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Hint: $$E(X^2)=2\int_0^1\int_0^tE(W(t)^2W(s)^2)\mathrm ds\mathrm dt$$ Can you compute $E(W(t)^2W(s)^2)$ for $s\lt t$?

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Here's a different argument. Define $$X_t = \int_0^t f(s,\omega) dW_s \\ Y_t = X_t^4 \\ Z_t = X_t^2.$$ Apply the Ito formula to $Y_t$ and $Z_t$: $$Y_t = 4 \int_0^t f(s,\omega) X_s^3 dW_s + 6 \int_0^t f(s,\omega)^2 X_s^2 ds \\ Z_t = 2 \int_0^t f(s,\omega) X_s dW_s + \int_0^t f(s,\omega)^2 ds$$ Now substitute and take expectations, thereby canceling the ...

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The answer is certainly: yes. By definition of a martingale, $\mathsf E M_t = \mathsf E M_0$ (just using the tower property) and since $M_0 = 0$ in your case, all follows. Note that Ito integral may not be a martingale unless some integrability conditions are satisfied, in general it may be only a local martingale - so you need to see which integrability ...

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Suppose that $(X_t,Y_t)_t$ is a solution to the given system. Then in particular, we can write $$X_t-X_0 = - \underbrace{\int_0^t Y_s \, dB_s}_{\text{local martingale}} - \frac{1}{2} \underbrace{\int_0^t X_s \, ds}_{\text{bounded variation, continuous}}.$$ Applying the optional stopping theorem yields that $$(t,\omega) \mapsto \left( \int_0^{t \wedge ... 1 When you are not sure, there is not harm in treating entries as separate variables to apply Ito's formula. The only thing you have to take care of in that case are quadratic covariations. 2 Applying Itô's formula for the Itô process$$X_t := \int_0^t f(s) \, dW_s$$and f(x) := x^4 gives$$X_t^4 = 4 \int_0^t X_s f(s) \, dW_s + \frac{4 \cdot 3}{2} \int_0^t X_s^2 f(s)^2 \, ds.$$Taking expecation yields$$\begin{align*} \mathbb{E}(X_t^4) &= 6 \int_0^t \mathbb{E}(X_s^2 f(s)^2) \, ds \\ &\leq 6 \mathbb{E} \left(\sup_{s \leq t} |X_s|^2 ...

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Let $H:x\mapsto F(\min(x,\tau))$, then $\mathrm dH(x)=\mathbf 1_{x\leqslant\tau}\,\mathrm dF(x)$ hence (3) reads $$g(t)=h(t)+\int_0^\infty g(t-x)\,\mathrm dH(x).$$ It follows that $$\tilde{g}(s)=\frac{\tilde{h}(s)}{1-\tilde{H}(s)},$$ where $$\tilde{H}(s)=\int_0^\infty\mathrm e^{-sx}\,\mathrm dH(x)=\int_0^\tau\mathrm e^{-sx}\,\mathrm dF(x).$$

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It follows from Itô's formula that the solution to the SDE $$dM_t = M_t \sigma_t dW_t$$ equals $$M_t = M_0 \exp \left( \int_0^t \sigma_s \, dW_s - \frac{1}{2} \int_0^t \sigma_s^2 \, ds \right).$$ By assumption, $M_0 \geq 0$. Hence, \begin{align*} \sqrt{M_T} &= \sqrt{M_0} \exp \left( \frac{1}{2} \int_0^T \sigma_s \, dW_s - \frac{1}{4} \int_0^T ... 2 Basically, you can simply apply Itô's formula:\begin{align*} f(W_t^1, W_t^2)-f(0,0) &= \int_0^t f_x(W_s^1,W_s^2) \, dW_s^1 + \int_0^t f_y(W_s^1,W_s^2) \, dW_s^1+ \\ &\quad \frac{1}{2} \bigg( \int_0^t f_{xx} (W_s^1,W_s^2) \, \underbrace{dW_s^1 dW_s^1}_{ds} + 2 \int_0^t f_{xy}(W_s^1,W_s^2) \, \underbrace{dW_s^1 dW_s^2}_{\varrho \, ds} \\ &\quad ...

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Two ways to fix this: Any martingale with respect to a nice filtration (i.e. a filtration satisfying the usual conditions) has a càdlàg modification. This means that we can assume without loss of generality that the process has càdlàg sample paths. In the case of Brownian motion, the completion of the canonical filtration satisfies indeed the usual ...

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For #1: Let's call your isomorphism $T$, the isometry from $\mathcal{H}_K$ to $L^2(P)$ which maps $K_t$ to $B_t$. Yes, $T$ really is a stochastic integral: $Tf$ is the Itô integral, not of $f$, but of $f'$. So $$Tf = \int_0^1 f'(t)\,dB_t$$ or in other words, for deterministic $g \in L^2([0,1])$, $$\int_0^1 g(t)\,dB_t = T\left(\int_0^\cdot ... 1 I like "Brownian Motion Calculus" by Ubbo F. Wiersema (2008). I found it more approachable than other books I've seen. But it depends on your level of mathematical sophistication (I'm a mathematical hick). 0 partial answer here 1)Measure, Integral and Probability by Capinski 2)Probability Essentials by Jacods 3)Knowing the odds by Walsh are all good but more elemetary than gut 0 No. Let \Omega = [0,1]. Let C_1 = [1/4,1/2] and let C_2 = [1/3,2/3]. Then$$ \sigma(C_1) = \{\emptyset, [1/4,1/2], [0,1/4) \cup (1/2,1], [0,1]\}$$and$$ \sigma(C_2) = \{\emptyset, [1/3,2/3], [0,1/3) \cup (2/3,1], [0,1]\}$$so that \sigma(C_1) \cap \sigma (C_2) = \{\emptyset, [0,1]\}. On the other hand, C_1 \cap C_2 = [1/3,1/2] so that$$ \sigma ...

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I like the book Brownian Motion - An Introduction to Stochastic Processes by René Schilling & Lothar Partzsch pretty much. It does not only cover stochastic differential equations (in particular, several possibilites are presented how to solve SDEs, e.g. by transforming them into linear SDEs), but also contains a very interesting and detailed exposition ...

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