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Partial answer: According to a theorem (Gil-Pelaez), if $0$ is a point of continuity of $F_X$, then $$ F_X(0) = \frac{1}{2} - \frac{1}{\pi} \int_0^{\infty} \frac{\mathrm{Im}[\phi_X(t)]}{t} dt, $$ so, since a random variable $X$ is a.s. non-negative, then $F_X(0) = 0$, $\int_0^{\infty} \frac{\mathrm{Im}[\phi_X(t)]}{t} dt = \pi / 2$ is the condition for a.s. ...


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You're saying the pair $(X,Y)$ has the same distribution as the pair $(\bar X,\bar Y)$ and $\bar X,\bar Y$ are independpendent and you want to prove $X,Y$ are independent. \begin{align} & \Pr(X\in A\ \&\ Y\in B) \\[10pt] = {} & \Pr((X,Y)\in A\times B) \\[10pt] = {} & \Pr((\bar X,\bar Y)\in A\times B) & & \text{(since the joint ...


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Denote by $\mathbb{P}_X$ the distribution of a random variable $X$ and by "$\stackrel{d}{=}$" equality in distribution. Since $\tilde{X}$ and $\tilde{Y}$ are independent, the distribution of $\tilde{Z}$ equals $$\mathbb{P}_{\bar{Z}} = \mathbb{P}_{\tilde{X}} \otimes \mathbb{P}_{\tilde{Y}}.$$ Moreover, $\tilde{X} \stackrel{d}{=}X$ and $\tilde{Y} ...


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First, $k=1,\dots,n$ (otherwise, the total prob. does not sum up to 1). Then $$\varphi_{X_n}(t)=\mathbb{E}[e^{\mathrm{i}tX_n}]= \sum_{k=1}^{n}e^{\mathrm{i}t\frac{k}{n}}\cdot\frac{1} {n}= \frac{\exp\left(\mathrm{i}t\cdot\frac{n+1}{n}\right)-\exp\left(\mathrm{i}t\cdot\frac{1}{n}\right)}{\left(\exp\left(\mathrm{i}t\cdot\frac{1}{n}\right)-1\right)\cdot n} ...


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Think about what can the Jordan normal form of the matrix be. The nullity is the number of $0$-blocks (each $0$-block contains a single eigenvector). The size of these blocks are either $1$ or $2$ (because $x^2\,|\,m_A(x)$) that sum up to $6$. For this, the only possibility for the sizes of $0$-blocks is $2,2,1,1$. The other eigenvalues ($\pm\sqrt2,\ ...


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\begin{align} \operatorname{E}(e^{itY}\mid X=x) & = \operatorname{E}(e^{it\sqrt{x} \, Z})\quad\text{where }Z\sim N(0,1), \\[10pt] & = \varphi_Z(t\sqrt x) = \exp \left( \frac{-1}2 t^2 x \right). \end{align} \begin{align} \operatorname{E}(e^{itY}) & = \operatorname{E} \left( \operatorname{E}(e^{itY}\mid X) \right) = \operatorname{E}\left( ...


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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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Let $I = \overline {E^c} \cap [0,1]$ (the bar denotes topological closure). Being closed an bounded, $I$ is compact. Note that since $|E|=0$, then $\Big| I \cap [\frac 1 n, \frac 2 n] \Big|= \frac 1 n$. It is known that if $f_n \to f$ uniformly on a compact $X$, then $\int \limits _X f_n \to \int \limits _X f$. In your case, applying this on $I$ gives $1 \to ...


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Hint: Look, for each $n\geq 1$, at $\phi_n$, the characteristic function of the $n$ i.i.d. random variables guaranteed by the hypothesis that $\chi$ is infinitely divisible. Letting $\phi$ be the characteristic function of $\chi$, you get for all $t\in \mathbb{R}$ $$ \chi_n(t) = \chi(t)^{\frac{1}{n}}$$ Now, show that for each $t\in\mathbb{R}$, ...


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By definition, the characteristic function of a discrete random variable $X$ with mass distribution $f_X(x)$ over support $\mathcal X$ is: $$\varphi(t) =\mathsf E(\mathsf e^{itX})= \sum_{x\in\mathcal X} f_X(x) e^{itx}$$ The characteristic function of a binomial distribution is: $$\begin{align} \varphi(t) & = ((1-\color{red}{p})+\color{red}{p}\mathsf ...



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