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3

As @Did already pointed out, it is much easier to show that $$a W_t - \frac{1}{2} a^2 t \to - \infty \qquad \text{almost surely as t \to \infty.}$$ Recall that the process $$B_t := \begin{cases} t W_{\frac{1}{t}}, & t>0, \\ 0, & t=0 \end{cases}$$ defines a Brownian motion; in particular $$\lim_{t \to 0} t W_{\frac{1}{t}} = \lim_{t \to 0} ... 2 Here is an interpretation of the original question. Assume the states are S_i, each representing the event that there are exactly i non-empty boxes. Clearly, we have$$p_{ij} = \begin{cases}i/k & \text{if }i = j \\ 1-i/k & \text{if } j = i + 1 \\ 0 & \text{otherwise}.\end{cases},$$where i,j\in\{0,1,\dots, k\}. This gives us a Markov chain ... 2 You can get the variance by the following trick:$$\begin{align*} \mathbb{E}\left[\left(\int_0^T W_t^2\,dt\right)^2\right] &= \mathbb{E}\left[\int_0^T W_t^2\,dt \int_0^T W_s^2\,ds\right] \\ &= \mathbb{E}\left[\int_0^T \int_0^T W_t^2 W_s^2 \,dt\,ds\right] \\ &= \int_0^T \int_0^T \mathbb{E}[W_t^2 W_s^2]\,dt\,ds. \end{align*}$$Now writing W_t = ... 2 Lemma: Let I=[a,b) be an interval and f,g: [a,b) \to E functions which are continuous from the right. Then$$f=g \iff \forall x \in \mathbb{Q} \cap [a,b): f(x) = g(x).$$Proof: "\Rightarrow" is obvious. For "\Leftarrow" we fix x \in [a,b). Then there exists a sequence (x_n)_{n \in \mathbb{N}} \subseteq I \cap \mathbb{Q} such that x_n ... 2 Well, let's explore the expectation directly:$$E\left[e^{e^y}\right] = \int_{-\infty}^\infty e^{e^x} e^{-(x-\mu)^2/(2\sigma^2)}\, dx = \int_{-\infty}^\infty \exp\left( e^x-\frac{(x-\mu)^2}{2\sigma^2}\right)\, dx.$$Now take x \to +\infty; it is clear that e^x \to \infty much faster than -x^2 \to -\infty, so it dominates in the exponential. ... 2 We have to check that the integral$$\int_0^{+\infty}\exp\left(e^y-(y-\mu)^2/(2\sigma^2)\right)\mathrm dy$$is divergent. There is a huge problem at +\infty because \lim_{y \to +\infty}e^y-(y-\mu)^2/(2\sigma^2)=+\infty. 2 I'd recommend going with J.R. Norris' Markov Chains book since it focuses specifically on Markov chains with excellent presentation. There is also some queuing theory in these notes and these notes as well as books like Bertsekas' Data Networks (since you're in engineering). Assumussen's Applied Probability and Queues is also a nice book. Frank Kelly also ... 2 For fixed k \in \mathbb{N} define a bounded stopping time \tau_k by$$\tau_k := \inf\{t \geq 0; M_t > x \} \wedge k = \tau \wedge k.$$Then, by the optional stopping theorem, we have$$\mathbb{E}(M_{\tau_k} \mid \mathcal{F}_0) = M_0. \tag{1}$$On the other hand, since (M_t)_{t \geq 0} is a continuous martingale, we know that$$M_{\tau_k} = ...

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Property $1$ says nothing about independence. It only tells you about the distribution of the increments $B_t-B_s$. If not for property $2$, you may have that $B_t-B_s$ is dependent on $B_s-B_u$ for some $u<s$. Property $1$ only tells you that $B_t-B_s$ has the distribution $N(0,t-s)$, and that $B_s-B_u$ has the distribution $N(0,s-u)$. Given property ...

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Since $$\mathbb E[e^{W_T}|\mathcal F_t]=\mathbb E[e^{W_T-W_t}e^{W_t}|\mathcal F_t]=e^{W_t}\mathbb E[e^{W_T-W_t}|\mathcal F_t]=e^{W_t}\mathbb E[e^{W_T-W_t}]=e^{W_t+\frac{1}{2}(T-t)}\,,$$ an application of Itô's lemma gives $\mathbb E[e^{W_T}|\mathcal F_t]=\int_{0}^{t}e^{W_u+\frac{1}{2}(T-u)}\mathrm dW_u\ (\text{for }0\leqslant t\leqslant T)$, which is of the ...

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Consider the collection $\mathcal{C}$ consisting of all sets of the form $\{ f: (f(t_1),...,f(t_n)) \in B_1 \times \cdots \times B_n \} = \bigcap_{i=1}^n \pi_{t_i}^{-1} (B_i)$, where $n \in \mathbb{N}$, $t_i \in [0,1]$, and $B_i \in \mathcal{B}(\mathbb{R})$. First notice that $\mathcal{C} \subset \mathcal{B}(S)$. To show this note that if $t \in [0,1]$ and ...

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For $i\leqslant t-1$, we have $\mathbb E\left[X_i\mid \mathcal F_{t-1}\right]=X_i$, hence the sequence $(Z'_t)_{t\geqslant 1}$, where $Z'_t=\sum_{i=1}^t\alpha^{-i}X_i$ is a martingale with respect to the filtration $(\mathcal F_t)_{t\geqslant 1}$ as long as $\mathbb E\left[X_t\mid\mathcal F_{t-1}\right]=0$ (which is seemed to be assumed in the first ...

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Starting with the definition, $\newcommand{\prob}{\mathbb{P}}$ $$\prob(X_n = x_n | X_{n-1} = x_{n-1},…,X_0 = x_0 ) = \prob(X_n = x_n | X_{n-1} = x_{n-1})$$ Then let $j_1...j_N$ be a labelling of $\{0,1,2,3,...,t_n\}\setminus \{t_1,t_2,...,t_n\}$. You just need to sum over the values you don't specify in the conditioning: \begin{align} &\prob(X_{t_n ...

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(a) Your answer is right until the last step. Because $N(t)$ has a Poisson distribution with parameter $\lambda t$, it should be: $$P(N(t)=1)\cdot P(N(t)=1)\cdot P(N(t)=1) = (\lambda t \;e^{-\lambda t})^3.$$ (b) Your answer is right. (c) \begin{eqnarray*} E(T_2\mid T_1\lt T_2\lt T_3) &=& \int_{t_2}{t_2 P(T_2=t_2\mid T_1\lt T_2\lt T_3)\;dt_2} \\ ...

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The distribution is also Poisson. Since this process is a Compound Poisson process. It's mean is $\lambda$TE[T] and it's variance is $\lambda$TE[$T^2$]. Here's a good reference: http://www.columbia.edu/~ww2040/3106F14/lec1023.pdf

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Write : \begin{align*} W_k&=bW_{k-1}+X_k=b\color{red}{(bW_{k-2}+X_{k-1})}+X_k\\ &=b^2W_{k-2}+bX_{k-1}+X_k\\ & \qquad\vdots\\ &=b^{k-1}W_{1}+b^{k-2}X_{2}+\cdots+X_k=b^{k-1}\color{red}{(bW_0+X_1)}+b^{k-2}X_{2}+\cdots+X_k\\ &=ab^{k}X_{0}+b^{k-1}X_{1}+b^{k-2}X_{2}+\cdots+X_k \end{align*} Now you don't have independence problem ! ...

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There are two parts to your question. For the second part, assuming X(t) and Y(t') are independent, requires the cross spectral density to be zero. The cross spectral density is the Fourier transform of the cross correlation function. The cross correlation is the ensemble average of the time-shifted product of X(t) and Y(t'), and if these are independent ...

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It asks you to find the probability of at least $2$ men arriving within the first inter-arrival time for the women. Therefore, letting $T_W$ be the women's inter-arrival time, and $N_M(t)$ be the number of arrivals of men within time $t$, \begin{eqnarray*} P(\text{$\geq 2$ men before first woman}) &=& \int_{t=0}^{\infty}{P(T_W=t) (1- P(N_M(t)=0) - ...

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Repeating what you get, we have $$P_{i+1} - P_i =(P_1 - P_0) \prod_{k=1}^i \frac{q_k}{p_k} = P_1 \prod_{k=1}^i \frac{q_k}{p_k}$$ since $P_0 = 0$ Then we have $$P_n = \sum_{i=0}^{n-1}(P_{i+1}-P_i) + P_0 = P_1 \sum_{i=0}^{n-1}\prod_{k=1}^i \frac{q_k}{p_k}, \forall n = 1,2,\cdots, N$$ At last, use the fact $P_N = 1$ to see \$P_1 = ...

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