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{{definitioncard|Characteristic function|Let <math>(\Omega,\A,\p)</math> be a probability space. Let <math>X</math> be a r.v. with values in <math>\R^d</math>, i.e. <math>X:(\Omega,\A,\p)\to\R^d</math>. Then we can look at the characteristic function of <math>X</math>, which is given by the Fourier transform | |||
<math display="block"> | |||
\Phi_X:\R^d\to\C,\xi\mapsto \Phi_X(\xi)=\E\left[e^{i\langle \xi,X\rangle}\right]=\int_{\R^d}e^{i\langle \xi, x\rangle}d\p_X(x) | |||
</math> | |||
where <math>\langle\xi,X\rangle=\sum_{k=1}^d\xi_kX_k</math>.}} | |||
{{alert-info | | |||
For <math>d=1</math> and <math>\xi\in\R</math>, we get | |||
<math display="block"> | |||
\Phi_X(\xi)=\E\left[e^{i\xi X}\right]=\int_\R e^{i\xi x}d\p_X(x). | |||
</math> | |||
}} | |||
{{alert-info | | |||
<math>\Phi_X(\xi)</math> is continuous on <math>\R^d</math> and bounded. For boundedness note | |||
<math display="block"> | |||
\vert\Phi_X(\xi)\vert=\left\vert\E\left[e^{i\langle\xi,X\rangle}\right]\right\vert\leq \E\left[\underbrace{\left\vert e^{i\langle\xi,X\rangle}\right\vert}_{=1}\right]\leq 1. | |||
</math> | |||
Moreover, we know that <math>e^{i\langle\xi,x\rangle}</math> is a continuous function of <math>\xi\in\R^d</math> for every <math>x\in\R^d</math>. | |||
<math display="block"> | |||
\left\vert e^{i\langle\xi,X\rangle}\right\vert\leq 1,\E[1]=1 < \infty. | |||
</math> | |||
So it follows that <math>\Phi_X(\xi)</math> is a continous function of <math>\xi</math>. | |||
}} | |||
{{proofcard|Theorem|thm-1|The characteristic function uniquely characterizes probability distributions, meaning that for two r.v.'s <math>X</math> and <math>Y</math> satisfying | |||
<math display="block"> | |||
\Phi_X(\xi)=\Phi_Y(\xi), | |||
</math> | |||
for all <math>\xi\in\R^d</math>, we get that | |||
<math display="block"> | |||
\p_X=\p_Y. | |||
</math> | |||
|No proof here.}} | |||
{{proofcard|Lemma|lem-1|Let <math>X</math> be a r.v. which is <math>\mathcal{N}(0,\sigma^2)</math> distributed. Then | |||
<math display="block"> | |||
\Phi_X(\xi)=\exp\left(-\frac{\sigma^2\xi^2}{2}\right),\xi\in\R. | |||
</math> | |||
|According to the formula we get | |||
<math display="block"> | |||
\Phi_X(\xi)=\int_{-\infty}^{\infty}e^{i\xi x}e^{-\frac{x^2}{2\sigma^2}}\frac{dx}{\sigma\sqrt{2\pi}}. | |||
</math> | |||
Assume for simplicity <math>\sigma=1</math> (change of variables: <math>\xi=\frac{x}{\sigma}</math>). Therefore we have | |||
<math display="block"> | |||
\Phi_X(\xi)=\int_{-\infty}^\infty\frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}\cos(\xi x)dx+\underbrace{i\int_{-\infty}^\infty\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\sin(\xi x)dx}_{0,\text{by parity}}=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\cos(\xi x)dx. | |||
</math> | |||
Hence we have | |||
<math display="block"> | |||
\frac{d\Phi_X}{d\xi}(\xi)=-\int_{-\infty}^{\infty}\frac{1}{\sqrt{2\pi}}xe^{-\frac{x^2}{2}}\sin(\xi x)dx. | |||
</math> | |||
We have used the fact that <math>e^{-\frac{x^2}{2}}\cos(\xi x)</math> is <math>C^\infty</math> in both variables and that <math>\left\vert x\sin(\xi x)e^{-\frac{x^2}{2}}\right\vert\leq \vert x\vert e^{-\frac{x^2}{2}}</math>, which is integrable on <math>\R</math>. Using integration by parts we get | |||
<math display="block"> | |||
\frac{d\Phi_X}{d\xi}(\xi)=\underbrace{\left[\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\sin(\xi x)\right]_{-\infty}^\infty}_{=0}-\xi\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}}\cos(\xi x)dx. | |||
</math> | |||
So we have the following Cauchy-problem | |||
<math display="block"> | |||
\begin{cases}\frac{d\Phi_X}{d\xi}(\xi)=-\xi \Phi_X(\xi)\\ \Phi_X(0)=1\end{cases} | |||
</math> | |||
Solving the differential equation, we get | |||
<math display="block"> | |||
\Phi_X(\xi)=e^{-\frac{-\xi^2}{2}}. | |||
</math>}} | |||
{{proofcard|Proposition|prop-1|Let <math>X=(X_1,...,X_d)\in\R^d</math> such that <math>\E[\vert X\vert^2] < \infty</math>, where <math>\vert\cdot\vert</math> denotes the euclidean norm. Then | |||
<math display="block"> | |||
\lim_{\vert\xi\vert\to0}\Phi_X(\xi)=1+i\sum_{j=1}^d\E[X_j]\xi_j-\frac{1}{2}\sum_{j=1}^d\sum_{k=1}^d\xi_j\xi_k\E[X_jX_k]+o(\vert\xi\vert^2), | |||
</math> | |||
|Note that we can write | |||
<math display="block"> | |||
\frac{\partial\Phi_X(\xi)}{\partial\xi_j}=i\E\left[X_je^{i\langle\xi,X\rangle}\right]. | |||
</math> | |||
This follows from the differentiation under the integral sign with <math>\left\vert X_je^{i\langle \xi,X\rangle}\right\vert\leq \vert X_j\vert</math>, which is integrable. Since | |||
<math display="block"> | |||
\E[\vert X_iX_k\vert]\leq \E[\vert X_j\vert^2]^\frac{1}{2}\E[\vert X_k\vert^2]^{\frac{1}{2}} < \infty, | |||
</math> | |||
we have | |||
<math display="block"> | |||
\frac{\partial^2\Phi_X(\xi)}{\partial\xi_j\partial \xi_k}=-\E\left[\underbrace{X_jX_ke^{i\langle\xi,X\rangle}}_{\leq \vert X_jX_k\vert\in L^1}\right]. | |||
</math> | |||
Taking <math>\xi=0</math> we get that <math>\frac{\partial\Phi_X}{\partial\xi_j}(0)=i\E[X_j]</math> and <math>\frac{\partial^2\Phi_X}{\partial\xi_j\partial\xi_k}(0)=-\E[X_jX_k]</math>. we see that the equation in the proposition is the taylor-expansion at order 2 near 0 of the <math>C^2</math>-function <math>\Phi_X(\xi)</math>.}} | |||
{{alert-info | | |||
From the proof we see that when <math>d=1</math>, we have | |||
<math display="block"> | |||
\E[\vert X\vert] < \infty\Longrightarrow\E[X]=i\Phi_X'(0)\text{and}\E[X^2] < \infty\Longrightarrow\E[X^2]=-\Phi_X''(0). | |||
</math> | |||
}} | |||
==General references== | |||
{{cite arXiv|last=Moshayedi|first=Nima|year=2020|title=Lectures on Probability Theory|eprint=2010.16280|class=math.PR}} |
Latest revision as of 01:53, 8 May 2024
Let [math](\Omega,\A,\p)[/math] be a probability space. Let [math]X[/math] be a r.v. with values in [math]\R^d[/math], i.e. [math]X:(\Omega,\A,\p)\to\R^d[/math]. Then we can look at the characteristic function of [math]X[/math], which is given by the Fourier transform
For [math]d=1[/math] and [math]\xi\in\R[/math], we get
[math]\Phi_X(\xi)[/math] is continuous on [math]\R^d[/math] and bounded. For boundedness note
The characteristic function uniquely characterizes probability distributions, meaning that for two r.v.'s [math]X[/math] and [math]Y[/math] satisfying
No proof here.
Let [math]X[/math] be a r.v. which is [math]\mathcal{N}(0,\sigma^2)[/math] distributed. Then
According to the formula we get
Let [math]X=(X_1,...,X_d)\in\R^d[/math] such that [math]\E[\vert X\vert^2] \lt \infty[/math], where [math]\vert\cdot\vert[/math] denotes the euclidean norm. Then
Note that we can write
From the proof we see that when [math]d=1[/math], we have
General references
Moshayedi, Nima (2020). "Lectures on Probability Theory". arXiv:2010.16280 [math.PR].