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Any càdlàg function $f: [0,T] \to \mathbb{R}$ has at most finitely many jumps with jump size $>\epsilon$ for any (fixed) $\epsilon>0$, see e.g. this answer. The estimate "$\leq [X,X]_t$" is not used to conclude that $\log V_t$ is of bounded variation, but to show that the series $\sum_{s \leq t} (\log(1+U_s)-U_s)$ is absolutely convergent. Recall ...

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This is only an answer for a part of the question To answer your second question $2)$: No, in general it's not true that a function bounded by a function of finite variation is itself of finite variation, for example take $$f(x)=\begin{cases} 0, x=0 \\ \sin(\frac{1}{x}), \text{otherwise}\end{cases}$$ which is not of bounded variation although $f(x)$ is ...

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No, we cannot apply the dominated convergence theorem in this way (see all the comments above). For fixed $R>0$ denote by $$\tau := \inf\{t>0; |B_t| \geq R\}$$ the exit time from $(-R,R)$. Moreover, we denote by $$H \bullet B(T) := (H \bullet B)(T) := \int_0^T H(s) \, dB_s$$ the stochastic integral of $H$. By Markov's inequality and Itô's ...

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Define $\tau_L = \inf\{t \geq 0, |B_t|>L\}$. Note that $\tau_L \uparrow \infty$ and therefore $\Bbb{P}(\tau_L<n) \xrightarrow[L \to \infty]{} 0$ by the bounded convergence theorem. That is sufficient because once you prove $$\lim \sum_{j=1}^n f(B_{t^{(n)}_j})(t^{(n)}_{j+1} \wedge T - t^{(n)}_{j}\wedge T) \xrightarrow[n\to ... 1 There is not one answer to this question. Your question is really one about modeling, rather than being strictly about mathematics, so the best answer depends on what you're trying to model. Two answers that come to mind for me are as follows. One would be to instead consider applying small iid normally distributed perturbations every \Delta t and ... 3 E^x is the expectation under the path measure of W(\cdot) started at W(0) = x. Example: E^x[W(t)] = E^x[W(t)-W(0)+W(0)] = E^x[W(t)-W(0)] + x = x since W(t)-W(0) \sim \mathcal{N}(0,t). In plain English, E^x is just ordinary expectation with the understanding that the starting point of BM is a variable, x. 1 Fubini's theorem on [0,t]\times\Omega gives$$\mathbb{E}\left[\int^{t}_{0} W^2_s\, ds\right] =\int_\Omega \int^t_0 W_s^2(\omega)\,ds\,\mathbb{P}(d\omega) =\int^t_0\int_\Omega W_s^2(\omega)\,\mathbb{P}(d\omega)\,ds = \int^{t}_{0} \mathbb{E}[W^2_s]\, ds.$$1 Let g be a predictable function such that$$\mathbb{E} \left( \int_0^t \!\! \int |g(s,y)| \, \nu(dy) \, ds \right)<\infty,$$then one can show that the integral$$\int_0^t \!\! \int g(s,y) \, N(dy,ds)$$is well-defined and$$\mathbb{E} \left( \int_0^t \!\! \int g(s,y) \, N(dy,ds) \right) = \mathbb{E} \left( \int_0^t \!\! \int g(s,y) \, \nu(dy) \, ...

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The Itô isometry is important because it allows you to extend the stochastic integrals that aren't in general path-by-path well defined to any function in $L^2$. Remember that the Brownian motion is of unbounded variation and the integral in Stieltjes sense is not readily available. If you would like to see a path-by-path construction of the integral you ...

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To keep notation simple, we write $S_t$ instead of $S_t^i$, $\sigma_t^j$ instead of $\sigma_{t}^{ij}$ and so on. That's okay, because $i$ is a fixed number throughout this calculation. Suppose $(X_t)_{t \geq 0}$ is an Itô process of the form $$dX_t = b(t) \, dt + \eta(t) \, dW_t$$ where $\eta = (\eta_1,\ldots,\eta_m)$ and $(W_t)_{t \geq 0}$ is an ...

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