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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

To use Ito's formula to calculate a stochastic integral, you want to find an Ito process whose differential is the integrand. When calculating $\int_0^t B_s dB_s$, you initially guess (by intuition from regular calculus) that the process might be $\frac{1}{2} B_t^2$. So you use Ito's formula to calculate its differential and get $d \left ( \frac{1}{2} B_t^2 ...


3

As @Did already pointed out, it is much easier to show that $$a W_t - \frac{1}{2} a^2 t \to - \infty \qquad \text{almost surely as $t \to \infty$.}$$ Recall that the process $$B_t := \begin{cases} t W_{\frac{1}{t}}, & t>0, \\ 0, & t=0 \end{cases}$$ defines a Brownian motion; in particular $$\lim_{t \to 0} t W_{\frac{1}{t}} = \lim_{t \to 0} ...


3

Assume that $g\colon \mathbf R_+\to\mathbf R_+^*$ is a function such that $$\lim_{t\to\infty}\frac{g(t)}{\sqrt{t\log\log t} } =0.$$ Then we have $$\limsup_{t\to +\infty }\frac{B(t) }{g(t)}=+\infty.$$ Indeed, since $\limsup_{t\to\infty} \frac{B(t)}{\sqrt{2t\log\log t}} = 1$, we have $$\lim_{R\to \infty}\sup_{t\geqslant R}\frac{B(t)}{\sqrt{2t\log\log ...


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

For any $0 = t_0 < \ldots < t_n$, we know that $$B_{t_n}-B_{t_{n-1}}, B_{t_{n-1}}-B_{t_{n-2}},\dots,B_{t_1}-\underbrace{B_{t_0}}_{0}$$ are independent random variables. Choosing $t_0 = 0$, $t_1 = 1$, $t_2 = 2$, $t_3 = 3$, $t_4 = \pi$, $t_5 = 4$, we get that $$B_4-B_{\pi},B_{\pi}-B_3, B_3-B_2,B_2-B_1,B_1 \tag{1}$$ are independent. This implies in ...


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

Levy's characterisation and the use of It$ô$'s lemma give the same answer. I think the independence argument in your approach is not quite enough to conclude that $b$ must be zero, unfortunately (please see my comment below). Levy's Characterisation: The aim here is to show that there is a choice of $a$ and $b$ such that ...


2

Let $A$ be the event that $\int_0^\infty e^{B_t} \,dt < +\infty$. By the Kolomorgov 0-1 law, $P(A) = 0$ or $1$. Now let $B$ be the event that $\int_0^\infty e^{-B_t}\,dt < +\infty$. By symmetry, $P(A) = P(B)$. Moreover, on $A \cap B$ we have $$\int_0^\infty (e^{B_t} + e^{-B_t}) \,dt < +\infty$$ which is absurd since $e^x + e^{-x} \ge 1$ for all ...


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

Denote by $$p(t,x) := \frac{1}{\sqrt{2\pi t}} \exp \left(- \frac{x^2}{2t} \right), \qquad x \in \mathbb{R},$$ the density of the normal distribution with mean $0$ and variance $t$. As you already noted, this function solves the heat kernel equation, i.e. $$\frac{\partial}{\partial t} p(t,x) = \frac{1}{2} \frac{\partial^2}{\partial x^2} p(t,x).$$ For $f ...


2

Since $\cosh(x) \ge 1$ for all real $x$ you have $\frac{e^{t/2}}{\cosh X_t} \le e^{t/2}$. So on any finite time interval you have a bounded local martingale, which is a martingale. Alternatively, note that $\frac{\sinh(x)}{(\cosh(x))^2} \le \frac{1}{2}$ for all $x$. So for any $T$, $$\int_0^T E\left[\left|-\frac{e^{t/2} \sinh ...


2

Hint: Let $X$ be a (non-trivial) random variable such that $X$ is independent from $\mathcal{F}^W_{\infty} := \sigma(\mathcal{F}_t^W; t \geq 0)$. Consider the filtration defined by $$\mathcal{F}_t := \sigma(\mathcal{F}_t^W, X).$$


2

Obviously, $$\ln X = \int_0^T \sigma(s) \, dW_s \qquad \text{for} \quad \sigma(s) := 1_{[0,T/2]}(s) + 1_{[0,T]}(s). \tag{1}$$ Define an Itô process $(Y_t)_{t \geq 0}$ by $$Y_t := \int_0^t \sigma(s) \, dW_s - \frac{1}{2} \int_0^t \sigma(s)^2 \, ds. \tag{2}$$ Then, by Itô's formula, $$\begin{align*} \exp(Y_t)-1 &= \int_0^t \exp(Y_s) \, dY_s + ...


1

Step 1: The assumption holds for all $f \in C_b^2$, i.e. functions which are twice differentiable and have bounded continuous derivatives. To this end, pick a function $\chi \in C^2_b$ such that $0 \leq \chi \leq 1$, $\chi(x) = 1$ for $x \in (-1,1)$ and $\chi(x)=0$ for all $|x|>2$. Then the function $$f_n(x) := f(x) \cdot \chi \left( \frac{x}{n} ...


1

Part (i): Use $$M_t := \frac{W_{t \wedge \tau(-b)} + b}{b}.$$ How to come up with this choice? Well, we are interested in $$\sup_{0 \leq t \leq \tau(-b)} W_t = \sup_{t \geq 0} W_{t \wedge \tau(-b)}.$$ We know that $(W_{t \wedge \tau(-b)})_{t \geq 0}$ is a martingale and that $W_{t \wedge \tau(-b)} \to -b$ as $t \to \infty$. This means that, in order to get ...


1

Consider the application $$ F(\xi) : F(\xi)(T)=\xi_0+\int_0^Tf(\xi(t))dt+W(T) $$ $$ \left[ F(\xi) - F(\zeta)\right] (T) = \int_0^T \left[f(\xi(t)) - f(\zeta(t))\right] dt $$ under the same assumptions as in the Cauchy Lipschitz theorem, you can prove that (for the right space and right norm) $F$ is Lipschitz with constant $<1$ (the proof is exactly ...


1

For notational convenience, define $f_{n,t}:=\cos(n-1/2)\pi t$ and choose any finite set of $k$-distinct natural numbers $n_1, \ldots, n_k$. Then, the Itô integrals $\{\int_0^1 f_{n_1,t}~\text dW_t,\ \ldots,\ \int_0^1 f_{n_k,t}$$\text dW_t\}$ are jointly normal. Consequently, these intergrals are independent if any pair of integrals has zero correlation. So, ...


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

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

first you can get $b$ simply be computing the time zero expectation: $$\mathbb{E}(e^{5B_t})= e^{0.5 \times 25 \times t}.$$ So $b = 12.5.$ With this value, your final equation $$\exp\{\frac{25(t-s)}{2}+5B_s-bt\} = \exp\{5B_s-bs\}$$ holds and we are done.


1

Yeah, that would be a way to do it. May I suggest the following way of ordering the terms in the covariance calculation? $$\text{cov}(X,Y) = \text{cov}(B_1+(B_3-B_2),B_1-(B_3-B_2))$$ $$=\text{cov}(B_1,B_1)-\text{cov}(B_1,B_3-B_2)+\text{cov}(B_3-B_2,B_1)-\text{cov}(B_3-B_2,B_3-B_2)$$ $$=\text{var}(B_1)-\text{var}(B_3-B_2)$$ $$=1-1=0.$$ Underway I use that ...


1

First consider the case $n=2$. By definition, the joint distribution of $(X_{t_1},X_{t_2})$ is given by $$p_{t_1}(0,x_1) p_{t_2-t_1}(x_1,x_2) \, dx_1 \, dx_2 = q_{t_1}(x_1) q_{t_2-t_1}(x_2-x_1) \, dx_1 \, dx_2$$ where $$q_t(x) := \frac{1}{\sqrt{2\pi t}} \exp \left(- \frac{x^2}{2t} \right), \qquad x \in \mathbb{R}.$$ Consequently, $$\begin{align*} ...


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

I suppose I'll relieve my own worries then! Of course most credits should go to @Saz, who was tremendously helpful in the comments. The problem here was basically that of modes of convergence, I think. While $X_t = t^H X_1$ holds, it holds only in distribution. That is the reason that when we taking expectations of these variables on both sides, we get ...


1

(Not a proof, but here's some intuition at least) Standard Brownian motion has mean 0 and variance t for $0 \le t \lt \infty$ Thus, on average the integral becomes... $$\int_0^{\infty} e^0 \ ds=\int_0^{\infty} 1 \ ds$$ Which clearly diverges. What you should aim for proving is that Brownian motion crosses between negative and positive values in such a ...


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

You have that $X = \frac{B_1 - B_3 + B_2}{\sqrt{2}}$. Since $B$ is a Brownian motion, you know that $B_3-B_2$ and $B_1$ are independent. Thus $$\text{var}(X) = \frac{1}{2}\text{var}(B_1-B_3+B_2)=\frac{1}{2}\text{var}(B_1-(B_3-B_2))=\frac{1}{2}(\text{var}(B_1)+\text{var}(B_3-B_2)).$$ By definition, the increments $B_s-B_t$ are normally distributed with ...


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 ...



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