= Brownian exit-time skeleton
{c}
{title2=$\sigma_a^{k+1}=\inf\{t\geq\sigma_a^k:|B_t-B_{\sigma_a^k}|=a\}$}
Successive exits from intervals of radius $a$ around the current position embed a <simple symmetric random walk> into <Brownian motion>. The <Strong Markov property> makes the durations independent with common law $a^2\sigma_1$; the displacements divided by $a$ are independent symmetric signs. The displacement sequence is independent across exits; no <independence> between a given displacement and its duration is needed for these conclusions.
Back to article page
