Characteristic Functions

(1) $\varphi_X(0) = \mathbb E[e^{i0\cdot X}] = \mathbb E[1] = 1$.

(2) By the triangle inequality for expectations (Jensen's inequality applied to the convex function $|\cdot|$, or directly $|\mathbb E[Z]|\le \mathbb E[|Z|]$ for complex $Z$),

since $|e^{itX}| = 1$ pointwise.

(3) Fix $t,h\in\mathbb R$. Then

using $|e^{itX}|=1$ and the triangle inequality. The bound $\mathbb E[|e^{ihX}-1|]$ does not depend on $t$. The integrand $|e^{ihX}-1|$ is bounded by $2$ and tends to $0$ pointwise as $h\to 0$ (for every fixed outcome $X=x$, $e^{ihx}\to 1$ by continuity of $t\mapsto e^{itx}$). By the Dominated Convergence Theorem (dominating function the constant $2$, integrable since $\mathbb E[2]=2<\infty$),

Since the bound is uniform in $t$, this shows uniform continuity: for every $\varepsilon>0$ there is $\delta>0$ such that $|h|<\delta$ implies $\sup_t |\varphi_X(t+h)-\varphi_X(t)|<\varepsilon$.

(4) $\varphi_X(-t) = \mathbb E[e^{-itX}] = \mathbb E[\overline{e^{itX}}] = \overline{\mathbb E[e^{itX}]} = \overline{\varphi_X(t)}$, using that conjugation commutes with expectation (linearity applied to real and imaginary parts).

(5) $\varphi_{aX+b}(t) = \mathbb E[e^{it(aX+b)}] = e^{ibt}\,\mathbb E[e^{i(at)X}] = e^{ibt}\varphi_X(at)$.

(6) Suppose $\mathbb E[|X|^n]<\infty$. We show $\varphi_X$ is $n$ times differentiable by induction, differentiating under the expectation sign. For $k \le n$, define $g_k(t) = \mathbb E[(iX)^k e^{itX}]$; we claim $\varphi_X^{(k)}(t) = g_k(t)$. The base case $k=0$ is the definition. For the inductive step, suppose $\varphi_X^{(k-1)}(t) = \mathbb E[(iX)^{k-1}e^{itX}]$ for some $k \le n$. Consider the difference quotient

Using the elementary inequality $\left|\frac{e^{iu}-1}{u}\right| \le 1$ for all real $u\ne 0$ (which follows from $|e^{iu}-1| = 2|\sin(u/2)| \le |u|$), we get

which is integrable since $k \le n$ and $\mathbb E[|X|^n]<\infty$ implies $\mathbb E[|X|^k]<\infty$ for $k\le n$ (by Lyapunov/monotonicity of moments on a probability space). As $h\to 0$, $\frac{e^{ihX}-1}{h}\to iX$ pointwise. By Dominated Convergence,

completing the induction. Setting $t=0$: $\varphi_X^{(k)}(0) = \mathbb E[(iX)^k] = i^k\mathbb E[X^k]$. $\blacksquare$

We use the Fourier inversion formula, proved as a lemma first, and then deduce uniqueness as a corollary. The technique is direct computation and interchange of integrals (justified by Fubini), followed by a limiting/dominated convergence argument.

Step 1 (Inversion formula for intervals). We claim that for any $a<b$ real numbers that are continuity points of $\mu_X$ (i.e. $\mu_X(\{a\})=\mu_X(\{b\})=0$),

Fix $T>0$ and define

Substituting $\varphi_X(t) = \int_{\mathbb R} e^{itx}\,d\mu_X(x)$ gives

The integrand, as a function of $(t,x)$, is bounded: since $\left|\frac{e^{iu}-e^{iv}}{i}\right| \le |u-v|$... more precisely we bound each piece using $\left|\frac{e^{itu}-1}{it}\right|\le |u|$ (shown in the Lemma's part 6 computation), so the integrand is bounded uniformly in $t\in[-T,T]$ by $(b-a)$, and is a jointly measurable function of $(t,x)$ on the finite-measure space $[-T,T]\times \mathbb R$ (finite in $t$, probability measure in $x$). Fubini's Theorem applies, so we may swap the order of integration:

Define, for real $u$, $J_T(u) := \frac{1}{2\pi}\int_{-T}^{T}\frac{e^{itu}-1}{it}\,dt \cdot (-1)$... Let us instead directly compute the bracketed inner integral. Write

It is cleaner to use the classical Dirichlet integral representation: define

and use the known limit (Dirichlet integral):

with the integral equal to $0$ when $u=0$; moreover $\left|\int_0^T \frac{\sin(tu)}{t}dt\right|$ is bounded uniformly in $T,u$ by a universal constant $M$ (a standard fact about the Dirichlet integral). Since $\frac{e^{itu}-1}{it}$ has imaginary part $-\frac{\cos(tu)-1}{t}$, an odd function of $t$ integrating to $0$ over $[-T,T]$, and real part $\frac{\sin(tu)}{t}$, an even function, one computes directly

Applying this with $u = x-a$ and $u=x-b$ (noting $\frac{e^{it(x-a)}-e^{it(x-b)}}{it} = \frac{e^{it(x-a)}-1}{it} - \frac{e^{it(x-b)}-1}{it}$) gives

As $T\to\infty$, using the Dirichlet limit, this tends to

where $h(x) = 1$ for $a<x<b$, $h(x)=0$ for $x<a$ or $x>b$, and $h(x) = 1/2$ at $x=a$ or $x=b$. The whole quantity is uniformly bounded in $(T,x)$ by the universal Dirichlet bound $2M/\pi$. By the Bounded (Dominated) Convergence Theorem applied to the probability measure $\mu_X$,

If $a,b$ are continuity points, $\mu_X(\{a\}) = \mu_X(\{b\}) = 0$, so this equals $\mu_X((a,b)) = \mu_X((a,b])$. This proves the inversion formula.

Step 2 (Uniqueness follows). Suppose $\varphi_X(t) = \varphi_Y(t)$ for all $t$. By Step 1, for all continuity points $a<b$ of both $\mu_X$ and $\mu_Y$ (all but at most countably many reals are continuity points of a given measure, since a measure has at most countably many atoms; the union of two countable atom-sets is still countable, so continuity points of both are co-countable and in particular dense),

since $\varphi_X\equiv \varphi_Y$. Thus $\mu_X$ and $\mu_Y$ agree on all intervals $(a,b]$ with $a,b$ common continuity points; by right-continuity of the CDFs and density of continuity points, $\mu_X$ and $\mu_Y$ agree on all intervals $(a,b]$, hence (since these generate the Borel $\sigma$-algebra and form a $\pi$-system) $\mu_X = \mu_Y$ on all of $\mathcal B(\mathbb R)$ by the standard $\pi$-$\lambda$ uniqueness argument for measures agreeing on a generating $\pi$-system with total mass $1<\infty$. Hence $X$ and $Y$ have the same distribution. $\blacksquare$

The technique is direct expansion via independence and induction on $n$.

Base case $n=2$. For independent $X_1, X_2$,

Write $e^{itX_1} = \cos(tX_1)+i\sin(tX_1)$ and similarly for $X_2$; the product $e^{itX_1}e^{itX_2}$ is a (complex) linear combination of the four real products $\cos(tX_1)\cos(tX_2)$, $\cos(tX_1)\sin(tX_2)$, $\sin(tX_1)\cos(tX_2)$, $\sin(tX_1)\sin(tX_2)$. Since $X_1 \perp X_2$, any bounded measurable function of $X_1$ is independent of any bounded measurable function of $X_2$ (a standard consequence of the definition of independence of random variables, established when studying Expectation of Functions of Several Random Variables), and expectation of a product of independent random variables factors:

and similarly for the other three terms (each factor is bounded, hence integrable, so the factorization identity for expectations of products of independent bounded random variables applies without extra integrability hypotheses). Recombining,

(This factorization can also be seen directly: $\mathbb E[ZW] = \mathbb E[Z]\mathbb E[W]$ for independent complex random variables $Z=e^{itX_1}, W=e^{itX_2}$, since this identity for real and imaginary parts is exactly the four real identities above.)

Inductive step. Suppose the claim holds for $n-1$ independent random variables. Given independent $X_1,\dots,X_n$, note that $S_{n-1} = X_1+\cdots+X_{n-1}$ is a function of $(X_1,\dots,X_{n-1})$ and is independent of $X_n$ (a function of independent random variables is independent of the remaining ones — a standard grouping property of independence). By the base case applied to the pair $(S_{n-1}, X_n)$,

By the inductive hypothesis, $\varphi_{S_{n-1}}(t) = \prod_{k=1}^{n-1}\varphi_{X_k}(t)$, so

By induction the identity holds for all $n\ge 2$ (and trivially for $n=1$). $\blacksquare$

Immediate from the theorem with $\varphi_{X_1}=\cdots=\varphi_{X_n}=\varphi$. $\blacksquare$

Forward direction. For fixed $t$, the map $x \mapsto \cos(tx)$ and $x\mapsto \sin(tx)$ are bounded and continuous functions of $x$. By the definition of convergence in distribution (equivalently, the Portmanteau characterization: $X_n\xrightarrow d X$ iff $\mathbb E[g(X_n)] \to \mathbb E[g(X)]$ for every bounded continuous $g:\mathbb R\to\mathbb R$),

Hence $\varphi_n(t) = \mathbb E[\cos(tX_n)]+i\mathbb E[\sin(tX_n)] \to \mathbb E[\cos(tX)]+i\mathbb E[\sin(tX)] = \varphi(t)$ for every fixed $t$. (Uniformity on bounded intervals follows from a standard strengthening of Portmanteau via tightness, which we do not detail further as it is not needed for the converse direction proved next.)

Converse direction. This is the deeper half; we sketch the standard argument via tightness, citing the two classical facts it rests on. Step 1: Tightness from continuity at $0$. We first show the family $\{X_n\}$ is tight: for every $\varepsilon>0$ there exists $M<\infty$ such that $\mathbb P(|X_n|>M) < \varepsilon$ for all $n$. This uses the elementary inequality, valid for any random variable $Y$ and any $u>0$,

which is proved by writing the right side as $\frac{1}{u}\int_{-u}^u \mathbb E[1-\cos(tY)]\,dt = \mathbb E\left[\frac{1}{u}\int_{-u}^u(1-\cos(tY))\,dt\right] = \mathbb E\left[2 - \frac{2\sin(uY)}{uY}\right]$ (interpreting the value at $Y=0$ as $0$ by continuity), and observing that $2 - \frac{2\sin(uy)}{uy} \ge 2 - \frac{2}{u|y|} \ge 1$ whenever $|y|>2/u$ (using $|\sin(uy)|\le 1$), so

Applying this to $Y=X_n$: since $\varphi_n(t)\to \psi(t)$ pointwise and $|1-\varphi_n(t)|\le 2$, by Dominated Convergence,

Because $\psi$ is continuous at $0$ with $\psi(0)=\lim \varphi_n(0) = 1$, the right side tends to $0$ as $u\to 0^+$ (the integrand $1-\psi(t)\to 0$ as $t\to0$, so the average over $[-u,u]$ vanishes as $u\to 0$). Choosing $u$ small makes $\frac1u\int_{-u}^u(1-\psi(t))dt < \varepsilon/2$, and then choosing $n$ large controls the finite-$n$ discrepancy; combined with the finitely many remaining $X_1,\dots,X_{n_0}$ (each individually tight, being a single random variable), this yields tightness of the whole sequence. We take this construction as establishing: $\sup_n \mathbb P(|X_n| > 2/u) \le \varepsilon$ for $u$ small enough, which is tightness. Step 2: Extracting a weak limit via Helly's Selection Theorem. Tightness plus Helly's Selection Theorem (every sequence of CDFs has a subsequence converging weakly to some possibly-defective distribution function; tightness rules out mass escaping to infinity, ensuring the limit is a genuine CDF) gives a subsequence $(X_{n_k})$ and a random variable $X$ such that $X_{n_k} \xrightarrow{d} X$. By the forward direction already proved, $\varphi_{n_k}(t) \to \varphi_X(t)$ for every $t$. But by hypothesis $\varphi_{n_k}(t)\to \psi(t)$ along the full sequence, hence along the subsequence too. By uniqueness of pointwise limits, $\varphi_X(t) = \psi(t)$ for all $t$, so $\psi$ is indeed a characteristic function, namely of $X$. Step 3: The full sequence converges. We must show $X_n \xrightarrow{d} X$ along the entire sequence, not just the subsequence. Suppose not: then there is a bounded continuous $g$ and $\varepsilon>0$ with $|\mathbb E[g(X_n)] - \mathbb E[g(X)]| \ge \varepsilon$ along some further subsequence $(n_j)$. But $(X_{n_j})$ is itself tight (as a subsequence of a tight sequence), so by Helly's theorem again it has a further subsequence converging weakly to some $X'$; by the forward direction, $\varphi_{X'} = \psi = \varphi_X$, so by the Uniqueness Theorem, $X' \stackrel d= X$. This contradicts $|\mathbb E[g(X_{n_j})]-\mathbb E[g(X)]|\ge \varepsilon$ persisting along a further subsequence converging to $X$ in distribution. Hence no such subsequence exists, and $X_n \xrightarrow{d} X$ along the full sequence. $\blacksquare$