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Note that \begin{align*} \int_t^T X_s ds \int_t^T Y_s ds = \int_t^T\int_t^T X_s Y_u ds du. \end{align*} Then, by assuming the Fubini, \begin{align*} E\left(\int_t^T X_s ds \int_t^T Y_s ds\right) = \int_t^T\int_t^T E\left( X_s Y_u\right) ds du, \end{align*} which is generally not equal to $\int_t^TE\left( X_s Y_s\right) ds$.


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It is easy to check that \begin{align*} d\left(e^{-\frac{\alpha^2}{2}t - \alpha W_t } S_t \right) = e^{-\frac{\alpha^2}{2}t - \alpha W_t } f(t) dW_t. \end{align*} Then, $S_t$ can be solved subsequently.


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You could try a Monte Carlo approach. Basically, you can simulate a large number of strong solutions and then evaluate the sample mean and variance of the specific instant of interest. Depending on the structure of the diffusion coefficient, it is possible to perform exact simulation. In this case, no approximation error will be propagated to your ...


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Let $Y_t = \int_0^ts\,dB_s$. Note that $Y$ is a martingale and hence a semimartingale. Then consider the process $X_t = f(t,Y_t)$ with $f(t,x) = e^{\frac{t^3}{6}+x}$. Since the partial derivatives $f_t,f_x,f_{xx}$ all exist and are continuous, Ito's lemma applies in this situation. $$f(t,Y_t) = f(0,Y_0) + \int_0^tf_t(s,Y_s)\,ds + \int_0^tf_x(s,Y_s)\,dY_s + ...


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Are you sure that's the answer you want? The differential equation you show, $bp + \frac{1}{2}\sigma^2 \frac{dp}{dx}=0$ should not have a solution that's a Gaussian. Are you sure you don't mean to have $\int \frac{b}{\sigma^2} dx$?


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$$ bp+\frac{1}{2}\sigma^2\partial_{x}p = 0 $$ $$ 2bp+\sigma^2\partial_{x}p = 0 $$ $$ \sigma^2\partial_{x}p = -2bp $$ $$ \frac{1}{p}\,\partial_{x}p = -\frac{2b}{\sigma^2} $$ $$ \partial_{x}\left(\ln p\right) = -\frac{2b}{\sigma^2} $$ $$ \ln p(x) = \ln p(A) - \int_{A}^{x}\frac{2b}{\sigma^2}\,dx' $$ where $A$ is a constant $$ p = p(A)\,\exp\left(- ...


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By assumption, $$P[|H\cdot M|^*_t\ge c]\le P[|H\cdot U|^*_t\ge c/2]+P[|H\cdot V|^*_t\ge c/2]\le 18/c(\|U_t\|_1+\|V_t\|_1)$$ The $M=U-V$ decomposition as described by Writing a martingale as the difference of two non-negative martingales satisfies $\|M_t\|_1=\|U_t\|_1+\|V_t\|_1$.


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Your attempt: How do you conclude that $$\mathbb{E} \left( \int_0^t \frac{1}{1-s} \, dW_s \right) = \mathbb{E} \left( \frac{W_t}{1-t} \right)$$ ...? To me it looks as you used something of the form $$\int_0^t f(s) \, dW_s = f(t) W_t,$$ but this identity is, in general, not correct. Your professor's answer: Your professor uses the following well-known ...


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Itô's formula states that $$\mathbb{E} \left( \left| \int_0^T f(t_n) \, dW(t_n) \right|^2 \right) = \mathbb{E} \left( \int_0^T |f(t_n)|^2 \, dt_n \right)$$ for any "nice" function $f$. Applying this for $$f(t_n) := \int_0^{t_n} \dots \int_0^{t_2} h(t_1,\ldots,t_n) \, dW_{t_1} \dots dW_{t_{n-1}}$$ yields $$\mathbb{E} \left( \left| \int_0^T \int_0^{t_n} ...



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