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I would decompose the integration interval defining $X_\infty$ into the part that is measurable with respect to $\cal F_t$, i.e., $X_t,$ and the rest, which is $\int_t^\infty H_s\ dB_s.$ Since Brownian motion is a martingale the second term has zero conditional expectation w.r.t. $\cal F_t$ so the first term is the conditional expectation.

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The expectations is easy to calculate, using Fubini's theorem, which applies since the integrand is positive: \begin{align} E\left[\int_0^t B(s)^2 ds\right] = \int_0^tE[B(s)^2]ds = \int_0^ts\,ds = \frac{t^2}{2} \end{align}

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Your SDE is an example of a linear SDE $$dY_t = (\alpha(t)+\beta(t)Y_t)dt+(\gamma(t)+\delta(t)Y_t)dW_t,$$ where $\alpha$, $\beta$, $\gamma$ and $\delta$ denote deterministic processes and $W$ denotes a one-dimensional Wiener process. In your case, processes $\beta$ and $\delta$ are zero processes and $\gamma$ is a constant process and $\alpha(t) = ... 0 Intuitively, the symbol$\text{d}W_{t}$may be interpreted as an infinitesimal increment of a (one-dimensional) Wiener process,$W$. $$\text{d}W_{t} = W_{t+\text{d}t}-W_{t}\ .$$ Increments of a Wiener process are normal distributed random variables whose expected value is zero and whose variance is equal to the time-increment. Thus the symbol$\text{d}W_{t}$... 1 You don't have the right argument, the real reason why you can get the third line is because first$B$and$Z$are orthogonal so that $$\mathbb{E}^{\mathbb{Q}^{t} }\bigg[\bigg(\int_0^t\sigma_r^t(s)dB_s^t \bigg)\cdot\bigg(\int_0^t\mathbb{E}^{\mathbb{Q}^{t}}_s[D_s^{Z^t}\mathbb{1}_{S_t > K}]dZ_s^t \bigg)\bigg] =0$$ and because : ... 0 For the equality you wrote there is some "intuition" based on the Itô integral. Namely, write this integral as $$\int_{0}^T B(T) d B(t) = \int_0^T B(t) dB(t) + \int_0^T \big(B(T)- B(t)\big) dB(t).$$ The first integral is a usual Itô integral equal to$(B(T)^2 - T)/2$. The second is a kind of "purely anticipative" integral: the integrand is ... 1 It seems like you need some further clarification. What you want to do is to fix a point$x$and a time$t$, and consider them as known. Now your solution is $$f(t,x) = E[ h(X_T) | x,t]$$ where$X_uis a stochastic process satisfying \begin{aligned} &\mathrm{d} X_u = 2u \, \mathrm{d}u + u^2\, \mathrm{d}W_u \quad \text{when} \quad t ... 1 As you correctly write, the integral\int_0^t b(X_s) dX_s$is well defined (as an extended Itô integral) provided that$\int_0^t b(X_s)^2 ds<\infty$almost surely. So the question is: When is the integral$\int_0^t f(X_s) ds$finite almost surely? The answer to the latter question is well known: this is the case if$^*f\in ...

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Consider a function $f: [0,T] \times \Omega \to \mathbb{R}$ which is also in your "good conditions" space, denoted $\mathcal{V}$. Consider the operator $I : \mathcal{V} \to L^2(\Omega, \mathcal{F}_T, P)$ given by $$I_T(f) := \int_0^T f(s,\omega) \, dB_t(\omega).$$ Ito's isometry tells us $$E\left( \left( \int_0^T f(s) \, dB_t \right)^2 \right) = \int_0^T ... 0 Hi this kinds of SDE are called McKean-Vlasov SDE, you can have a look here and there and to the references therein. Best regards. 0 First write$$ \int_{0}^{\infty}e^{-\frac{4}{5}x}e^{-isx}dx =\frac{1}{\frac{4}{5}+is}. $$Then$$ \begin{align} \int_{0}^{\infty}x^{2}e^{-\frac{4}{5}x}e^{-isx}dx & =-\frac{d^{2}}{ds^{2}}\int_{0}^{\infty}e^{-\frac{4}{5}x}e^{-isx}dx \\ & = -\frac{d^2}{ds^2}\frac{1}{\frac{4}{5}+is} \\ & = ...

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Hint, too long for a comment. It's really bad practice to use the same variable in both the spatial and Fourier domain, and will lead to massive confusion. Switching to more usual notation, note that the result is $\frac{d^k}{d \omega^k} \hat{f}(\omega) = \mathcal{F}[{i^k x^k f(x)}](\omega)$, where $\hat{f}(\omega) = \mathcal{F}[f(x)](\omega)$. You know ...

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Define \begin{align*} \beta_t = \int_0^t \frac{dB_t^2 + X_t dB_t^1}{\sqrt{1+X_t^2}}. \end{align*} Then \begin{align*} \langle\beta, \beta\rangle_t = t. \end{align*} By Levy's characterization, $\{\beta_t \mid t \ge 0\}$ is a Brownian motion.

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Intuitive argument: by symmetry we should have $P(X_1 \ge 0 \mid W_1 = 0) = \frac{1}{2}$. However, $P(f(1, W_1) \ge 0 \mid W_1 = 0)$ is either $0$ or $1$ depending on whether $f(1,0) \ge 0$. Of course this is not really a proof because I conditioned on an event of probability 0. So let's try to use the same idea in an actual proof. Consider the ...

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