Uniqueness of Distributions from Characteristic Functions

Why Uniqueness Matters

A characteristic function packages a probability law into a function of a real argument. If two random variables have the same characteristic function, they give the same value to every frequency $t$, but it is not immediate that this forces their probability distributions to agree. The key is to add a small independent Gaussian noise: the resulting distributions have smooth densities that can be expressed directly in terms of the characteristic functions. As the noise becomes small, those smoothed distributions recover the original laws.

Definition: For a real-valued random variable $X$, its characteristic function is $\phi_X(t)=E[e^{itX}]$, $t\in\mathbb R$. Its distribution function is $F_X(x)=P(X\le x)$. Two random variables have the same distribution if their distribution functions agree for every $x\in\mathbb R$.

The result in this tutorial is specific and fundamental: if $\phi_X(t)=\phi_Y(t)$ for every real $t$, then $X$ and $Y$ have the same distribution. The proof will not assume that either law has a density or has finite moments.

Gaussian Smoothing

Let $\mu$ be any probability distribution on $\mathbb R$. For $\varepsilon>0$, define the centered Gaussian density with standard deviation $\varepsilon$ by

g_\varepsilon(x)=\frac{1}{\sqrt{2\pi}\,\varepsilon} \exp\left(-\frac{x^2}{2\varepsilon^2}\right).

Take a random variable $X$ with law $\mu$, and take $Z\sim N(0,1)$ independent of $X$. The law of $X+\varepsilon Z$ is the Gaussian-smoothed law of $\mu$. It has density

p_{\mu,\varepsilon}(x) =\int_{\mathbb R} g_\varepsilon(x-u)\,\mu(du).

This formula follows by conditioning on $X=u$: given that value, $X+\varepsilon Z$ has density $g_\varepsilon(x-u)$. Integrating those conditional densities over the law of $X$ gives the displayed mixture. The construction is valid for every probability law $\mu$, including discrete and singular laws.

The Gaussian Fourier identity, with the convention $\phi_X(t)=E[e^{itX}]$, is

g_\varepsilon(z)=\frac{1}{2\pi}\int_{\mathbb R} e^{-itz}e^{-\varepsilon^2t^2/2}\,dt.

It is the usual characteristic-function formula for a centered normal distribution, written as an integral for its density. Substituting this identity into the mixture formula gives the useful representation below. The interchange of integrals is justified because $\int_{\mathbb R}e^{-\varepsilon^2t^2/2}\,dt$ is finite and $|e^{itu}|=1$.

Smoothed-density formula: If $\phi_\mu(t)=\int e^{itu}\,\mu(du)$, then for every $\varepsilon>0$ and $x\in\mathbb R$, $$ p_{\mu,\varepsilon}(x) =\frac{1}{2\pi}\int_{\mathbb R} e^{-itx}\phi_\mu(t)e^{-\varepsilon^2t^2/2}\,dt. $$

Indeed, after substituting $z=x-u$, the factor $e^{-it(x-u)}$ separates into $e^{-itx}e^{itu}$. Integrating $e^{itu}$ against $\mu$ yields $\phi_\mu(t)$. The Gaussian factor also makes the final integral absolutely convergent, since $|\phi_\mu(t)|\le 1$.

Worked Example: Smoothing a Two-Point Distribution

Let $B$ take the value $0$ with probability $2/3$ and the value $1$ with probability $1/3$. Its characteristic function is $$ \phi_B(t)=\frac{2}{3}+\frac{1}{3}e^{it}, $$ because $E[e^{itB}]=(2/3)e^0+(1/3)e^{it}$. For independent $Z\sim N(0,1)$, the density of $B+\varepsilon Z$ is the mixture $$ p_\varepsilon(x)=\frac{2}{3}g_\varepsilon(x)+\frac{1}{3}g_\varepsilon(x-1). $$ Each term is a shifted Gaussian density, weighted by the probability of the corresponding value of $B$. The smoothed distribution function at $x=1/2$ is $$ P(B+\varepsilon Z\le 1/2) =\frac{2}{3}\Phi\left(\frac{1}{2\varepsilon}\right) +\frac{1}{3}\Phi\left(-\frac{1}{2\varepsilon}\right), $$ where $\Phi$ is the standard normal distribution function. As $\varepsilon\downarrow0$, the two normal probabilities tend to $1$ and $0$, respectively, so the expression tends to $2/3=P(B\le1/2)$. This illustrates recovery of the original distribution at a continuity point.

Recovering the Original Distribution

The smoothing noise vanishes as $\varepsilon$ decreases, but convergence of the smoothed distribution functions is guaranteed at continuity points, not necessarily at every point. The following argument establishes exactly the convergence needed for uniqueness.

Lemma: Let $X$ be real-valued and let $Z\sim N(0,1)$ be independent of $X$. At every continuity point $x$ of $F_X$, $$ P(X+\varepsilon Z\le x)\longrightarrow F_X(x) \quad\text{as }\varepsilon\downarrow0. $$

Proof. Fix $\delta>0$. On the event $\{X\le x-\delta,\ |\varepsilon Z|\le\delta\}$, we have $X+\varepsilon Z\le x$. Conversely, if $X+\varepsilon Z\le x$ and $|\varepsilon Z|\le\delta$, then $X\le x+\delta$. These event inclusions give

F_X(x-\delta)-P(|\varepsilon Z|>\delta) \le P(X+\varepsilon Z\le x) \le F_X(x+\delta)+P(|\varepsilon Z|>\delta).

For fixed $\delta$, the error probability tends to zero. For example, Markov's inequality applied to $Z^2$ gives $P(|\varepsilon Z|>\delta)\le \varepsilon^2/\delta^2$, since $E[Z^2]=1$. Taking lower and upper limits as $\varepsilon\downarrow0$, and then letting $\delta\downarrow0$, gives the claimed limit: at a continuity point, both $F_X(x-\delta)$ and $F_X(x+\delta)$ tend to $F_X(x)$. $\square$

The Uniqueness Theorem

Theorem (Uniqueness of characteristic functions): If $X$ and $Y$ are real-valued random variables and $$ \phi_X(t)=\phi_Y(t)\quad\text{for every }t\in\mathbb R, $$ then $X$ and $Y$ have the same distribution.

Proof. Write $\mu$ and $\nu$ for the laws of $X$ and $Y$. For each $\varepsilon>0$, smooth each law by adding an independent $N(0,\varepsilon^2)$ random variable. The smoothed-density formula shows that, for every $x$,

p_{\mu,\varepsilon}(x) =\frac{1}{2\pi}\int_{\mathbb R}e^{-itx} \phi_X(t)e^{-\varepsilon^2t^2/2}\,dt =\frac{1}{2\pi}\int_{\mathbb R}e^{-itx} \phi_Y(t)e^{-\varepsilon^2t^2/2}\,dt =p_{\nu,\varepsilon}(x).

Thus the smoothed laws have identical densities and hence identical distribution functions for every $\varepsilon>0$. Now take any $x$ at which both $F_X$ and $F_Y$ are continuous. Applying the lemma to both variables and letting $\varepsilon\downarrow0$ in the equality of smoothed distribution functions gives $F_X(x)=F_Y(x)$.

It remains to extend equality from common continuity points to every real $x$. A distribution function is nondecreasing and has at most countably many discontinuities; each discontinuity is a positive jump, and there can be only countably many positive jumps because their total is at most one. Therefore the union of the discontinuity sets of $F_X$ and $F_Y$ is countable. For any $x$, choose common continuity points $x_k$ with $x<x_k<x+1/k$. Then $x_k\downarrow x$ can be arranged, and right-continuity gives $$ F_X(x)=\lim_{k\to\infty}F_X(x_k) =\lim_{k\to\infty}F_Y(x_k)=F_Y(x). $$ The distribution functions agree everywhere, so the laws are equal. $\square$

Worked Example: Comparing Two Normal Laws

Suppose $X\sim N(2,9)$ and $Y\sim N(2,9)$. Their characteristic functions, using variance as the second normal parameter, are $$ \phi_X(t)=\exp\left(2it-\frac{9t^2}{2}\right) \quad\text{and}\quad \phi_Y(t)=\exp\left(2it-\frac{9t^2}{2}\right). $$ They agree for every $t$, so the uniqueness theorem confirms that the two variables have the same distribution.

More generally, for $X\sim N(\mu_1,\sigma_1^2)$ and $Y\sim N(\mu_2,\sigma_2^2)$ with positive variances, equality of characteristic functions implies equality of their moduli: $$ e^{-\sigma_1^2t^2/2}=e^{-\sigma_2^2t^2/2}\quad\text{for every }t. $$ Taking any nonzero $t$ gives $\sigma_1^2=\sigma_2^2$. The remaining equality is $e^{it(\mu_1-\mu_2)}=1$ for all $t$, which forces $\mu_1=\mu_2$. Thus the parameters must match as well.

Worked Example: Distinguishing a Uniform Law from a Triangular Law

Let $U$ have the uniform distribution on $[-1,1]$, with density $1/2$ there. For $t\ne0$, $$ \phi_U(t)=\frac12\int_{-1}^{1}e^{itx}\,dx =\frac{e^{it}-e^{-it}}{2it} =\frac{\sin t}{t}. $$ Now let $T$ have density $1-|x|$ on $[-1,1]$ and zero elsewhere. This is a probability density because $$ \int_{-1}^{1}(1-|x|)\,dx =2\int_0^1(1-x)\,dx=1. $$ By symmetry, $\phi_T(t)=2\int_0^1(1-x)\cos(tx)\,dx$. For $t\ne0$, integration gives $$ \int_0^1(1-x)\cos(tx)\,dx=\frac{1-\cos t}{t^2}, \qquad \phi_T(t)=\frac{2(1-\cos t)}{t^2}. $$ At $t=\pi$, $\phi_U(\pi)=0$, whereas $\phi_T(\pi)=4/\pi^2$. The characteristic functions differ, so the distributions cannot be equal. This example uses the contrapositive of uniqueness: unequal characteristic functions rule out equality of laws.

Common Pitfall and Practical Meaning

A tempting but incorrect shortcut is to say that $X+\varepsilon Z$ approaches $X$, and therefore its distribution function approaches $F_X$ at every point. The Bernoulli example shows why the qualification about continuity points matters. At the atom $x=0$, $$ P(B+\varepsilon Z\le0) =\frac{2}{3}\Phi(0)+\frac{1}{3}\Phi(-1/\varepsilon) \longrightarrow \frac{1}{3}, $$ whereas $P(B\le0)=2/3$. The added symmetric noise places asymptotically half of the mass at zero on each side of zero. The correct recovery argument uses continuity points, then right-continuity to determine the distribution function everywhere.

Uniqueness makes characteristic functions useful even when a law has no density or its moments do not exist. Once a characteristic function has been computed, it identifies at most one probability distribution. The theorem does not itself perform a distributional identification or provide a general inversion formula; it guarantees that any valid identification based on matching characteristic functions is conclusive.

Key takeaway: Equal characteristic functions give equal Gaussian-smoothed densities at every positive smoothing scale. Those smoothed laws converge back to their original distribution functions at continuity points, which is enough to establish equality of the original laws everywhere.

Check Your Understanding

Try to answer each question using the definitions and arguments above.

Why does Gaussian smoothing produce a density even if the original law is discrete?
In the smoothed-density formula, what ensures that the Fourier integral is absolutely convergent?
Why does the recovery lemma require $x$ to be a continuity point of the original distribution function?
Two random variables have characteristic functions that agree for every real $t$. What does the uniqueness theorem conclude?
For the Bernoulli variable in the example, why does the smoothed distribution function at zero not converge to $P(B\le0)$?