Why a Characteristic Function Identifies a Law
A characteristic function is more than a convenient way to encode moments or simplify sums: it determines the entire distribution. This means that if a complicated random variable has the same characteristic function as a familiar one, its law is the familiar law, even when finding its density directly would be difficult. The key point is that equality must hold for every real argument.
Let \(X\) be a real-valued random variable on a probability space \((\Omega,\mathcal{F},\mathbb{P})\). Its characteristic function is \(\varphi_X(t)=\mathbb{E}[e^{itX}]\), for \(t\in\mathbb{R}\). Write \(F_X(x)=\mathbb{P}(X\le x)\) for its cumulative distribution function. Two real-valued random variables have the same distribution when their cumulative distribution functions agree at every real \(x\).
The converse follows directly because \(\mathbb{E}[e^{itX}]\) depends only on the distribution of \(X\). The forward direction is the substantive claim. We prove it by adding a small independent Gaussian noise to each variable. The resulting variables have smooth densities that can be expressed directly in terms of the characteristic functions. When the noise shrinks to zero, their distribution functions recover the original ones.
Gaussian Smoothing
Proof. Define \(I(u)=\int_{-\infty}^{\infty}e^{-\sigma^2t^2/2}e^{-itu}\,dt\). The integral is absolutely convergent, and differentiating under the integral is justified because \(|t|e^{-\sigma^2t^2/2}\) is integrable. Integration by parts, with the Gaussian factor vanishing at both endpoints, gives $$ \int_{-\infty}^{\infty}t e^{-\sigma^2t^2/2}e^{-itu}\,dt =-\frac{iu}{\sigma^2}I(u). $$ Consequently, \(I'(u)=-i\int t e^{-\sigma^2t^2/2}e^{-itu}\,dt=-uI(u)/\sigma^2\). Solving this differential equation gives \(I(u)=I(0)e^{-u^2/(2\sigma^2)}\). The Gaussian integral, obtained by squaring the integral and using polar coordinates, gives \(I(0)=\sqrt{2\pi}/\sigma\). Substitution yields the displayed identity. \(\square\)
Now let \(Z\) be a standard normal random variable independent of \(X\); if necessary, form the product of the original probability space with a space carrying \(Z\). For \(\sigma>0\), the random variable \(X+\sigma Z\) has a density. Conditional on \(X\), its density at \(x\) is the Gaussian density evaluated at \(x-X\). Averaging over \(X\) gives $$ f_{X+\sigma Z}(x)=\mathbb{E}\left[\frac{1}{\sqrt{2\pi}\sigma}e^{-(x-X)^2/(2\sigma^2)}\right] =\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-itx}\varphi_X(t)e^{-\sigma^2t^2/2}\,dt. $$ For the second equality, apply the Gaussian Fourier identity with \(u=x-X\) and interchange expectation and integration. This interchange is valid because the absolute value of the integrand before taking expectation is bounded by \(e^{-\sigma^2t^2/2}\), whose integral is finite.
Proof of the forward direction of the theorem. Assume \(\varphi_X(t)=\varphi_Y(t)\) for every \(t\in\mathbb{R}\). For each \(\sigma>0\), the density formula just proved gives \(f_{X+\sigma Z}(x)=f_{Y+\sigma Z}(x)\) at every \(x\), using independent standard normal noise for each variable if they are on different probability spaces. Thus the smoothed variables have the same distribution.
We next show that their distribution functions approach the original distribution functions at continuity points. For any \(\varepsilon>0\), the event \(\{X\le x-\varepsilon,\ |\sigma Z|\le\varepsilon\}\) implies \(X+\sigma Z\le x\). Also, if \(X+\sigma Z\le x\) and \(|\sigma Z|\le\varepsilon\), then \(X\le x+\varepsilon\). Therefore $$ F_X(x-\varepsilon)-\mathbb{P}(|\sigma Z|>\varepsilon) \le \mathbb{P}(X+\sigma Z\le x) \le F_X(x+\varepsilon)+\mathbb{P}(|\sigma Z|>\varepsilon). $$ Since \(\mathbb{E}[(\sigma Z)^2]=\sigma^2\), Markov's inequality applied to \((\sigma Z)^2\) gives \(\mathbb{P}(|\sigma Z|>\varepsilon)\le \sigma^2/\varepsilon^2\). If \(x\) is a continuity point of \(F_X\), first let \(\sigma\) decrease to zero in the bounds and then let \(\varepsilon\) decrease to zero. The result is \(\mathbb{P}(X+\sigma Z\le x)\to F_X(x)\). The same argument holds for \(Y\).
It follows that \(F_X(x)=F_Y(x)\) at every point where both distribution functions are continuous, because the smoothed distribution functions agree for every \(\sigma>0\). Such points are dense: each discontinuity of a distribution function is a positive jump, and there are at most countably many positive jumps, since the sum of the jumps over any finite collection of points is at most \(1\). For any real \(x\), choose common continuity points \(x_n\) with \(x<x_n<x+1/n\). Right-continuity of distribution functions gives \(F_X(x)=\lim_nF_X(x_n)=\lim_nF_Y(x_n)=F_Y(x)\). Thus the distributions agree everywhere, proving the forward direction. \(\square\)
Identifying Distributions by Matching
In practice, the theorem is used by calculating a characteristic function and recognizing a known expression. A candidate law must itself be a probability distribution, and the match must hold for every real \(t\), not just at selected values. The Gaussian Fourier identity above gives the familiar formula \(\varphi_X(t)=e^{i\mu t-vt^2/2}\) for a normal random variable with mean \(\mu\) and variance \(v>0\).
Worked Example: Recognizing a Normal Distribution
Suppose a real-valued random variable \(X\) has characteristic function \(\varphi_X(t)=e^{3it-8t^2}\) for every \(t\in\mathbb{R}\). Identify its distribution.
Compare the exponent with the characteristic function of a normal distribution: $$ e^{i\mu t-vt^2/2}. $$ Matching the coefficient of \(it\) gives \(\mu=3\). Matching the coefficient of \(t^2\) gives \(v/2=8\), so \(v=16\). Thus the candidate law is \(N(3,16)\). Its characteristic function is $$ e^{i(3)t-(16)t^2/2}=e^{3it-8t^2}, $$ which agrees with \(\varphi_X(t)\) for every real \(t\). The uniqueness theorem therefore gives \(X\sim N(3,16)\).
Check: at \(t=0\), both characteristic functions equal \(e^0=1\), as required. At \(t=1\), both equal \(e^{3i-8}\). The match is not limited to these checks: the exponent calculation establishes equality for all \(t\).
Worked Example: Recovering a Three-Point Distribution
Suppose \(X\) has characteristic function \(\varphi_X(t)=\tfrac13+\tfrac23\cos t\) for every real \(t\). Identify its distribution.
Consider a variable \(Y\) with probability \(1/3\) at each of \(-1,0,1\). These probabilities are nonnegative and sum to \(1/3+1/3+1/3=1\). Its characteristic function is $$ \varphi_Y(t)=\tfrac13 e^{-it}+\tfrac13+\tfrac13 e^{it} =\tfrac13+\tfrac13(e^{-it}+e^{it}) =\tfrac13+\tfrac23\cos t. $$ Thus \(\varphi_X(t)=\varphi_Y(t)\) for every real \(t\), so uniqueness gives \(\mathbb{P}(X=-1)=\mathbb{P}(X=0)=\mathbb{P}(X=1)=1/3\).
Check: the proposed probabilities sum to \(1\), and their characteristic function reproduces the given expression exactly. In particular, at \(t=\pi\), the expression is \(1/3+(2/3)(-1)=-1/3\), which also equals \(\tfrac13e^{-i\pi}+\tfrac13+\tfrac13e^{i\pi}=-1/3\).
Worked Example: A Sum with a Cumbersome Density Calculation
Let \(U_1,\ldots,U_5\) be independent standard normal random variables on a probability space, and set \(X=U_1+2U_2+3U_3+4U_4+5U_5\). Identify the distribution of \(X\) without computing its density by repeated convolution.
The characteristic function of a standard normal variable is \(e^{-t^2/2}\). By independence, the characteristic function of a sum of independent variables is the product of their characteristic functions. Therefore $$ \begin{aligned} \varphi_X(t) &=e^{-t^2/2}e^{-(2t)^2/2}e^{-(3t)^2/2}e^{-(4t)^2/2}e^{-(5t)^2/2}\\ &=\exp\left(-\frac{t^2}{2}(1+4+9+16+25)\right) =e^{-55t^2/2}. \end{aligned} $$ Here the squared coefficients add to \(1+4+9+16+25=55\). The last expression is the characteristic function of \(N(0,55)\), so uniqueness gives \(X\sim N(0,55)\). This avoids deriving the density through four successive convolution integrals.
Check: the candidate characteristic function at \(t=0\) is \(e^0=1\). Substituting the five coefficients in the product gives the exponent \(-t^2/2-4t^2/2-9t^2/2-16t^2/2-25t^2/2=-55t^2/2\), exactly the exponent for \(N(0,55)\).
Common Pitfall: Matching Too Little
A characteristic function must match at every real argument. Equality at one argument gives only one equation, and equality of absolute values discards information about location. For a concrete example, let \(X=1\) with probability \(1\) and \(Y=-1\) with probability \(1\). Then \(\varphi_X(t)=e^{it}\) and \(\varphi_Y(t)=e^{-it}\). Both have modulus \(1\) for every \(t\), but the distributions are different: \(X\) is always \(1\), whereas \(Y\) is always \(-1\). At \(t=0\), their characteristic functions agree, since both equal \(1\); that single match does not identify a law. The uniqueness theorem applies only when the characteristic functions themselves agree for all real \(t\).
The theorem is useful whenever a transform calculation is easier than a direct probability calculation. It turns a verified identity into a distributional conclusion, but it does not replace the need to compute the transform correctly or to check that the comparison law is valid.
Check Your Understanding
For each question, use equality of characteristic functions only when it holds for every real argument.
- If \(\varphi_X(t)=e^{-2it-9t^2/2}\) for all real \(t\), what normal distribution does uniqueness identify?
- Find the characteristic function of a variable that takes values \(-2\) and \(2\), each with probability \(1/2\), and state what a matching characteristic function would identify.
- Why does equality of characteristic functions imply equality of the densities after adding independent \(N(0,\sigma^2)\) noise for \(\sigma>0\)?
- In the smoothing proof, why is \(\mathbb{P}(|\sigma Z|>\varepsilon)\le \sigma^2/\varepsilon^2\) useful when \(\sigma\) decreases to zero?
- Give two distinct distributions whose characteristic functions agree at \(t=0\), and explain why this does not contradict uniqueness.