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## Hot answers tagged brownian-motion

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Using that $$M_t := \exp(i \xi B_t + \frac{t}{2} \xi^2)$$ is a martingale, one can show that for any $\lambda>0$ there exists $\beta>0$ such that $$\mathbb{E}e^{\beta T_{\lambda}}<\infty.$$ By Markov's inequality, this implies in particular that $$\mathbb{P}(T_{\lambda} \geq t) \leq e^{-\beta t} \underbrace{\mathbb{E}(e^{\beta ... 4 Typically a Brownian motion is defined to have continuous sample paths. If you take some countable, dense subset S\subset [0,T], you then have$$\sup_{t\in S} B_t = \sup_{t\in [0,T]} B_t$$by continuity of B_t. 3 You need to use the fact that B_t depends continuously on t. That shows that the sup is the same as the sup restricted to rational t. 3 Since the augmented filtration is right-continuous, we may assume that (M_t)_{t \geq 0} has càdlàg sample paths. Since (M_t)_{t \geq 0} is a local martingale, there is a sequence of stopping times (\tau_k) such that \tau_k \uparrow \infty and (M_{t \wedge \tau_k})_{t \geq 0} is a martingale. Set$$Y := M_{T \wedge \tau_k}$$for fixed k \in ... 3 To a large extent, it depends on whether you are interested in learning important properties and facts or whether you want to do proofs and eventually original research, whether you are interested primarily in theory or in applications. Many of the properties and much of what is done in applications is accessible without measure theory and many important ... 2 In 1971, John Hawkes established the exact modulus of continuity of L_t:$$ \lim_{\epsilon\to 0}\sup_{0\le t\le 1,0<s<\epsilon}{L_{t+s}-L_t\over \sqrt{s\cdot\log(1/s)}}=c, $$almost surely, for a certain explicitly given constant c. (See "A lower lipschitz condition for the stable subordinator" in Zeitschrit f. Warsch., vol. 17 (1971) 23-32.) In ... 2 Since B_2-B_1 is independent of B_1,$$ P(B_2>0\mid B_1=x) = P(B_2-B_1>-x\mid B_1=x) = P(B_2-B_1>-x) = \Phi(x). $$Therefore,$$ P(B_2>0,B_1>0) = \int_0^{+\infty} \Phi(x) \varphi(x)dx = \frac12\Phi^2(x)\Big|_{x=0}^{+\infty} = \frac38, $$whence$$ P(B_2>0\mid B_1>0) =\frac34. $$2 A Process X_t is called standard Brownian motion iff: X_0 = 0. X_t is almost surely continuous. All finite-dimensional distributions of X are joint Gaussian. \operatorname{cov}(X_s, X_t) = s \wedge t and E[X_s] = 0 for all s, t > 0. The processes B and C inherit the properties (1) and (3) directly from A [if you additionally define ... 2 By Ito's formula:$$W_t^2=2 \int_0^t W_s dW_s + \int_0^t ds = 2 \int_0^t W_s dW_s + t.Now the Ito integral (of a square integrable, independent-of-future function) is a martingale, so W_t^2-t is a martingale. This procedure works for a wide class of diffusion processes. 2 For many stochastic processes (but certainly not all) you can find such a function X_t. It turns out that it will be a finite variation process, called the bracket process, or quadratic variation process. In this particular case, one can use the markov property in place of Ito's formula: \begin{align} \Bbb E[ W_t^2 | \mathcal{F}_s] &= \Bbb E\left[ ... 2 If A,B>0\int_{0}^{+\infty}\exp\left(-A^2 x^2-\frac{B^2}{x^2}\right)\,dx =\sqrt{\frac{B}{A}}\int_{0}^{+\infty}\exp\left(-ABx^2-\frac{AB}{x^2}\right)\,dx\tag{1}$$hence it is enough to compute:$$ I(C)=\int_{0}^{+\infty}\exp\left(-C^2 x^2-\frac{C^2}{x^2}\right)\,dx. \tag{2}$$By splitting the integration range as (0,1)\cup(1,+\infty) and setting ... 1 This is not a Brownian motion indeed, as noted above. The pdf of Z_t=A_tB_t is$$ f(z)=\int\frac{dadb}{2\pi t}\delta(z-ab)e^{-\frac12\frac{a^2+b^2}{t}} $$where the integral is over the whole plane. You can integrate this directly to obtain the Bessel functions of second kind of order zero and with imaginary argument (from Maple)$$ ...

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Not at all. In fact, the Wiener measure of a singleton is just zero. The Wiener measure of a set is just the probability that a Wiener process trajectory is a member of that set. Thus for instance the set of functions which are differentiable at some point in $\mathbb{R}_+$ has Wiener measure zero. Similarly the set of monotone functions has Wiener measure ...

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My bet would be that the definition is in fact through the quadratic covariation: $\langle w_t,\bar w_t\rangle = \rho t$ for all $t$. Then, as a result from Ito lemma you get: $$\mathrm dw_t\bar w_t = w_t\mathrm d\bar w_t + \bar w_t\mathrm dw_t + \mathrm d\langle w_t,\bar w_t\rangle$$ which gives you that $\Bbb Ew_t\bar w_t = \rho t$. I am not sure ...

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Since $|X_t|^2=\sum_{i=1}^d (B_t^{(i)})^2$ where $B_t^{(i)}$ are a family of iid Brownian motions, and since $(B_t^{(i)})^2-t$ is a martingale, it follows also that $|X_t|^2-td$ is also a martingale. Thus $\langle X_t\rangle=td$.

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