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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 ... 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 $$\mathbb{E} \left( \sup_{t \in [0,K]} ... 2 At stationarity, one asks that$$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)} ... 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 ... 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 ... 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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$\Omega = \{a,b,c\}$, 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 ...

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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 $$\max_{s \in [0,t]} g(s) = \max_{s \in [0,t] \cap ... 1 Take any path P from (0,0) to (n,a-b). By adding a +1 step from (-1,-1) to (0,0) we now have a path P^{'} of length n+1 from (-1,-1) to (n,a-b). We now shift path P^{'} one unit up and one unit right, giving us a path P^{''} from (0,0) to (n+1,a-b+1). Now, for i\geq 1,\; P has all its X_i\geq 0\; iff \;P^{''} has all its ... 1 (a) Well, the idea of your proof is correct, but there a several things which I would like to point out: There is some abuse of notation in it. So, for example, every probabilist will know what you mean by$$\{(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) ...

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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, $$\liminf_p\left(\mathbb E[X_p]\right)^{1/p} +\frac 1j\leqslant ... 1 Let 0 \leq t_1<\ldots<t_n. Since (B_t)_{t \geq 0} is a Gaussian process, we know that (B_{t_1},\ldots,B_{t_n}) is Gaussian. This implies in particular that \sum_{j=1}^n B_{t_j} is Gaussian. Since a Gaussian random variable is uniquely characterized by its mean and variance, it remains to calculate those two. As \mathbb{E}B_t=0 for any t ... 1 \mathcal{F}_{t_i} is a member of a filtration which is a set of \sigma algebras indexed by time. To be measurable with respect to \mathcal{F}_{t_i} means to only depend on the information available up to and including time t_i. Conditioning probability on it means given the information available up to and including t_i, what is the probability. 1 The Wikipedia article you cite provides everything you need to evaluate the analytical solution of the Ornstein–Uhlenbeck process. However, for a beginner, I agree that it may not be very clear. 1. Simulating SDEs You should first be familiar with how to simulate this process using the Euler–Maruyama method. The stochastic differential equation (SDE) ... 1 Set \tau_j := (\lfloor 2^j \rfloor+1)/2^j. Then \tau_j is a discrete stopping time and \tau_j \downarrow \tau as j \to \infty. Fix F \in \mathcal{F}_{\tau+}. Then, by the right-continuity and the dominated convergence theorem,$$\mathbb{E}(e^{\imath \, \xi (Y_t-Y_s)} 1_F) = \lim_{j \to \infty} \mathbb{E}(e^{\imath \, \xi ...

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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 ...

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