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The answer is indeed yes, as every adapted measurable real-valued process admits a progressively measurable modification. This is classical Theorem due to Paul-André Meyer (cf. Paul A. Meyer. Probabil …

As your reflected Brownian motion is nothing else then the absolute value of a Brownian motion you have $$ \{t \in [0,T] : Y_t = 0\} = \{t \in [0,T] : \vert W_t \vert= 0\} = \{t \in [0,T] : W_t …

One possible approach is to use the fact that the density process $\left. Z_t =\frac{d\mathcal{Q}}{d\mathcal{P}} \right\vert_{\mathcal{F}_t}$ for every equivalent local martingale measure $\mathcal{Q} …

Note that for fixed $t_0$, we have by the martingale representation theorem that $$Y_{t_0} = \int_0^{t_0} I\bigl( s \in [0,t_0-1) \bigr) \, dB_s.$$ In particular that for $t_0\leq 1$ the indicator y …

A Grigelionis process is a special semimartingale with absolutely continuous integral characteristics (in time). This is insofar a generalization of a Lévy process, as Lévy processes can be characteri …

The problem as stated has no solution except in the special case where it is optimal not to consume at all. To see this, note that the payoff depends on the consumption rate process $c(t)$ only throu …

I think question 1) is reasonably answered by Wolfgang Loehr in his comment. To get a counterexample for your claims in 2), just set $a_u=W_u^2-u$ for your Brownian motion. Ito's formula gives you the …

I am not sure if I understand your question correctly. A typical Brownian BSDE has the form $$dY_t = f(\omega, t, Y_t, Z_t)dt - Z_t dW_t$$ with terminal condition $$Y_T = \xi \in \mathcal{F}^{W}_T$$ w …

No, that both processes have the same one-dimensional marginals is not sufficient. In contrary, when $X$ is an arbitrary elliptic Itô-process, you can always find a Markov process with the same margin …

No, but under some regularity conditions you might represent $a(t)$ in terms of Malliavin calculus by means of the Clark-Ocone formula (see e.g. the Lecture notes by Eulalia Nualart, Section 1.5.3.)

First, a martingale is always only specified with respect to a filtration, and so is thus a local martingale. You do not specify any filtration in your problem, so I assume you mean the natural filtra …

Here an answer for the case with determinist drift as mentioned in the edit. (Note: I fail to see why to use different notations for $F$ and $\tilde{F}$ as it is the same process) As $$ dF_t = \mu_t …

The proof is not correct, as without additional integrability condition you will not be able to conclude that $g_n \in L^2$ for $n$ large enough, and therefore the $L^2$ convergence argument fails. As …

You can solve this by reducing it to a problem of Brownian motion: Define the scale function $\varsigma(x) = \int_{X_0}^x e^{-2\int_{X_0}^y \frac{b(z)}{\sigma^2(z)} dz} dy$ the process $M_t = \var …

Hi lpdbw, I think this is a very interesting questions, here at least a partial answer; It depends heavily on the little word "strong" in parentheseis. Assuming that $(X_t)$ is strong Markov, the an …