# Tag Info

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For a rigorous proof, let $\{N_t\}$ be the Poisson process related to crew check. Accordingly, denote $X_1, X_2, \ldots$ to be the independent random exponential random variables representing the time between events and $S_n = X_1 + \cdots + X_n$. For fixed $t > 0$, define $X^{(t)} = S_{N_t + 1} - t$. For a machine that starts working at time $0$, if a ...

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Consider the case of an i.i.d. sequence $(\varepsilon_j)_{j\in\mathbb Z}$ and define $X_j := j\varepsilon_j$. The sequence $\left(X_j\right)_{j\in\mathbb Z}$ is independent hence $\phi$-mixing but not stationary, even in the wide sense.

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For the case $a+b\le x$ where $0 \le x \le 1$ the region is a triangle with base and height $x$ which has the area $x^2/2.$ To use an integral as you did, the limits on $a$ should be from $0$ to $x$ rather than $0$ to $1$ as you have it. When $1 \le x \le 2$ a good way is to subtract the area of the triangle above the line $a+b=x$ from $1.$

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No. I think you mean $\mathbb E$ instead of $\mathbb P$ It does not make sense to take the intersection of a collection of events (e.g. $\mathcal F_{t-1}$) and a collection of sample points (e.g. $(B_t = 1)$). Perhaps you meant $\sigma((B_t = 1))$. Careful about the extension you're trying to make here. You seem to be thinking we can do something like: ...

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For the first one you are right of course (finite stopping times), For the second question then yes again but be aware that some authors do not allow a stopping times to be infinite to avoid any problems in that case, some others always express their claims by conditioning with respect to the event $T<\infty$ on $X_T$. For the last one well then $X_T$ ...

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The representation of Lévy processes by the means of its characteristic function in the Lévy-Khintchine Representation gives you by unicity of such representation the tool to get convergence results (i.e. for a sequence of processes $X^n_t\to X_t$). But your claim with $t\to \infty$ is simply wrong. For example for a Brownian motion it has no meaning, as ...

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$\mathscr G_m = \sigma(V_1,\ldots,V_m)$ is strictly larger than $\mathscr F_m = \sigma(Z_1,\ldots,Z_m)$. ($\mathscr G_m$ has $2^{2^m}$ measurable sets while $\mathscr F_m$ only has $2^{m+1}$) But $\mathscr F_m$ is a subset of $\mathscr G_m$ and so $E[X \mid \mathscr F_m] = E[E[X \mid \mathscr G_m] \mid \mathscr F_m]$. Now, with $X$ being independant of ...

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Notice that $u=(S_n+n)/2$ is $Beta-Binomial(n,1,1)=U[0,n]$. Hence $$Y|S_n\sim Beta(1+\frac{n+S_n}2,1+\frac{n-S_n}2)=Beta(1+u,1+d)$$ with average $\hat y_n=\frac {1+u}{2+u+d}=\frac{1+u}{n+2}$ and hence $$S_{n+1}-S_n\sim B(P((Y|S_n)>X_{n+1}))= B(\hat y_n)$$ where $B(p)$ is Bernoulli on $\{-1,1\}$ with $P(B(p)=1)=p$.

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Regarding the necessity of indexing with the rational numbers in the interval $[0,t]$ it is very important, since otherwise you cannot now whether the uncountable intersection that is $\{T_{x}>t\}$ is actually an event; by definition this set is an event if is an element of a sigma-algebra and sigma-algebras are closed under countable set operations.

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A linear SDE $$\text{d}Y_{t} = (\alpha(t)+\beta(t)Y_{t})\,\text{d}t+(\gamma(t)+\delta(t)Y_{t})\,\text{d}W_{t}$$ has an explicit (strong) solution which can be found on Wikipedia . Here $\alpha$, $\beta$, $\gamma$ and $\delta$ denote deterministic functions and $W$ a one-dimensional Wiener process. The solution, $Y\ ,$ is expressed in terms of the stochastic ...

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Since the stochastic integral process $$Y(t)=\int_{0}^{t}X_{s}\,\text{d}W_{s}\ , \quad t \geq 0$$ is a martingale and $\tau_{b}$ is an (almost-surely) bounded stopping time (adapted to the same filtration as process $Y$) then by the Optional Stopping Theorem the expected value of the random variable $Y(\tau_{b})$ is equal to that of the random variable ...

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First of all, it is not true that $\mathbb{E}_x \tau_x=\mathbb{E}_y \tau_x$. For example consider a chain with two states: $\mathcal{S}=\{x,y\}$ with the transition matrix $\pi(i,j)=1_{\{i\neq j\}}$ (i.e. changing state with probability $1$ at every step). In this case $\mathbb{E}_x \tau_x=2>1=\mathbb{E}_y \tau_x$. One way to show what you want is by ...

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Your SDE is an example of a linear SDE $$dY_t = (\alpha(t)+\beta(t)Y_t)dt+(\gamma(t)+\delta(t)Y_t)dW_t,$$ where $\alpha$, $\beta$, $\gamma$ and $\delta$ denote deterministic processes and $W$ denotes a one-dimensional Wiener process. In your case, processes $\beta$ and $\delta$ are zero processes and $\gamma$ is a constant process and $\alpha(t) = ... 0 Intuitively, the symbol$\text{d}W_{t}$may be interpreted as an infinitesimal increment of a (one-dimensional) Wiener process,$W$. $$\text{d}W_{t} = W_{t+\text{d}t}-W_{t}\ .$$ Increments of a Wiener process are normal distributed random variables whose expected value is zero and whose variance is equal to the time-increment. Thus the symbol$\text{d}W_{t}$... 1 Try to use the fact that$B_t/\sqrt{t}$has the same distribution as$B_1$and the fact that$\sup_{s\le t }B_s/\sqrt{s}$is monotonic w.r.t.$t$[Warning: more details below] For any constant$C>0$, let $$A = (\lim_{t\to 0} \sup_{0<s\le t}\frac{B_s}{\sqrt{s}}<C )$$ We want to show that$P(A)=0$. Since$A\in {\mathcal F}_{0+}$, we only have to ... 0 Based on Lost1's comment: In the first place, to have conditional expectation, we need integrability. A product of integrable random variables is not necessarily integrable: Let$X, Y \in \mathscr L^{1}(\Omega, \mathscr F, \mathbb P)$. Consider$X$and$X - Y$w/$Xhaving an infinite second moment but finite first moment. Then $$E[X(X-Y)] = E[X^2] - ... 3 Your question is quite interesting, and deserves more attention than it has been getting. Whether or not a (time homogeneous) Markov process (X_t)_{t\geq 0} with state space E is strong Markov depends on a precise definition. I hope that my explanation below substracts from, rather than adds to, your confusion. The standard definition of the strong ... 1 The transition matrix has the "block cycle structure"$$P=\pmatrix{0&A_0&0&0&\cdots&0\cr 0&0&A_1&0&\cdots&0\cr 0&0&0&A_2&\cdots&0\cr \vdots&&&\ddots&&\vdots\cr A_{d-1}&0&0&0&\cdots&0}$$For any dth root of ... 1 No it's not a Lévy process because Lévy processes must have the property (which is almost equivalent to a definition(*)) to be continuous in probability, in other words almost surely at a time s (for example t+h) the paths of the process are continuous. As you know that the process is almost surly discontinuous your process is not Lévy. For a reference look ... 0 You are right, this probability is not given without any reason. Since n_i is the number of transactions involving i items, the rate at which n_i goes to n_i+1 is the rate at which orders involving i products arrive. The total rate at which orders arrive is \lambda, and the probability that any order contains i items is 1/M, so that ... 0 Since it seems like what you wrote is a linear SDE of the form dX_t=(a+b.X_t)dt+(c+d.X_t)dW_t, I think a solution is available (no ?) Define$$\epsilon _t := \exp\left[\int_0^t b-\frac{d^2}{2} ds + \int_0^td. dW_s\right]$$, then$$X_t=\epsilon _t.\left[X_0+\int_0^t\frac{a-c.d}{\epsilon _s}ds + \int_0^t\frac{c}{\epsilon _s}dW_s \right]$$Also since in that ... 0 The probability of ever reaching zero does not depend on the probability to not move at all. Thus you can instead consider the walk with a 2/3 chance to go right and a 1/3 chance to go left. There is a nice general solution to this problem. Let n \in \mathbb{N}, then the probability to hit 0 before hitting n starting at 0<x<n satisfies: ... 0 Let P1 be the probability it ever reaches 0, starting at 1, and P2 the probability it ever reaches 0, starting at 2. To reach 0 from 2, it has to reach 1 from 2, then 0 from 1. Reaching 1 from 2 is the same as reaching 0 from 1. So the probability P2=P1^2 Now, start at 1, and make one move. The chances are 1/3 you are at 0 (success!) and 2/3 you are at 2. ... 0 Suppose it reaches 0 in n steps. We can calculate the probability using the multinomial distribution. Out of n steps some k will be +1, k+1 will be -1, and n-2k-1 will be 0. Hence, P(S_n=0) = \sum_{k=0}^{(n-1)/2} (1/2)^k (1/4)^{n-k}\frac{n!}{k! (k+1)! (n-2k-1)!} 0 In interest rate modeling, this is known as the Brennan and Schwartz model. My understanding is that, there is no close form solution, and numerical methods are used in practice. 1 Since p>\frac{1}{2}, we can write p= \frac{1}{2}+q for some q>0. Now$$\frac{(n-1)^p}{\sqrt{n}} = \sqrt{\frac{n-1}{n}} (n-1)^q \geq \frac{1}{2} (n-1)^q, \qquad n \in \mathbb{N},$$implies$$\mathbb{P} \left(B_1 > C \frac{(n-1)^p}{\sqrt{n}} \right) \leq \mathbb{P} \left(B_1 > \frac{C}{2} (n-1)^q\right).$$Choose k \in \mathbb{N} ... 1 You want to know something about \sum_iX_i. However, X_i has a different distribution for different values of i, so this is difficult. Estimating \sum_i X_i+Y_i is easier, since X_i+Y_i has a Bin(n,p) distribution for all i. So Y_i is just an auxiliary variable to show that X_i is dominated by a Bin(n,p) variable. I would say that ... 0 In general, a measure \nu:2^S\to [0,\infty)^S is said to be invariant under the stochastic kernel P which specifies the transition probabilities of X, i.e. P_{ij} = \mathbb P(X_{n+1}=j\mid X_n=i) for i,j\in S if$$\nu = \nu P. $$Now, since in this case we are also assuming$$\sum_{i\in S}P_{ij} = 1, \ j\in S $$as well as the usual assumption ... 1 HINT: The stationary distribution \pi satisfies the relationship \pi P=\pi, which is equivalent to showing \forall j\in S :$$\pi_j = \sum_{i \in S}\pi_i p_{ij}You can also write out \pi explicitly as a row vector where every entry is \frac{1}{5}, and P is a matrix with entries p_{ij} in row i, column j. Expanding this out and using ... 0 Here, Page 45, example 4.1.3. 2 You are mistaken in saying \Pr(t<T) = \beta e^{-\beta t}. That should say \Pr(t<T) = e^{-\beta t} (for t\ge0), so \Pr(T<t) = 1- e^{-\beta t} (for t\ge0), and hence the density is f_T(t)= \beta e^{-\beta t} for g\ge0. \begin{align} \Pr(S<T) & = \operatorname{E}(\Pr(S<T\mid S)) \\[10pt] & = \operatorname{E}(e^{-\beta S}) ... 2 Fix a positive integer n. By the martingale property and L^1 convergence of \{X_n\} we have for m\geqslant n,\mathbb E[|X_n - \mathbb E\left[X\mid \mathcal F_n]|\right] = \mathbb E\left[| \mathbb E[X_m-X\mid \mathcal F_n]|\right]\leqslant \mathbb E\left[|X_m-X|\right]\stackrel{m\to\infty}\longrightarrow 0, $$so that$$X_n=\mathbb E[X\mid \mathcal ... 0 Along the lines of my last comment in your previous question, you could do a reverse convolution type thingy: Suppose we have a renewal processN(t)$with rate$\lambda>0$. Let$X$be a random variable with the same distribution as the inter-arrival times of$N(t)$. Then$E[X]=1/\lambda$. Fix$\epsilon>0$. We want to add another process$A(t)$... 0 The jump process$\left(N_{\int_0^t f(s)\ \mathsf d s}\right)_{t\geqslant 0}$is a non-homogeneous Poisson process with rate$f(s)$. If we define the time transformation $$u = F(t) = \int_0^t f(s)\ \mathsf ds,$$ then the process$\left(M_u\right)_{u\geqslant 0}$is homogeneous with rate$1$. Therefore it follows that $$\left\langle N_{\int_0^\cdot f(s)\ ... 1 You don't have the right argument, the real reason why you can get the third line is because first B and Z are orthogonal so that$$\mathbb{E}^{\mathbb{Q}^{t} }\bigg[\bigg(\int_0^t\sigma_r^t(s)dB_s^t \bigg)\cdot\bigg(\int_0^t\mathbb{E}^{\mathbb{Q}^{t}}_s[D_s^{Z^t}\mathbb{1}_{S_t > K}]dZ_s^t \bigg)\bigg] =0$$and because : ... 1 A direct proof would be to observe that for any real number a,$$ 0 \le \sum^n_{k=1} (x_k - a)^2 p_k = \sum^n_{k=1} x_k^2 p_k - 2 a \sum^n_{k=1} x_k p_k + a^2 \, . $$Then set a = \sum^n_{k=1} x_k p_k to obtain 0 \le W_1 - W_2. This is actually the same as the Cauchy–Schwarz inequality, applied to the vectors$$ (x_1 \sqrt{p_1}, \ldots, x_n ... 1 There is the following general result: Let$f: [0,\infty) \to \mathbb{R}$be a continuous function and$g: [0,\infty) \to \mathbb{R}$be a function of bounded variation (on compact intervals). Then $$\langle f, g \rangle_t = 0$$ for all$t \geq 0$, i.e. $$\langle f,g \rangle_t = \lim_{|\Pi| \to 0} \sum_{t_j \in \Pi} (f(t_{j+1})-f(t_j)) ... 0 In the case where the two component renewal processes have the same interrenewal distribution F, the interarrival distribution of the superimposed process has density$$g(x) = -\frac1\mu\frac{\mathsf d}{\mathsf dx}\left[\bar F(x)\int_x^\infty\bar F(u)\ \mathsf d u\right], $$as derived in this answer. Here \bar F=1-F is the survival function of the ... 0 For the equality you wrote there is some "intuition" based on the Itô integral. Namely, write this integral as$$ \int_{0}^T B(T) d B(t) = \int_0^T B(t) dB(t) + \int_0^T \big(B(T)- B(t)\big) dB(t). $$The first integral is a usual Itô integral equal to (B(T)^2 - T)/2. The second is a kind of "purely anticipative" integral: the integrand is ... 0 Let g be the density of the interarrival times in the superimposed process, and F the interrenewal distribution of a component renewal process. Let U be the limiting backward recurrence time of the superimposed process, and U_i the same for the i^{\mathrm{th}} component renewal process. Then for x>0,$$\mathbb P(U>x)=\prod_{i=1}^n \mathbb ... 3 The original renewal process has independent and identically disributed (i.i.d.) inter-renewal times$\{T_1, T_2, T_3, ...\}$. Thus, starting from time 0, renewals occur at times$\{T_1, T_1+T_2, T_1+T_2+T_3, ...\}$. Now fix a probability$p >0$. Independently place each renewal time to$P_1$with prob$p$. So we get new inter-renewal times$\{Z_1, ...

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If $\Bbb E[1_A|\mathscr F_t]=1$, then $\Bbb E[1_A]=1$, which is the same as $1_A=1$. In this case $\Bbb E[1_A|\mathscr F_s]=1$ (almost surely) for each $s$. Likewise, if $\Bbb E[1_A|\mathscr F_t]=0$, then $\Bbb E[1_A]=0$, which is the same as $1_A=0$. In this case $\Bbb E[1_A|\mathscr F_s]=0$ (almost surely) for each $s$. If $\Bbb E[1_A|\mathscr F_t]=p$, ...

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Here is a simple explanation: It serves to exclude process with jumps at fixed (nonrandom) times. See P68 of R. Cont/P.Tankov

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I interpret it the way that Henry does in the comments on Graham Kemp's answer: Heuristically, any change in infinitesimal time is almost surely infinitesimal (i.e., is infinitesimal with probability one).

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For Poisson process, the quadratic variation is just $N(t)$. For $Z_t$, the quadratic variation is just $N(\int_0^tf(X_s)ds)$, if $f$ is positive. Without positive assumption, the 'time' $\int_0^tf(X_s)ds$ can be negative

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In simple terms, property 4 states that there are no step discontinuities in the cumulative density function. $$\neg\exists \epsilon > 0:\;\lim\limits_{h\searrow 0}\;\mathsf P\big(\big\lvert X_{t+h}{-}X_t\big\rvert> \epsilon\big) > 0$$

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Hint: Your argument shows that $\Bbb P(\tau>km)\le(1-\epsilon)\Bbb P(\tau>(k-1)m)$ for $k=1,2,\ldots$. (Take your $n$ equal to $k-1$.) Proceed recursively.

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Note that by continuity we have $\omega \in \Omega$, $$X_p(\omega) \ne X_q(\omega) \ \forall p \ne q$$ Define: $$A_n:= (X_n > \max\{X_1, X_2, ... X_n\})$$ Observe that: $$A_1:= (X_1 > \max\{X_1\}) = (X_1 > \max\{X_1\}) = \emptyset$$ $$A_2:= (X_2 > \max\{X_1, X_2\})$$ $$= (X_2 > X_1) = \emptyset \ \text{if} \ (X_1 > X_2)$$ $$= (X_2 ... 1 This may be a bit of overkill but the Lindeberg-Feller Central Limit Theorem (for triangular array) gives you the answer rather immediately. The condition that Var[|E|]={n \choose 2} p_n (1-p_n) \to \infty is both a necessary and sufficiently condition that \frac{|E|-{n \choose 2}p_n}{\sqrt{{n \choose 2}p_n (1-p_n)}} converges in distribution to a ... 2 Since the Poisson process has stationary increments, i.e.$$\mathbb{E}(|X_t-X_s|^{\alpha}) = \mathbb{E}(|X_{|t-s|}|^{\alpha}), \qquad s,t \geq 0,$$it suffices to consider the case s=0. By definition X_t is Poisson distributed with parameter \lambda t, and therefore$$\mathbb{E}(|X_t|^{\alpha}) \geq \mathbb{E}(1 \cdot 1_{\{X_t=1\}}) = \mathbb{P}(X_t ...

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