New answers tagged

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For Brownian motion it is true that $B=C$. This means that if $f\in B$ there exists $g\in C$ such that $f(\omega,t)=g(\omega,t)$ for $\Bbb R\times\mu$-a.e. $(\omega,t)\in\Omega\times[0,T]$. This does not mean that $t\mapsto f(\cdot,t)$ is a predictable process. The matter is discussed in some detail in Chapter 3 of Introduction to Stochastic Integration by ...


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Here is the idea (you need to fill in some details) $R_j = o(|\Delta t_j|^2 + |\Delta X_j|^2)$ means $$R_j \le r(|\Delta t_j|^2 + |\Delta X_j|^2)(|\Delta t_j|^2 + |\Delta X_j|^2)$$, where $r(y)\to 0 \ as \ y\to 0$ (using the condition that $g$ is twice continuously differentiable, you can show such $r$ exists and does not depends on $j$). Now $$|\sum_j R_j| ...


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@webbster: I'm going to flip this around and deal instead with right-limited functions. So let $h:[0,1]\to\Bbb R$ have right limits ate each point of $[0,1)$. Extend $h$ to all of $[0,\infty)$ by setting $h(t)=h(1)$ for $t\ge 1$. It suffices to show that for a fixed $\epsilon>0$, the set $B:=\{t\in[0,1): |h(t)-h(t+)|\ge\epsilon\}$ is countable. For ...


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Choose $$f(t, B_t) = \frac{B_t^3}{3}$$ so that we have $$ \frac{B_t^3}{3} = \int_0^t B_s^2 dB_s + \int_0^t 0 ds + \frac{1}{2}\int_0^t 2B_s ds $$ $$ \to E[\frac{B_t^3}{3}] = E[\int_0^t B_s^2 dB_s + \int_0^t 0 ds + \frac{1}{2}\int_0^t 2B_s ds] $$ $$ \to E[\frac{B_t^3}{3}] = E[\int_0^t B_s^2 dB_s] + \frac{1}{2}\int_0^t 2E[B_s] ds $$ $$ \to ...


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Let's show that $\mathbb{E}[\int_0^t (e^{Z_s}(1-X_su_s))^{2}ds] < \infty $ By Cauchy-Schwarz inequality: $\mathbb{E}[\int_0^t (e^{Z_s}(1-X_su_s))^{2}ds] \leq \mathbb{E}[\int_0^t e^{4 Z_s}ds]^{\frac{1}{2}} \mathbb{E}[\int_0^t (1-X_su_s)^{4}ds]^{\frac{1}{2}} $ Because $u$ is bounded, it is clear that $\mathbb{E}[\int_0^t (1-X_su_s)^{4}ds] < ...


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It depends very much on what you know about stochastic integrals. If you know Novikov's condition, then this might be your first choice. If you know local martingales, then you can use the fact that $(e^{Z_t})_t$ is a local martingale and the boundedness of $u$ to conclude that $e^{Z_t} \in L^2$. In my answer, I'll present a solution which does not require ...


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Let's try to prove this rigorously. Lemma 1 Let $E$ be a topological space $t\ge 0$ and $t_0^{(n)},\ldots,t_n^{(n)}\ge 0$ with $$0=t_0^{(n)}<\cdots<t_n^{(n)}=t$$ for some $n\in\mathbb N$ $F:[0,t]\times H\to E$ be continuous $(X_t)_{t\ge 0}$ be a left-continuous $H$-valued stochastic process on $(\Omega,\mathcal A,\operatorname P)$ ...


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Well, \begin{align} S_n&:=\sum_{i=1}^n\int_{t_{i-1}}^{t_i}L_i\Phi_0\;{\rm d}W_s\\ &=\sum_{i=1}^n\int_{t_{i-1}}^{t_i}F_x(t_{i-1},X_{t_{i-1}})\Phi_0\;{\rm d}W_s\\ &=\int_0^t\bigg[\sum_{i=1}^nF_x(t_{i-1},X_{t_{i-1}})\Phi_0\mathbf{1}_{(t_{i-1},t_‌​i]}(s)\bigg]\;{\rm d}W_s. \end{align} Now if $F(s,x)$ is continuous for $(s,x)\in[0,t]\times H$ and ...


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Apply Ito's formula for: $$ f(x_1,x_2)=e^{x_1}\cos(x_2) $$ $$ f_{x_1}(x_1,x_2)=e^{x_1}\cos(x_2)\quad f_{x_2}(x_1,x_2)=-e^{x_1}\sin(x_2) $$ $$ f_{x_1x_1}(x_1,x_2)=e^{x_1}\cos(x_2)\quad f_{x_1x_2}(x_1,x_2)=-e^{x_1}\sin(x_2) $$ $$ f_{x_2x_2}(x_1,x_2)=e^{x_1}\cos(x_2) $$ Then: $$ \mathrm{d}f(B_1,B_2)=f_{x_1}(B_1,B_2)\mathrm{d}B_1+f_{x_2}(B_1,B_2)\mathrm{d}B_2+ ...


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Let me split this answer into two parts: Part 1 Let $U$ and $H$ be arbitrary Hilbert spaces, $L\in\mathcal L(U,H)$ and $x\in H$. As Q. Huang noted in his answer, the authors of Stochastic Differential Equations in Infinite Dimensions$^3$ "define" $$(L^\ast x)u:=\langle Lu,x\rangle\;\;\;\text{for }u\in U\;.\tag 7$$ I hate that, it's awful. Why? Well, cause ...


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There is exactly a definition of the term $\int_0^t\langle\Phi_s{\rm d}W_s,F_x(s,X_s)\rangle$. For $\Phi_s$ taking values in $\mathfrak L_2(Q^\frac{1}{2}U,H)$ and satisfying the condition that the integral of $\Phi_s$-'s square-norm in $\mathfrak L_2(Q^\frac{1}{2}U,H)$ is a.s. finite (just called the "Energe Condition" privately), and for $\Psi_s$ a ...


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Without an additional regularisation (say with colored noise), the integral you are trying to compute is strongly divergent. The integrand is dominated by $\xi^2=\left(\frac{dW}{dt}\right)^2$ which is positive and of order $\frac{1}{dt}$ and there is thus no hope to give it a meaning in the stochastic integral sense. You are just integrating infinite ...


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Your stochastic integral formula is correct, as are your values for the variances of the two integrals on the right side. This implies that the covariance of those integrals is $t^3/6$. To see this directly, let $M_t$ denote the first stochastic integral on the right; this is a martingale. Therefore, $$ \Bbb E\left[M_t\cdot\int_0^t ...


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First of all, note that $\mathbb{E}(B_T^3)=0$, i.e. we have to find $F_t$ such that $$B_T^3 = \int_0^T F_s \, dB_s. \tag{1}$$ It follows from Itô's formula that $$B_t^3 = 3 \int_0^t B_s^2 \, dB_s + 3 \int_0^t B_s \, ds \tag{2}$$ (please compare this with what you wrote in your question; there are so many typos in there that I'm really not sure whether ...


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Actually, I have managed to find a solution to the equation, I post it for people who followed the topic and if it could interest some students : $ -v_t + sup_{u}(-\frac{1}{2}u^2v_{xx} -x^2 + \frac{1}{2}u^2) =0$ $v(1,x)= x^2$ This system can be rewritten : $ -v_t + -x^2+ sup_{u}(\frac{1}{2}u^2(1-v_{xx})) =0$ $v(1,x)= x^2$ We make the assumption that we ...


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I think you can use the Wiener integral to solve that problem ? $\int_{0}^{t}h_rdB_r$~ $ \mathscr{N}(0,\int_{0}^{t}h_r^2dr)$ as h is continuous adapted to the filtration (Wiener integral) Then using the 4th moment of a central normal distribution, $E[\int_{0}^{t}h_rdB_r] = 3*(\int_{0}^{t}h_r^2dr)^2 < 3C^4t^2$ using the fact that h is bounded.


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It means that $$ lim_{k \to \infty}\Bbb E\left\{\left[\sum_{i = 0}^{k - 1} {X_{t_{i+1}} + X_{t_i}\over 2} \left( W_{t_{i+1}} - W_{t_i} \right) - \int_0^T X_t \circ dW_t\right]^2\right\}=0. $$


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Consider a general linear SDE $$dX_t = (a_1 X_t+a_2)dt + (b_1 X_t+b_2)dW_t.\qquad (*),$$ with $X_0 = x_0$ and where $W_t$ denotes standard Brownian motion. Define $$\phi_t = \exp \left \{a_1 t - \frac{b_1^2}{2}t + b_1W_t \right \},$$ then $(*)$ has solution $$X_t = \phi_t \left(x_0+(a_2-b_1b_2)\int_{0}^{t} \phi_s^{-1} ds + b_2 \int_{0}^{t} \phi_s^{-1} dW_s ...


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Since the initial condition is $X(0) = 0$ you can try the test function $$X(t) = a(t)\int_{0}^{t}b(s)dB(s),$$ and computing $dX(t)$ implies that you obtain the following ODE, which can be solved by integrating both sides \begin{align} &\frac{a'(t)}{a(t)}dt = \frac{-2dt}{1-t} \\ &\Leftrightarrow \ln|a(t)| = 2 \ln|1-t| + C \\ &\Leftrightarrow a(t) ...


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Hint: Apply the polarization identity in $\mathbb{R}$: $$xy = \frac{1}{4}((x+y)^2-(x-y)^2),$$ to $x = \displaystyle \int_{0}^{t} \cos(u) dB_u$ and $y = \displaystyle \int_{0}^{t} \sin(u) dB_u$, together with the Itô-isometry. This will lead to the result.


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To prove the desired identity, it is enough to show that for each fixed $x$, $$ H_n(x)=\frac{d^n}{d\alpha^n}e^{\alpha x-\frac{1}{2}\alpha^2}\Big|_{\alpha=0}$$ By completing the square, we may write $$ e^{\alpha x-\frac{1}{2}\alpha^2}=e^{\frac{x^2}{2}}e^{-\frac{(x-\alpha)^2}{2}}$$ and hence $$\frac{d^n}{d\alpha^n}e^{\alpha ...



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