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3

The mapping $\omega \mapsto X_{\xi(\omega)}(\omega)$ is measurable if the corresponding process is jointly measurable, i.e. if $$(t,\omega) \mapsto X(t,\omega) \tag{1}$$ is $\mathcal{B}([0,T]) \otimes \mathcal{A} /\mathcal{B}(\mathbb{R})$-measurable for $T>0$. This can be proved in the following way: Let $f$ be a continuously differentiable function, ...


2

Let $(X_t,\mathcal{F}_t)_{t \geq 0}$ be an (càdlàg) $L^2$-martingale, i.e. a martingale which satisfies $$\sup_{t < \infty} \mathbb{E}(X_t^2)<\infty.$$ Then it follows from the Doob-Meyer decomposition that there exists a unique increasing previsible process $(A_t)_{t \geq 0}$ such that $A_0=0$ and $$(X_t^2-A_t,\mathcal{F}_t)_{t \geq 0} \,\, ...


2

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 $||f||$, $||f'||$, $\sup_{t \in [a,b]}|W(t)|$.


1

For simplicity, I'll assume that $(M_t)_{t \geq 0}$ is a continuous martingale. The argumentation is similar for a local martingale, but more technical. Set $$I_n(t) := \int_0^t I_{n-1}(s) \, dM_s, \qquad n \in \mathbb{N}, t \geq 0, \tag{1}$$ and suppose that $$(n-1) I_{n-2}(t) = I_{n-2}(t) M_t - I_{n-3}(t) \langle M \rangle_t, \qquad t \geq 0. \tag{2}$$ ...


1

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 ?


1

Such a process does not exist. Two possible argumentations: Since a stochastic integral with respect to Brownian motion is a martingale, the expectation is constant, i.e. $$\mathbb{E} \left( \int_0^t X_s \, dB_s \right) = \mathbb{E} \left( \int_0^t X_s \, dW_s \right) = 0.$$ (To be more precise: Here, we need some integrability assumptions on $(X_t)_{t ...



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