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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.

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

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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{...

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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.}{\...

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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.

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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 $t>\epsilon$. ...

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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), $$where \tau_k is the kth jump time. It is well known that, given N(t) = n, the jump times \tau_1,\dots,\tau_n are distributed as order statistics U_{(1)},\dots, U_{(n)} of an iid U[0,t] sample U_1,\... 1 In fact, there is a deeper relation between the Laplacian and Brownian motion. Let (M, g=\langle\cdot, \cdot\rangle) be a smooth Riemannian manifold without boundary. The Laplace-Beltrami operator is defined as the contraction of the covariant derivative of the differential of any smooth function on M$$\forall f \in C^\infty(M): \Delta_M f := \mathrm{...

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