Limit of the stochastic process at time 0 - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T22:28:48Z http://mathoverflow.net/feeds/question/113894 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/113894/limit-of-the-stochastic-process-at-time-0 Limit of the stochastic process at time 0 Kamil 2012-11-20T02:53:00Z 2012-11-20T18:30:58Z <p>This is not a homework question so please be kind not to remove it right away. I am working on some research but have to justify the following argument: Assume $S_t$ is a continuous stochastic process, don't want to make an assumption about distribution, think about something like a smooth function of Brownian motion. I define another process $$Y_t=\frac{1}{t} \int_0^t S_u du$$ Now I am interested in the limit of $Y_t$ as $t$ approaches zero. I would like to know in what sense the argument would hold, the guess is that the limit is $S(0)$. Please suggest a solution or the way to approach this problem. </p> http://mathoverflow.net/questions/113894/limit-of-the-stochastic-process-at-time-0/113971#113971 Answer by Guillaume for Limit of the stochastic process at time 0 Guillaume 2012-11-20T18:30:58Z 2012-11-20T18:30:58Z <p>You can prove it reasoning $\omega$ by $\omega$. I mean, if you are working in a probability space $\Omega$, the continuous path $(S_t)_{t\geqslant 0}$ depends on $\omega\in \Omega$. But for all $\omega$, $t\mapsto S_t(\omega)$ is continuous, so the following convergence holds : $$Y_t(\omega) \rightarrow S_0(\omega)$$ </p> <p>It is enough to conclude that $Y_t$ converges almost surely to $S_0$ as $t$ goes to zero.</p>