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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

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 ...

5

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 : $$\varphi_S(t) = \mathbb{E}\left(\mathrm{e}^{i t U_1}\cdot ... 4 No, any characteristic function that is equal to 1 on an interval around 0 must be equal to 1 everywhere. This can easily be deduced from the the fact that |\phi(t)|\leq 1 and the inequality 1-\cos(2t)\leq 4(1-\cos(t)) which allows you to conclude 1-\Re \phi(2t)\leq 4[1-\Re \phi(t)] which is essentially a statement that says the behavior of \phi(t) ... 4 Since S_n=\sum\limits_{k=1}^nX_k converges almost surely to S=\sum\limits_{k=1}^\infty X_k, \varphi_{S_n}\to\varphi_S pointwise by dominated convergence. By independence, \varphi_{S_n}=\prod\limits_{k=1}^n\varphi_k. Thus, \prod\limits_{k=1}^n\varphi_k converges pointwise when n\to\infty and its limit is both \prod\limits_{k=1}^\infty\varphi_k ... 4 Factoid 1: If a characteristic function is infinitely differentiable at zero, all the moments of the corresponding random variable are finite. Factoid 2: If all the moments of a random variable are finite, the corresponding characteristic function is infinitely differentiable everywhere on the real line. Factoid 3: The function t\mapsto|\cos(t)| is ... 4 I should choose for induction: E\left[\left(\sum_{i=1}^{n}X_{i}\right)^{3}\right]=E\left[\left(\sum_{i=1}^{n-1}X_{i}\right)^{3}+3X_{n}\left(\sum_{i=1}^{n-1}X_{i}\right)^{2}+3X_{n}^{2}\sum_{i=1}^{n-1}X_{i}+X_{n}^{3}\right]. By induction: E\left[\left(\sum_{i=1}^{n-1}X_{i}\right)^{3}\right]=\sum_{i=1}^{n-1}E\left[X_{i}^{3}\right]. Further we have: ... 3 Let A be independent of X with P(A=1)=P(A=0)=\frac{1}{2}. Then$$ E\left[e^{it\{AX+(1-A)(-X)\}}\right]=\frac{1}{2}E\left[e^{itX}\right]+\frac{1}{2}E\left[e^{it(-X)}\right]=\frac{\phi(t)+\phi(-t)}{2}, $$but using that \cos is even and \sin is odd, we obtain$$ \phi(-t)=E\left[e^{i(-t)X}\right]=E[\cos(-tX)]+iE[\sin(-tX)]=E[\cos(tX)]-iE[\sin(tX)] $$... 3 Let me just state the theorem I linked to in my comment, so that this question does not go unanswered. Bochner's theorem If \varphi:\mathbb{R}^d\to \mathbb C is a complex-valued function with \varphi(0)=1, continuous at 0 and nonnegative-definite in the sense that for n\geq 1 we have that$$ ...

3

The characteristic function is \begin{align} \frac12 \int_{-\infty}^{\infty} dx\, e^{-|x|} e^{i t x} &= \frac12 \int_{-\infty}^0 dx \, e^{x+i t x} + \frac12 \int_0^{\infty} dx \, e^{-x + i t x}\\ &= \frac12 \int_0^{\infty} dx \, e^{-(1+i t) x} + \frac12 \int_0^{\infty} dx \, e^{-(1-i t) x} \\ &= \frac12 \frac{1}{1+i t} + \frac12 \frac{1}{1-i ... 3 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 ... 3 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) = ... 3 Let  P  denote the probability distribution of  X  and  \mu  the Lebesgue measure on  \mathbb{R} . Then \begin{align*} \mathsf{E}[g(X)] &= \int_{\mathbb{R}} g(x) \, d{P(x)} \\ &= \int_{\mathbb{R}} \left[ \int_{\mathbb{R}} G(t) e^{i t x} \, d{\mu(t)} \right] d{P(x)} \\ &= \int_{\mathbb{R}} \left[ \int_{\mathbb{R}} G(t) e^{i t ... 2 By Bochner's theorem, a function \phi : \mathbb{R} \to \mathbb{C} is the characteristic function of a probability measure if and only if \phi is positive definite, \phi(0) = 1, and \phi is continuous at the origin. Since these properties are conserved under convex combination, your second statement is true whenever \alpha_i are non-negative. ... 2 Yes, for example when \mu_n is normally distributed with mean 0 and standard deviation n^2: the sequence of characteristic functions converges pointwise to the map \varphi such that \varphi(t)=0 if t\neq 0 and \varphi(0)=1. But such a sequence of measure cannot converge weakly (it's not tight). However, if the sequence of characteristic ... 2 Denote the PDE asa(x, y) \frac{\partial u}{\partial x} + b(x, y) \frac{\partial u}{\partial y} = 0, \tag{1}$$where$$a(x, y) = y \text{ and } b(x, y) = x. \tag{2}$$Is this PDE nonlinear? A PDE is nonlinear if the solution space of the homogeneous equation (with RHS = 0) is not a vector space. Suppose that u and v are solutions of the PDE ... 2 If X is a random variable with density f (with respect to Lebesgue measure on the real line) and G is a well-behaved function, then$$E[G(X)]=\int_{\Bbb R}G(x)f(x)d\lambda(x).$$In the particular case of the exponential law, this gives$$\phi(t)=\int_0^{+\infty}e^{itx}e^{-\lambda x}\lambda dx.$$If X is a random variable with values in the set of ... 2 You are almost there.$$\phi(t) = E[e^{itX}] = \sum_{j = 0}^{\infty} e^{itj} (1 - P)^j P = P \sum_{j = 0}^{\infty} [e^{it} (1 - P) ]^j $$And now, if you don't know about the geometric series ( \sum_{k=0}^\infty a^k ), it's time to learn about it. 2 (1) The problem lies in the definition of e^{itx} when x=\pm \infty, as \lim_{x\to +\infty} e^{itx} is not well defined. We are not bothered by this when X\in\mathbb R almost surely, because we can define \widetilde X a real valued random variable equal to X almost everywhere. In particular, it won't change the integral. (2) More generally, if ... 2 Note that (a+bi)+(a-bi)=2a so by splitting the integral at 0$$ \frac12\int_{-\infty}^\infty e^{itx}e^{-|x|}dx = \int_0^\infty \mathfrak{Re}(e^{itx}e^{-x})dx  =\int_0^\infty \cos(tx)e^{-x}dx  =e^{-x}\frac{t\sin(tx)-\cos(tx)}{t^2+1}\biggr|_{x=0}^{\infty}  =\frac{1}{1+t^2}. $$I will leave the calculation of the integral (by parts twice) to ... 2 Since \varphi is the characteristic function of an infinitely divisible distribution we have that$$ \varphi(t)=\varphi_n(t)^n,\quad n\in\mathbb{N},\tag{1} $$for a sequence of characteristic functions \varphi_n. Now we use that |\varphi_n|^2 is also a characteristic function for each n, and thus by (1) we have that |\varphi|^{2/n} is a ... 2 Just to iron out the details in my comment, since others have also posted complete answers, \phi(t) is infinitely differentiable at 0; in fact, \left.\frac{d^n \phi}{dt^n} \right|_{t = 0} = 0 and hence if \phi(t) is the cf of some random variable X it must be that E[X^n] = 0 for all n. In particular, \mbox{Var}(X) = 0 and E[X] = 0 so that ... 2 Any random variable Z:(\Omega,\mathcal F)\to(\mathbb R^n,\mathcal B(\mathbb R^n)) such that \mathrm E(\mathrm e^{\mathrm i\langle u,Z\rangle})=\mathrm e^{-\kappa\|u\|^2} for every u in \mathbb R^n and for some positive \kappa, is centered normal with variance-covariance matrix 2\kappa I. A quick way to see this is to note that the function ... 2 The answer is yes if the series$$\sum_{m=1}^{+\infty}\frac{t^m}{m!}\mathbb E|X|^m\quad\mbox{ and }\quad\sum_{n=1}^{+\infty}\frac{t^n}{n!}\mathbb E|Y|^n$$are convergent for each t (it follows from dominated convergence theorem which gives the splitting equality mentioned in the OP). It's in particular the case when X and Y are bounded. 2 Let X_1,X_2,\dots be a sequence of i.i.d. random variables with characteristic function \phi. Let N be a Poisson random variable with parameter \lambda, independent with the X_i. Then the random variable$$ Z = \sum_{i=1}^N X_i \overset{\text{def}}{=} \sum_{i=1}^\infty X_i\,1_{N \geq i} $$has characteristic function e^{\lambda(\phi(t)-1)}. See ... 2 Simply because the characteristic function of a random variable Y is defined as$$ \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$$Remember: When a random variable X has a ... 2 One way to do this is by a simple induction on the number of terms after proving it works when n=2.$$ E((X_1+X_2)^3) = E(X_1^3)+3E(X_1^2X_2)+3E(X_1X_2^2)+E(X_2^3). $$Because of independence this becomes$$ E(X_1^3)+3E(X_1^2)E(X_2)+3E(X_1)E(X_2^2)+E(X_2^3). $$Then the middle two terms are 0 because each has a factor that is 0. (But this doesn't ... 2 Choosing the z axis along k and denoting |k| by q, we have$$ \begin{align} \int_{\mathbb R^3}\chi_{|x|\lt r}(x)\exp(-\mathrm ikx)\,\mathrm dx &=\int_0^r R^2\,\mathrm dR\int_0^\pi\sin\theta\,\mathrm d\theta\int_0^{2\pi}\mathrm d\phi\exp(-\mathrm iqR\cos\theta) \\ &=2\pi\int_0^r R^2\,\mathrm dR\int_0^\pi\sin\theta\,\mathrm ...

2

You're off to a good start. Next note that $$\chi_{[0,1]}(x-y) = \begin{cases} 1, & \text{for x-1 \le y \le x} \\ 0, & \text{otherwise}\end{cases}.$$ Draw some graphs! Hence, if $0< x < 1$, then $\chi_{[0,1]}(x-y)$ on the whole of $[0,1]$. Thus the integral is $0$. If $0 < x < 1$, then $\chi_{[0,1]}(x-y) = 1$ on $[0,x]$ and $=0$ on ...

2

Change variables $x-y=s$, thus the new domain is $(x,x-1)$. You have $$\int_0^1\Xi(x-y)dy=\int_{x-1}^x\Xi(s)ds.$$ Now, if $x-1>1$ and if $x<0$ the integral vanishes. Moreover, you have $$\int_{x-1}^x\Xi(s)ds=\text{length}\left((x-1,x)\cap(0,1)\right).$$ If $0<x<1$ $$\text{length}\left((x-1,x)\cap(0,1)\right)=x.$$ Otherwise, if $1<x<2$ ...

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