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Let $T$ be the last time before $1$ a Brownian motion visits $0$. Explain why $X(t)=B(t+T)-B(T)=B(t+T)$ is not a Brownian motion. This problem is from Introduction to Stochastic Calculus with Applications, by Klebaner,Exercise 3.15, and I can't understand the solution provided by the book.

Any hints, thanks.

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    X(t)has the same sign in a small interval (0, something) contradicting the law o$f$ the iterated logaritm, for example, which says it must oscillate.2012-11-16

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Let $(B_t)_{t \geq 0}$ a Brownian Motion and $\xi_t := \sup \{s \leq t; B_s = 0\}$. Then we know (by arc-sine law, see René L. Schilling/Lothar Partzsch: "Brownian Motion - An Introduction to stochastic processes", Theorem 6.19)

$\mathbb{P}[\xi_t < s] = \frac{2}{\pi} \arcsin \sqrt{\frac{s}{t}}$

for all $s \leq t$. Thus

$\mathbb{P}[\exists t \in (0,\varepsilon): B_t = 0] = 1- \mathbb{P}[\forall t \in (0,\varepsilon): B_t \not= 0] = 1- \mathbb{P}[\xi_{\varepsilon} = 0]=1$

for all $\varepsilon>0$ which means that you can (a.s.) find for every $w \in \Omega$ a sequence $(t_n)_n$ such that $t_n \to 0$, $B(t_n,w)=0$.

And this means that $(X_t)_{t \geq 0}$ can't be a Brownian motion: Let $w \in \Omega$, then we have $X_{t}(w) \not= 0$ for all $0 (by definition of $T$). So there can't exist a sequence $(t_n)_{n \in \mathbb{N}}$ such that $t_n \to 0$, $X_{t_n}(w)=0$.