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Jochen Wengenroth's "Wahrscheinlichkeitstheorie" (de Gruyter, 2008), Theorem 5.6, p. 90 (continuity & uniqueness). Concerning uniqueness, see also the paragraph on p. 84 immediately following the proof of the Portmanteau theorem (theorem 5.2, p. 83). In order to be able to follow theorem 5.6's proof, one is advised to read the entire chapter 5 up to the ...


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The general property of characteristic functions that you need follows from properties of exponents and independence: $$E[e^{it \frac{\sum_{i=1}^n X_i}{n}}]=E[e^{it/n X_1} e^{it/n X_2} \cdots e^{it/n X_n}] = E[e^{it/n X_1}] \cdots E[e^{it/n X_n}] = \phi(t/n)^n$$ where $\phi$ is the characteristic function of each of the $X_i$. Now you just need to check ...


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Denote $a_n:=\mathbb E[M_n]$ and $b_n:=\mathbb E[S_n^+ ]/n$. The LHS is $$\sum_{n=0}^\infty r^na_n-\sum_{n=1}^\infty r^na_{n-1}=a_0+\sum_{n=1}^\infty r^n(a_n-a_{n-1}).$$ By identification of the coefficients of power series, we derive that for each $n\geqslant 1$, $$a_n-a_{n-1}=b_n,\quad a_0=0, $$ from which the wanted formula follows.


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If $E$ is a set such that $f_n\to 0$ uniformly in $E^c$, then $E$ has a positive measure. Indeed, notice that $$\sup_{x\in E^c}f_n(x)=\begin{cases}n&\mbox{if }[1/n,2/n]\cap E^c\neq\emptyset, \\ 0&\mbox{otherwise}. \end{cases} $$ Therefore, we should have $[1/n,2/n]\cap E^c=\emptyset $ for some $n$. This implies that $$\frac ...


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Believe it or not, this question does make sense. We have to make sense of what it means for an algorithm to take a real number as input. Such an algorithm works as a normal algorithm, except that it can, at any time, ask questions of the form "is the input in the rational interval (a,b)?" (see oracle machine and effective Polish space). You can view such ...


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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 ...


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The $c$ doesn't have to do anything with the quadratic. We know that $s_n=s_{n-1}+s_{n-2}$, thus $s_n-s_{n-1}-s_{n-2}=0$. Now replace $s_n$ with $x^2$, $s_{n-1}$ with $x$ and $s_{n-2}$ with 1 to get the characteristic equation.


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i) By definition of $\mu_n$, if $f\colon [0,1]\to\mathbf C$ is a measurable function, then $$\int f(x)\mathrm d\mu_n=\frac 1n\sum_{j=0}^{n-1}f\left(\frac jn\right),$$ therefore, with the choice $f(x)=e^{itx}$, $$\zeta_n(t)=\frac 1n\sum_{j=0}^{n-1}e^{ijt /n},$$ which can be computed using the formula for a geometric series if $e^{it /n}\neq 1$. ii) You have ...


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If $Y_j,1\leqslant j\leqslant n$ are independent random variable with respective characteristic functions $\varphi_j$, then the characteristic function of $\sum_{j=1}^nY_j$ is given by $\prod_{j=1}^n\varphi_j(w)$. Applying this to $Y_j:= n^{-1/2} x_j$, we obtain that the characteristic function $\phi_n$ of $n^{-1/2}\sum_{j=1}^nx_j$ is equal to ...


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I guess the formula comes from the summation of $e^{ijt}$, but this is not valid if $e^{it}=0$. Let us call for $t\notin 2\pi\mathbf Z$, $$\varphi_n(t):=\frac{1}{n+1} \frac{1- \exp\left( \left(n+2 \right)it \right)}{1-\exp(it)}.$$ The limit $\lim_{t\to 0}\varphi_n(t)$ exits an is equal to $\frac{n+2}{n+1}$; by periodicity $\lim_{t\to ...


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Mathematica suggests the equation is solvable: $c(t,r) = n W\left(\frac{e^{\frac{c_1\left(\frac{1}{3} \left(r^3-3 K t\right)\right)}{n}-\frac{\text{Da} t}{n}}}{n}\right)$ where $W$ denotes ProductLog function.


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The characteristic function is defined as: $$ \varphi(t) = \mathbb{E}[e^{itX}] \tag{1}$$ and assuming either $X\in L^2$ or $\varphi\in C^2$ we have: $$ \frac{d^2}{dt^2}\,\varphi(t) = \mathbb{E}[-X^2 e^{itX}]\tag{2} $$ by the linearity of the expected value, hence $\varphi''(0)=0$ implies $\mathbb{E}[X^2]=0$. As you already noticed, that leads to a ...


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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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The basic case is that the Fourier transform of the indicator function of an interval centered at zero is a sinc function. So by this calculation and linearity, your question essentially reduces to determining how the Fourier transform is affected by translation. It is easy to calculate this by a change of variable: you find that $\widehat{f(x-x_0)}(\xi) = ...


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So the characteristic function of $\text{B}(n,\lambda/n)$ is $$ ((1-\lambda/n)+\lambda/n e^{it})^{n} = \left( 1 + \frac{1}{n} \lambda\left( e^{it}-1 \right) \right)^n. $$ Now use that $$ \lim_{n \to \infty} \left( 1 + \frac{x}{n} \right)^n = e^x. $$ Then the convergence and uniqueness theorems for characteristic functions imply that the distribution is ...


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You misunderstood how the arrangement should look. For $n=2$, you have the lines (hyperplanes) $x_1=0$, $x_2=0$, and $x_1=x_2$. So your hyperplane arrangement looks like Remember, the rank of the central arrangement $\mathcal{B}$ is the dimension of the space spanned by the normal vectors of the hyperplanes in $\mathcal{B}$. Now it should make sense why ...



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