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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 ...

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 ...

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) $$for any 'nice' function f. This is the law of the unconscious statistician for discrete random variables. 5 I should choose for induction: E\left[\left(\sum_{i=1}^{n}X_{i}\right)^{3}\right]=E\left[\left(\sum_{i=1}^{n-1}X_{i}\right)^{3}+3X_{n}\left(\sum_{i=1}^{n-1}X_{i}\right)^{2}+3X_{n}^{2}\sum_{i=1}^{n-1}X_{i}+X_{n}^{3}\right]. By induction: E\left[\left(\sum_{i=1}^{n-1}X_{i}\right)^{3}\right]=\sum_{i=1}^{n-1}E\left[X_{i}^{3}\right]. Further we have: ... 4 It's probably better to use the cumulative distribution function, since it behaves better with respect to the maximum of an independent sequence. We have$$\mu\{n(1-M_n)\leqslant t\}=\mu\{1-t/n\leqslant M_n\}=1-(1-t/n)^n$$and we recognize a well-known limit. 4 The characteristic function is an expectation:$$ \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

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

$$\mathbf 1_{(x,x+a]}(y)=1\iff x\lt y\leqslant x+a\iff y-a\leqslant x\lt y\iff \mathbf 1_{[y-a,y)}(x)=1$$

4

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 ...

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 ...

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 $\varphi(\theta) = ... 3$A=2J-I$, where$J$is the all-one matrix with$J^2=3J$. Therefoer $$A^2-4A-5I=(2J-I)^2-4(2J-I)-5I=4J^2-12J=0.$$ 3 If you would like to sped up the calculation (supposing you are doing it on your own) you can use this approach. Let's take the equation $$A^2 - 4A - 5I = 0.$$ And now forget about the matrices. What we have is simple quadratic equation in$x$written as $$x^2 - 4x - 5 = 0,$$ now just find the roots $$(x-5)(x+1) = 0.$$ Let's get back to that matrix ... 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 1we have that $$... 3 Let A be independent of X with P(A=1)=P(A=0)=\frac{1}{2}. Then$$ 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 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 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 ... 3 In the approach with characteristic functions, we have to use the fact that \varphi_{S_n}(t)=\prod_{j=1}^n\varphi_{X_j}(t). here is an other approach: we have, noting by [n] the set \{1,\dots,n\},\mathbb E\left(\sum_{i=1}^nX_i\right)^3=\sum_{(i,j,k)\in [n]^3}\mathbb E(X_iX_jX_k).$$The set [n]^3 can be divided into the (i,j,k) such that: ... 3 The characteristic function is$$\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 ... 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 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 Yes, thanks to Lévy's continuity theorem, the pointwise limit is a characteristic function as soon as it is continuous at$0$. 2 Hint: Split to two cases,$w \in A$and$w \notin A$. In each case, think in which (if any) of the subsets$A_n$you can expect to find$w$. 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 It's true because$\phi_n(-\omega)=\overline{\phi_n(\omega)}$(as long as$\omega$is real not complex). Thus, you get $$|\phi_n(\omega)|^2 =\phi_n(\omega)\overline{\phi_n(\omega)} =\phi_n(\omega)\phi_n(-\omega).$$ 2 Define$w=y-Cx$, then $$E[\mathrm e^{\mathrm iu^Tw}|\mathcal F^x] =E[\mathrm e^{\mathrm iu^Ty}|\mathcal F^x]\,\mathrm e^{-\mathrm iu^TCx} = \mathrm e^{-u^T \tilde{Q}u/2}$$ is independent of$x$. Thus$w$is independent of$x$and centered normal with covariance$\tilde{Q}. 2 We have \begin{align} \frac{e^{-ita}-e^{-itb}}{it}e^{itx} &= \frac{e^{it(x-a)}-e^{it(x-b)}}{it}\\ &= \frac{\cos\left( t(x-a)\right) - \cos \left(t(x-b)\right)}{it} + \frac{\sin \left(t(x-a)\right) - \sin \left(t(x-b)\right)}{t} \end{align} by the addition theorem for the exponential function and Euler's formulae^{iz} = \cos z + i\sin z\$. The ...

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