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If you consider the process $Y_t=|X_t|,$ Ito's formula gives $$ dY_t=\sqrt{2}dB_t+\frac{dt}{Y_t}-l(\theta_t)Y_t\,dt, $$ where $\theta_t$ is an argument of $X_t$, $l(\theta_t)\in \{1,2,3\}$ is given b …

The easiest way to show this is to check that if $\hat{u}$ is a bounded solution to your boundary value problem, then $\hat{u}(t-s,B_{s\wedge\tau}+x)$ is a martingale, where $\tau$ is the minimum of t …

(I will work with the Brownian bridge instead). This is to explain why there is no such RKHS of the form $$ W=\left\{f(x)=\sum_{k=1}^\infty \hat{f}_k\sin(\pi k x):\sum_{k=1}^\infty w_k\hat{f}^2_k<\inf …

There are several very different cases here. The first case is when $\mathbb{E}(X_i)=\mathbb{E}(Y_i)=0$, which in your case means that the variables are symmetric. In this case, $P(\sum_{i=0}^{kn} X_i …

It is transient. Consider the process $Y_t=R_t-X_t$; it is a Brownian motion on the positive half-line with a constant negative drift $\nu-\mu$. The probability that it does not hit the origin up to t …

Let $\varphi$ be a smooth function such that $\varphi(x)=0$ for $x\leq 0$ and $\varphi(x)=1$ for $x\geq 1$, and $0\leq\varphi(x)\leq 1$ else. If $\tau_1$ and $\tau_2$ are is the first times $B_t$ hits …

Fix $A>0$; we will be proving that $X_t\stackrel{\mathcal D}= B_t$ on $[0;A].$ Let $T^{(m)}$ be an a.s. increasing sequence of stopping times such that $T^{(m)}\to\infty$ almost surely, and for each $ …

For any $\theta>0$, if you substitute $M(t)=e^{\theta t}N(t),$ then the equation becomes $$ N(t)=\int_0^\infty N(t-\tau)ae^{-\theta\tau}f(\tau)d\tau. $$ If $a>1$, then there is, by continuity, exactl …

A Gaussian process can be thought of as a "standard Gaussian" on its Cameron-Martin space $\mathcal{M}$. That is, given an orthonormal basis $\psi_1,\psi_2, \dots$ of the Cameron-Martin space, the pro …

Another construction, which does not use the structure of $\mathbb{R}$ and works for a sigma-finite measure $\nu$ on arbitrary measurable space $\Omega$, is as follows: let $\Omega=\bigcup_i E_i$ with …

What fails for other lattices is that there seems to be no parafermionic observable with properties as nice as for hexagonal lattice; specifically, there is no analog of Lemma 1. In the definition of …

For $n\in\mathbb{Z}_{\geq 0}$ and $0\leq i< 2^n$, denote $$ X_{n,i}=\sup_{t\in[i2^{-n}, (i+1)2^{-n}]}{|W_t-W_{i2^n}|}. $$ Let $n$ be such that $2^{-{n}}<|\Delta t|\leq 2^{-{n+1}}$. Then, $$\sup_{s,t\i …

Using the representation in terms of i. i. d. Gaussians $\xi_1,\xi_2,\dots,$ $$ B_t=\sqrt{2}\sum_{n=1}^\infty (-1)^{n+1}\xi_n\frac{\sin \pi \left(n-\frac12\right)t}{\pi \left(n-\frac12\right)}, $$ we …

How about $\min\{n\cdot \mathrm{dist}(t,X),1\}$, where $X$ is your favorite point process, say, the Poisson process? For an example with Markov property, consider a continuous time Markov chain with …

The rate is exponential. First, let me show that $|X(t)|$ (which is a reflected BM with constant drift towards the origin) has exponential rate of convergence. I do it by a coupling argument: you run …