# Finite dimensional distributions of brownian motion changed at a random point

In the book on Brownian Motion by Yuval Peres, he makes the following claim:

Suppose that $\{ B(t); \ t \geq 0 \}$ is a Brownian Motion and $U$ is an independent random variable, uniformly distributed on $[0, 1]$. Then the process $\{ \tilde{B} (t); \ t \geq 0 \}$ defined as, $$\tilde{B}(t) = \begin{cases} B(t) & \mbox{ if t \neq U} \\ 0 & \mbox{ if t = U} \end{cases}$$

has the same finite-dimensional distributions as a Brownian motion, but is discontinuous if $B(U) \neq 0$, i.e. with probability one, and hence this process is not a Brownian motion.

I'm having difficulty in proving that the finite dimensional joint distribution ( basically the joint distribution of $( \ \tilde{B}(t_1), \ \tilde{B}(t_2), \tilde{B}(t_3), \dots \tilde{B}(t_n) \ )$ ) is the same as the finite dimensional distribution of $B(t)$. Can you please give me some clues.

• Can you show that $B_t$ has the same distribution as $\bar{B}_t$ for fixed $t$?
– saz
Commented Feb 4, 2018 at 12:14
• For each $t$, $B(t) =\tilde B(t)$ almost surely. Commented Feb 4, 2018 at 12:22
• Yes I understand the $t, \ B(t) = \tilde B(t)$ almost surely, but if random variables are the same a.s, does it imply their joint distributions are the same a.s. Also how'd you prove discontinuity? Commented Feb 4, 2018 at 12:51

Since

$$\{B_t \neq \tilde{B}_t\} \subseteq \{U=t\}$$

for any $t \geq 0$ we have

$$\bigcup_{j=1}^n \{B_{t_j} \neq \tilde{B}_{t_j}\} \subseteq \bigcup_{j=1}^n \{U=t_j\},$$

and so

$$\mathbb{P}(\exists j \in \{1,\ldots,n\}: B_{t_j} \neq \tilde{B}_{t_j}) \leq \sum_{j=1}^n \mathbb{P}(U=t_j)=0;$$

this implies readily that $(B_t)_{t \geq 0}$ and $(\tilde{B}_t)_{t \geq 0}$ have the same finite dimensional distribution.

Since $(B_t)_{t \geq 0}$ has continuous sample paths, we know that $$\lim_{\substack{s \to U(\omega) \\ s \neq U(\omega)}} \tilde{B}_s(\omega) = \lim_{\substack{s \to U(\omega) \\ s \neq U(\omega)}} B_s(\omega) = B_U(\omega)$$

for almost all $\omega \in \Omega$. This shows that

$$\mathbb{P}(\{\omega; t \mapsto \tilde{B}_t(\omega) \, \, \text{is discontinuous at t=U(\omega)}\}) = \mathbb{P}(B_U \neq 0) = 1.$$

In particular, each sample path $t \mapsto \tilde{B}_t(\omega)$ has at least one discontinuity with probability 1.

• @VoB Since $U$ and $(B_t)_{t \geq 0}$ are independent, we have $$\mathbb{P}(B_U \neq 0) = \int \mathbb{P}(B_u \neq 0) \, d\mathbb{P}_U(u) = \int 1 \, d\mathbb{P}_U(u)=1,$$ where $\mathbb{P}_U$ denotes the distribution of $U$.
– saz
Commented Dec 5, 2019 at 7:26
• Are these intermediate steps correct? :$\mathbb{P}(B_U \neq 0) = \int \mathbb{P}(B_u \neq 0 | U = u) \, d\mathbb{P}_U(u) = \int \mathbb{P}(B_u \neq 0) \, d\mathbb{P}_U(u).$ Commented Apr 14, 2021 at 7:50
• @Error404 Yeah, it's okay.
– saz
Commented Apr 14, 2021 at 8:36
• Thanks! Also does $\, d\mathbb{P}_U(u) = \mathbb{P}(u = U)$? I am asking this because $\mathbb{P}(B_U \neq 0)= \text{"sum over" [0,1] of} \; \mathbb{P}(B_u \neq 0, u=U) = \text{"sum over" [0,1] of} \; \mathbb{P}(B_u \neq 0| u = U)\mathbb{P}(u=U).$ Commented Apr 14, 2021 at 14:12