-1\) and \(2/3\) at \(2\), write down the real and imaginary parts of its characteristic function.
Which identity rules out \(q(t)=1+it\) as a characteristic function, and for which values of \(t\) does the violation occur?
Do normalization, the modulus bound, and conjugate symmetry alone prove that a proposed function is a characteristic function? Explain.