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Yes, there is a very strong relation between the (distribution of the) jumps of a Lévy process and its Lévy measure. In fact, the Lévy measure describes exactly the jump behaviour of the Lévy process. To this end, define the jump counting measure $$N([0,t] \times B) := |\{0 \leq s \leq t; \Delta X_s \in B\}| \tag{1}$$ where $\Delta X_s := X_s-X_{s-}$ ...

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$Y|X=x$ has the same distribution of $f(x,U)$ for some measurable $f$ (take $f(x,.)$ the inverse of the CDF of $Y|X=x$ for an 'explicit' choice). Then $(X,Y), (X,f(X,Y))$ have the same distributions.

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Number the people from $1$ to $9,$ with $1,2,3$ Turks, $4,5,6$ Brits, and $7,8,9$ French. Let $T_{ij}$ be the event that $j$ sits immediately to the right of $i.$ We need to exclude $18$ events: $T_{12},$ $T_{21},$ $T_{13},$ $T_{31},$ $T_{23},$ $T_{32},$ $T_{45},$ $T_{54},$ and so on. Since we will need to use the principle of inclusion-exclusion, we need ...

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The probability model of quantum mechanics is different from the Kolmogorov model, is due to von Neumann, and actually predates the Kolmogorov model. The basic model goes something like this: events are represented by the lattice of projectors on a Hilbert space. The elementary outcomes are the one-dimenstional projectors. Probabilities are assigned to ...

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The problem seems to be taylored for inclusion-exclusion formula... Let $I=\{1,2,3,4,5\}$. For every $i$ in $I$, let $A_i$ denote the event that $i$ does not appear before $6$ appears for the third time. For every $J\subseteq I$, let $A_J=\bigcap\limits_{i\in J}A_i$. One asks for $1-P(A)$ where $A=\bigcup\limits_{i\in I}A_i$ is the event that one ...

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I am not quite sure about how to use the uniform continuity to prove this conclusion. Then don't. Maybe we should apply some techniques in Fourier analysis. Hmmm... And maybe we should look for explicit simple counterexamples? Try $X$ such that $P(X=1)=1$, and $s=0$, $t\to0$.

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I think that it is an "historical" reference; see : Nicolas Bourbaki, Elements of the History of Mathematics (French ed 1984), Ch.16 : Metric spaces, page 165 : the notion of metric space was introduced in 1906 by M.Fréchet, and developed a few years later by F. Hausdorff in his "Mengenlehre". See : Maurice Fréchet (2 September 1878 – 4 June 1973) ...

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Define for fixed $s,t\in\mathbb R_+$ the random variable $$Y_N:=\frac{t}{\sqrt{\pi}}X_0+\sqrt{\frac 2{\pi}}\sum_{n=1}^NX_n\frac{\sin(nt)-\sin(ns)}n.$$ For a fixed $N$, the random variable $Y_N$ has Gaussian distribution has a linear combination of the Gaussian vector $(X_0,\dots,X_N)$. By the results mentioned in the OP, the sequence $(Y_N)_{N\geqslant ... 2 Since$K_\epsilon$is closed it is complete (as a subspace of$S$). Choose some$n$. Then the${{{\overline K }_{j,n}}}$($j=1,...,k_n$) cover$K_\epsilon$. Select$x_j \in K_\epsilon \cap {{{\overline K }_{j,n}}}$. Then the$x_j$form a${2 \over n}$-net for$K_\epsilon$. Hence for any$\delta>0$we can find a finite$\delta$-net for$K_\epsilon$and ... 2 I'd bet that there is a simpler way, but I couldn't simplify it. Some slight changes in notation; and to aid readability, I write in lowercase bold the (column) vectors, in uppercase the matrices. Let $$\underbrace{V}_{n\times \ell}=\pmatrix{{\bf v}_1^t\\ {\bf v}_2^t\\ \cdots\\ {\bf v}_n^t\\ }= ({\bf u}_1 \, {\bf u}_2 \cdots {\bf u}_\ell)$$ where${\bf ...

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A little correction to the previous answer by janak: Let Y have cdf $G(y)$ then: \begin{align}G(y)&=P(Y\leq y)\\ &=P(X^2\leq y)\\ &=P(-\sqrt{y} \leq X\leq \sqrt{y})\\ &=P(X\leq \sqrt{y})-P(X\leq -\sqrt{y})\\ &=F(\sqrt{y})-F(-\sqrt{y})\end{align}

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I believe you have that the second moment of $\displaystyle d_i$ is bounded (right?). Now use Cramer-Wold device: Choose $\lambda_1$ and $\lambda_2$ such that $\lambda_1^2+\lambda_2^2=1$. Then since the second moment of sine and cosine are bounded and by virtue of independence you assumed we have that $\frac{1}{\sqrt{n}}(\lambda_1 X+\lambda_2 Y) \sim ... 0 Write for a positive$R$, $$\mathbb P(|f(X_n)|\gt R)=\mathbb P(|f(X_n)|\gt R,|X_n|\leqslant M)+\mathbb P(|X_n|\gt M).$$ For a fixed$\varepsilon$, take$M$such that$\mathbb P(|X_n|\gt M)\lt\varepsilon$. Once this$M$is chosen, notice that the function$f$is bounded on$[-M,M]$, hence for$R$greater than$\sup_{t\in[-M,M]}|f(t)|, we obtain \mathbb ... 3 Let Y has cdf G(y) and also assume X is a continuous random variable. Then G(y)=P(Y\leq y)=P(X^2\leq y)=P(X\leq \sqrt{y})-P(X<-\sqrt{y})=F(\sqrt{y})-F(-\sqrt{y}) 0 We have a sample of n Bernouli trials with a success probability of p except for the k-1 trials subsequent to a success in which failure is certain. Call these auto-fails. Define a k length success string, as one success followed by k-1 auto-fails which are automatic failures. For there to be X successes in the sample, there must be X of ... 2 Let X_0:=0, \ X_n:=X_{n-1}-(n-1)+Y_{n}, where Y_n is independent, \mathcal{F}_{n} measurable and P(Y_n=0)=1/n, P(Y_n=n)=1-1/n. Then E[Y_n]=n-1, so E[X_n|\mathcal{F}_{n-1}]=X_{n-1}. Since X_n\rightarrow\infty is a 0-1 event, we just need to show that it occurs with positive probability to conclude X_n\rightarrow\infty almost surely. If ... 3 Let X\sim\mathcal{Geo}(p), Y\sim\mathcal{Geo}(q), X\perp Y \begin{align} \Pr(X\geq k) & = (1-p)^{k-1} & \impliedby X\sim \mathcal{Geo}(p) \tag{1} \\[2ex] \Pr(Y\geq k) & = (1-q)^{k-1}& \impliedby Y\sim \mathcal{Geo}(q)\tag{2} \\[2ex] \Pr(\min(X,Y)\geq k) & = \Pr(X\geq k,Y\geq k) \\[1ex] & = \Pr(X\geq k)\Pr(Y\geq k) & \impliedby ... 4 Let the parameters of the two geometrics be \alpha and \beta. So these are the probabilities of "success," and the geometrics give the number of trials until the first success. Let Z=\min(X,Y).We have Z\ge z if and only if X\ge z and Y\ge z. The probability that X is \ge z is the probability of z-1 "failures" in a row. This probability is ... 0 If X\sim -Y then X+Y \sim -Y-X. It follows that \mathsf P(X+Y < - t) = \mathsf P(-Y-X < - t).\therefore \mathsf P(X+Y< -t) = \mathsf P(X+Y> t)$$Basically, if X and -Y are distributed identically then the sum X+Y is distributed symmetrically about 0. 0 For one direction, each interval of the form (-\infty,b] could be represented as the countable union of intervals of the other type, namely (-\infty,b]=\cup_{n=0}^\infty [b-n-1,b-n]. For the other direction, each [a,b] equals (-\infty,b]\setminus(-\infty,a), and (-\infty,a) in turn equals \cup_{n=1}^\infty (-\infty,a-\frac1n]. That is, ... 1 \varphi_X(t) is periodic (with period \pi/2). So, we may consider a small neighborhood around zero, say (-\delta,\delta) As I mentioned in the comments a_n=2^n is not enough to get transience. So, let a_n=4^n For t\in(-\delta,\delta) and some constant c$$1-\varphi_X(t)=\sum_{n=1}^\infty \frac{1-\cos(t4^n)}{2^n}\ge c\sqrt{|t|}Finally, ... 0 One method is to integrate wrt x to find the CDF and then differentiate it back wrt y. Another is to use the "change of variable transformation", which involves one differentiation. \begin{align} \\ f_Y(y) & = f_X{\circ}x(y)\; \Bigg|\frac{\mathrm d x(y)}{\mathrm d y}\Bigg|\tag{1: change of variable} \\[1ex] f_X(x) & = 3x^2\tag{2 given} \\[1ex] ... 0 \begin{align} \mathsf {Cov}(X,Y) & = \mathsf{Cov}(X,X^2) \\[1ex] & = \mathsf E((X-\mathsf E(X))\;(X^2-\mathsf E(X^2))) \\[1ex] & = \mathsf E(X^3-X^2\mathsf E(X)-X\mathsf E(X^2)+\mathsf E(X)\mathsf E(X^2)) & \text{ by expansion} \\[1ex] & = \mathsf E(X^3)-\mathsf E(X^2)\mathsf E(X)-\mathsf E(X)\mathsf E(X^2)+\mathsf E(X)\mathsf E(X^2) ... 0 If the sequence (T_k) is i.i.d., then, for each integer n\geqslant1 and each permutation \pi on \{1,2,\ldots,n\}, by symmetry of the distribution of (T_k)_{1\leqslant k\leqslant n}, the eventA_\pi=[T_{\pi(1)}\lt T_{\pi(2)}\lt\cdots\lt T_{\pi(n)}\lt t]$$has the same probability. Furthermore, the union of these n! events is ... 0$$F_Y(y)=P(Y\leq y)=P(X^2\leq y)=P(X\leq\sqrt{y})=F_X(\sqrt{y})=\int_0^\sqrt{y}3t^2\,dt=y^{3/2}\Longrightarrow\frac{3}{2}\sqrt{y}=f_Y(y).$$This holds for 0<\sqrt{y}<1. 0 We know it is E(X^3) so:$$E(X^3)=\int_{-1}^{1}x^3f(x)=0$$So it is correct 2 Your mistake is in the last implication. Just because X<\infty a.s. does not mean that E(X) < \infty. You can only say anything if you make the stronger assumption that X<M<\infty a.s. In this case E(X) < M < \infty. For instance, you can have X whose probability density is 1/x^2 for x \geq 1 and 0 otherwise. Then ... 2 Toss two coins, set \xi = 1 if the first coin is a head, 0 otherwise, \eta = 1 if the second coin is a head, 0 otherwise, \zeta = 1 if the coins show different symbols, 0 if they are the same. Then \xi, \eta and \zeta are pairwise independent (hence pairwise uncorrelated) random variables with E(\xi) = E(\eta) = E(\zeta) = \frac 12 but ... 0 Since F is continuous and (P_n)_{n\in\mathbb{N}} weakly converges to P, we've got$$\lim F_n(x)=F(x)\;\;\;\text{for all }x\in\mathbb{R}\tag{1}$$Let N\in\mathbb{N},$$x_k:=\inf\left\{x\in\overline{\mathbb{R}}:F(x)\ge\frac{k}{N}\right\}\;\;\;\text{for }k=0,\ldots, N$$and$$R_n:=\max_{1\le k<N}\left|F_n(x_k)-F(x_k)\right|$$(1)\Rightarrow ... 1 Actually you need the opposite result, i.e. that Y^- is stochastically larger than X^-. Note that if P(X> t)\ge P(Y> t) then for t> 0 you have that$$\begin{align*}P(X^-\ge t)&=P(\max\{-X,0\}\ge t)=1-P(\max\{-X,0\}<t)\overset{t>0}=\\&=1-P(-X< t)=1-P(X> -t)\\&\le 1-P(Y>-t)=1-P(-Y< t)=P(\max\{-Y,0\}\ge t)=P(Y^-\ge ... 2 The mode is the most likely value to occur. Poisson takes only discrete values, so you are given that $$P(X=1)=P(X=2)$$ which implies that $$e^{-λ}\frac{λ^1}{1!}=e^{-λ}\frac{λ^2}{2!} \iff λ=\frac{λ^2}{2} \iff λ^2-2λ=0 \iff λ(λ-2)=0$$ soλ=0$which is rejected (since$λ>0$) or$λ=2$which is acceptable. Thus$X \sim$Poisson$(λ=2)$and therefore ... 0 Define$I:=\left\{ \left(r,s\right)\in\mathbb{N}\times\mathbb{N}\mid0\leq r\leq s\leq5\right\} $. Let$E_{r,s}$denote the following event: Before the first number$6$occurs exactly$r$distinct numbers in$\{1,2,3,4,5\}$occur, before the second number$6$occurs exactly$s$distinct numbers in$\{1,2,3,4,5\}$occur, and before the third number ... 2 Well, certainly it is a useful result for coupling. Let us define$M_\epsilon:=\mu+\nu$, then$M_\epsilon$is a coupling of$\Bbb P$and$\Bbb Q$by$(1)$, and $$M\left(d(x,y) \geq \alpha+\epsilon\right)\leq\beta+\epsilon.$$ The latter fact is very useful: say, you have two random variables$\xi\sim \Bbb P$and$\eta\sim \Bbb Q$, and you would like to ... 2 What type of distribution is this? I do not recognize it, but the wording of the question seems to imply it is identifiable. Indeed it is! Hint: The distribution puts weight$p_k$on each integer$k$between$0$and$n$, where $$p_k=\binom{n}{k}a^k(1-a)^{n-k},\qquad a=\frac{t}{t+s}.$$ Does this ring a bell? Second hint: The LaTeX encoding of the first ... 1 I don't know regarding the distribution, but support is easy to check: you know that$X_t$can take any value in$\Bbb N = \{0,1,2,\dots\}$unconditionally. You also know that$X_t$is non-decreasing, so if$X_{t+s} = n$then$X_t\in [0;n]$almost surely. I'm pretty sure that you've used this fact while computing the conditional distribution. An educated ... 1 Let$B_N = \bigcup_{n \ge N} A_n$. Then$A_n \ \text{i.o.} = \bigcap_{N=1}^\infty B_N$. Note that$B_{N+1} \subseteq B_N$. Then countable additivity implies$P(\bigcap_{N=1}^\infty B_N) = \lim_{N \to \infty} P(B_N)$. 0 Notice that for each$R$,$\mathbb P\{Y_n\leqslant R\}=(\mathbb P\{X_1\leqslant R\})^n$. Since the sequence of sets$(\{Y_n\leqslant R\})$is non-increasing, we have $$\mathbb P\left\{\sup_kX_k \leqslant R\right\}=\lim_{n\to\infty} (\mathbb P\{X_1\leqslant R\})^n.$$ We thus have that$\mathbb P\left\{\sup_kX_k \leqslant R\right\}=0$for each$R$as long as ... 0 As mentioned in the comments, the attempt is good up to a typo and the verification that the variance of$2X_i$is$1$in order to be sure that the limiting process has variance$1$. 0 Hint: For every real number$t$, define$X_t=\max(X,t)-t$and$Y_t=\max(Y,t)-t$, then$X_t$and$Y_t$are nonnegative random variables and$X_t$is stochastically larger than$Y_t$(can you show this?) hence$E(\max(X,t))\geqslant E(\max(Y,t))$. Consider the limit when$t\to-\infty$, assuming that$X$and$Y$are integrable. 1 Hint: $$\Bbb Q = \bigcup_{q\in\Bbb Q} \{q\}$$ and$P(\bigcup A_n)\le \sum P(A_n)$if$(A_n)$is enumerable. 2 Let$(Y_t)_{t \in [0,1]}$be a Lévy process and denote by$\Delta Y_t := Y_t-Y_{t-}$the jump height at time$t$. The aim is to show that$(Y_t)_{t \in [0,1]}$has only finitely many jumps with jump size larger than$r$, i.e. that $$\{t \in [0,1]; |\Delta Y_t(\omega)|>r\}$$ is a finite set for (almost) all$\omega \in \Omega$. This follows from the ... 1 I think a complete characterization might be difficult, so I have collected some postitive and negative results that should cover most practical cases. Any compact topological group has a finite Haar measure. If the group is Hausdorff and infinite this measure is atomless. This is applicable to e.g. the Cantor space, which can be construed as the product ... 3 Let$\left(\Omega,\mathcal{A},P\right)$be a probability space. Then a random variable on it is a function$X:\Omega\rightarrow\mathbb{R}$such that$X^{-1}\left(B\right)=\left\{ \omega\in\Omega\mid X\left(\omega\right)\in B\right\} \in\mathcal{A}$for each Borelset$B$. Denoting the collection of Borelsets on$\mathbb{R}$by$\mathcal{B}$we state that ... 2 Your proof is correct. More generally: If$E(X_n)\to C$and the series$\sum\limits_n\mathrm{var}(X_n)$converges, then$X_n\to C$almost surely. Hint: There exists some positive sequence$(\alpha_n)$such that$\alpha_n\to0$and$\sum\limits_n\frac1{\alpha_n^2}\mathrm{var}(X_n)$converges. Consider the events$A_n=[|X_n-E(X_n)|\geqslant\alpha_n]$. 1 To help you prove that$Y$is a random variable, we need to know what kind of definition of random variables you are familiar with. Regarding the distribution function:$F(t) = P(X \leq t)$. Let$F_Y$be the distribution function of$Y$. Then $$F_Y(t) = P(Y \leq t) = P(aX + b \leq t) = P\left(X \leq \frac{t-b}{a}\right) = F\left(\frac{t-b}{a}\right)$$ if$a ...

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The union of the $\sigma$-algebras has no reason to be a $\sigma$-algebra; $\sigma(X_1,\dots,X_n)$ is the $\sigma$-algebra generated by these unions. Take $n=2$. Then $X_2$ can take the values $1$ on the set $A_1:=\{Y_1=1/2\}\cap \{Y_2=1/2\}$, $0$ on $A_0:=(\{Y_1=1/2\}\cap \{Y_2=-1/2\})\cup(\{Y_1=-1/2\}\cap \{Y_2=1/2\})$ and $-1$ on ...

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The following proof shows the claim for $p \leq 2$: First of all, since $\mathbb{P}(\ldots) \leq 1$, it suffices to prove the claim for $\lambda \geq 1$. Moreover, since $\lambda^{-2} \leq \lambda^{-p}$ for all $p \leq 2$ and $\lambda \geq 1$, we can restrict ourselves to the case $p=2$. It follows from Etemadi's inequality that $$\mathbb{P} \left( ... 2 I learnt about RN derivative from "Real Analysis" by Folland, and would advise you to check it out there (Chapter 3) as it may answer your coming questions. In particular, Theorem 3.5 answers your Q1. It state that If \nu is a finite signed measure and \mu is a positive measure, then \nu\ll \mu iff for any \varepsilon >0  there exists \delta ... 3 The game where one looses \1 when the toss is heads and one wins \2 when the toss is tails corresponds to a random walk with positive bias. These have positive probability to never hit \0. The game where one looses \1 when the toss is heads and one doubles one's fortune when the toss is tails, is at least as favorable as the first one at every step such ... -1 You can always verify your results by taking the inverse laplace transform of your function. Here:$$\mathcal{L}^{-1}(e^{-sc})=\delta(t-c)\text{ where }\delta\text{ is Dirac's Delta Function}

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