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

Proof ▶

Lemma 4.3

The characteristic function is a continuous function.

Proof ▶

Lemma 4.4

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

Proof ▶

Lemma 4.5

For \(a \in \mathbb {R}\), the characteristic function of \(a X\) is \(t \mapsto \phi _X(at)\).

Proof ▶

Lemma 4.6

If two random variables \(X, Y : \Omega \to S\) are independent, then \(X+Y\) has characteristic function \(\phi _{X+Y} = \phi _X \phi _Y\).

Proof ▶

Lemma 4.7

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

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

Proof ▶

TODO

Lemma 4.8

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

Lemma 4.9

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

Proof ▶

TODO

4.2 Convergence of characteristic functions and weak convergence of measures

Lemma 4.10

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

Proof ▶

TODO

Lemma 4.11

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 ▶

TODO

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 (todo: why?).

Lemma 4.12

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 4.13

Two probability measures on a TODO space are equal iff they have the same characteristic function.

Proof ▶

The sub-algebra \(\mathcal M\) of exponential polynomials is separating by Lemma ??. Equality of characteristic functions implies equality on \(\mathcal M\) by Lemma 4.12, which then implies equality of the measures since \(\mathcal M\) is separating.

Theorem 4.14 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 4.11 that the family of measures is tight. Then apply Lemma 3.13 to the subalbegra \(\mathcal M\): it suffices to show that \(\mu _n[f] \to \mu [f]\) for all \(f \in \mathcal M\). This is obtained with Lemma 4.12.