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Here is an example in my textbook to illustrate why we need the continuous sample path in the definition of Brownian motion. Let $(B_t)$ be a Brownian motion and $U$ be a uniform random variable on $[0,1]$. Define $(\tilde B_t)$ by

$$ \tilde B_t = \begin{cases} B_t & \mbox{if } t\neq U, \\ 0 & \mbox{if } t = U. \end{cases} $$

Then it is claimed that $\tilde B_t$ has the same finite-dimensional distribution as $B_t$ but with discontinuous sample path if $B(U)\neq 0$. I do not understand this example. To be more specific,

  1. What does it mean by $t \neq U$ or $t=U$, please? Here, $t$ is a deterministic time and $U$ is a random variable. How can we talk about whether the two are equal or not?
  2. Since I do not get the definition of $\tilde B_t$, I cannot see why it has the same finite dimensional distribution and discontinuous sample path. Could anyone explain this with more detail, please? Thank you!
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Could you please tell in which textbook you found this example? – Cm7F7Bb Aug 20 '14 at 10:49
up vote 4 down vote accepted

Re 1., the expanded definition (maybe clearer) of the random process $\tilde B$ is that, for every $\omega$ in $\Omega$, $\tilde B_{U(\omega)}(\omega)=0$ and $\tilde B_t(\omega)=B_t(\omega)$ if $t\ne U(\omega)$. In short, $$ \tilde B_t=B_t\,\mathbf 1_{U\ne t}. $$ Re 2., the almost sure path discontinuity of $\tilde B$ should be obvious. To wit, the event that $\tilde B$ is continuous is $[B_U=0]$, the notation being again a shortcut, this time to the event $$ [\omega\in\Omega\mid B_{U(\omega)}(\omega)=0]. $$ Since $P(B_t=0)=0$ for every $t$ and $U$ is independent of $B$, one gets indeed that $P(B_U=0)=0$.

The equality of the finite marginal distributions of $B$ and $\tilde B$ follows from the remark that, for every finite subset $T$ of the time interval, the event $$ A=[\omega\in\Omega\mid\forall t\in T, B_t(\omega)=\tilde B_t(\omega)], $$ has full probability, since $P(U=t)=0$ for every fixed $t$ and $$ \Omega\setminus A\subseteq[U\in T]=\bigcup_{t\in T}[U=t]. $$

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$U$ is a random variable, but its value is determined before the Brownian motion begins to evolve, we can tell if t equals $U$ or not.

Look at the finite dimensional distribution $(\tilde{B}_{t_1}, \cdots, \tilde{B}_{t_n})$ of $\tilde{B}_t$, we have $$(\tilde{B}_{t_1}, \cdots, \tilde{B}_{t_n}) = ({B}_{t_1}, \cdots, {B}_{t_n})$$ on the event $I = \{U\neq t_i,\cdots, U\neq t_n\}$.

But $P(I) = 1$, which means $(\tilde{B}_{t_1}, \cdots, \tilde{B}_{t_n}) = ({B}_{t_1}, \cdots, {B}_{t_n})$ almost surely. So we have $E[f(\tilde{B}_{t_1}, \cdots, \tilde{B}_{t_n})] = E[f({B}_{t_1}, \cdots, {B}_{t_n})]$ for any measurable function $f$

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