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12

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}. $$Using \phi_\alpha(t)=\mathrm e^{-c|t|^\alpha} for ... 6 Below, there is a discrete example because I misread the word continuous. However, if X has the density f given by$$f(x)=\begin{cases}0&\mbox{ if }|x|\leqslant 2,\\ \frac C{x^2\log |x|}&\mbox{ if }|x|>2,\end{cases}$$then X is not integrable and the characteristic function is given by ... 6 To prove these are characteristic functions, let us use random variables. This yields simpler, and more intuitive, proofs. In the first case, assume that \phi_1(t)=\mathrm E(\mathrm e^{itX_1}) and \phi_2(t)=\mathrm E(\mathrm e^{itX_2}) for some random variables X_1 and X_2 defined on the same probability space and introduce a Bernoulli random ... 6$$T=ZX+(1-Z)Y\implies E(\mathrm e^{\mathrm itT})=pE(\mathrm e^{\mathrm itX})+(1-p)E(\mathrm e^{\mathrm itY})$$5 I should choose for induction: E\left[\left(\sum_{i=1}^{n}X_{i}\right)^{3}\right]=E\left[\left(\sum_{i=1}^{n-1}X_{i}\right)^{3}+3X_{n}\left(\sum_{i=1}^{n-1}X_{i}\right)^{2}+3X_{n}^{2}\sum_{i=1}^{n-1}X_{i}+X_{n}^{3}\right]. By induction: E\left[\left(\sum_{i=1}^{n-1}X_{i}\right)^{3}\right]=\sum_{i=1}^{n-1}E\left[X_{i}^{3}\right]. Further we have: ... 5 Since you seem to be turning around this question and some of its variants again and again, let us try to answer it (almost) completely. First, as mentioned partially by the text you are reading, to know the characteristic function of every normal random vector, it is enough to know the characteristic function of a standard one-dimensional normal random ... 5$$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 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 IfX$is a random variable with values in$\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 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 ...

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

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

Hint: The Cayley-Hamilton theorem implies that $A^n$ is a linear combination of $I,A,A^2,\ldots,A^{n-1}$.

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 : $$\varphi_S(t) = \mathbb{E}\left(\mathrm{e}^{i t U_1}\cdot ... 4 No. For example this won't work for P(Z=a)=P(Z=b)=1/2, a\not= b. Since the convolution can only make measures less singular, X_1 must have purely discrete distribution also, but then it's easy to see that \sum w_j X_j simply takes too many values (with positive probability). 4 Any random variable Z:(\Omega,\mathcal F)\to(\mathbb R^n,\mathcal B(\mathbb R^n)) such that \mathrm E(\mathrm e^{\mathrm i\langle u,Z\rangle})=\mathrm e^{-\kappa\|u\|^2} for every u in \mathbb R^n and for some positive \kappa, is centered normal with variance-covariance matrix 2\kappa I. A quick way to see this is to note that the function ... 4 This is a consequence of Levy's Inversion Formula (aka Fourier Inversion Theorem). If \varphi is the CF of X and \int_{\mathbb{R}}|\varphi(\theta)|d\theta < \infty then X is absolutely continuous with density$$ f(x)= \frac{1}{2\pi}\int_\mathbb{R}e^{-i\theta x}\varphi(\theta)d\theta. $$(Here we are using the definition \varphi(\theta) = ... 4 The probability density function is non-negative. Hence, |f(x)|=f(x) and$$ \int |f(x)|\mathrm dx=\int f(x)\mathrm dx=1. $$It follows that f\in L^1. 4 Factoid 1: If a characteristic function is infinitely differentiable at zero, all the moments of the corresponding random variable are finite. Factoid 2: If all the moments of a random variable are finite, the corresponding characteristic function is infinitely differentiable everywhere on the real line. Factoid 3: The function t\mapsto|\cos(t)| is ... 4 No, any characteristic function that is equal to 1 on an interval around 0 must be equal to 1 everywhere. This can easily be deduced from the the fact that |\phi(t)|\leq 1 and the inequality 1-\cos(2t)\leq 4(1-\cos(t)) which allows you to conclude 1-\Re \phi(2t)\leq 4[1-\Re \phi(t)] which is essentially a statement that says the behavior of \phi(t) ... 4 This is a weakened form of Bernstein's Theorem (weakened by unnecessarily assuming identical distributions having finite variances), so the proof is shorter. Here's a sketch, adapted from "On three characterizations of the normal distribution" by M. P. Quine: Define \quad U = X+Y, \quad V=(X-Y)^2 and characteristic functions$$\phi(t) = E e^{itX}\\ ...

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

The letters $a,b$ and $c$ are just letters. I think your problem is with the quadratic formula itself. Think like this: $$\color{red}{\rm stuff}\,x^2 + \color{blue}{\rm stuff}\,x + \color{green}{\rm stuff} = 0 \implies x = \frac{-\color{blue}{\rm stuff} \pm \sqrt{(\color{blue}{\rm stuff})^2 - 4\,\color{red}{\rm stuff}\,\color{green}{\rm ... 3 Let Z_n=a_n(X_n-X). We are told that \frac{1}{a_n}\to 0 so, in particular, \frac{1}{a_n}\overset{P}{\to}0. So by Slutsky's Theorem, X_n-X=\frac{1}{a_n}\times Z_n\overset{d}{\to}0\times Z=0. But convergence in distribution to constant is equivalent to convergence in probability to that same constant, so we conclude that X_n-X\overset{P}{\to}0. 3 Just to iron out the details in my comment, since others have also posted complete answers, \phi(t) is infinitely differentiable at 0; in fact, \left.\frac{d^n \phi}{dt^n} \right|_{t = 0} = 0 and hence if \phi(t) is the cf of some random variable X it must be that E[X^n] = 0 for all n. In particular, \mbox{Var}(X) = 0 and E[X] = 0 so that ... 3 Let A be independent of X with P(A=1)=P(A=0)=\frac{1}{2}. Then$$ 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)] $$... 3 By Bochner's theorem, a function \phi : \mathbb{R} \to \mathbb{C} is the characteristic function of a probability measure if and only if \phi is positive definite, \phi(0) = 1, and \phi is continuous at the origin. Since these properties are conserved under convex combination, your second statement is true whenever \alpha_i are non-negative. ... 3 There is a mistake in the computation of the characteristic function. By definition,$$\varphi_Z(t) = \mathbb{E}e^{\imath \, t Z}, \qquad t \in \mathbb{R}.$$Therefore,$$\begin{align*} \varphi_{\frac{Y}{\sqrt{\lambda}}}(t) &= \mathbb{E} \exp \left( \imath \frac{t}{\sqrt{\lambda}} Y \right) = \exp \left( \lambda \left[ \exp \left(- ...

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