Tag Info

Hot answers tagged

6

From the stochastic integral analogue of integration by part, which is a corollary of Ito's Lemma, $$d(W_t\ln t) = \frac{W_t}{t}dt+(\ln t)dW_t,$$ so $$I:=\int_0^t \frac{W_s}{s}ds=W_t\ln t-\int_0^t (\ln s)dW_s=\int_0^t \ln\frac{t}{s}dW_s$$ as $W_t\ln t\to 0$ in probability, as $t\to 0^+$. Since the random variable $I$ is a linear combination of independent ...


3

I think both of you and your book are right! The point here is $d(S_t)d(e^{-rt})=0$. The reason is $dt\cdot dW=0$ and $dt\cdot dt=0$.


3

Edit: This answer shows the identity $$\mathbb{E} \left( \left| \sum_{i=0}^{n-1} \int_{t_i^n}^{t_{i+1}^n} (W(t_i^n)-W(t)) \, dt \right|^2 \right) = \mathbb{E} \left( \sum_{i=0}^{n-1} \left[\int_{t_i^n}^{t_{i+1}^n} (W(t_i^n)-W(t)) \, dt\right]^2 \right). \tag{1}$$ Fix $i<j$. Then, by the tower property, $$\begin{align*} ...


2

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 ...


2

We have that $$ V = W_t^2 - t$$ hence (written with $t$ and $x$, where we plug in $S_t = W_t$ for $x$), we have $$ V(x,t) = x^2 - t$$ Now \begin{align*} \frac{\partial V}{\partial t} &= -1\\ \frac{\partial V}{\partial x} &= 2x\\ \frac{\partial^2 V}{\partial x^2} &= 2 \end{align*} Hence (note that $S_t = W_t$) \begin{align*} ...


2

It holds that $$\mathbb{E}^x(F) = \int F(w) \, d\mathbb{P}_x(w) = \int F(x+w) \, d\mathbb{P}(w) \tag{1}$$ for any measurable function $F: (C[0,\infty),\mathcal{B}(C[0,\infty)) \to [0,\infty)$. This follows from the fact that $(1)$ holds for simple functions and we can extend the equality to all measurable non-negative functions (by the monotone convergence ...


2

Since $$\exp \left( - \frac{a^2}{2t} \right) \uparrow 1 \qquad \text{as $t \to \infty$},$$ there exists $R>0$ such that $$\exp \left( - \frac{a^2}{2t} \right) \geq \frac{1}{2}$$ for all $t \geq R$. Consequently, $$\begin{align*} \mathbb{E}(\tau_a) &\geq \int_R^{\infty} \frac{at}{\sqrt{2\pi t^3}} \exp \left(- \frac{a^2}{2t} \right) \, dt \\ ...


2

This is a non-rigorous answer. If $t \mapsto R_t$ is a differentiable function from $\mathbb R$ to $SO(n)$, then its derivative satisfies $$ \frac{dR_t}{dt} = R_t A_t ,$$ where $A_t$ is an anti-symmetric matrix. I could rewrite this non-rigorously as $$ dR_t = R_t A_t \, dt = R_t (\exp(A_t \, dt) - I) ,$$ where $\exp(A)$ denotes the matrix exponential of a ...


2

My writeup of this in a previous project, with slightly different notation, looks like this: $$u(x) = \Bbb E[f(x+W_{\tau_{\partial D},x})] \\ = \Bbb E[\Bbb E[f(x+W_{\tau_{\partial D},x})|W_{\tau_r}]] \\ = \Bbb E[u(x+W_{\tau_r})] \\ = \int_{\partial B(x,r)} u(y) dS(y).$$ In the first step, we use the tower property, inserting a conditional expectation in ...


2

In order to prove measurability of $\Lambda(m)$, it suffices to show that the processes $$(t,\omega) \mapsto \max_{(t-1/m)^+ \leq s \leq t} W_s(\omega) \qquad \quad (t,\omega) \mapsto \min_{t \leq s \leq t+1/m} W_s(\omega)$$ are progressively measurable. Since any continuous (adapted) process is progressively measurable, we are done if we can show that ...


2

Ito's formula: http://en.wikipedia.org/wiki/Itō's_lemma $Y = X^2$ 1.) find $dY$ using the chain rule to order $(dX)^2$ with a taylor expansion and substitute for $dX$ $$dY=\frac {\partial Y}{\partial t}dt + \frac {\partial Y}{\partial X}dX + \frac 12 \frac {\partial^2 Y}{\partial X^2}(dX)^2$$ $$=2X(a(t)Xdt+2dW) + \frac 12*2(a(t)Xdt+2dW)^2 = ...


2

Ito's formula says that given $$ dX_t = \mu_t dt + \sigma_t dW_t $$ and a $C^2$-function $f \colon \mathbf R^2 \to \mathbf R$, we have for $Y_t = f(t, X_t)$ that $$ dY_t = \left(\frac{\partial f}{\partial t}(t, X_t) + \mu_t \frac{\partial f}{\partial x}(t, X_t) + \frac{\sigma_t^2}{2}\frac{\partial^2 f}{\partial x^2}(t, X_t)\right)\, dt + \sigma_t ...


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

By definition, $$U_t(X)-U_t(Y) = (X_0-Y_0) + \int_0^t (b(s,X_s)-b(s,Y_s)) \, ds + \int_0^t (\sigma(s,X_s)-\sigma(s,Y_s)) \, dW_s$$ and therefore $$\begin{align*}&|U_t(X)-U_t(Y)|^2 \leq \\ & \color{red}{3 |X_0-Y_0|^2} + 3 \left( \int_0^t |b(s,X_s)-b(s,Y_s)|ds\right)^2 + 3\left(\int_0^t |\sigma(s,X_s)-\sigma(s,Y_s)|dW_s\right)^2. \end{align*}$$ The ...


1

I would do this. Show $$\begin{align}P_t (x,A) &{:=} \int_A g_t (x,y)dy \\ P_0 (x,A) &{:=} \delta_x (A), \end{align}$$ where $g_t$ density of the folded normal distribution, is a Markov kernel (Chapmann K. Equation, etc.). Then you know there exists a unique Markov process. That's the reflected Brownian motion. Have someone any comments? Is that ...


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 ...


1

Heuritically, $d(S_t)d(e^{-rt})=-re^{-rt}dt(dS_t)\sim O(dt)^{3/2}$. Thus, it is not considered in the SDE for $S_te^{-rt}$.


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) ...


1

Hints: (This answer does not use Borel Cantelli lemma; instead it is based on basic martingale techniques.) Show that for any fixed $\xi>0$, the process $$M_t^{\xi} := \exp \left( \xi B_t - \frac{1}{2} \xi^2 t \right), \qquad t \geq 0,$$ defines a martingale. Fix $T>0$. For $b>0$ we define a stopping time by $\tau_b := \inf\{t>0; B_t \geq b\}$. ...


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

Obviously, $$|M_t| \leq |B_t|^3 + |\alpha| \int_0^t |B_s| \, ds.$$ Since $(B_t)_{t \geq 0}$ is a Brownian motion, we have in particular $B_t \sim N(0,t)$ and $B_t \sim \sqrt{t} B_1$. Consequently, $$\mathbb{E}(|B_t|^3) <\infty \qquad \text{and} \qquad \mathbb{E}(|B_s|) = \sqrt{s} \mathbb{E}(|B_1|).$$ Using Tonelli's theorem, we get ...


1

Your calculations are correct, but the claim is not. Instead of $\tau_a \stackrel{d}{=} \sqrt{a} \tau_1$ it should read $\tau_a \stackrel{d}{=} a^2 \tau_1$. References: Revuz/Yor: Continuous martingales and Brownian motion, Proposition III.3.10 Schilling/Partzsch: Brownian motion - An introduction to stochastic processes, Problem 6.6


1

the answer to this is quite simple if you look carefully to the definition of H, you should notice that $h_{i-1}$ is an $\mathcal{F}_{t_{i-1}}$-measurable random variable. From this and the following elementary properties on conditional expectation we get the result : If X is $\mathcal{F}$-measurable and for any bounded random variable $Y$, we have a.s. : ...


1

Integrating any locally bounded predictable process (in particular a continuous adapted process) with respect to a local martingale (resp locally square integrable martingale) results in a new process which is a local martingale (resp locally square integrable). A proof of this can be found in Jacod & Shirjaev (Chapter 1; Theorem 4.31). (It basically ...


1

For any $0 \leq s_0 < \ldots < s_m \leq T =T + t_0 < \ldots < T+t_n$ the random variables $$\underbrace{B_{t_n+T}-B_{t_{n-1}+T}}_{B'_{t_n}-B'_{t_{n-1}}},\ldots,\underbrace{B_{t_1+T}-B_{t_{0}+T}}_{B'_{t_1}-B'_{t_{0}}}, B_{s_m}-B_{s_{m-1}},\ldots,B_{s_1}-B_{s_0} \tag{1}$$ are independent. Recall that for any two measurable mappings $f: ...


1

Define a martingale by $$M_t := \begin{cases} \mathbb{E}(\phi(W_T) \mid \mathcal{F}_t), & t \leq T, \\ \phi(W_T), & t>T. \end{cases}$$ Then, by the martingale representation theorem, there exists a representation of the form $$M_t = c+ \int_0^t \beta_s \, dW_s.$$ For $t=T$, this proves the claim. Remark: One possibility to prove the mentioned ...


1

First of all, note that the strong Markov property of Brownian motion does not state that $(B_t)_{t \in [0,\tau]}$ and $(B_{t+\tau})_{t \geq 0}$ are independent, but that $(B_t)_{t \in [0,\tau]}$ and $(B_{t+\tau}-B_{\tau})_{t \geq 0}$ are independent. This independence yields in fact $$\mathbb{E}(1_A \cdot 1_B) = \mathbb{E}(1_A) \mathbb{E}(1_B)$$ for any ...


1

First consider the case that $\sigma$ is a simple function, i.e. $$\sigma(s) =\sum_{j=1}^n 1_{[t_{j-1},t_j)} \xi_j \tag{1}$$ where $(\xi_j)_{j=1,\ldots,n}$ are random variables independent from the Brownian motion $W$. Without loss of generality, we may assume that $t_k = t$ for some $k \in \{1,\ldots,n\}$ (otherwise we add the point to the partition). ...


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.



Only top voted, non community-wiki answers of a minimum length are eligible