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The set $Z_{\mathcal{P^\ast}}$ is never compact except in the case where it is a singleton (or empty). This is for the general case with $S=(S^1,S^2,\ldots,S^d)$ being an $\mathbb{R}^d$-valued semimar …

Yes, it is possible to extend the state space with respect to which $Y$ is a Feller process. Then, $X$ will be a dense open subset of the extension $\hat X$. Furthermore, for any initial distribution …

No, it is not true that a process W satisfying the properties (1), (3) and (4) has to be a Brownian motion. We can construct a counter-example as follows. This construction is rather contrived, and I …

Yes, in this case it is true that $p$ is a proper martingale! Note that your integrand $\exp\left(-\int_0^tr_u du\right)$ is an adapted, continuous, and decreasing process bounded by 1. So, the follow …

No, $\rho$ need not be a proper martingale. To guarantee that $p_t=\int_0^ta_sd\rho_s$ is a martingale for all predictable $0\le a_t\le 1$ you need the additional property that $\sup_{s\le t}\rho_s$ i …

Here's a proof of the statement for $f=0$, so that $X=W$ is a Wiener process. (The proof with general $f$ is a bit more involved, and I give this further below). I'll base the proof on the following s …

No, it is not true for simple examples such as standard Brownian motion or a sequence of independent random variables. Suppose $W$ is a standard Brownian motion on the interval $[0,T]$. The covarian …

Yes, the Bochner integral does agree with the Lebesgue integral of the sample paths of the process. We can prove this in a slightly more general situation than that asked for in the question. For a p …

As suggested in my comment, here's a simple fact which applies to any probability measure $\mu$ on (the Borel σ-algebra of) a second countable topological space $X$. There is a unique minimal closed s …

As suggested in the question, $P_t$ need not be a well defined operator on $L^\infty(W,\mu)$. That is, $F$ can be zero $\mu$-almost everywhere, but $P_tF$ is nonzero on a set of positive $\mu$-measure …

To further elaborate on my comment, it is a theorem that if $X^1,X^2,\ldots,X^n$ are Lévy processes with respect to a common filtration, all starting from zero, then they are independent if and only i …

No, that is not true. Consider the following, defined on a filtered probability space $(\Omega,\mathcal{F},\{\mathcal{F}_t\}\_{t\in[0,T]},\mathbb{P})$. $W$ is a standard Brownian motion. $U$ is an $ …

Consider the following continuous Markov process X, starting from position x if x = 0 then Xt = 0 for all times. if x ≠ 0 then X is a standard Brownian motion starting from x. This is not strong M …

You cannot define a Lévy process by the individual distributions of its increments, except in the trivial case of a deterministic process Xt − X0 = bt with constant b. In fact, you can't identify it b …