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3

Check out Girsanov's paper from 1962 (here or here) for the proof of why his counterexample works; he gives the counterexample: $$ dX_t = \frac{|X_t|^{\alpha}}{1+|X_t|^{\alpha}}dB_t, \quad 0\le \alpha < \frac{1}{2} $$ where $B_t$ is Brownian motion/standard Wiener process.


3

See https://en.wikipedia.org/wiki/Renewal_theory#The_elementary_renewal_theorem The elementary renewal theorem


2

We assume that \begin{align*} d\left(\! \begin{array}{c} S^1(t)\\ S^2(t) \end{array} \!\right) =\textrm{diag}\left(S^1(t), S^2(t)\right)\bigg[\left(\! \begin{array}{c} r\\ r \end{array} \!\right)dt + \left(\! \begin{array}{cc} \sigma_{1,1} &\sigma_{1,2}\\ \sigma_{2,1} &\sigma_{2,2} \end{array} \!\right)d \left(\! \begin{array}{c} W_t^1\\ W_t^2 \end{...


2

As it was hinted in the comments, the result is not true unless you assume uniform convergence somewhere. Here is a simple example. Let $X_n = X$ have non-degenerate Bernoulli distribution and $Y_{un} = X \mathbf{1}_{u\ge n}$. Then $P(Y_{un}\neq X_n)\to 0$, $u\to\infty$, and $e^{itY_{un}} \to 1$, $n\to\infty$, in all senses. However, $e^{itX_n} = e^{itX}\...


2

To be clear, you want to find $P_n(t)$ for a birth-death process with constant birth rate $q(i,i+1)=\lambda$ and constant death rate $q(i,i-1)=\mu$? This particular birth and death process is exactly the $M/M/1$ queue. As you can see, under "Transient solution", there is a solution for the probability mass function dependent on time for a particular state.


1

Thanks for @Did's help. Now I have got the answer. Because $X_t-X_{s+1/n}$ is independent of ${\mathcal{ F_{s+1/n} }}$ and $\mathcal{ F_{s+} }\subset\mathcal{F_{s+1/n}}\ $ , $X_t-X_{s+1/n}\ $ is independent of ${\mathcal{ F_{s+} }}$ . Now the question is to prove "Assume that $\left\{Y_n\right\}$ is independent of $\mathcal{G}$ and that $Y_n\overset{a.s.}{\...


1

Here is another answer that is more straightforward. $\mu(t,x)=1,\,\sigma(t,x)=2\sqrt{x}$. For any $\alpha\ge 0$, $\tau_\alpha:=\inf\{t\ge\alpha: B_t=0\}$, by Ito's Lemma, $$X_t(\alpha)=\begin{cases} 0, & t<\tau_\alpha \\ B_t^2, &t\ge\tau_\alpha \end{cases} $$ is a solution of the above stochastic differential equation.


1

By the stationarity of the increments, this is equivalent to $$\min_{s \in [0,t]} \mathbb{P}(|X_s| \leq \epsilon)>0$$ for all $t>0$. Without any additional assumptions this does, in general, not hold true. Consider for example $X_t := t$ (i.e. a deterministic drift process), then $$\mathbb{P}(|X_t| \leq \epsilon) = 0$$ for all $t>\epsilon$. ...


1

You're right, this is no hard, although is quite tricky. The only thing you need is self-similarity. First note that $x^qa^{-q} = q\cdot\sup_{\lambda>0}(\lambda x - p^{-1}\lambda^p a^p)$. Then $$ \sup_{t\ge 0} \left(\frac{X_t}{1+t^{p/2}}\right)^q = q\cdot\sup_{t\ge 0}\sup_{\lambda>0}\left(\lambda X_t - p^{-1}\lambda^p (1+t^{p/2})^p\right)\\ = \sup_{\...


1

You'll want to have a look at "A Signed Measure on Path Space Related to Wiener Measure" by K. Hochberg http://www.jstor.org/stable/2243147 Hocberg constructs a Markovian signed measure (of unbounded variation) on the space of continuous paths, associated with the operator $Lu:={\partial^4u\over\partial x^4}$. The transition density $p$ is the fundamental ...


1

This is not a complete answer, just a correction for the case of Poisson process. Clearly, given $N(t) = n$, $$ K(t) = \sum_{k=1}^n (t-\tau_k), $$ where $\tau_k$ is the $k$th jump time. It is well known that, given $N(t) = n$, the jump times $\tau_1,\dots,\tau_n$ are distributed as order statistics $U_{(1)},\dots, U_{(n)}$ of an iid $U[0,t]$ sample $U_1,\...


1

In fact, there is a deeper relation between the Laplacian and Brownian motion. Let $(M, g=\langle\cdot, \cdot\rangle)$ be a smooth Riemannian manifold without boundary. The Laplace-Beltrami operator is defined as the contraction of the covariant derivative of the differential of any smooth function on $M$ $$\forall f \in C^\infty(M): \Delta_M f := \mathrm{...



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