The 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
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].