New answers tagged

1

I doubt that this identity holds true. Just consider $f:=1$, then the assertion reads $$B_t = 0,$$ which is obviously not correct. As @Did pointed out in a comment, Itô's formula shows that the identity $$\int_0^t f(s) \, dB_s = f(t) B_t - \int_0^t f'(s) B_s \, ds$$ holds for $f \in C^1(\mathbb{R})$.


1

If $\{B_t:t\in\mathbb R_+\}$ is a standard Brownian motion, then $$X_t := X_0\exp\left(\left(\mu-\frac12\sigma^2\right)t+\sigma B_t\right) $$ (where $X_0$, $\mu$, and $\sigma$ are constants) defines a geometric Brownian motion. It is clear that if $X_0=0$ then $X_t$ is identically zero, and similarly if $\sigma=0$ then $X_t = X_0 e^{\mu t}$ with probability ...


4

Actually, the answer to this lies in quite a bit more advanced topic called rough path theory (beware: PDF). A rough path is a way of "enhancing" a $\alpha$-Hölder continuous path with some extra information. A rough path is an ordered pair, $\textbf{X}=(X, \mathbb{X})$ where $X\colon [0,T]\to V$ where $V$ is some Banach space (typically $\Bbb{R}$) and a ...


0

So, not that I want to answer my own question, but I seem to have been able to work out the first two moments at least. Lemmas/Facts Let $\{B^i\}$ and $\{W^i\}$ be standard independent Brownian motions. $\mathbb{E}[B_t^i] = 0$, $\;\mathbb{E}[(B_t^i)^2] = t$, $\;(B_{t+k}^i-B_{t}^i)\sim\mathcal{N}(0,k)$. [1] If $\theta\sim\mathcal{N}(0,\sigma^2)$, $\;\...


0

Section 5.4 on p. 13 of this document, seems to confirm the notion that we can extend the class of allowable integrands to locally square-integrable processes (or at least when we restrict to predictable processes, instead of general progressively measurable processes). Not only that, but apparently we can also expand the class of integrators to include ...


3

As Did mentioned, $[W_1,W_2](t)=0$. Let $\{t_i\}_{i=1}^{n}$ be a partition of $[0,t]$. We want to consider $$A_n=\sum_{i=0}^{n-1}(\,W_1(t_{i+1})-W_1(t_{i})\,)(\,W_2(t_{i+1})-W_2(t_{i})\,)$$ Using independent of $W_1$ and $W_2$ , thus $\mathbb{E}[A_n]=0$. Since increment of Wiener process are independent, the variance of sum is sum of variance , and we have ...


0

For fixed $t$, the function $f:s\mapsto 1/\sqrt{t-s+1}$ is an element of $L^2[0,t]$, so the stochastic integral $X_t:=\int_0^t{1\over\sqrt{t-s+1}}\,dW(s)=\int_0^t f(s)\,dW(s)$ is defined and is normally distributed with mean $0$ and variance $$ \int_0^t{1\over t-s+1}\,ds=\log(t+1). $$ If $0<t<u$ then the covariance of $X_t$ and $X_u$ is $$ \int_0^t{1\...


0

I think I have figured out a way to prove that $v(x)$ is differentiable. The idea is to use the sort of techniques used in viscosity solution for optimal control. I will give a sketch of the argument below. My proof hardly makes use of the structure given above. In fact, I think, for a piecewise Lipschitz, bounded $\mu,\sigma$ and $f$, the same reasoning ...


0

Note that \begin{align*} X_t = X_0 e^{\alpha t} + \sigma\int_0^t e^{\alpha(t-s)} dW_s. \end{align*} Therefore \begin{align*} E(X_t) = X_0 e^{\alpha t}, \end{align*} and \begin{align*} E\left(X_t^2\right) &= E\left(\left(X_t-E(X_t)\right)^2\right) + \left(E\left(X_t\right) \right)^2\\ &=\sigma^2 \int_0^te^{2\alpha(t-s)} ds + X_0^2 e^{2\alpha t}\\ &...


-1

$$X_t = exp(\int_0^t {f(\omega,s)dB_s}-\frac 12 \int_0^t{f(\omega,s)^2 ds})$$ and $Y_t = \ln(X_t)$ then $$Y_t = \int_0^t {f(\omega,s)dB_s}-\frac 12 \int_0^t{f(\omega,s)^2 ds}$$ then we have $dY_t=f(\omega,t)dB_t - \frac 12 f^2(\omega,t)dt$ . so by Ito formula we have $$dX_t = d(e^{Y_t}) = e^{Y_t}dY_t + \frac 12 e^{Y_t}(dY_t)^2=X_t f(\omega,t)dB_t - \frac ...


2

Set $f(x)=-\frac{1}{x}-\tan^{-1}x$. we have $$f'(x)=\frac{1}{x^2}-\frac{1}{1+x^2}=\frac{1}{x^2+x^4}$$ and $$f''(x)=-\frac{2x+4x^3}{(x^2+x^4)^2}$$ By application of Ito's lemma we have $$f(B_t)=f(B_1)+\int_{1}^{t}f'(B_s)dB_s+\frac{1}{2}\int_{1}^{t}f''(B_s)ds$$ therefore $$-\left(\frac{1}{B_t}+\tan^{-1}(B_t)\right)=-\left(\frac{1}{B_1}+\tan^{-1}(B_1)\right)+\...


0

Note that \begin{align*} dJ_c(s) = dB_s -cJ_c(s) ds. \end{align*} Moreover, $d\langle J_c, J_c\rangle_t = dt$. Then, by Ito's lemma, \begin{align*} \int_0^1J_c(s) dJ_c(s) &= \int_0^1 d\left(J_c^2(s) -s \right)\\ &= J_c^2(1) - 1. \end{align*} Since \begin{align*} J_c(1) &= \int_0^1 e^{-c(1-s)}dB_s \end{align*} is normal with zero mean and the ...



Top 50 recent answers are included