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3

I think everything you're doing is correct, but perhaps it would benefit from exploiting some cancellations so things work out more smoothly. In your second line, with six terms, I claim if you take the first, fourth, fifth, and sixth terms and sum them, you get $$\boxed{ \lambda^2 s^2 -\lambda s -2\lambda t X_s}.$$ I used this grouping because lots of ...


2

You have : $$\mathbb{P}(A_4=k|N_{10}= 1000)=\frac{\mathbb{P}(A_4=k,N_{10}=1000)}{\mathbb{P}(N_{10}=1000)}$$ $$\mathbb{P}(N_{10}=1000)=\frac{(10\times 100)^{1000}}{1000!}e^{-10\times 100}$$ You have : $$A_t = \sum_{i=1}^{N_t} \xi_i$$ where $\xi_i$ are independent and identically distributed $$\mathbb{P}(\xi_i=1)=\mathbb{P}(\xi_i=0)=\frac{1}{2}$$ ...


2

Note that $$ \left( \frac{N_t}{t}-\lambda \right)^2= \frac{1}{t^2} (N_t-t\lambda)^2 \leq \frac{1}{\sigma^2} (N_t-t\lambda)^2$$ for all $t \in [\sigma,\tau]$. Hence, $$\mathbb{E} \left[ \sup_{t \in [\sigma,\tau]} \left( \frac{N_t}{t}-\lambda \right)^2 \right] \leq \frac{1}{\sigma^2} \mathbb{E} \left[ \sup_{t \in [\sigma,\tau]} (N_t-t\lambda)^2 \right].$$ ...


1

This isn't an answer per se, rather some thoughts to consider. If $\{X_n:n=1,2,\ldots\}$ is a Markov chain and we define the holding times $\{T_n:n=1,2,\ldots\}$ by $T_n\sim\mathsf{Exp}(\lambda_{X_n})$, then the sequence $\{T_n\}$ only defines a Poisson process if $\lambda_{X_n}$ is equal for all $n$. Otherwise, the intensity would be a stochastic process ...


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Let's try to prove this rigorously. Lemma 1 Let $E$ be a topological space $t\ge 0$ and $t_0^{(n)},\ldots,t_n^{(n)}\ge 0$ with $$0=t_0^{(n)}<\cdots<t_n^{(n)}=t$$ for some $n\in\mathbb N$ $F:[0,t]\times H\to E$ be continuous $(X_t)_{t\ge 0}$ be a left-continuous $H$-valued stochastic process on $(\Omega,\mathcal A,\operatorname P)$ ...


1

It depends very much on what you know about stochastic integrals. If you know Novikov's condition, then this might be your first choice. If you know local martingales, then you can use the fact that $(e^{Z_t})_t$ is a local martingale and the boundedness of $u$ to conclude that $e^{Z_t} \in L^2$. In my answer, I'll present a solution which does not require ...


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Let's show that $\mathbb{E}[\int_0^t (e^{Z_s}(1-X_su_s))^{2}ds] < \infty $ By Cauchy-Schwarz inequality: $\mathbb{E}[\int_0^t (e^{Z_s}(1-X_su_s))^{2}ds] \leq \mathbb{E}[\int_0^t e^{4 Z_s}ds]^{\frac{1}{2}} \mathbb{E}[\int_0^t (1-X_su_s)^{4}ds]^{\frac{1}{2}} $ Because $u$ is bounded, it is clear that $\mathbb{E}[\int_0^t (1-X_su_s)^{4}ds] < ...


1

If $f: (0,\infty) \times \Omega \to \mathbb{R}$ is progressively measurable and $$\mathbb{E} \left( \int_0^t |f(s)|^2 \, ds \right)<\infty \quad \text{for all $t \geq 0$}$$ then $$M_t := \int_0^t f(s) \, dB_s, \qquad t \geq 0,$$ is a martingale. This implies in particular $$\mathbb{E}(M_t) = \mathbb{E}(M_0) = 0.$$ Since $f(s,\omega) := ...


1

There are many questions here, and some of them should perhaps be split into their own posts. But to address the title question, consider the Ornstein–Uhlenbeck process, which can be defined by the SDE $dX_t = dB_t - X_t \,dt$, or in various other equivalent ways. (In Wikipedia's notation, I am using the parameters $\mu = 0$ and $\theta = \sigma = ...


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The property you're looking for, namely that $X_t$ is $\mathcal{B}([0,t])\otimes\mathcal{F}_t$-measurable for all $t$, is called progressive measurability. Now your $X$ may not be progressively measurable, but as an adapted (and measurable) process, it has a progressively measurable modification $Y$, and it is easy to see that modification does not change ...


1

There is exactly a definition of the term $\int_0^t\langle\Phi_s{\rm d}W_s,F_x(s,X_s)\rangle$. For $\Phi_s$ taking values in $\mathfrak L_2(Q^\frac{1}{2}U,H)$ and satisfying the condition that the integral of $\Phi_s$-'s square-norm in $\mathfrak L_2(Q^\frac{1}{2}U,H)$ is a.s. finite (just called the "Energe Condition" privately), and for $\Psi_s$ a ...


1

Let $D:=\{(t,\omega):f(t,B_t(\omega)\not=0\}$ and $C=\{(t,x):f(x,t)\not=0\}$. Observe that $D=\varphi^{-1}(C)$, where $\varphi(t,\omega) =(t,B_t(\omega))$ (mapping $[0,T]\times\Omega$ into $[0,T]\times\Bbb R$). Consequently, writing $\lambda$ for Lebesgue measure on $[0,T]$, $$ 0=\lambda\otimes\Bbb P(D) =\int\int 1_{C}(t,x){1\over\sqrt{2\pi ...


1

For any measurable set $A \subseteq [0,T] \times \Omega$ it holds that $$\begin{align*} \{(t,\omega); (t,B_t(\omega)) \in A\} &= \int_0^T \! \int 1_A(t,B_t) \, d\mathbb{P} \, dt \\ &= \int_0^T \! \int 1_A(t,x) \frac{1}{\sqrt{2\pi t}} e^{-x^2/2t} \, dx \, dt. \end{align*}$$ Since $e^{-x^2/2t}$ is strictly positive, this implies $$(\lambda|_{[0,T]} ...


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The values of $N(0), R(0), a$ do not matter, $\frac {R(t)}{N(t)} \to 1$ with probability $1$. Intuitively, when few people know the rumor, the number who know it doubles every time step. This will overtake the linear growth rate of the population. When most of the population knows the rumor, everybody around hears it pretty quickly and the only ones who ...



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