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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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It would maybe helpful for some people if you cited the solution and told precisely where your problems are with that solution. – Julian Kuelshammer Nov 16 '12 at 15:50
It would be helpful if you could summarize the solution given by the book. (Who is the author?) – Per Manne Nov 16 '12 at 15:50
X(t)has the same sign in a small interval (0, something) contradicting the law of the iterated logaritm, for example, which says it must oscillate. – mike Nov 16 '12 at 16:08
up vote 0 down vote accepted

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 $0<t \leq 1-T(w)$ (by definition of $T$). So there can't exist a sequence $(t_n)_{n \in \mathbb{N}}$ such that $t_n \to 0$, $X_{t_n}(w)=0$.

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thanks for you help – RequiemInDm Nov 17 '12 at 2:37

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