# Tag Info

## Hot answers tagged markov-process

3

Your question is quite interesting, and deserves more attention than it has been getting. Whether or not a (time homogeneous) Markov process $(X_t)_{t\geq 0}$ with state space $E$ is strong Markov depends on a precise definition. I hope that my explanation below substracts from, rather than adds to, your confusion. The standard definition of the strong ...

3

Let $T:=\inf\{t:X_t=0\}$, and suppose that $X_0=x\not=0$. Fix $t>0$. Then $\Bbb P^x[X_t=0]=0$ because $X_t\sim\mathscr N(x,t)$. On the other hand, if $X$ were a strong Markov process, then on the event $\{T<t\}$ we would have $X_t=0$ because the post-$T$ process $\{X_{T+s}:s\ge 0\}$ would start in state $0$, and so stay in state $0$ for all time by ...

2

1. The uniform distribution is invariant Let $\theta_0$ have a uniform distribution in the circle, so that its pdf is $f(\theta)= \frac 1{2\pi}$ for $\theta \in [0,2\pi]$ and $0$ otherwise. If $\theta_0$ moves according to the law of Brownian motion during a time interval $[0,t]$, then its new distribution over the reals can be calculated by applying the ...

1

There are two way to think about this problem. First, let $N_t$ be the number of jumps by the time $t$ (clearly you only need to consider $N_t\ge n$) and let $\bar n=n-1/2$. Then you are interested in $$EV(n)=E(\sum_{i=1}^{N_t}(\frac 1 2 P(S_{i-1}=n-1)+\frac 1 2P(S_{i-1}=n)))=E(\sum_{i=n}^{N_t}\frac 1 2\binom {i-1}{\lfloor\frac {i-n}2\rfloor})$$ Now you can ...

Only top voted, non community-wiki answers of a minimum length are eligible