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2

Let $B_{n}=\cup_{i=1}^{n} E_{i}$. Notice, that $\lim\limits_{n \to \infty} B_{n} =E$ (The details are left to you!, but as a hint, $B_{n}$ is ncreasing!). $$\mu (E)=\mu(\lim B_{n})= \lim \mu(B_{n})=\lim \mu(\cup_{i=1}^{n} E_{i})= \lim \sum\limits_{i=1}^{n} \mu (E_{i}) =\sum\limits_{i=1 }^{\infty}\mu(E_{i}).$$ The seccond lime is because $B_{n}$ is increasing ...


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Recall that the arrival times in a Poisson process with rate $\lambda$ follow an Exponential distribution with mean $1/\lambda$. Since inter-arrival times are independent, and you have two Poisson process, then it sounds like you are asking for $$P(T<S)$$ where $T = X_1+X_2$ is the sum of two waiting times for one process, and $S = Y_1+Y_2$ is the sum ...


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$$|f|1\{|f|>n\}\le |f|\in L^1.$$ Thus, $$\lim_{n\to\infty}\int|f|1\{|f|>n\}d\mu=\int \lim_{n\to\infty}|f|1\{|f|>n\}d\mu.$$


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Let $W_n=n^{-1/2}Y_n$. Then $$n^{-1/2}E[Y_n\mid Y_n>0]=E[W_n\mid W_n>0]\le \frac{\sqrt{E|W_n|^2}}{P\{W_n>0\}}=\frac{\sigma}{P\{W_n>0\}},$$ and $P\{W_n>0\}\to 1/2$ (by the CLT), i.e. for any $\epsilon>0$, $P\{W_n>0\}\ge \frac{1}{2}-\epsilon$ for $n$ large enough.


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Multivariate standard normal density function is in the form $f(x)=\exp(-\|x\|^2)$ where $x$ is the vector. Observe that $f(x)$ doesn't care about $\frac{x}{\|x\|}$ and only cares $\|x\|$, hence conditioned on any slice $\|x\|=a$, $f(x)$ is a constant which means all directions are equally-likely (i.e. uniform distribution). Now, the conditional ...


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Fix $\epsilon>0$. Since $(X_t)_{t \geq 0}$ satisfies the SLLN, there exists for almost all $\omega \in \Omega$ a constant $S=S(\omega)>0$ such that $$\left| \frac{X_s(\omega)}{s} - \mathbb{E}(X_1) \right| \leq \epsilon \quad \text{for all $s \geq S$}.$$ As $U_t \uparrow \infty$, we have $U_t(\omega) \geq S$ for $t \geq T=T(\omega)$ sufficiently ...



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