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21

I will give two answers: Do it without complex numbers, notice that $$\begin{eqnarray} \mathcal{F}(\omega) = \int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x &=& \int_{0}^\infty \frac{1}{\sqrt{2\pi}} \mathrm{e}^{-\frac{x^2}{2}} \mathrm{e}^{j \omega x} \mathrm{d} x + \int_{-\infty}^0 ... 13 Theorem (Kac's theorem) Let X,Y \in L^1 \mathbb{R}^d-valued random variables. Then the following statements are equivalent. X,Y are independent \forall \eta,\xi \in \mathbb{R}^d: \mathbb{E}e^{\imath \, (X,Y) \cdot (\xi,\eta)} = \mathbb{E}e^{\imath \, X \cdot \xi} \cdot \mathbb{E}e^{\imath \, Y \cdot \eta} Proof: (1) \Rightarrow ... 10 Denote by \Phi(t) = \mathbb{E}e^{\imath \, t \cdot X} the characteristic function of X. We have$$X = \frac{1}{2} \big((X+Y)+(X-Y) \big).$$Thus,$$\begin{align*} \Phi(t) &= \mathbb{E}e^{\imath \, \frac{t}{2} (X+Y)} \cdot \mathbb{E}e^{\imath \, \frac{t}{2} (X-Y)}= \left( \mathbb{E}e^{\imath \, \frac{t}{2} X} \right)^2 \cdot \mathbb{E}e^{\imath \, ...

7

If $\phi$ is a characteristic function, then, for every real values of $s$ and $t$, $K(t,s)\geqslant0$ where $K(t,s)$ is the determinant $$K(t,s)=\det\begin{pmatrix}\phi(0) & \phi(t) & \phi(t+s) \\ \phi(-t) & \phi(0) & \phi(s) \\ \phi(-t-s) & \phi(-s) & \phi(0)\end{pmatrix}.$$ Using $\phi_\alpha(t)=\mathrm e^{-c|t|^\alpha}$ for ...

7

This is a consequence of Levy's Inversion Formula (aka Fourier Inversion Theorem). If $\varphi$ is the CF of $X$ and $\int_{\mathbb{R}}|\varphi(\theta)|d\theta < \infty$ then $X$ is absolutely continuous with density $$f(x)= \frac{1}{2\pi}\int_\mathbb{R}e^{-i\theta x}\varphi(\theta)d\theta.$$ (Here we are using the definition $\varphi(\theta) = ... 7 Finally, the semi-explicit (but quite useable) answer for$K \neq 0$, which I jot down quickly before coming back with clean derivations and proofs. $$\psi[\alpha, \beta,\sigma,K]=\sigma \frac{\Gamma \left(\frac{\alpha -1}{\alpha }\right) \left(\left(1+i \beta \tan \left(\frac{\pi \alpha }{2}\right)\right)^{1/\alpha }+\left(1-i \beta \tan ... 6 To prove these are characteristic functions, let us use random variables. This yields simpler, and more intuitive, proofs. In the first case, assume that \phi_1(t)=\mathrm E(\mathrm e^{itX_1}) and \phi_2(t)=\mathrm E(\mathrm e^{itX_2}) for some random variables X_1 and X_2 defined on the same probability space and introduce a Bernoulli random ... 6 As user75064 already pointed out, the answer is "no". But there is the following result: Let X,Y \in L^1 \mathbb{R}-valued random variables. Then the following statements are equivalent. X,Y are independent \forall \eta,\xi \in \mathbb{R}: \mathbb{E}e^{\imath \, (X,Y) \cdot (\xi,\eta)} = \mathbb{E}e^{\imath \, X \cdot \xi} \cdot ... 6 Below, there is a discrete example because I misread the word continuous. However, if X has the density f given by$$f(x)=\begin{cases}0&\mbox{ if }|x|\leqslant 2,\\ \frac C{x^2\log |x|}&\mbox{ if }|x|>2,\end{cases}$$then X is not integrable and the characteristic function is given by ... 6$$T=ZX+(1-Z)Y\implies E(\mathrm e^{\mathrm itT})=pE(\mathrm e^{\mathrm itX})+(1-p)E(\mathrm e^{\mathrm itY})$$6$$\sum_{k,\ell}\xi_k\bar{\xi_{\ell}}\varphi_X(t_k-t_{\ell})=E\left(\sum_{k,\ell}\xi_k\bar{\xi_{\ell}}e^{i(t_k-t_{\ell})X}\right)=E\left(\left|\sum_k\xi_ke^{it_kX}\right|^2\right)\geqslant0$$5 Any random variable Z:(\Omega,\mathcal F)\to(\mathbb R^n,\mathcal B(\mathbb R^n)) such that \mathrm E(\mathrm e^{\mathrm i\langle u,Z\rangle})=\mathrm e^{-\kappa\|u\|^2} for every u in \mathbb R^n and for some positive \kappa, is centered normal with variance-covariance matrix 2\kappa I. A quick way to see this is to note that the function ... 5 Factoid 1: If a characteristic function is infinitely differentiable at zero, all the moments of the corresponding random variable are finite. Factoid 2: If all the moments of a random variable are finite, the corresponding characteristic function is infinitely differentiable everywhere on the real line. Factoid 3: The function t\mapsto|\cos(t)| is ... 5 No. The property in your post is called subindependence, and it is strictly weaker than independence. (Note that some people use the term "subindependent" as a synonym for "uncorrelated".) In addition to the references given in Wikipedia, you can find an example in this short note. Unfortunately it's behind a paywall. The example consists of two random ... 5 I should choose for induction: E\left[\left(\sum_{i=1}^{n}X_{i}\right)^{3}\right]=E\left[\left(\sum_{i=1}^{n-1}X_{i}\right)^{3}+3X_{n}\left(\sum_{i=1}^{n-1}X_{i}\right)^{2}+3X_{n}^{2}\sum_{i=1}^{n-1}X_{i}+X_{n}^{3}\right]. By induction: E\left[\left(\sum_{i=1}^{n-1}X_{i}\right)^{3}\right]=\sum_{i=1}^{n-1}E\left[X_{i}^{3}\right]. Further we have: ... 5 It's probably better to use the cumulative distribution function, since it behaves better with respect to the maximum of an independent sequence. We have$$\mu\{n(1-M_n)\leqslant t\}=\mu\{1-t/n\leqslant M_n\}=1-(1-t/n)^n$$and we recognize a well-known limit. 5$$A^2 = \left( \begin{array}{ccc} 9 & 8 & 8 \\ 8 & 9 & 8 \\ 8 & 8 & 9 \\ \end{array} \right)-4A = \left( \begin{array}{ccc} -4 & -8 & -8 \\ -8 & -4 & -8 \\ -8 & -8 & -4 \\ \end{array} \right)A^2 - 4A = \left( \begin{array}{ccc} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 5 \\ ... 5 If$X$is a random variable with values in$\mathbb{N}$, then $${\rm E}[f(X)]=\sum_{k=0}^\infty f(k)P(X=k)$$ for any 'nice' function$f$. This is the law of the unconscious statistician for discrete random variables. 5 Since you seem to be turning around this question and some of its variants again and again, let us try to answer it (almost) completely. First, as mentioned partially by the text you are reading, to know the characteristic function of every normal random vector, it is enough to know the characteristic function of a standard one-dimensional normal random ... 5 Suppose from the start that I have three vector bundles$E_1, E_2, E_3$and an open cover of the base space on which all three of these trivialize. Then if the transition functions for$E_i$are denoted$\phi_{i,U}: U \to GL_n$, then the tensor product$E_1 \otimes E_2$is the vector bundle given by transition functions$\phi_1 \otimes \phi_2$, where I mean ... 4 The characteristic function is an expectation: $$\varphi_S(t) = \mathbb{E}\left(\exp(i S t)\right) = \mathbb{E}\left(\exp\left(i \left(U_1 + U_2 + \cdots + U_n \right) t\right)\right)$$ Now, if$U_i\$ is independent, the expectation factors into product of expectations, because : $$\varphi_S(t) = \mathbb{E}\left(\mathrm{e}^{i t U_1}\cdot ... 4 No, any characteristic function that is equal to 1 on an interval around 0 must be equal to 1 everywhere. This can easily be deduced from the the fact that |\phi(t)|\leq 1 and the inequality 1-\cos(2t)\leq 4(1-\cos(t)) which allows you to conclude 1-\Re \phi(2t)\leq 4[1-\Re \phi(t)] which is essentially a statement that says the behavior of \phi(t) ... 4 Simply because the characteristic function of a random variable Y is defined as$$ \phi(t) = E(e^{itY}) $$And so with Y= X^2 and using the density of X \sim \mathcal{N}(0,1) we have$$ \phi(t) = E(e^{itY}) = E(e^{itX^2}) = \int_{-\infty}^\infty e^{itx^2} \frac{1}{\sqrt{2\pi}} e^{-\frac{x^2}{2}} dx$$Remember: When a random variable X has a ... 4 Since S_n=\sum\limits_{k=1}^nX_k converges almost surely to S=\sum\limits_{k=1}^\infty X_k, \varphi_{S_n}\to\varphi_S pointwise by dominated convergence. By independence, \varphi_{S_n}=\prod\limits_{k=1}^n\varphi_k. Thus, \prod\limits_{k=1}^n\varphi_k converges pointwise when n\to\infty and its limit is both \prod\limits_{k=1}^\infty\varphi_k ... 4 A step function is a linear combination of charateristic functions of intervals. Since any interval is measurable, any step function is simple. On the other hand, the characteristic function of Cantor's set is simple, but not a step function. Cantor's function is neither simple, nor a step function. 4 If X_n\to 0 in distribution, then any \varepsilon does the job. The converse is harder. Here it's the proof of Levy's continuity theorem which will be used. Denoting by \varphi_n the characteristic function of X_n and \mu_n its distribution, we indeed have the equality ... 4$$\mathbf 1_{(x,x+a]}(y)=1\iff x\lt y\leqslant x+a\iff y-a\leqslant x\lt y\iff \mathbf 1_{[y-a,y)}(x)=1$$4 The probability density function is non-negative. Hence, |f(x)|=f(x) and$$ \int |f(x)|\mathrm dx=\int f(x)\mathrm dx=1. $$It follows that f\in L^1. 4 Hint: The Cayley-Hamilton theorem implies that A^n is a linear combination of I,A,A^2,\ldots,A^{n-1}. 4 This is a weakened form of Bernstein's Theorem (weakened by unnecessarily assuming identical distributions having finite variances), so the proof is shorter. Here's a sketch, adapted from "On three characterizations of the normal distribution" by M. P. Quine: Define \quad U = X+Y, \quad V=(X-Y)^2 and characteristic functions$$\phi(t) = E e^{itX}\\ ...

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