# Tag Info

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, $x$.

3

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 ...

1

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 ...

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 $$\mathbb{E} \left( \int_0^t \!\! \int g(s,y) \, N(dy,ds) \right) = \mathbb{E} \left( \int_0^t \!\! \int g(s,y) \, \nu(dy) \, ... 1 No, we cannot apply the dominated convergence theorem in this way (see all the comments above). For fixed R>0 denote by$$\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$$the stochastic integral of H. By Markov's inequality and Itô's ... 1 There is not one answer to this question. Your question is really one about modeling, rather than being strictly about mathematics, so the best answer depends on what you're trying to model. Two answers that come to mind for me are as follows. One would be to instead consider applying small iid normally distributed perturbations every \Delta t and ... 1 Fubini's theorem on [0,t]\times\Omega gives$$\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

Are you sure about your definition of $\mathcal H_0^{2,c}$? Shouldn't this be the class of $L^2$-bounded continuous martingales? This is a indeed a Hilbert space with the scalar product $\langle L,N\rangle =E[L_\infty N_\infty]$. This makes sense because $L$ and $N$ are uniformly integrable and therefore $L_\infty=\lim L_t$ exists by the martingale ...

Only top voted, non community-wiki answers of a minimum length are eligible