# Tag Info

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The key point is to write $B_T^2$ in a clever way: \begin{align*} B_T^2 &= \sum_{j=1}^n (B_{t_j}^2-B_{t_{j-1}}^2) \\ &= \sum_{j=1}^n \big( (B_{t_j}-B_{t_j^{\ast}})+B_{t_j^{\ast}} \big)^2 - \big( (B_{t_{j-1}}-B_{t_j^{\ast}})+B_{t_j^{\ast}} \big)^2 \\ &= 2 \sum_{j=1}^n B_{t_j^{\ast}} (B_{t_j}-B_{t_{j-1}}) + \sum_{j=1}^n ... 1 The proof is in two steps: it is true for elementary functions: you just have to prove that for every Y_i mesurable with respect to F_{t_{i-1}}, E\left[\sum_{i=0}^{N-1} Y_i (B_{t_i} - B_{t_{i-1}})\right] =0 $$(it is quite easy). then, write an element f as an L^2 limit of such elementary processes f_n. Then,$$ E \left[\int_S^T f_n dB - ...

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I'm sorry, I was very stupid. We can just apply Ito's formula and get: \begin{align*} & \int_a^b f(t)dW_t=f(b)W_b-f(a)W_a-\int_a^b W_t f'(t)dt. \end{align*} This yields that indeed we can find the desired pathwise upper bound of $\int_a^b f(t)dW_t$ in terms of $||f||$, $||f'||$, $\sup_{t \in [a,b]}|W(t)|$.

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In the second case it is not necessarily the case that $dX_t=\mu_t\,dt+\sigma_t\,dB_t$. But the second case should imply the first. What is $dX_t\cdot dX_t$ for the first case ?

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In order to solve the SDE $$dX_t = \alpha X_t \, dt + \sqrt{2} \, dB_t \qquad X_0 = x \tag{1}$$ we consider the corresponding ordinary differential equation $$dx(t) = \alpha x(t) \, dt, \qquad x(0)=c.$$ It is not difficult to see that its unique solution equals $$x(t) = c \, e^{\alpha t}.$$ Now the idea is to use an analogue of the variation of ...

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Your stochastic differential equation is for geometric Brownian motion. Using your notation, the analytic solution for geometric Brownian motion under Itō's interpretation is $$X_t = X_0\space\text{exp}\left(\left(\mu-\frac{\sigma^2}{2}\right)t + \sigma B_t\right)$$ This is not what you're using in your Matlab code for Xtrue. To simulate your system, you ...

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