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| Boilerpipe Text | $U$ 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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# [A counter example of Brownian Motion](https://math.stackexchange.com/questions/903892/a-counter-example-of-brownian-motion)
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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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\$\\begingroup\$ Could you please tell in which textbook you found this example? \$\\endgroup\$
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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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