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Continuity and the local martingale property should (hopefully) be clear. For the quadratic variation/covariation, look at the quantities $(dW_1)^2$, $(dW_2)^2$, $dW_1dW_2$. The first two should be $dt$, and the third $0$. (Use the fact that $dB_1dB_2=\rho_t$). Then integrate to get the required answer.

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One of the characterizations of Multidimensional Brownian Motion $B_i(t)$ is that the covariations satisfy: $$[B_i, B_j]_t = \delta_{ij}t$$

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A partial answer, but perhaps others will contribute... Quoting from http://www.researchgate.net/post/What_is_the_difference_between_essential_boundary_conditions_and_natural_boundary_conditions "The essential boundary conditions are imposed on the functions in the space where the minimzation of the energy functional is made or the weak formulation is ...

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example Without assuming some "joint" measurability of $X_t(\omega)$ in $t$ and $\omega$ you are out of luck. We use the Continuum Hypothesis. There is a set $A \subseteq [0,1] \times [0,1]$ such that: $\qquad$For each $t \in [0,1]$,$\qquad \{\omega \in [0,1] : (t,\omega) \in A\}$ is countable, $\qquad$For each $\omega \in [0,1]$,$\qquad \{t \in ... 1 All of these statements can be false. What follows is more or less the standard counterexample of a local martingale that is not a martingale. I've stolen the details from an MO post of mine which constructs something slightly different. (To avoid search-and-replace I'm keeping my process called$Y$instead of$\beta$.) Set$T=1$. Let$r(t)$be any ... 0 For any Wiener process$(B_t)_t$, we have$B_t \sim N(0,t)$; in particular$\mathbb{P}(B_t < 0) = \frac{1}{2}$. Since$X_t= W_t^2 \geq 0$is non-negative, we have$\mathbb{P}(X_t<0)=0$and therefore$(X_t)_{t \geq 0}$cannot be a Wiener process. 1 Model: Set$A:=0$,$B:=50$. Let$\xi_j \sim \frac{1}{6} \delta_{-1} + \frac{1}{3} \delta_0 + \frac{1}{2} \delta_1$,$j \geq 1$, be independent identically distributed random variables. Then $$X_n := k + \sum_{j=1}^n \xi_j$$ is the position of the person after$n$steps if the person starts at$X_0=k$. We would like to calculate the probability that the ... 0$\mathscr{F}_{t_i}$represents a sigma algebra, which is a fancy term for "available information". In this case your only available information for computing the probability is the portfolio value$C(t_i)$. 1$\mathcal{F}_{t_i}$is a member of a filtration which is a set of$\sigma$algebras indexed by time. To be measurable with respect to$\mathcal{F}_{t_i}$means to only depend on the information available up to and including time$t_i$. Conditioning probability on it means given the information available up to and including$t_i$, what is the probability. -1 Independence is equivalent to zero covariance when normals are jointly normal. So just compute the expectation of the product and see if it's zero. 1 I am not sure this qualifies for an alternative proof anyway, define $$A_j:=\left \{ \liminf_p X_p^{ 1/p}\geqslant \liminf_p\left(\mathbb E[X_p]\right)^{1/p} +\frac 1j\right\}.$$ We have to prove that for each$j$,$\mathbb P(A_j)=0$. If not, then by Markov's inequality and Fatou's lemma, $$\liminf_p\left(\mathbb E[X_p]\right)^{1/p} +\frac 1j\leqslant ... 0 This is a simple example of the Orstein Uhlenbeck Process. Everything is described on the wiki: Ornstein Uhlenbeck Process 0 The requirement is simply that f is integrable. An ito integral is approximated by \sum_{s_i} f(W_{s_i}, s_i) (W_{s_{i+1}} - W_{s_i}). Since f(W_{s_i}, s_i) is independent of that interval (by definition of brownian motion), the expectation of each summand is zero. The result follows. 2 Yes, the stopped compensated Poisson process is uniformly integrable: By Doob's maximal inequality, we have$$\mathbb{E} \left( \sup_{t \in [0,K]} |\bar{N}(t \wedge T)|^2 \right) \leq 4 \mathbb{E}(|\bar{N}(K \wedge T)|^2)$$for any K>0. Since (\bar{N}_t^2-\lambda t)_{t \geq 0} is a martingale, we obtain$$\mathbb{E} \left( \sup_{t \in [0,K]} ... 1 I find the following example easier to understand: Consider any probability space$(\Omega,\mathcal{A},\mathbb{P})$. Set$\mathcal{F}_t := \mathcal{A}$for all$t \geq 0$and let$A \subseteq [0,T]$be a set which is not Borel-measurable. The process $$X_t(\omega) := 1_A(t)$$ takes only values in$\{0,1\}$and for each fixed$t \geq 0$, we have either ... 2 At stationarity, one asks that $$X\stackrel{\text{law}}{=}X\cdot\mathbf 1_{X<\theta}+\delta,$$ where the random variables$\delta$and$X$in the RHS are independent. Using the PDF$f$and$g$of$X$and$\delta$, this translates as the condition that, for every$x>0$, $$\ \qquad \qquad f(x)=P(X>\theta)\cdot g(x)+\int_0^{\min(\theta,x)} ... 1 If I is uncountable and A_i \in \mathcal{F}_t, then it does in general not follow that \bigcap_{i \in I} A_i \in \mathcal{F}_t. We only know that \mathcal{F}_t is stable under countable intersections. Recall the following lemma: Let g: [0,t] \to \mathbb{R} be a continuous function. Then$$\max_{s \in [0,t]} g(s) = \max_{s \in [0,t] \cap ... 0 You need to use the density function of the "running maximum" of the Brownian motion. This density function is well known, see for example here. 2 I'm also studying a course where things like this appear, so I'll present an attempt but corrections are welcome. I know stochastic calculus doesn't always play by the rules. I get that: $$\mathbb{E}[W_T \cdot \int_0^Tf(s)dW_s] = \int_0^Tf(s)ds$$ If$\mathbf{P} = \{t_0 = 0, t_1,t_2,\dots,t_n = T\} $is a tagged partition,$||\mathbf{P}|| = ...

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Let $0 \leq t_1<\ldots<t_n$. Since $(B_t)_{t \geq 0}$ is a Gaussian process, we know that $(B_{t_1},\ldots,B_{t_n})$ is Gaussian. This implies in particular that $\sum_{j=1}^n B_{t_j}$ is Gaussian. Since a Gaussian random variable is uniquely characterized by its mean and variance, it remains to calculate those two. As $\mathbb{E}B_t=0$ for any $t ... 0 Mathematica claims that there is an analytical solution for this: $$\sum_{x=0}^\infty \sum_{y=0}^\infty P(\lambda_x,x) P(\lambda_y,y)\frac{x}{x+y} =\frac{\lambda_x+e^{\lambda_x+\lambda_y}\lambda_y}{\lambda_x+\lambda_y}$$ although I have not yet managed to prove it. 0 Notice that $$X_t = \exp\{-\alpha t\}\left( X_0 + W_{\sigma^2( \exp\{2\alpha t\}-1)/2\alpha}\right) =\exp\{-\alpha t \} X_0 + \sigma \exp\{ - \alpha t \} \int_0^t \exp\{\alpha s \} d W_s.$$ Now we can simply note that the process$X_t$is a Gaussian process it being a linear combination of the Gaussian process$W_t$. Its mean and variance are now easily ... 1 Set$\tau_j := (\lfloor 2^j \rfloor+1)/2^j$. Then$\tau_j$is a discrete stopping time and$\tau_j \downarrow \tau$as$j \to \infty$. Fix$F \in \mathcal{F}_{\tau+}$. Then, by the right-continuity and the dominated convergence theorem, $$\mathbb{E}(e^{\imath \, \xi (Y_t-Y_s)} 1_F) = \lim_{j \to \infty} \mathbb{E}(e^{\imath \, \xi ... 2 Take any path P from (0,0) to (n,a-b). By adding a +1 step from (-1,-1) to (0,0) we now have a path P^{'} of length n+1 from (-1,-1) to (n,a-b). We now shift path P^{'} one unit up and one unit right, giving us a path P^{''} from (0,0) to (n+1,a-b+1). Now, for i\geq 1,\; P has all its X_i\geq 0\; iff \;P^{''} has all its ... 0 Okay, I had a look into the paper and the first thing to notice is that it is easy to proof that E(X)<=x. So let us first do that and see what happens: Define (just like in the paper) : X=\lim \inf_{p \rightarrow \infty} X_p^{\frac{1}{p}}, x= \lim \inf_{p \rightarrow \infty} E(X_p)^{\frac{1}{p}} and notice that Fatou's Lemma states that E(X) \leq ... 1 \Omega = \{a,b,c\}, three atoms each of probability 1/3. X(a)=X(b)=0, X(c)=1. \mathcal F = \big\{\Omega, \varnothing,\{a,b\}, \{c\}\big\} \mathcal G = \big\{\Omega,\varnothing,\{a\}, \{b,c\}\big\}. Compute E[E[X|\mathcal G]|\mathcal F]: it has value 1/4 at a and b, value 1/2 at c. Complute E[X|\mathcal F \cap \mathcal G]: it is ... 1 The Wikipedia article you cite provides everything you need to evaluate the analytical solution of the Ornstein–Uhlenbeck process. However, for a beginner, I agree that it may not be very clear. 1. Simulating SDEs You should first be familiar with how to simulate this process using the Euler–Maruyama method. The stochastic differential equation (SDE) ... 2 We should focus on simulating \int_0^t e^{\theta (s)}\, \mathrm{d}W_s. If f(s)=e^{\theta (s)} is continuously differentiable, you could use the fact that$$\sum_{i=1}^{[tn]}f(s_i^*)\Big(W(s_i)-W(s_{i-1})\Big)\to\int_0^tf(s)dW(s)$$in quadratic mean, for s_i^* \in [s_{i-1},s_i]. Note that you should use$$W(s_i)-W(s_{i-1}) \sim N(0,s_i-s_{i-1})$$and ... -1 For s\le n,$$\mathbb{E}(M_{n+1}\mid \mathcal{F}_s)=\mathbb{E}((X_n+Y_{n+1})^2-(n+1)\mid \mathcal{F}_s)\\=X_n^2+2X_n\mathbb{E}(Y_{n+1})+\mathbb{E}(Y_{n+1}^2)-(n+1)\\=X_n^2-n=M_n$$the steps have used the fact that Y_n is an i.i.d. \mathcal{N}(0,1) sequence. The possible mistake that you are making in your approach is the one rightly pointed out by @saz ... 0 This is much easier than your comments suggest:$$\mathbb{E}(M_t^2)=\mathbb{E}(\langle M \rangle_t)=t<\infty$$0 Normally, we define (\Omega, \sigma-\text{algebra}, P) as a probability space and here we define (X_p)_{p \geq 0} on that space with (X_p)\geq 0. We write then E(X_p) = \displaystyle\int_\Omega X_p dP. Now define Y_p:=X_p^{1/p}. So the usual statement of Fatou's Lemma becomes then E(\liminf Y_p)\le\liminf E(Y_p). Now note that the difference ... 0 This looks a lot like Fatou's lemma: http://en.wikipedia.org/wiki/Fatou%27s_lemma Maybe by defining f_p =X_p^(1/p) and letting \mu be the Lebesgue measure, you can already obtain the result. 0 I guess we are talking about continuous time: stochastic basis (\mathcal F_t)_{t \in (0,\infty)}, say. Write \sigma = \inf_n \tau_n. Can we verify that \{\sigma \le 5\} \in \mathcal F_5 ??? Well, suppose B \in \bigcap_{t>5}\mathcal F_t but B \not\in \mathcal F_5. Then for each n,$$ \tau_n(\omega) = \begin{cases} 5+\frac{1}{n},\qquad ... 1 (a) Well, the idea of your proof is correct, but there a several things which I would like to point out: There is some abuse of notation in it. So, for example, every probabilist will know what you mean by $$\{(t,\Omega); h(t) \in B\},$$ but (at least from my point of view) that's not a nice way to put it. Just write $$\{(t,\omega); \omega \in \Omega, h(t) ... 2 Let's look at \mathbb{E}[X_{n+1}^2-(n+1)|F_n] (take s=n). We need show it equals to X_n^2-n. Since X_{n+1}=X_n+ Y_{n+1} It should be$$ \mathbb{E}[X_{n}^2|F_n]+2\mathbb{E}[X_{n}Y_{n+1}|F_n]+\mathbb{E}[Y_{n+1}^2|F_n]-\mathbb{E}[n+1|F_n] $$, Since X_{n} is measurable w.r.t. F_n, and Y_{n+1} is independent of F_n (for which we can "take out ... 2 For s\le n,$$\mathbb{E}(M_{n+1}\mid \mathcal{F}_s)=\mathbb{E}((X_n+Y_{n+1})^2-(n+1)\mid \mathcal{F}_s)\\=X_n^2+2X_n\mathbb{E}(Y_{n+1})+\mathbb{E}(Y_{n+1}^2)-(n+1)\\=X_n^2-n=M_n$$the steps have used the fact that Y_n is an i.i.d. \mathcal{N}(0,1) sequence. The possible mistake that you are making in your approach is the one rightly pointed out by @saz ... 4 Let \mathbb{P}(X_{n+1}=X_{n}+1)=p and \mathbb{P}(X_{n+1}=X_{n}-1)=1-p=q. Since X_{2n}=0, we must have taken n upward and n downward steps, as \#\{steps up\}+\#\{steps down\}=2n and \#\{steps up\}-\#\{steps down\}=0. So the probability is C_{n}p^nq^n where C_n is the number of ways it can be done. We will prove that \displaystyle ... 1 Yes, they are independent; for a proof see e.g. N. Ikeda, S. Watanabe: Stochastic Differential Equations and Diffusion Processes, Theorem II.6.3. Note that a similar question (for the case of Poisson processes, not Poisson random measures) has been discussed on mathoverflow. 1 Hints: It suffices to consider the case that \mathbb{E}(X_t^2)<\infty for all t \geq 0. Show that \mathbb{E}(X_t^2) = t \mathbb{E}(X_1^2) for any t \in \mathbb{Q} \cap [0,\infty) using the additivity you have already shown. (Hint: write t= \frac{m}{n} for m,n \in \mathbb{N}.) Fix arbitrary t \in [0,\infty) and a sequence (t_n)_{n \in ... 1 If you believe that the system of two balls is ergodic (which is true for strictly convex scatterers) and mixing fast (which nobody is able to prove by the moment but is also probably true), (very) roughly speaking you can think of it as of two independent (dicrete) random walks. Do you know the answer for this simplified case of random walks? What about ... 0 Part(a) Condition on X_n = x:$$ E[X_{n+1}|X_n = x] = \sum yP(x, y) = Ax + B $$Taking expectation of both sides leads to the result. Part(b) Write EX_n + \frac{B}{A-1} = A\left(EX_{n-1} + \frac{B}{A-1}\right) given A\neq 1. The rest is easy. 0 Note that$$P(X_1=y)=\sum_{x\in\{0,1,\cdots,\ d\}}p_{xy}P(X_0=x)$$where p_{xy} is the transition probability from state x to state y of the Ehrenfest chain and is given by p_{x,x-1}=x/d,p_{x,x+1}=(d-x)/d and p_{xy}=0 else.So, ... 0 Everything is based around an application of the definition of conditional probability and the additive rule for probabilities of disjoint events, plus a few additional steps. The solution for Part (b), for instance, also includes the use of distribution and the assertion that all (C_i\cap D) are all mutually exclusive (disjoint), if all C_i are such. 2 Here is one argument that would work, assuming that one already knows the following facts: For any s>0, the process (B_{s+t}-B_s)_{t \geq 0} is a Brownian motion. The time inversion (tB_{\frac{1}{t}}) of a Brownian motion is a Brownian motion (defined to start at 0 at t=0). \limsup_{t \to \infty} B_t >0 (in fact, the lim sup is ... 1 Basically, the construction is divided into three steps: Step 1: Constructing a consistent set of finite dimensional distributions. Consider any starting point x\in\mathbb{R} and a set of times 0<t_1<t_2<\cdots<t_n<T. Define a measure on finite dimensional space \mathbb{R}^n as$$ \nu_{t_1,\cdots,t_n}(F_1,\cdots,F_n) ... 2 For part (i), we don't even need right-continuity or stationary increments from$X$. Notice that if$t>s$, then for any (bounded-measurable) function$f: E \to \mathbb{R}$, we can write $$E[f(X_t)|\mathcal{F}_s] = E[f(X_t-X_s+X_s)|\mathcal{F}_s] = E[g(X_t-X_s,X_s)|\mathcal{F}_s]$$ where$g(x,y) = f(x+y)$. Now, we know that$X_s$is ... 1 Let$(X_t)_{t \geq 0}$be a weak solution of the given SDE, i.e. there exists a Brownian motion$(B_t)_{t \geq 0}$such that $$X_t = X_0 + \int_0^t b(s,X_s) \, ds + \int_0^t \sigma(s,X_s) \, dB_s.$$ Using the elementary estimate $$(a+b+c)^p \leq 3^p (a^p+b^p+c^p), \qquad a,b,c \geq 0,$$ we find $$|X_t|^p \leq 3^p |X_0|^p + 3^p \left| \int_0^t b(s,X_s) ... 0 The inequalities, \;t-u\lt S_k\leq t\lt S_{k+1}\leq t+v,\; mean two things: Up to time t we need exactly k events but we cannot have it such that those k events occured up to time t-u and then none between t-u and t. Between times t and t+v we have at least 1 event. Therefore, \begin{eqnarray*} P(t-u\lt S_k\leq t &\cap& t\lt ... 0 Here are formulas for the expectation and variance for a Compound Poisson Process. Define random variables: \begin{eqnarray*} X &=& \text{Total amount spent in one 10-hour day} \\ N(t) &=& \text{Number of customers up to time t} \\ Y_i &=& \text{Amount spent by i^{th} customer (even if he buys nothing).} \\ \end{eqnarray*} Now ... 0 I agree with you that (2) is equivalent to \psi-irreducibility and that it is not clear how the authors derive (2) from the given assumptions. In the proof, the key point is that \varphi(\bar{A}(k))>0 for k sufficiently large where$$\bar{A}(k) := \left\{y; \sum_{n=1}^k P^n(y,A) >k^{-1} \right\}.$\$ Here is possibility to prove this ...

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