4 The characteristic function

4.1 Definition and first properties

Definition 4.1 Characteristic function

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Let \(\mu \) be a measure on a real inner product space \(E\). The characteristic function of \(\mu \), denoted by \(\hat{\mu }\), is the function \(E \to \mathbb {C}\) defined by

\begin{align*} \hat{\mu }(t) = \int _x e^{i \langle t, x \rangle } d\mu (x) \: . \end{align*}

The characteristic function of a random variable \(X\) is defined as the characteristic function of \(\mathcal L(X)\).

Lemma 4.2

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For all \(t\), \(\Vert \hat{\mu }(t)\Vert \le 1\).

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Lemma 4.3

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If \(\mu [|X|^n] {\lt} \infty \) then the characteristic function of \(\mu \) is continuously differentiable of order \(n\).

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Lemma 4.4

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The characteristic function is a continuous function.

Proof ▶

Lemma 4.5

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\(\hat{\mu }(-t) = \overline{\hat{\mu }(t)}\).

Proof ▶

Lemma 4.6

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For \(a \in \mathbb {R}\) and \(X\) a random variable with law \(\mu \), the characteristic function of \(a X\) is \(t \mapsto \hat{\mu }(at)\).

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Lemma 4.7

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If two random variables \(X, Y : \Omega \to S\) with laws \(\mu \) and \(\nu \) are independent, then \(X+Y\) has characteristic function \(\hat{\mu } + \hat{\nu }\).

Proof ▶

Lemma 4.8

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\begin{align*} \int _{-r}^r e^{i t} \, d t = 2 \sin r \: . \end{align*}

Proof ▶

Lemma 4.9

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\begin{align*} \int _{-r}^r \hat{\mu }(t) dt = 2 r \int \frac{\sin (r x)}{r x} \, d\mu (x) \end{align*}

Proof ▶

We will use Fubini’s theorem to exchange the order of integration.

\begin{align*} \int _{-r}^r \hat{\mu }(t) \, dt & = \int _{-r}^r \int e^{i t x} \, d\mu (x) \, dt \\ & = \int \left( \int _{-r}^r e^{i t x} \, dt \right) \, d\mu (x) \\ & = \int 2 \frac{\sin (r x)}{x} \, d\mu (x) \\ & = 2 r \int \frac{\sin (r x)}{r x} \, d\mu (x) \end{align*}

Lemma 4.10

For \(\mu \) a probability measure on \(\mathbb {R}\) and \(r {\gt} 0\),

\begin{align*} \mu [-r, r] & \le 2 r \int _{-1/r}^{1/r} |\hat{\mu }(t)| dt \: . \end{align*}

Proof ▶

TODO

Lemma 4.11

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For \(\mu \) a probability measure on \(\mathbb {R}\) and \(r {\gt} 0\),

\begin{align*} \mu \left\{ x \mid |x| {\gt} r\right\} & \le \frac{r}{2} \int _{-2/r}^{2/r} (1 - \hat{\mu }(t))dt \: . \end{align*}

Proof ▶

We will use Lemma 4.9 with \(r\) replaced by \(2/r\).

\begin{align*} \frac{r}{2} \int _{-2/r}^{2/r} (1 - \hat{\mu }(t))dt & = 2 - \frac{r}{2} \int _{-2/r}^{2/r} \hat{\mu }(t)dt \\ & = 2 - 2 \int \frac{\sin (2 x / r)}{2 x / r} \, d\mu (x) \\ & = 2 \int \left(1 - \frac{\sin (2 x / r)}{2 x / r}\right) \, d\mu (x) \\ & \ge 2 \int _{|x| \ge r} \left(1 - \frac{\sin (2 x / r)}{2 x / r}\right) \, d\mu (x) \\ & \ge \mu \{ x \mid |x| {\gt} r\} \: . \end{align*}

In the end we used that \(\sin x \le x /2\) for \(x \ge 2\).

Lemma 4.12

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For \(\mu \) a probability measure on an inner product space \(E\) and \(r {\gt} 0\), \(a \in E\),

\begin{align*} \mu \left\{ x \mid \vert \langle a, x\rangle \vert {\gt} r\right\} & \le \frac{r}{2} \int _{-2/r}^{2/r} (1 - \hat{\mu }(t \cdot a))dt \: . \end{align*}

Proof ▶

TODO (see code)

Lemma 4.13

Let \(X\) be a random variable with law \(\mu \). Then for any \(s, t\),

\begin{align*} \vert \hat{\mu }(s) - \hat{\mu }(t) \vert \le 2 \mu \left[ \left\vert (s - t) X\right\vert \wedge 1\right] \: . \end{align*}

Proof ▶

\begin{align*} \vert \hat{\mu }(s) - \hat{\mu }(t) \vert & = \vert \mu \left[e^{isX} - e^{itX}\right] \vert \\ & \le \mu \left[\vert e^{isX} - e^{itX} \vert \right] \\ & = \mu \left[\vert 1 - e^{i(t - s)X} \vert \right] \le 2 \mu \left[ \left\vert (s - t) X\right\vert \wedge 1\right] \: . \end{align*}

Lemma 4.14

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The Gaussian distribution \(\mathcal N(m, \sigma ^2)\) has characteristic function \(\phi (t) = e^{itm - \sigma ^2 t^2 /2}\).

Proof ▶

4.2 Convergence of characteristic functions and weak convergence of measures

Lemma 4.15

If two probability measures \(\mu , \nu \) have same characteristic function, then for all exponential polynomial \(f \in \mathcal M\), \(\mu [f] = \nu [f]\).

Proof ▶

Lemma 4.16

Two finite measures on a complete separable metric space are equal iff they have the same characteristic function.

Proof ▶

See Mathlib PR #19783.

The sub-algebra \(\mathcal M\) of exponential polynomials is separating by Lemma 3.10. Equality of characteristic functions implies equality on \(\mathcal M\) by Lemma 4.15, which then implies equality of the measures since \(\mathcal M\) is separating.

Lemma 4.17

A family of measures \((\mu _a)\) on a finite dimensional inner product space is tight iff the family of functions \((\hat{\mu }_a)\) is equicontinuous at 0.

Proof ▶

TODO: equicontinuous implies tight.

For the other direction, by Lemma 4.13, for all \(a\),

\begin{align*} \vert \hat{\mu }_a(s) - \hat{\mu }_a(t) \vert \le 2 \mu _a\left[ \left\vert (s - t) X\right\vert \wedge 1\right] \: . \end{align*}

Since \((\mu _a)\) is tight, the right-hand side converges to 0 as \(s - t \to 0\) uniformly in \(a\). We obtain equicontinuity at 0 of the family \((\hat{\mu }_a)\).

Lemma 4.18

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Let \((\mu _n)_{n \in \mathbb {N}}\) be measures on \(\mathbb {R}^d\) with characteristic functions \((\hat{\mu }_n)\). If \(\hat{\mu }_n\) converges pointwise to a function \(f\) which is continuous at 0, then \((\mu _n)\) is tight.

Proof ▶

By Lemma 4.12 and dominated convergence, for all \(a \in \mathbb {R}^d\) and \(r {\gt} 0\),

\begin{align*} \lim _{n \to +\infty } \mu _n \left\{ x \mid |\langle a, x\rangle | \ge r\right\} & \le \lim _{n \to +\infty } \frac{r}{2} \int _{-2/r}^{2/r} (1 - \hat{\mu }_n(\langle a, t\rangle ))dt \\ & = \frac{r}{2} \int _{-2/r}^{2/r} (1 - f(\langle a, t\rangle ))dt \end{align*}

The right-hand side converges to 0 as \(r \to +\infty \) by continuity of \(f\) at 0. This shows that the family is tight.

Remark: the finite dimension is necessary. Counterexample: \(\ell ^2\) with \(\mu _n\) the law of \(X_n = \sum _{k=1}^n \zeta _k e_k\) where \(e_k = (0, \ldots , 0, 1, 0 \ldots )\) and the \(\zeta _k\) are i.i.d. \(\mathcal N(0,1)\). Then \(\hat{\mu }_n(t) \to e^{- \Vert t \Vert ^2 / 2}\) for all \(t \in \ell ^2\) but \((\mu _n)\) is not tight.

Theorem 4.19 Convergence of characteristic functions and weak convergence of measures

Let \(\mu , \mu _1, \mu _2, \ldots \) be probability measures on \(\mathbb {R}^d\) with characteristic functions \(\hat{\mu }, \hat{\mu }_1, \hat{\mu }_2, \ldots \). Then \(\mu _n \xrightarrow {w} \mu \) iff for all \(t\), \(\hat{\mu }_n(t) \to \hat{\mu }(t)\).

Proof ▶

For all \(t\), \(x \mapsto e^{i \langle t, x \rangle }\) is a bounded continuous function. Hence by the definition of convergence in distribution, \(\mu _n \xrightarrow {w} \mu \implies \hat{\mu }_n(t) \to \hat{\mu }(t)\) for all \(t\).

Let’s now prove the other direction. If the characteristic functions converge pointwise, we get from Lemma 4.18 that the family of measures is tight. Then apply Lemma 3.11 to the subalgebra \(\mathcal M\): it suffices to show that \(\mu _n[f] \to \mu [f]\) for all \(f \in \mathcal M\). This is obtained with Lemma 4.15.