# Random “start afresh” of Brownian Motion [duplicate]

In the book "Brownian Motion and Stochastic Calculus" by Karatzas and Shreve I encountered Problem 6.1 which the authors refer to as "not be hard to show":

Let $$\{B_t, \cal{F}_t;t\geq 0\}$$ be a standard, one-dimensional Brownian motion. Give an example of a random time $$S$$ with $$P[0\leq S<\infty]=1$$, such that with $$W_t:=B_{S+t}-B_S$$, the process $$W=\{W_t, \cal{F}^W_t;t\geq 0\}$$ is not a Brownian motion.

Surely, if S being a stopping time $$(W_t)_t$$ would still be a Brownian motion. I tried out some distributions for $$S$$ but none of them gave a sufficient answer.

Most likely I am missing a crucial point here and appreciate any kind of help.

## marked as duplicate by Davide Giraudo probability-theory StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 16 '18 at 15:23

Let $$S = \inf \{s\ge 0:B_{s+1}-B_s\ge 0\}$$. Since $$W_1\ge0$$ almost surely, $$W_t$$ cannot be a Brownian motion.