# Tag Info

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Hint: $X_2 = X_1 + Y$ where $Y$ is Poisson with parameter ... and independent of ...

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Changing slightly the notations and using the homogeneity of the problem, $$[N\geqslant n]=[\text{car}\ n\ \text{is fully served}]=[U_1+\cdots+U_n\leqslant2],$$ where the random variables $U_n$ are i.i.d. and uniform on $(0,1)$. The mean time elapsed between two cars are served is $1/\lambda$ hence the mean time before the station runs out of gas is $$... 0 Preliminary, let us do a slight change in notation: I will indicate with k \in \mathbb{Z} the discrete time, with K the k present in the summation and the period (if it exists) with N \in \mathbb{N}_0. Firstly, let us refresh the following concepts. Let \theta \in \mathbb{R}_+, k \in \mathbb{Z}, i = \sqrt{-1}, and consider the following ... 0$$ P_0(t)=\phi(0,t) \qquad P_1(t)=\frac{\partial\phi}{\partial s}(0,t)  P_n(t)=\frac1{n!}\,\frac{\partial^n\phi}{\partial s^n}(0,t) $$1 Each visitor is of type 1 with probability p=80\% and of type 2 with probability 1-p hence the number of visitors of type 1, from a given time instant, before a type 2 visitor arrives, is the number N of heads before the first tail in a heads-or-tails game of probabilities p and 1-p. This is the geometric distribution such that P(N=n)=p^n(1-p), ... 0 Thank you Bryon Schmuland, the answer of "The strong Markov property with an uncountable index set" does answer my question too. I was first put off why T should be finite, but since then \{T<\infty \}=\Omega, X_T is well defined for every \omega without having to look at the Limits. 0 It's a measure of (linear) dependence between now and one period back, albeit that the scale of X matters. So it's better to look at the autocorrelation instead. 1 Yes, except for a minor technical condition pertaining to the existence of moments, eg Cauchy distribution. 1 Yes, since X_i and X_j are independent for i \neq j, their covariance is 0, and hence the autocovariance function has value 0 for all i \neq j. Note that identical distribution is not needed for this result to hold; independence suffices. In fact, all that is needed is for X_i and X_j to be uncorrelated for all i \neq j, which is true when ... 2 This might be one of the rare cases I know where the CDF is the most convenient approach... For every n\geqslant0, let M_n=\max_{0\leqslant k\leqslant n}X_k, then, for every x,$$ [X_N\leqslant x]=\bigcup_{n\geqslant1}[x\geqslant X_n\gt X_0\geqslant M_{n-1}]=\bigcup_{n\geqslant1}[x\geqslant M_n,A_n], $$where$$ A_n=[X_n\gt X_0\geqslant M_{n-1}]. $$... 0 I do not know whether it is elegant or not, but what I'd try is as follows: first,$$ \mathbb{P}\{N=1\} = \mathbb{P}\{X_1 > X_0\} = \int_{\mathbb{R}} \mathbb{P}\{X_1 > x,\ X_0 \in dx\} = \int_{\mathbb{R}} dx f(x)\left(1-F(x)\right) the last step by independence. Then, \begin{align*} \mathbb{P}\{N=2\} &= \mathbb{P}\{X_2 > X_0,\ X_1 \leq ... 1 This is called thinning of a Poisson Process. Suppose that \Phi is the set of decay points in the real line following Poisson with intensity \lambda. Note that the expected value of D(t) is equal to \lambda t. The problem with your original idea, which I prefer too, is that the conditional distribution is indeed binomial distribution because you ... 0 It seems that that the term emergent might refer here to some behaviors or properties which appear when a number of simple entities (agents) operate in an environment, forming more complex behaviors as a collective than was expected in view of the simplicity of the mechanisms ruling them at an individual level. An example of this perspective in the context ... 1 Answer: I have given the answer for n random variables which can be simplified to two.F_{min}(x) = P(X_{min} \leq x)= 1- P(X_{min} > x)= 1-P(X_1 > x; X_2 > x;\cdots X_n > x)= 1 - [1 - F_1(x)][1 - F_2(x)]\cdots[1 - F_n(x)]$$For x > 0,$$ F_{i}(x) = 1 - e^{-\lambda x} 1- F_{i}(x) = e^{-\lambda x}F_{max}(x) = ...

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Hint: The hazard function of an absolutely continuous random variable $Z$ can be obtained by $$\lambda_Z(t)=-\frac{\mathrm d}{\mathrm d t}\log S_Z(t),$$ where $S_Z(t)=P(Z>t)$ denotes the survivor function.

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The theorem claims [that] $G$ is the common CDF of $X_{n}$ (...) but then claims [that] $G_{k}(t) = P(S_{k}\leq t)$. I don't see how both of those statements can be true. If $G$ is the CDF [of every] $X_{n}$ then $G(t) = P(X_n \leq t) = P(S_n - S_{n-1} \leq t)$ and this clearly does not equal $P(S_{k}\leq t)$. Why should it? Nobody said that $G=G_k$ ...

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First remark : the events "$S_n^X$ hits $A$ before $B$" are negligible. Let's be rigorous. Let $\tau_A := \inf \{n \geq 0: S_n \geq A\}$, and let $\tau_B := \inf \{n \geq 0: S_n \leq B\}$, both with values in $\mathbb{N} \cup \{+\infty\}$. Then you wan to prove that: $$\mathbb{P}_{(S_n^X)} (\tau_A < \tau_B) > \mathbb{P}_{(S_n^Y)} (\tau_A < ... 1 You went almost as far as one can go with this approach... Assume that all the coefficients \mu_k are different (otherwise things become more complicated to write down). Starting from your recursion relations, one can guess that, for every 0\leqslant j\leqslant i there exists some coefficients (\alpha_{ij}^k)_{j\leqslant k\leqslant i} such that, for ... 0 It is true for absolutely continuous distributions (but not for discrete distributions). The empirical distribution function is nothing else but the sample mean of Bernoulli random variables that has expected value F(x) and variance (1/n)F(x)\cdot[1-F(x)]. Therefore the Central Limit Theorem applies and we have$$\sqrt n\Big(\hat F_n(x) - F(x)\Big) ...

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Hints: $P(\text{extinction}\mid X_n\ne0)=P(\text{extinction}; X_n\ne0)/P(X_n\ne0)$ $P(X_n\ne0)\geqslant P(\text{survival})=1-P(\text{extinction})$ $P(\text{extinction}; X_n\ne0)=P(\text{extinction})-P(X_n=0)$ This yields (if I am not mistaken and to help you check your solution to come) the upper bound $$P(\text{extinction}\mid ... 0 We are given that X,Y\sim \mathrm{Exp}(1) so that the probability density function (pdf) of X is f(x)=e^{-x} and the pdf of Y is f(y)=e^{-y} For all z_1,z_2>0 we need to evaluate the joint density of X and Y with the appropriate limits as follows:- ... 1 In general, the sum of two compensated poisson processes is not a martingale. Example Let (N_t^1)_{t \geq 0} a Poisson process with intensity \lambda^1 =1. If we set N_t^2 := N_{t/2}^1, then (N_t^2)_{t \geq 0} is a Poisson process with intensity \lambda^2 = 1/2. For the joint filtration \mathcal{F}_t := \sigma(N_s^1,N_s^2; s \leq t) it holds ... 0$$g(x,t) = ∫_0^t \theta(u) \exp(-\alpha(t-u) ) dB(u) \\ = e^{-\alpha t} ∫_0^t \theta(u) \exp(\alpha u ) dB(u) $$Now use the fact that t \to e^{-\alpha t} is of finite variation and the Ito formula for a product: If X,Y are two Ito processes then so is XY:$$ dX_t = a_t dt + \sigma_t dB_t \\ dX_t = b_t dt + \sigma'_t dB_t \\ d(XY)_ t = X_t dY_t ...

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$\newcommand{\E}{\mathbb E}$ I will assume here that $S$ is independent of $N$. (Very often people writing about this neglect to state that assumption. And so it is here, if in fact that was intended, and if it wasn't then the nature of the dependence would need to be mentioned at least implicitly.) You can use the law of total expectation: \begin{align} ...

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Yes, it's the delta function. The formula means that for every $t$ the variable $f(t)$ has variance $R$ and for every $s\neq t$ variables $f(s)$ and $f(t)$ are uncorrelated. This corresponds to the more intuitve discrete-time white noise.

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Actually, there is nothing left to do. From $$dZ_t = b \exp \left( \frac{c^2}{2} t - c W_t \right) \, dt$$ it follows that $$Z_t = \underbrace{Z_0}_{0} + b \int_0^t \exp \left( \frac{c^2}{2} s - c W_s \right) \, ds.$$ Hence, $$X_t = Y_t \cdot Z_t = b \exp \left(- \frac{c^2}{2} t+c W_t \right) \cdot \int_0^t \exp \left( \frac{c^2}{2} s - c W_s \right) ... 1 Some hints to (b): In an incomplete market \Pi(C) is an interval because we put lower and upper bounds on the arbitrage free prices. For example the upper bound on the price of a non-replicable C is$$ \inf(E_{\mathbb{Q}}[X/(1+R)]:X\geq C, \: X \text{ is replicable}) $$because if it would trade at a higher price than this, you could do arbitrage by ... 1 What have you done so far, where are you stuck? The transition rate matrix for the standard M/M/1 queue model is given by$$Q=\begin{pmatrix} -\lambda & \lambda \\ \mu & -(\mu+\lambda) & \lambda \\ &\mu & -(\mu+\lambda) & \lambda \\ &&\mu & -(\mu+\lambda) & \lambda &\\ &&&&\ddots \end{pmatrix}.$$... 1 Hint It follows from Itô's formula that$$\exp(\theta t) \cdot B_t = \int_0^t \exp(\theta s) \, dB_s + \theta \int_0^t \exp(\theta s) B_s \, ds,$$i.e.$$ \int_0^t \exp(\theta s) \, dB_s = \exp(\theta t) \cdot B_t - \theta \int_0^t \exp(\theta s) B_s \, ds.$$0 Hmmm ... it seems that your book is confusingly using X to denote both the process and the state space. Let me change the state space to S to make things simpler. This is the description of a stochastic process; it is just a collection of random variables. Instead of just one or two, you have a random variable for every t in some index set T:$$ ...

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Apply Ito's lemma to $f(x,t) = t - x^2$ $$\mathrm{d} f(W_t,t) = \mathrm{d}t - 2 W_t \mathrm{d}W_t - \frac{1}{2} 2 \mathrm{d}t = 2 W_t \mathrm{d}W_t$$ It is a property of an Ito integral $I_t = \int_0^t a(s, \omega) \mathrm{d}X_s$ where $X_t$ is a martingale and $a(t,\omega)$ is adapted that $I_t$ is a martingale. QED EDIT: you need a few more technical ...

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Let $$W_t := \sigma B_t - \mu \cdot t$$ be a Brownian motion with drift $\mu>0$ and $$T_x := \inf\{t \geq 0; W_t \geq x\}.$$ Using the stationarity and independence of the increments of the Brownian motion $(B_t)_{t \geq 0}$, it is not difficult to see that $\left(\exp\left[\frac{2\mu W_t}{\sigma^2}\right]\right)_{t \geq 0}$ is a martingale. In ...

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Recall that an SDE of the form $$dX_t = b(X_t) \, dt + \sigma(X_t) \, dW_t$$ has a unique solution whenenver the coefficients $b$, $\sigma$ are globally Lipschitz continuous. In particular, we see that the SDE $$dX_t = (1+X_t) \, dt+ X_t \, dW_t$$ has a unique solution. Now let $(Y_t)_t$ the (unique) solution of the SDE $$dY_t = Y_t \, dt + Y_t \, dW_t ... 1 The conditional distribution of (S_k)_{1\leqslant k\leqslant n} is uniform, that is, its density is$$n!\,t^{-n}\,\mathbf 1_D,$$where$$ D=\{(s_k)_{1\leqslant k\leqslant n}\mid 0\leqslant s_1\leqslant\cdots\leqslant s_n\leqslant t\}. $$One is interested in the conditional distribution of (T_k)_{1\leqslant k\leqslant n} where T_1=S_1 and ... 0 The probability that 11 is a subsequence of \sigma is the probability that a Markov chain on the state space \{0,1,2\} starting from the state 0 visits the state 2 at or before time k, if the transition probabilities are 1-x for the transitions 0\to1 and 1\to2, and x for the transitions 0\to0 and 1\to0, where x=1-1/n. The usual ... 0 The results follow from the following identities, using the structure of the process (V_i). Note that$$X_i=\sum_{n\geqslant1}\beta^{n-1}V_{i-n} $$hence$$ \mathrm{var}(X)=\sum_{n\geqslant1}\beta^{2(n-1)}\mathrm{var}(V)=\frac{\mathrm{var}(V)}{1-\beta^2}$$and that, for every j\geqslant0,$$X_i=\beta^jX_{i-j}+\sum_{n=1}^j\beta^{n-1}V_{i-n} $$hence$$ ...

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Yes, your approach is correct. Your calculation simplifies (a bit) if you use the symmetry, i.e. the fact that $$\text{var} \left( \sum_{i=1}^n W_{t_i} \right) = \sum_{i=1}^n \text{var}(W_{t_i}) + 2 \sum_{i=1}^n \sum_{j=1}^{i-1} \text{cov}(W_{t_i},W_{t_j}).$$ An alternative approach is the following: The vector $(W_{t_1},\ldots,W_{t_n})$ is a Gaussian ...

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$$Y_n=1/(1+X_n)\implies P(|Y_n|\geqslant2)\leqslant P(|X_n|\geqslant\tfrac12)$$

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I have never heard of bounded in probability before this so I googled and am going off of this pdf. Here is a screenshot of the relevant definition(I assume the bent equals sign is meant to be $Θ$?): It felt unlikely that $\frac{1}{1+X_n}$ will be converging to 0 in probability so I decided to try to prove the other case.$\newcommand{\prob}{\mathbb{P}}$ ...

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EDIT: Sorry I don't know why I was so sloppy with capital versus lowercase letters, I intended to make capitals random variables but I was sloppy and did it inconsistently in some places. I might edit it at some point. This follows the structure of the derivation of the regular B-S equation by Thayer Watkins. I guess the risk-free interest rate is $0$? Let ...

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For simplicity (of notation), we assume $n=2$, i.e. that $(B_s)_{s \geq 0}$ is a $2$-dimensional Brownian motion, and $t=1$. For $k,j \in \mathbb{Z}$ and $m \in \mathbb{N}$ set $$A_{k,j}^m := \left[ \frac{k}{2^m}, \frac{k+1}{2^m} \right) \times \left[ \frac{j}{2^m}, \frac{j+1}{2^m} \right)$$ and $$B_{k,j}^m := \left[ \frac{k-1}{2^m}, \frac{k+2}{2^m} ... 1 Note that$$ G(z_1,z_2)=E(z_1^{X_1-X_2}z_2^{2X_1+X_2-N})=z_2^{-N}E((z_1z_2^2)^{X_1}(z_1^{-1}z_2)^{X_2}). $$Since X_1 and X_2 are independent,$$ G(z_1,z_2)=z_2^{-N}g_1(z_1z_2^2)g_2(z_1^{-1}z_2). $$This does not use the constraints. 2 If (X_1,X_2,X_3) is independent,$$ G(z_1,z_2)=E(z_1^{X_1-X_2}z_2^{X_1-X_3})=E((z_1z_2)^{X_1}z_1^{-X_2}z_2^{-X_3}), $$hence$$ G(z_1,z_2)=g_1(z_1z_2)g_2(z_1^{-1})g_3(z_2^{-1}). $$2 In general, a point process is a random variable N from some probability space (\Omega,\mathcal{F},P) to a space of counting measures on {\bf R}, say (M,\mathcal{M}). So each N(\omega) is a measure which gives mass to points$$ \ldots < X_{-2}(\omega) < X_{-1}(\omega) < X_0(\omega) < X_1(\omega) < X_2(\omega) < \ldots $$of {\bf ... 0 The integral to solve is: E$$[S_s]$=$\int_{-\infty}^{\infty}$$S_s$$exp((\mu-\frac{1}2\sigma^2)s+\sigma W_s)$$exp$$(W_s^2/2t)$$dW$$/(\sqrt(2\pi t))$From this integral the expression$S_s$$exp((\mu-\frac{1}2\sigma^2)s can be factored out. The remaining expression inside the integral is exp(\sigma W_s)$$exp$$(W_s^2/2t)=exp(\sigma W_s+$$W_s^2/2t)$This ... 1 From Wikipedia: For any disjoint bounded subsets$B_1,\ldots,B_n$and non-negative integers$k_1,\ldots,k_n$we have that $$\Pr[\xi(B_i) = k_i, 1 \leq i \leq n] = \prod_i e^{-\lambda \|B_i\|}\frac{(\lambda \|B_i\|)^{k_i}}{k_i!}.$$ The constant$\lambda$is called the intensity of the Poisson point process. Note that the Poisson point process is ... 8 I will complement saz's answerby proving that$L$(to use saz's terminology) is a random variable. EDIT: A simpler and more powerful proof was given later by saz here (see his remark at the end). Introduction & proof outline 0 Following is a proof that the measure of the path of an$n$-dimensional ($n \geq 1$) Brownian motion is itself a measurable ... 0 If$\mu$is predictable and Lebesgue integrable, this converges to$0$. This is an Ito process with$\sigma =0$. Quadratic Variation of Ito process 1 If you start in j, then you automatically get one count of being in the target state. Now pick your next transition k. Do a weighted average of the expected number of visits from k to j, where you weigh by the probability of going to a particular k first. Let$S_{j}$be a random variable that counts the number of times one visits state$j$. $$S_{j} = ... 1 Suppose that$$dX_t = - \mu X_t \, dt + \sigma \, dW_t \tag{1}$$and set Y_t := e^{\mu t} X_t. Itô's formula states that$$df(t,X_t)= \frac{\partial}{\partial x} f(t,X_t) \, dX_t + \frac{\partial}{\partial t} f(t,X_t) \, dt + \frac{1}{2} \frac{\partial^2}{\partial x^2} f(t,X_t) \, d\langle X \rangle_t. \tag{2}$$Here, we have$f(t,x) = e^{\mu t} \cdot ...

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