<ruby id="d9npn"></ruby>

      <sub id="d9npn"><progress id="d9npn"></progress></sub>

      <nobr id="d9npn"></nobr>

      <rp id="d9npn"><big id="d9npn"><th id="d9npn"></th></big></rp>

      <th id="d9npn"><meter id="d9npn"></meter></th>

      New answers tagged

      1

      For bounded and continuous $f$, the map $\omega\mapsto\int_0^T f(\omega(r))\,dr$, from $D([0,T]; \Bbb R^d)$ to $\Bbb R$ is continuous and bounded. See, for example, https://math.stackexchange.com/questions/271738/is-integration-a-continuous-functional-on-the-skorohod-space


      0

      Since $(X_t(\omega))_[0,T]$ is a cadlag process it is progressively measurable, in particular $(\omega,t) \to X_t(\omega)$ is measurable. Let $\tilde f(y,t) := f(y)$, then $\tilde f$ is bounded and continuous again. W.l.o.g. assume $T = 1$. Then $\mu_n \otimes \lambda \to \mu \otimes \lambda$ weakly. It immediately follows that $$\lim_{n \to \infty} \int \...


      0

      (i,ii) Not quite. A trivial counterexample is $\ell(x, y) \equiv 0$. A simple non-trivial counterexample would be something like $$\ell(x,y) = \begin{cases} 1 / (x^{1/2} y^{3/2}) & \text{if $0 < x \leqslant 1$, $y < x$} \\ 0 & \text{if $0 < x \leqslant 1$, $y \geqslant x$} \\ e^{-y} & \text{otherwise.} \end{cases} $$ Then $L_t$ started ...


      6

      Suppose the sidelengths of your hexagon are $k, \ell, m$. A standard way of looking at this is to turn the picture by 30 degrees to the left and to view the yellow / cyan tiles as $m$ functions $f_i \colon \{1,\ldots,k+\ell\}$ into $\mathbb{Z}$ with the following constraints: $f_i(1) = i$, $f_i(k+\ell) = i-k+\ell$, $f_{i+1}(n) \ge f_i(n)+1$ and $f_i(n+1)-f_i(...


      1

      Short answer: If the RKHS $\mathcal{H}_C$ associated to the kernel of your GP is infinite dimensional, then sample paths are in $\mathcal{H}_C$ with probability 0. Nevertheless, one can show that sample paths are contained in a "larger" RKHS, and Steinwart showed how to construct such an RKHS as a power of the original RKHS. Sections 4.2 and 4.3 of ...


      0

      Q1: I have not seen such a process, but I can easily imagine one: Let $Z_t$ be any non-negative Lévy process with positive drift, and let $X_t = \int_{t-\epsilon}^t Z_s ds$. Q2: No, unless we know something more about $X_t$. Imagine the process $X_t$ as above, with $Z_t$ having frequent extremely large jumps. As $a \to \infty$, the random variable $T_{a+b} -...


      1

      Q1: Have you seen an analogous process to $X _t$ anywhere else? $$$$ I have not. Are you sure they exist ? If $\epsilon = 0$ the increments must be from positive, infinitely divisible distribution, none of which has moments of all orders. The strictly monotone condition also bothers me. It seems to only work with b = 1. If b > 1 then $X_t > t^b$ ...


      0

      (This is not a complete answer, rather an attempt to describe the distribution of the integral.) The integral $\int_\rho^t W(t) dW(t)$ is simply $\tfrac{1}{2} ((W(1))^2 - 1) - \tfrac{1}{2} ((W(\rho))^2 - \rho)$. Let us handle the other term: $$I_\rho = \int_\rho^1 W(t-\rho) dW(t).$$ Let us approximate $W(t-\rho)$ by an elementary process $X_n(t)$, which is ...


      1

      A useful inequality is \begin{equation} \tag{1} \frac{1}{\sqrt{2 \pi}} \frac{x}{x^2+1} \mathrm{e}^{-x^2/2} \le \frac{1}{\sqrt{2\pi}} \int_x^\infty \mathrm{e}^{-y^2/2} \, \mathrm{d} y \le \frac{1}{\sqrt{2 \pi}} \frac{1}{x} \mathrm{e}^{-x^2/2} \end{equation} for $x >0$, which can be shown by elementary arguments (see below). For example, one can prove the ...


      1

      Let $\tau$ be a Markov time, and define the usual $\sigma$-algebras: $$\mathcal F^{<\tau} = \sigma\{X_t^{-1}(E) \cap \{t < \tau\} : t \geq 0, \, E \text{ — Borel}\}$$ and $$\begin{aligned} \mathcal F_{\geqslant\tau} & = \sigma\{X_{\tau + t}^{-1}(E) : t \geq 0, \, E \text{ — Borel}\} \\ & = \sigma\{X_t^{-1}(E) \cap \{t \geqslant \tau\} : t \geq ...


      0

      This question is not research level and would be better suited on MathStackExchange. Positivity here just means $f\ge 0$ $\Rightarrow$ $Tf\ge 0$ (where $f\ge 0$ is defined as $f(x)\ge 0$ for each $x\in X$). The example is verified using Proposition 2.2: If $f(\eta)=\min f(X)$ then $g=f-f(\eta)1\ge 0$ and since $T(1)=1$ we get $$ (T-I)(f)(\eta)=T(f)(\eta)-...


      1

      Strong Markov property says that under the condition that $\forall i, y(\tau_i)=y_i$, on each $[\tau_i,\tau_{i-1}]$ the process $X_t$ is just a brownian motion starting at $(i,y_i)$ and ending at $(i-1,y_{i-1})$. This evolution only depend on the starting point and the ending point and not at all what append before $\tau_i$ or after $\tau_{i-1}$. Then $X_t$ ...


      2

      Let $X_i:=X^i$, $S_k:=\sum_1^n X_i$, $T_k:=S_k/\sqrt k$, $|\cdot|:=\|\cdot\|_2$, $n\in\{1,2,\dots\}$, and $m:=\lceil\log_2 n\rceil$, so that $2^m\ge n$. Then \begin{equation} \max_{k=1}^n|T_k|\le\max_{k=1}^{2^m}|T_k| =\max_{j=0}^m\max_{2^{j-1}<k\le 2^j}|T_k| \le\max_{j=0}^m 2^{-(j-1)/2}\max_{2^{j-1}<k \le 2^j}|S_k|. \end{equation} Hence, for any ...


      Top 50 recent answers are included

      特码生肖图
      <ruby id="d9npn"></ruby>

          <sub id="d9npn"><progress id="d9npn"></progress></sub>

          <nobr id="d9npn"></nobr>

          <rp id="d9npn"><big id="d9npn"><th id="d9npn"></th></big></rp>

          <th id="d9npn"><meter id="d9npn"></meter></th>

          <ruby id="d9npn"></ruby>

              <sub id="d9npn"><progress id="d9npn"></progress></sub>

              <nobr id="d9npn"></nobr>

              <rp id="d9npn"><big id="d9npn"><th id="d9npn"></th></big></rp>

              <th id="d9npn"><meter id="d9npn"></meter></th>

              彩票 极速赛车 我乐时时彩计划网 郑州红灯区 拉萨酒店小姐服务 pk10计划软件手机版苹果 三分pk10玩法 欧美人体艺术 北京pk10计划在线计划 90后漂亮女优的翻译是 计划软件免费版