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2

Fix $\omega \in \Omega$. Since the mapping $s \mapsto X_s(\omega)$ is càdlàg, we have $$\sup_{s \in [0,t]} |X_s(\omega)| < \infty$$ for any $t \geq 0$. As $f$ is continuous, this implies in particular that $$M:=\sup_{s \in [0,t]} |f(X_s(\omega))|<\infty.$$ Hence, $$\left| Y_t(\omega) - Y_r(\omega) \right| = \left| \int_r^t f(X_s(\omega)) \, ds ...


2

I think you misspoke somewhat in your first statement: first hitting times are stopping times, but there are many other kinds of stopping times. What you've written is a last exit time, which is not a stopping time, because as you've said we cannot know whether $\tau \leq t$ by observing the process up to time $t$. This creates some difficulties in ...


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$\newcommand{\var}{\operatorname{var}}\newcommand{\E}{\operatorname{E}}$The distribution of $S_N$ is called a "compound Poisson distribution". $$ \var(S_N \mid N=n) = \var(S_n) = n\sigma^2. $$ Therefore $$ \var(S_N\mid N) = N\sigma^2, $$ so $$ \E(\var(S_N\mid N)) = \sigma^2\E(N) = \sigma^2\lambda. $$ Next we have $$ \E(S_N\mid N=n) = \E(S_n) = n\mu. $$ ...


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Consider the process $$X_s = W^{(1)}_s-W^{(2)}_s$$ At any given time, this is the difference of two mean zero, independent, normals with variance $s$. That means $X_s$ is a mean zero normal with variance $2s$. So the question is, what is the expected value of the absolute value of a normal random variable. This is called a Folded Normal Distribution. One ...


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You just need to look at more stopping times. For any $t\geq 0$, you have $${E}[\mathbf{1}_{\{\tau_0<T\}}X(T \wedge (\tau_0+t))] ={E}[\mathbf{1}_{\{\tau_0<T\}}E[X(T\wedge (\tau_0+t))|F_{\tau_0}]] \le {E}[\mathbf{1}_{\{\tau_0<T\}}X(\tau_0)]=0.$$ This gives $$P\biggl(X(T\wedge(\tau_0+t))=0\mbox{ for all }t\in\mathbb{Q}\cap[0,T]; \tau_0<T\biggr) ...


2

You can generalize your derivation from: $$E[\mathbf{1}_{\{\tau_0<T\}}X(T)]=E[\mathbf{1}_{\{\tau_0<T\}}E[X(T)|F_{\tau_0}]]\le E[\mathbf{1}_{\{\tau_0<T\}}X(\tau_0)]=0$$ to: $$E[\mathbf{1}_{\{\tau_0=s\}}X(t)]=E[\mathbf{1}_{\{\tau_0=s\}}E[X(t)|F_{\tau_0}]]\le E[\mathbf{1}_{\{\tau_0=s\}}X(\tau_0)]=0$$ where $t > s$. As you mentioned, using the fact ...


1

Let $\epsilon > 0$ and consider the stopping time $$\tau_\epsilon = \inf\{t \in [\tau_0, T] : X_t \ge \epsilon\} \wedge T.$$ Since $\tau_0 \le \tau_\epsilon$ almost surely, optional stopping gives $E[X_{\tau_\epsilon}] \le E[X_{\tau_0}]$. Now let consider the events $\{\tau_0 < T\}$ and $\{\tau_0 = T\}$ and write $$E[X_{\tau_\epsilon} ; \tau_0 < T] ...


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You could try a Monte Carlo approach. Basically, you can simulate a large number of strong solutions and then evaluate the sample mean and variance of the specific instant of interest. Depending on the structure of the diffusion coefficient, it is possible to perform exact simulation. In this case, no approximation error will be propagated to your ...


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The player meets monsters at the rate $\lambda$, and with probability $p$ they are winning monsters, so the player meets winning monsters at the rate $\lambda p$. We can extend the game up to $t=1$, with monsters appearing after the player's death meeting a dead player. Then, since the Poisson monsters appear uniformly in time, the probability distribution ...


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Yes. A Drunkards Walk follows a similar principle to Binomial Distribution; but rather than successes (and failures) in a certain number of trials, you are counting steps forward and backwards. $\mathsf P(\eta_t=m)$ is the probability that in $t$ steps of $\pm 1$ units each, you will have moved $m$ units in total; consisting of $x$ steps forward, and $y$ ...


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Your first solution looks reasonable. I'm not sure I understand what your second solution even really means, with the expected value of the conditional variance and so on. Maybe it works, I'm just not sure how to parse it. Here is a different idea: write $$\text{Var}(S_N)=E[S_N^2]-(E[S_N])^2=E[S_N^2]-\mu^2 \lambda^2$$ using part 1. Now $$E[S_N^2]=\sum_n ...


1

Let's redefine $x^*$ according to my comment above: $$ x^* = \inf\left\{ y : \int_0^T 1_{\{x(t)\leq y\}} dt \geq \tau\right\} $$ Assume $x(t)$ takes values in a state space of nonnegative integers, so $x(t) \in \{0, 1, 2, \ldots\}$ for all $t$. Assume that $x(0)=0$. Then the infimum value is achieved by a particular integer $x^* \in \{0, 1, 2, \ldots\}$ ...


1

Let $A = \left(\begin{array}{rr} 0 & 1\\ -1 & 0 \end{array} \right)$, $X_t = \left(\begin{array}{r} X_1(t)\\ X_2(t) \end{array} \right)$, and $B_t = \left(\begin{array}{r} B_1(t)\\ B_2(t) \end{array} \right)$. Then \begin{align*} dX_t = AXdt + \alpha dB_t. \end{align*} Note that \begin{align*} d\left(e^{-At} X_t \right) &= -Ae^{-At} X_t dt + ...


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$\tau$ here strikes me as a Stopping Time. In that case, it is not a process but a random variable a) defined on the entire space of paths, b) measurable with respect to the filtration $\mathcal{F}_t$ in the sense that: $$\{\tau \leq t\}$$ is $\mathcal{F}_t$ measurable for all $t$. It is strange for me to think of it as a standard adapted process. Say for ...


1

I claim that $\Pi$ is Borel measurable. Define the family $$ \mathscr{A} := \left\{\text{Borel measurable $E\subseteq\mathcal{M}(\mathbb{R})$}: \text{$\Pi^{-1}(E)$ is Borel measurable}\right\} \;. $$ It is enough to verify that $\mathscr{A}$ is a $\sigma$-algebra, and $\mathscr{A}$ contains a countable sub-family $\mathscr{C}$ that generates the weak ...



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