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



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