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3

Set $$M_t := f(X_t) - \int_0^t Af(X_s) \, ds$$ and write $P_t f(x) := \mathbb{E}^x f(X_t)$ for the semigroup associated with $(X_t)_{t \geq 0}$. Using the definition of the generator, it is not difficult to see that $$\frac{d}{dt} (\mathbb{E}^x f(X_t)) = \frac{d}{dt} P_t f(x) = P_t Af(x).$$ By the fundamental theorem of calculus, this implies $$P_t ...


3

Without loss of generality, we may assume $x>0$. By the very definition of the stopping time, we have $$\begin{align*} \mathbb{P}_x(\tau_0>t) &= \mathbb{P}_x \left( \left\{w \in C[0,1]; \inf_{s \in [0,t]} (w(s)-x) +x >0 \right\} \right) \end{align*}$$ Since $\mathbb{P}_x$ is defined by translational invariance (it is the image measure of the ...


1

Well, I managed to hand-build a dirty distribution that works for me. The idea is to paste two geometric distributions together, and to truncate them not to get out of $S$: give yourself two parameters $k \in [0,\frac{2}{3}]$ (overload for the probability of "staying here") and $\lambda \in \mathbb{R}^+$ (parameter for the geometric distribution). compute ...


1

Let $X_n$ be the corresponding Markov chain. Define $N(j)=\sum_{n=0}^\infty 1\{X_n=j\}$ (number of visits to $j$), $\tau_j=\inf\{n\ge 0:X_n=j\}$, and $(\theta_n(\omega))(m)=\omega(m+n)$ (shift operator). Then $$\mathbb{E}_iN(j)=\mathbb{E}_i\left[\sum_{n=0}^\infty 1\{X_n=j\}\right]=\sum_{n=0}^\infty \mathbb{E}_i[1\{X_n=j\}]=\sum_{n=0}^\infty p^n(i,j)$$ ...


1

Yes, one can go the way back if there are assumptions on the state space $(E,\mathcal{E})$. Given a transition semigroup you can define a transition probability function using indicator functions. And from the transition probability function you can construct a Markov process using the Kolmogorov extension theorem. Well, this works if $(E,\mathcal{E})$ ...


1

For the transition matrix, we require that P be row-stochastic, so for all i, $\sum\limits_{j}p_{ij} = 1$. In particular, for $i=0$, since $\sum\limits_{j=0}^{\infty}{(\frac{5}{6})^j} = \dfrac{1}{1 - \frac{5}{6}} = 6$ only the first definition (with the exponent of $j$ not $j-1$) makes sense. The exponent $j-1$ would only make sense if the middle ...


1

In some sense, there is an $x$ on the right-hand side because the equality $$\mathbb{E}_x(Y \circ \theta_s \mid \mathcal{F}_s^+)(\omega) = \mathbb{E}_{B_s(\omega)}(Y)$$ holds only $\mathbb{P}_x$-almost surely. For a function $f: \mathbb{R} \to \mathbb{R}$ the Markov property reads $$\mathbb{E}_x (f(B_{t+s}) \mid \mathcal{F}_s^+) = \mathbb{E}_{B_s} f(B_{t}), ...



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